# Khinchin families and Large powers

José L. Fernández  Departamento de Matemáticas, Universidad Autónoma de
Madrid, Spain<EMAIL_ADDRESS><EMAIL_ADDRESS>and
Víctor J. Maciá  Departamento de Matemáticas, Universidad Autónoma de Madrid,
Spain<EMAIL_ADDRESS>
###### Abstract.

The purpose of this paper is to present a wide picture of asymptotic results
of large powers of coefficients of power series, with non negative
coefficients, using as tools local central limit theorems for lattice random
variables and for continuous families of lattice random variables combined
with the theory of Khinchin families and of the Hayman class.

We also cast in this framework the asymptotic formula due to Otter and to
Meir-Moon of the coefficients of the solutions of Lagrange equations when the
data power series has nonnegative coefficients. Asymptotics of Lagrangian
probability distributions are also discussed.

###### Key words and phrases:

Khinchin families, Hayman admissible functions, asymptotic formulae, large
powers, Lagrange inversion, analytic combinatorics, local central limit
theorem

###### 2010 Mathematics Subject Classification:

30B10, 60F05, 05A16,60F99

Research of J. L. Fernández is supported by Fundación Akusmatika

Research of V. Maciá was partially funded by grant MTM2017-85934-C3-2-P2 of
Ministerio de Economía y Competitividad of Spain and European Research Council
Advanced Grant 834728.

##  1\. Introduction

The purpose of this paper is to present a wide perspective of the asymptotic
formulas for the coefficients of large powers of power series with non
negative coefficients.

The framework of the approach of this paper results from the combination of
local central limit theorems for both lattice random variables and for
continuous families of lattice random variables, on the one hand, with the
theory of Khinchin families and, particularly, of the Hayman class on the
other.

The saddle-point approximation, steepest descent or Laplace method, the usual
tools to obtain these asymptotic formulas for large powers, are somehow built
into the local central limit theorems and into the Hayman class.

We reserve $k$ to signify index of the coefficient and $n$ to denote the power
to which a power series $\psi$ is raised. The power $n$ tends to infinity.
Different results appear depending on how $k$ behaves with $n$, or actually,
on how $k/n$ behaves as $n\to\infty$.

Gardy, in [11], gives a nice presentation of the results for $k/n\asymp 1$ and
$k/n\to 0$ as $n\to\infty$, from the point of view of saddle-point
approximation. See also, De Angelis [7], for a quite direct approach.

If $\psi$ is the probability generating function of a random variable $X$ with
values in $\\{0,1,2,\ldots\\}$, then the $k$-th coefficient of $\psi^{n}$ is
the probability that the sum of $n$ independent copies of $X$ takes the value
$k$. See Daniels [6], particularly [6, Section 8], and Good [13], particularly
[13, Section 6], for saddle point approximations pertaining to this
application of large powers.

Plan of the paper. Section 2 reviews the basic theory of Khinchin families.
Section 3 is devoted to the local central theorem for continuous families of
lattice variables.

The results about large powers, within the tools and framework described in
the previous sections, are discussed in Section 4 and then in Sections 5, 6, 7
and 8.

We also cast in this framework the asymptotic formula due to Otter, [24] and
Meir-Moon, [19] for the coefficients of the solutions of Lagrange equations
when the data power series has nonnegative coefficients. The proof of the
Otter-Meir-Moon Theorem is the content of Section 9.

Section 10 discusses applications to probability generating functions and to
Lagrangian distributions.

Finally, the appendix 11 deals with the so-called uniformly Gaussian and
uniformly Hayman Khinchin families which are just needed in Section 7.

Notations.

Expectation and variance of a random variable $X$ are generically denoted
$\mathbf{E}(X)$ and $\mathbf{V}(X)$. Probability of an event $A$ is
generically denoted by $\mathbf{P}(A)$.

For random variables $X,Y$ the notation $X\overset{d}{=}Y$ signifies that $X$
and $Y$ have the same distribution: $\mathbf{P}(X\in B)=\mathbf{P}(Y\in B)$
for any Borel set $B\subset\mathbb{R}$.

For a sequence of random variables $(X_{n})_{n\geq 1}$ and a random variable
$Y$ we write

$X_{n}\stackrel{{\scriptstyle d}}{{\underset{n\to\infty}{\longrightarrow}}}Y$

to signify convergence in distribution.

A family $(Z_{s})_{s\in I}$ indexed in a real interval $I$ is continuous in
distribution if for any sequence $(s_{n})_{n\geq 1}$ of points in $I$ which
converges to a point $s^{\star}\in I$ it holds that $Z_{s_{n}}$ converges in
distribution to $Z_{s^{\star}}$.

The abbreviations egf and ogf mean, respectively, exponential generating
function of a labeled combinatorial class and ordinary generating function of
a combinatorial class. For general background on Combinatorics pertaining to
this paper we refer to the treatise [10].

The unit disk in the complex plane $\mathbb{C}$ is denoted by $\mathbb{D}$.
The disk of center $a\in\mathbb{C}$ and radius $R$ is denoted
$\mathbb{D}(a,R)$, while its closure is denoted $\mbox{\rm
cl}(\mathbb{D}(a,R))$.

For positive sequences $(a_{n})_{n\geq 1},(b_{n})_{n\geq 1}$, by $a_{n}\sim
b_{n}$ , as $n\to\infty$, we mean that $\lim_{n\to\infty}a_{n}/b_{n}=1$.

If $\psi$ and $f$ are probability generating functions, with
$\mathcal{L}(\psi,f)$ we denote the Lagrangian probability distribution with
generators $\psi$ and $f$, see Section 10.1.

##  2\. Khinchin families

We denote by $\mathcal{K}$ the class of nonconstant power series

$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$

with positive radius of convergence, which have nonnegative Taylor
coefficients and such that $a_{0}>0$. Since $f\in\mathcal{K}$ is nonconstant,
at least one coefficient other than $a_{0}$ is positive.

The Khinchin family of such a power series $f\in\mathcal{K}$ with radius of
convergence $R>0$ is the family of random variables $(X_{t})_{t\in[0,R)}$ with
values in $\\{0,1,\ldots\\}$ and with mass functions given by

$\mathbf{P}(X_{t}=n)=\frac{a_{n}t^{n}}{f(t)}\,,\quad\mbox{for each $n\geq 0$
and $t\in(0,R)$}\,.$

For $t=0$, we define $X_{0}\equiv 0$. Notice that $f(t)>0$ for each
$t\in[0,R)$.

Any Khinchin family is continuous in distribution in $[0,R)$. No hypothesis
upon joint distribution of the variables $X_{t}$ is considered; they are
families and not processes.

We refer to [3] and [4] for the basic theory of Khinchin families. Next we
describe the specific aspects of the theory to be used in the present paper;
for proofs, examples and applications we refer to [3].

####  2.0.1. Shifted Khinchin families

Sometimes it is convenient to consider power series
$g(z)=\sum_{n=0}^{\infty}b_{b}z^{n}$ with radius of convergence $R>0$ and
nonnegative coefficients with at least two positive coefficients, but which
may have $g(0)=0$. We say then that $g$ is in the shifted class
$\mathcal{K}^{s}$. Thus $g\in\mathcal{K}^{s}$ if there exists an integer
$l\geq 0$ such that

$\displaystyle g(z)/z^{l}\in\mathcal{K}\,,$

i.e., if there exists a power series $f(z)\in\mathcal{K}$ with radius of
convergence $R>0$ such that $g(z)=z^{l}f(z)$, for $z\in\mathbb{D}(0,R)$.

To $g\in\mathcal{K}^{s}$ we associate a Khinchin family $(Y_{t})_{t\
\in(0,R)}$ as above:

$\mathbf{P}(Y_{t}=n)=b_{n}t^{n}/g(t)\,,\quad\mbox{for $n\geq 0$ and
$t\in(0,R)$}\,.$

If $(Z_{t})_{t\in[0,R)}$ is the Khinchin family of $f\in\mathcal{K}$, then

$\displaystyle Y_{t}\stackrel{{\scriptstyle d}}{{=}}Z_{t}+l\,,$

including, $Y_{0}\equiv l$.

###  2.1. Basic properties

Along this section we let $f$ be a power series in $\mathcal{K}$ with radius
of convergence $R>0$ and Khinchin family $(X_{t})_{t\in[0,R)}$.

####  2.1.1. Mean and variance functions

For the mean and variance of $X_{t}$ we reserve the notation
$m(t)=\mathbf{E}(X_{t})$ and $\sigma^{2}(t)=\mathbf{V}(X_{t})$, for
$t\in[0,R)$. In terms of $f$, the mean and the variance of $X_{t}$ may be
written as

$m(t)=\frac{tf^{\prime}(t)}{f(t)},\qquad\sigma^{2}(t)=tm^{\prime}(t)\,,\quad\mbox{for
$t\in[0,R)$}\,.$

For each $t\in(0,R)$, the variable $X_{t}$ is not a constant, and so
$\sigma^{2}(t)>0$. Consequently, $m(t)$ is strictly increasing in $[0,R)$,
though, in general, $\sigma(t)$ is not increasing. We denote

(2.1) $M_{f}=\lim_{t\uparrow R}m(t)\,.$

For $g\in\mathcal{K}^{s}$, with Khinchin family $(Y_{t})$, we also write
$m_{g}(t)=\mathbf{E}(Y_{t})$ and $\sigma^{2}(t)=\mathbf{V}(Y_{t})$. If
$g(z)=z^{l}h(z)$, with $l\geq 1$ and $h\in\mathcal{K}$ we have that

$m_{g}(t)=l+m_{h}(t)\quad\mbox{and}\quad\sigma_{g}^{2}(t)=\sigma^{2}_{h}(t).$

The function $m_{g}$ is an increasing diffeomorphism from $[0,R)$ to
$[l,l+M_{h})$.

For finite radius of convergence $R>0$ and for any integer $k\geq 1$, the
function $f(t)$ and its derivatives $f^{(k)}(t)$ are all increasing functions
on the interval $[0,R)$. For $k\geq 1$, we denote with $f^{(k)}(R)$ the limit

$f^{(k)}(R)\triangleq\lim_{t\uparrow R}f^{(k)}(t)\,,$

including

$f(R)\triangleq\lim_{t\uparrow R}f(t)\,;$

these limits exist, although they could be $+\infty$.

####  2.1.2. Range of the mean

Since the mean function $m(t)$ is increasing, its range is given by
$[0,M_{f})$.

$\bullet$ The case where $M_{f}=\infty$ is particularly relevant and quite
general. If $M_{f}=\infty$, then $m(t)$ is a diffeomorphism from $[0,R)$ onto
$[0,+\infty)$ and, in particular, for each $n\geq 1$, there exists a unique
$t_{n}\in(0,R)$ such that $m(t_{n})=n$. These $t_{n}$ play an important role
in Hayman’s identity, Section 2.2 and in Hayman’s asymptotic formula, Section
2.6.

$\bullet$ The following Lemma A describes the quite specific cases where
$M_{f}<+\infty$. It is Lemma 2.2 of [3].

######  Lemma A.

For $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathcal{K}$ with radius of
convergence $R>0$, we have $M_{f}<\infty$ if and only if ($R<\infty$ and
$\sum_{n=0}^{\infty}na_{n}R^{n}<\infty$) or ($R=\infty$ and $f$ is a
polynomial).

In the first case, we have that
$M_{f}=\dfrac{\sum_{n=0}^{\infty}na_{n}R^{n}}{\sum_{n=0}^{\infty}a_{n}R^{n}}$,
while for a polynomial $f\in\mathcal{K}$, the second case, we have that
$M_{f}=\mbox{deg}(f)$.

####  2.1.3. Extension of the Khinchin family if $M_{f}<\infty$ (and
$R<\infty$)

We assume now that $M_{f}<\infty$ (and $R<+\infty$).

Thus we have that $\sum_{n=0}^{\infty}na_{n}R^{n}<\infty$ and also that
$f(R)=\sum_{n=0}^{\infty}a_{n}R^{n}<\infty$. The power series
$\sum_{n=0}^{\infty}a_{n}z^{n}$ defines a continuous (actually, $C^{1}$)
function on the whole closed disk $\mbox{\rm cl}(\mathbb{D}(0,R)$ which
extends $f$ from $\mathbb{D}(0,R)$. We let
$f(R)\triangleq\sum_{n=0}^{\infty}a_{n}R^{n}=\lim_{t\uparrow R}f(t)$ and

$\displaystyle f^{\prime}(R)$
$\displaystyle\triangleq\sum_{n=1}^{\infty}na_{n}R^{n{-}1}$
$\displaystyle=\lim_{t\uparrow R}f^{\prime}(t),$ $\displaystyle
f^{\prime\prime}(R)$
$\displaystyle\triangleq\sum_{n=2}^{\infty}n(n{-}1)a_{n}R^{n{-}2}$
$\displaystyle=\lim_{t\uparrow R}f^{\prime\prime}(t).$

In this case, we may extend the Khinchin family $(X_{t})_{t\in[0,R)}$ of $f$
to $t\in[0,R]$ by defining the variable $X_{R}$ by

$\mathbf{P}(X_{R}=n)=\frac{a_{n}R^{n}}{f(R)}\,,\quad\mbox{for each $n\geq
0$}\,.$

The extended family $(X_{t})_{t\in[0,R]}$ becomes continuous in distribution
in the closed interval $[0,R]$. Observe that $X_{R}$ (like any other $X_{t}$,
with $t\in(0,R)$) is nonconstant.

The variable $X_{R}$ has (finite) mean
$\mathbf{E}(X_{R})=Rf^{\prime}(R)/f(R)\triangleq m(R)$ and variance

$\mathbf{V}(X_{R})=\frac{\sum_{n=0}^{\infty}n^{2}a_{n}R^{n}}{f(R)}-\mathbf{E}(X_{R})^{2}\,.$

The variance $\mathbf{V}(X_{R})$ is nonzero since $X_{R}$ is nonconstant, but
it could be infinite. Actually, $\mathbf{V}(X_{R})$ is finite if and only if
$\sum_{n=0}^{\infty}n^{2}a_{n}R^{n}<+\infty$ if and only if
$f^{\prime\prime}(R)<+\infty$, and in any case

$\mathbf{V}(X_{R})=\lim_{t\uparrow R}\sigma^{2}(t)\,.$

If $\mathbf{V}(X_{R})$ is finite, we write
$\sigma^{2}(R)\triangleq\mathbf{V}(X_{R})$.

####  2.1.4. Normalization and characteristic functions

For each $t\in(0,R)$, we denote by $\breve{X}_{t}$ the normalization of
$X_{t}$:

$\breve{X}_{t}=\frac{X_{t}-m(t)}{\sigma(t)},\,\quad\mbox{for $t\in(0,R)$}\,.$

The characteristic function of $X_{t}$ may be written in terms of the power
series $f$ as:

$\mathbf{E}(e^{\imath\theta
X_{t}})=\frac{f(te^{\imath\theta})}{f(t)}\,,\quad\mbox{for $t\in(0,R)$ and
$\theta\in\mathbb{R}$}\,,$

while for its normalized version $\breve{X}_{t}$ we have

$\mathbf{E}(e^{\imath\theta\breve{X}_{t}})=\mathbf{E}(e^{\imath\theta
X_{t}/\sigma(t)})e^{-\imath\theta m(t)/\sigma(t)}\,,\quad\mbox{for $t\in(0,R)$
and $\theta\in\mathbb{R}$}\,,$

and so,

$\big{|}\mathbf{E}(e^{\imath\theta\breve{X}_{t}})\big{|}=\big{|}\mathbf{E}(e^{\imath\theta
X_{t}/\sigma(t)})\big{|}\,,\quad\mbox{for $t\in(0,R)$ and
$\theta\in\mathbb{R}$}\,.$

If $M_{f}<\infty$ (and $R<\infty$), we may define
$\breve{X}_{R}=(X_{R}-m(R))/\sigma(R)$, with the understanding that
$\breve{X}_{R}\equiv 0$, if $\sigma(R)=+\infty$. Recall that $\sigma(R)>0$.
The extended family $(\breve{X}_{t})_{t\in(0,R]}$ of normalized variables is
continuous (in distribution).

####  2.1.5. Scaling

For any power series $g(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathcal{K}$ of
radius $R>0$, we denote

$Q_{g}\triangleq\gcd\\{n\geq 1:a_{n}\neq 0\\}=\lim_{N\to\infty}\gcd\\{1\leq
n\leq N:a_{n}\neq 0\\}\,.$

If $Q_{g}>1$, then we can write $g(z)=f(z^{Q_{g}})$ for a certain companion
power series $f\in\mathcal{K}$ with has radius of convergence $R^{Q_{g}}$.
Observe that $Q_{f}=1$.

Let $(Y_{t})_{t\in[0,R)}$ be the Khinchin family of $g$ and
$(X_{t})_{t\in[0,R^{Q_{g}})}$ be the Khinchin family of $f$. We have that

$\displaystyle Y_{t}\stackrel{{\scriptstyle d}}{{=}}Q_{g}\cdot
X_{t^{Q_{g}}},\quad\text{ for any }t\in(0,R)).$

The mean and variance functions of $g$ and $f$ are related by

$\begin{aligned} m_{g}(t)=Q_{g}\cdot m_{f}(t^{Q_{g}})\,,\\\
\sigma^{2}_{g}(t)=Q_{g}^{2}\cdot\sigma^{2}_{f}(t^{Q_{g}})\,,\end{aligned}\qquad\mbox{for
$t\in(0,R)$}\,.$

####  2.1.6. Fulcrum $F$ of $f$

A power series $f$ in $\mathcal{K}$ does not vanish on the real interval
$[0,R)$, and so, it does not vanish in a simply connected region containing
that interval. We may consider $\ln f$, a branch of the logarithm of $f$ which
is real on $[0,R)$, and the function $F$, called the fulcrum of $f$, which is
defined and holomorphic in a region containing $(-\infty,\ln R)$ and it is
given by

$F(z)=\ln f(e^{z})\,.$

If $f$ does not vanish anywhere in the disk $\mathbb{D}(0,R)$, then the
fulcrum $F(z)$ of $f$ is defined in the whole half plane $\Re z<\ln R$.

In general, the fulcrum $F$ of $f$ is defined and holomorphic in the band-like
region $\\{s+\imath\theta:s<\ln R,|\theta|<\sqrt{2}/\sigma(e^{s})\\}$. See
Section 2.1.5 of [3]

For $t=e^{s}$, we have $m(t)=F^{\prime}(s)$ and
$\sigma^{2}(t)=F^{\prime\prime}(s)$, for $s<\ln R$.

###  2.2. Hayman’s identity

Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be a power series in $\mathcal{K}$.
Cauchy’s formula for the coefficient $a_{n}$ in terms of the characteristic
function of its Khinchin family $(X_{t})_{t\in[0,R)}$ reads

$a_{n}=\frac{f(t)}{2\pi t^{n}}\int_{|\theta|<\pi}\mathbf{E}(e^{\imath\theta
X_{t}})e^{-\imath\theta n}\,d\theta\,,\quad\mbox{for each $t\in(0,R)$ and
$n\geq 1$}\,.$

In terms of the characteristic function of the normalized variable
$\breve{X}_{t}$, it becomes

(2.2) $\displaystyle a_{n}=\frac{f(t)}{2\pi
t^{n}\sigma(t)}\int_{|\theta|<\pi\sigma(t)}\mathbf{E}(e^{\imath\theta\breve{X}_{t}})$
$\displaystyle e^{-\imath\theta(n-m(t))/\sigma(t)}\,d\theta\,,$
$\displaystyle\quad\mbox{for each $t\in(0,R)$ and $n\geq 1$}\,.$

$\bullet$ If $M_{f}=\infty$, we may take for each $n\geq 1$ the (unique)
radius $t_{n}\in(0,R)$ so that $m(t_{n})=n$, to write

(2.3) $a_{n}=\frac{f(t_{n})}{2\pi
t_{n}^{n}\sigma(t_{n})}\int_{|\theta|<\pi\sigma(t_{n})}\mathbf{E}(e^{\imath\theta\breve{X}_{t_{n}}})\,d\theta\,,\quad\mbox{for
each $n\geq 1$}\,,$

which we call Hayman’s identity.

Although this identity (2.3) is just Cauchy’s formula with an appropriate
choice of radius, it encapsulates the saddle point method.

If the Khinchin family $(X_{t})$ may be extended to the closed interval
$[0,R]$, then we have a similar identity with $t=R$.

$\bullet$ If $M_{f}<\infty$ and ($R<\infty$) the power series $f$ extends
continuously to the closed disk $\mbox{\rm cl}(\mathbb{D}(0,R))$ and the
Khinchin family $(X_{t})_{t\in[0,R)}$ extends to the closed interval $[0,R]$.
See Section 2.1.2 and the notations therein.

We may write

$a_{n}=\frac{f(R)}{2\pi R^{n}}\int_{|\theta|<\pi}\mathbf{E}(e^{\imath\theta
X_{R}})e^{-\imath\theta n}\,d\theta\,,\quad\mbox{for $n\geq 1$}\,.$

If, moreover, $\sigma^{2}(R)<\infty$, i.e., if
$\sum_{n=0}^{\infty}n^{2}a_{n}R^{n}<\infty$, then $\breve{X}_{R}$ given by
$\breve{X}_{R}=(X_{R}{-}m(R))/\sigma(R)$ is well defined and we also have that

(2.4) $a_{n}=\frac{f(R)}{2\pi
R^{n}\sigma(R)}\int_{|\theta|<\pi\sigma(R)}\mathbf{E}(e^{\imath\theta\breve{X}_{R}})\,e^{-\imath\theta(n-m(R))/\sigma(R)}\,d\theta\,,\quad\mbox{for
$n\geq 1$}\,.$

###  2.3. Basic moments of a Khinchin family near 0

Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be a power series in $\mathcal{K}$
with radius of convergence $R>0$. Let $(X_{t})_{t\in[0,R)}$ be its Khinchin
family and $m(t)$ and $\sigma^{2}(t)$ be, respectively, the mean and variance
of $X_{t}$, for $t\in[0,R)$.

Assume first that $a_{1}>0$. In this case, we have that

$m(t)=\frac{a_{1}}{a_{0}}t+O(t^{2})\quad\text{and}\quad\sigma^{2}(t)=\frac{a_{1}}{a_{0}}t+O(t^{2})\,,\quad\mbox{as
$t\downarrow 0$}\,,$

and, besides, that

$\mathbf{E}(|X_{t}-m(t)|^{3})=\frac{a_{1}}{a_{0}}t+O(t^{2})\quad\text{and}\quad\sqrt{t}\,\mathbf{E}(|\breve{X}_{t}|^{3})=\sqrt{\frac{a_{0}}{a_{1}}}+O(t)\,,\quad\mbox{as
$t\downarrow 0$}\,.$

In general, if for $k\geq 1$, we have $a_{k}>0$, but $a_{j}=0$, for $1\leq
j<k$, then

$m(t)=\frac{ka_{k}}{a_{0}}t^{k}+O(t^{k{+}1})\quad\text{and}\quad\sigma^{2}(t)=\frac{k^{2}a_{k}}{a_{0}}t^{k}+O(t^{k{+}1})\,,\quad\mbox{as
$t\downarrow 0$}\,,$

and, besides, that

$\mathbf{E}(|X_{t}-m(t)|^{3})=\frac{k^{3}a_{k}}{a_{0}}t^{k}+O(t^{k{+}1})\quad\text{and}\quad
t^{k/2}\,\mathbf{E}(|\breve{X}_{t}|^{3})=\sqrt{\frac{a_{0}}{a_{k}}}+O(t)\,,\quad\mbox{as
$t\downarrow 0$}\,.$

###  2.4. Gaussian Khinchin families

We introduce next Gaussian and strongly Gaussian power series.

######  Definition 1.

A power series $f$ in $\mathcal{K}$ of radius of convergence $R>0$ and its
Khinchin family $(X_{t})_{t\in[0,R)}$ are termed Gaussian if $\breve{X}_{t}$
converges in distribution to the standard normal, as $t\uparrow R$, or,
equivalently, if

$\lim_{t\uparrow
R}\mathbf{E}(e^{\imath\theta\breve{X}_{t}})=e^{-\theta^{2}/2}\,,\quad\mbox{for
each $\theta\in\mathbb{R}$}\,.$

######  Definition 2.

A power series $f\in\mathcal{K}$ and its Khinchin family $(X_{t})_{t\in[0,R)}$
are termed strongly Gaussian if the following two conditions are satisfied

$\lim_{t\uparrow R}\sigma(t)=+\infty\quad\mbox{and}\quad\lim_{t\uparrow
R}\int\limits_{|\theta|<\pi\sigma(t)}\Big{|}\mathbf{E}(e^{\imath\theta\breve{X}_{t}})-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

This notion of strongly Gaussian power series was introduced by Báez-Duarte in
[1]. Strongly Gaussian power series are Gaussian, see Theorem A.

We refer to [3] and [4] for examples, for criteria to verify whether a power
series is Gaussian or strongly Gaussian, and for a number of applications
(asymptotic formulas of coefficients) of these concepts for set constructions
in Analytic Combinatorics, including partitions.

We just mention (see [3]) that

  * •

the exponential $f(z)=e^{z}$ is strongly Gaussian and thus Gaussian,

  * •

polynomials are not Gaussian. In fact, if $(X_{t})_{t\in[0,\infty)}$ is the
Khinchin family of a polynomial of degree $N$, then, as $t \uparrow\infty$,
$X_{t}$ tends in distribution to the constant $N$ and $\breve{X}_{t}$ tends in
distribution to $0$.

######  Lemma B.

If $f$ is a Gaussian power series, then $M_{f}=+\infty$

###### Proof.

Assume that $M_{f}<+\infty$. We use the characterization of $M_{f}<+\infty$
given in Lemma A. If the radius of convergence $R$ of $f$ is $R=+\infty$, then
$f$ is a polynomial, which is not Gaussian as we just have mentioned.

If $R<\infty$, then (see Section 2.1.2) we may extend the Khinchin family to
include $t=R$ with $X_{R}$. Thus $X_{t}$ tends in distribution to $X_{R}$ as
$t\uparrow R$. If $\sigma(R)<\infty$, then $\breve{X}_{t}$ tends in
distribution to $\breve{X}_{R}=(X_{R}-m(R))/\sigma(R)$, while if
$\sigma(R)=\infty$, then $\breve{X}_{t}$ tends in distribution to the constant
$\breve{X}_{R}\equiv 0$. In both cases, the limit $\breve{X}_{R}$ is a
discrete random variable. ∎

###  2.5. Hayman’s central limit Theorem

Strongly Gaussian power series satisfy Theorem A, which is both a local and
global Central Limit Theorem. This results, for Hayman’s admissible power
series, comes from [16], but see [3] for a proof for the more general case of
strongly Gaussian power series.

######  Theorem A (Hayman’s central limit theorem).

If $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathcal{K}$ is a strongly
Gaussian power series with radius of convergence $R>0$, then

(2.5) $\lim_{t\uparrow
R}\sup\limits_{n\in\mathbb{Z}}\Big{|}\frac{a_{n}t^{n}}{f(t)}\sqrt{2\pi}\sigma(t)-e^{-(n-m(t))^{2}/(2\sigma^{2}(t))}\Big{|}=0\,.$

(In this statement $a_{n}=0$, for $n<0$.)

Besides,

$\lim_{t\uparrow R}\mathbf{P}(\breve{X}_{t}\leq b)=\Phi(b)\,,\quad\mbox{for
every $b\in\mathbb{R}$}\,.$

And so, $\breve{X}_{t}$ converges in distribution towards the standard normal
and $f$ is Gaussian.

In other terms, Theorem A claims that if for each $t\in[0,R)$ we define the
set

$\mathcal{V}_{t}=\Big{\\{}\frac{n-m(t)}{\sigma(t)}:n\in\mathbb{Z}\Big{\\}}\,,$

then

$\lim_{t\uparrow
R}\sup\limits_{x\in\mathcal{V}_{t}}\Big{|}\mathbf{P}(\breve{X}_{t}=x)\,\sigma(t)-\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}\Big{|}=0\,.$

###  2.6. Asymptotic formula of coefficients of strongly Gaussian power series

The coefficients of strongly Gaussian power series obey a neat asymptotic
formula:

######  Theorem B (Hayman’s asymptotic formula).

If $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathcal{K}$ is strongly Gaussian
then

(2.6)
$a_{n}\sim\frac{1}{\sqrt{2\pi}}\frac{f(t_{n})}{t_{n}^{n}\sigma(t_{n})}\,,\quad\mbox{as
$n\to\infty$}\,.$

In the asymptotic formula above, $t_{n}$ is given by $m(t_{n})=n$, for each
$n\geq 1$, which are uniquely defined because $M_{f}=+\infty$, see Lemma B.
This asymptotic formula follows readily from Theorem A, or alternatively, from
Hayman’s formula (2.3) and strong gaussianity.

###  2.7. Hayman class

The class of Hayman consists of power series $f$ in $\mathcal{K}$ which
satisfy some concrete and verifiable conditions which imply that $f$ is
strongly Gaussian, see Theorem C below.

######  Definition 3.

A power series $f\in\mathcal{K}$ is in the Hayman class (or is Hayman-
admissible or just $H$-admissible) if

(2.7) $\mbox{\rm variance condition:}\quad\lim_{t\uparrow
R}\sigma(t)=\infty\,,$

and for a certain function $h:[0,R)\rightarrow(0,\pi]$, which we refer to as a
cut between a major arc and a minor arc, the following conditions are
satisfied.

(2.8) major arc: $\displaystyle\lim_{t\uparrow R}\sup_{|\theta|\leq
h(t)\,\sigma(t)}\bigg{|}\mathbf{E}(e^{\imath\theta\breve{X}_{t}})e^{\theta^{2}/2}-1\bigg{|}=0\,,$
(2.9) minor arc: $\displaystyle\lim_{t\uparrow
R}\sigma(t)\,\sup_{h(t)\sigma(t)\leq|\theta|\leq\pi\sigma(t)}|\mathbf{E}(e^{\imath\theta\breve{X}_{t}})|=0\,.$

Some authors include in the Hayman class power series with a finite number of
negative coefficients. This is not the case in the present paper.

For $f$ in the Hayman class, the characteristic function of $\breve{X}_{t}$ is
uniformly approximated by $e^{-\theta^{2}/2}$ in the major arc, while it is
uniformly $o(1/\sigma(t))$ in the minor arc.

######  Theorem C.

Power series in the Hayman class are strongly Gaussian.

See [3] for a proof of Theorem C. We may consider membership in Hayman’s class
as a criterion for strong gaussianity.

##  3\. Local central limit theorem and continuous families

To obtain asymptotic results for the coefficients of large powers of power
series with nonnegative coefficients we will write down these coefficients in
terms of Khinchin families and Hayman’s identities.

To handle the Hayman’s identities of a Khinchin family we shall use the local
central limit Theorem for lattice variables in some integral form. This is
Theorem F, as applied to single lattice variable, and Theorem 3.3, when
dealing with continuous families of lattice variables. We provide a proof of
this last result; this requires some preliminary material.

###  3.1. Approximation and bound of characteristic functions

For a random variable $Y$ and integer $n\geq 1$, we have for
$\mathbf{E}\big{(}e^{\imath\theta Y/\sqrt{n}}\big{)}^{n}$ the following well
known approximation, Theorem D, and bound, Theorem E, involving the third
moment of $Y$.

######  Theorem D.

There exists a constant $C>0$, such that for any random variable $Y$ verifying
that $\mathbf{E}(Y)=0$, $\mathbf{E}(Y^{2})=1$ and
$\mathbf{E}(|Y|^{3})<\infty$, it holds that

$\Big{|}\mathbf{E}\big{(}e^{\imath\theta
Y/\sqrt{n}}\big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\leq
C\frac{\mathbf{E}(|Y|^{3})}{\sqrt{n}}\,,\quad\mbox{if
$|\theta|\leq\dfrac{\sqrt{n}}{\mathbf{E}(|Y|^{3})}$ and $n\geq 1$}\,.$

See, for instance, [14, Lemma 6.2, Chapter 7]. In particular,

(3.1) $\lim_{n\to\infty}\mathbf{E}\big{(}e^{\imath\theta
Y/\sqrt{n}}\big{)}^{n}=e^{-\theta^{2}/2}\,,\quad\mbox{for each
$\theta\in\mathbb{R}$}\,,$

which is the Central Limit Theorem for sums of independent identically
distributed copies of the variable $Y$ with the extra assumption of finite
third absolute moment.

######  Theorem E.

For any random variable $Y$ such that $\mathbf{E}(Y)=0$, $\mathbf{E}(Y^{2})=1$
and $\mathbf{E}(|Y|^{3})<\infty$, we have that

$\Big{|}\mathbf{E}\big{(}e^{\imath\theta Y/\sqrt{n}}\big{)}^{n}\Big{|}\leq
e^{-\theta^{2}/3},\quad\mbox{for any $\theta$ such that
$|\theta|\leq\dfrac{\sqrt{n}}{4\mathbf{E}(|Y|^{3})}$ and $n\geq 1$}\,.$

See, for instance, step 1 in the proof of [14, Lemma 6.2, Chapter 7].

###  3.2. Lattice distributed variables

A random variable $Z$ is said to be lattice distributed or a lattice random
variable if $Z$ is nonconstant and for a certain $a\in\mathbb{R}$ and $h>0$
one has that $\mathbf{P}(Z\in a+h\mathbb{Z})=1$. The largest $h$ such that
$\mathbf{P}(Z\in a+h\mathbb{Z})=1$ is called the gauge of the lattice variable
$Z$. For gauge $h$, the variable $(Z-a)/h$ takes integer values and has gauge
1.

Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be a power series in $\mathcal{K}$.
Let $\mathcal{I}=\\{n\geq 1:a_{n}\neq 0\\}$. Recall from Section 2.1.5 the
notation $Q_{f}=\mbox{gcd}\\{\mathcal{I}\\}$.

If $Q_{f}=1$, then each $X_{t}$ of the Khinchin family of $f$ is a lattice
variable of gauge 1 and each $\breve{X}_{t}$ is a lattice variable of gauge
$1/\sigma(t)$. The condition $Q_{f}=1$ is equivalent to the requirement that
for each $d\geq 1$ there is $n\in\mathcal{I}$ such that $d\nmid n$.

In general, if $Q_{f}\geq 1$, then each $X_{t}$ of the Khinchin family of $f$
is a lattice variable of gauge $Q_{f}$ and each $\breve{X}_{t}$ is a lattice
variable of gauge $Q_{f}/\sigma(t)$.

######  Lemma C.

If $Z$ is a lattice random variable with gauge $h$, then

$|\mathbf{E}(e^{\imath\theta Z})|<1\,,\quad\mbox{ for $0<\theta<2\pi/h$}\,.$

In particular, for each $0<\delta<\pi/h$, there exists
$\omega=\omega_{\delta}<1$ such that

$|\mathbf{E}(e^{\imath\theta Z})|\leq\omega\,,\quad\mbox{for
$|\theta-\pi/h|\leq\delta$}\,.$

See [8, Section 3.5]. The second part of the statement follows simply from
continuity.

Next result is a standard inversion lemma: the mass function of a variable $U$
is expressed in terms of its characteristic function.

######  Lemma D.

Let $U$ be a lattice random variable with gauge $h$, and so that for certain
$a\in\mathbb{R}$ we have that $\mathbf{P}(U\in a+h\mathbb{Z})=1$. Then

$\mathbf{P}(U=a+hk)=\frac{h}{2\pi}\int_{-\pi/h}^{\pi/h}\mathbf{E}(e^{\imath\theta
U})e^{-\imath\theta(a+kh)}\,d\theta\,,\quad\mbox{for each
$k\in\mathbb{Z}$}\,.$

In other terms, for each possible value $x$ of the variable $U$ we have that

$\mathbf{P}(U=x)=\frac{h}{2\pi}\int_{-\pi/h}^{\pi/h}\mathbf{E}(e^{\imath\theta
U})e^{-\imath\theta x}d\theta\,.$

###### Proof.

We may assume that $a=0$ and $h=1$. For each $k\in\mathbb{Z}$, let
$p_{k}=\mathbf{P}(U=k)$. Then, $\sum_{k\in\mathbb{Z}}p_{k}=1$ and

$\mathbf{E}(e^{\imath\theta U})=\sum_{k\in\mathbb{Z}}p_{k}e^{\imath
k\theta}\,,\quad\mbox{for each $\theta\in\mathbb{R}$}\,,$

and therefore, that
$p_{k}=\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}\mathbf{E}(e^{\imath\theta
U})e^{-\imath k\theta}d\theta\,,\quad\mbox{for each $k\in\mathbb{Z}$}\,.$ ∎

###  3.3. Local central limit Theorem for lattice variables: integral form

######  Theorem F.

Let $Z$ be a lattice random variable with gauge $h$ and such that
$\mathbf{E}(Z)=0$ and $\mathbf{E}(Z^{2})=1$. Then

$\lim_{n\to\infty}\int\limits_{|\theta|\leq\pi\sqrt{n}/h}\Big{|}\mathbf{E}(e^{\imath\theta
Z/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

From the convergence of this integral, the usual statement of the Local
Central Limit Theorem follows, see Section 3.3.1. In fact, Theorem F is what
is actually shown (as an intermediate step) in the usual proofs of the Local
Central Limit Theorem, see, for instance, [12, Section 43] and the proof of
[8, Theorem 3.5.3].

We give a proof below with the extra assumption that
$\mathbf{E}(|Y|^{3})<\infty$ anticipating the corresponding result, Theorem
3.3, for continuous families of lattice variables.

###### Proof.

The central limit Theorem, as in equation (3.1), gives us that, as
$n\to\infty$, the integrand converges towards 0 for each
$\theta\in\mathbb{R}$.

Denoting $\tau\triangleq 1/(2\mathbf{E}(|Z|^{3})$, Theorem E gives us the
bound

$|\mathbf{E}(e^{\imath\theta Z/\sqrt{n}})^{n}|\leq
e^{-\theta^{2}/6}\,,\quad\mbox{si $|\theta|\leq\tau\sqrt{n}$}\,.$

Thus, dominated convergence gives that

$\lim_{n\to\infty}\int_{|\theta|\leq\tau\sqrt{n}}\Big{|}\mathbf{E}(e^{\imath\theta
Z/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

Lemma C gives, in particular, a bound $\omega\in(0,1)$ such that

$|\mathbf{E}(e^{\imath\theta Z})|\leq\omega\,,\quad\mbox{si
$\tau\leq|\theta|\leq\pi/h$}\,,$

and so that

$|\mathbf{E}(e^{\imath\theta Z/\sqrt{n}})^{n}|\leq\omega^{n}\,,\quad\mbox{si
$\tau\sqrt{n}\leq|\theta|\leq\pi\sqrt{n}/h$}\,.$

The bound

$\int\limits_{\tau\sqrt{n}\leq|\theta|\leq\pi\sqrt{n}/h}\Big{|}\mathbf{E}(e^{\imath\theta
Z/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta\leq\int\limits_{\tau\sqrt{n}\leq|\theta|\leq\pi\sqrt{n}/h}\Big{(}\omega^{n}+e^{-\theta^{2}/2}\Big{)}\,d\theta\,,$

tends to 0 as $n\to\infty$. ∎

####  3.3.1. Local central limit theorem

Let $Y$ be a random variable such that $\mathbf{E}(Y)=0$ and
$\mathbf{E}(Y^{2})=1$. Let $Y_{1},Y_{2},\ldots$ be a sequence of independent
random variables all distributed like $Y$.

Assume that $Y$ is a lattice variable with gauge $h$ and such that
$\mathbf{P}(Y\in a+h\mathbb{Z})=1$, for some fixed $a\in\mathbb{R}$.

For each $n\geq 1$, we denote $S_{n}=(Y_{1}+\dots+Y_{n})/\sqrt{n}$. The
variable $S_{n}$ is likewise a lattice variable, with gauge $h/\sqrt{n}$.

Let $\mathcal{V}_{n}$ denote the set of possible values of $S_{n}$, i.e.,
$\mathcal{V}_{n}=\\{a\sqrt{n}+kh/\sqrt{n}:k\in\mathbb{Z}\\}$.

Applying Lemma D to the variable $S_{n}$ and using that
$\mathbf{E}(e^{\imath\theta S_{n}})=\mathbf{E}(e^{\imath\theta
Y/\sqrt{n}})^{n}$, we have that

$\mathbf{P}(S_{n}=x)=\frac{h}{2\pi\sqrt{n}}\int_{-\pi\sqrt{n}/h}^{\pi\sqrt{n}/h}\mathbf{E}(e^{\imath\theta
Y/\sqrt{n}})^{n}e^{-\imath\theta x}\,d\theta\,,\quad\mbox{for every
$x\in\mathcal{V}_{n}$}\,.$

Since

$\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-\theta^{2}/2}e^{-\imath\theta
x}\,d\theta\,,\quad\mbox{for every $x\in\mathbb{R}$}\,,$

we deduce that

$\displaystyle\sup_{x\in\mathcal{V}_{n}}\Big{|}\frac{\sqrt{n}}{h}\mathbf{P}(S_{n}=x)-\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}\Big{|}$
$\displaystyle\leq\frac{1}{2\pi}\int\limits_{|\theta|\leq\pi\sqrt{n}/h}\Big{|}\mathbf{E}(e^{\imath\theta
Y/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta$
$\displaystyle+\frac{1}{2\pi}\int\limits_{|\theta|>\pi\sqrt{n}/h}e^{-\theta^{2}/2}\,d\theta\,.$

Theorem F allows us to conclude then that

######  Corollary A (Local central limit theorem).

With the notations above,

$\lim_{n\to\infty}\sup_{x\in\mathcal{V}_{n}}\left|\frac{\sqrt{n}}{h}\mathbf{P}(S_{n}=x)-\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}\right|=0\,.$

###  3.4. Continuous families of lattice variables

Recall that a family $(Z_{s})_{s\in[a,b]}$ of random variables indexed in the
real interval $[a,b]$ is said to be continuous if it is continuous in
distribution in the sense that if a sequence $(s_{n})_{n\geq 1}$ with
$s_{n}\in[a,b]$ converges to $s\in[a,b]$ then $Z_{s_{n}}$ converges to $Z_{s}$
in distribution.

The family $(Z_{s})_{s\in[a,b]}$ is said to be bounded if
$\sup_{t\in[a,b]}\mathbf{E}\left(\left|Z_{t}\right|\right)<+\infty$.

######  Lemma 3.1.

If the family $(Z_{s})_{s\in[a,b]}$ is continuous and bounded, then the
function

$(s,\theta)\in[a,b]\times\mathbb{R}\longrightarrow\mathbf{E}(e^{\imath\theta
Z_{s}})\in\mathbb{D}\,$

is continuous.

We shall apply this lemma to the normalized version $\breve{X}_{t}$ of a
Khinchin family of $f\in\mathcal{K}$ when $t$ is restricted to a closed
interval $[a,b]\subset(0,R)$. Continuity is clear and regarding boundedness,
observe that
$\mathbf{E}(|\breve{X}_{t}|)\leq\mathbf{E}(|\breve{X}_{t}|^{2})^{1/2}=1$, for
$t\in[a,b]$. If $M_{f}<+\infty$ (and $R<+\infty$), and if, besides,
$\sigma(R)<\infty$ (see Section 2.1.3), then $(\breve{X}_{t})$ is continuous
and bounded for $t\in[a,R]$, (including $t=R$) for any $a\in(0,R)$.

The (un-normalized) Khinchin family $(X_{t})$ of any $f$ in $\mathcal{K}$ with
radius of convergence $R$ is continuous and bounded in any interval $[0,b]$
with $0<b<R$. If $M_{f}<+\infty$ (and $R<+\infty$) the family
$(X_{t})_{t\in[0,R]}$ extended to the closed interval $[0,R]$ is continuous
and bounded; observe that $\mathbf{E}(|X_{t}|)=\mathbf{E}(X_{t})\leq M_{f}$,
for $t\in[0,R]$.

###### Proof.

Take two sequences $(\theta_{n})_{n\geq 1}\subseteq\mathbb{R}$ and
$(s_{n})_{n\geq 1}\subseteq[a,b]$ which, respectively, converge to $\theta$
and $s$. First we write

(3.2) $\displaystyle\mathbf{E}(e^{i\theta_{n}Z_{s_{n}}})=\mathbf{E}(e^{i\theta
Z_{s_{n}}})+\mathbf{E}(e^{i\theta_{n}Z_{s_{n}}}-e^{i\theta Z_{s_{n}}}).$

By virtue of Levy’s convergence theorem, the first term in the right-hand side
of (3.2) converges, as $n\to\infty$, to $\mathbf{E}(e^{i\theta Z_{s}})$. We
have the bound

$\displaystyle\left|\mathbf{E}(e^{i\theta_{n}Z_{s_{n}}}-e^{i\theta
Z_{s_{n}}})\right|\leq\left|\theta_{n}-\theta\right|\mathbf{E}(|Z_{s_{n}}|),\quad\mbox{for
$n\geq 1$}\,,$

which follows from the numerical inequality: $|e^{ix}-e^{iy}|\leq|x-y|$, for
all $x,y\in\mathbb{R}$.

Now using that $(Z_{s})_{s\in[a,b]}$ is bounded, we conclude that

$(s,\theta)\in[a,b]\times\mathbb{R}\longrightarrow\mathbf{E}(e^{\imath\theta
Z_{s}})\in\mathbb{D}\,,$

is a continuous function. ∎

Let $(Z_{s})_{s\in[a,b]}$ be a continuous family of lattice variables. For
each $s\in[a,b]$, let $h(s)$ denote the gauge of $Z_{s}$. The gauge function
$h$ is upper semicontinuous:

$\limsup_{u\to s}h(u)\leq h(s)\,,\quad\mbox{for each $s\in[a,b]$}\,.$

In general, the gauge of a continuous family is not a continuous function. For
instance, for $s\in[0,1/2]$, let $Z_{s}$ take the values $0,1/2,1$ with
respective probabilities $1/2\leavevmode\nobreak\
{-}s,2s,1/2\leavevmode\nobreak\ {-}s$. The family $(Z_{s})$ is continuous, and
$h(s)=1/2$, for $s\neq 0$, but $h(0)=1$.

The next lemma is a counterpart of Lemma C for continuous and bounded
families; it follows by continuity from Lemma C and compactness.

######  Lemma 3.2.

Let $(Z_{s})_{s\in[a,b]}$ be a continuous and bounded family of lattice random
variables, with continuous gauge function $h(s)$.

Let $H>0$, be such that $h(s)\leq H$, for each $s\in[a,b]$.

Then,

$|\mathbf{E}(e^{\imath\theta Z_{s}})|<1,\quad\mbox{if $s\in[a,b]$ and
$0<|\theta|<2\pi/h(s)$}\,.$

In particular, for each $0<\tau<\pi/H$, there exists $\omega_{\tau}<1$ such
that

$|\mathbf{E}(e^{\imath\theta Z_{s}})|\leq\omega_{\tau},\quad\mbox{if
$s\in[a,b]$ and $\tau\leq|\theta|\leq\pi/h(s)$}\,.$

The next theorem is a uniform analogue of Theorem F for continuous families of
lattice variables:

######  Theorem 3.3.

Let $(Z_{s})_{s\in[a,b]}$ be a continuous family of lattice variables with
gauge function $h(s)$ continuous in $[a,b]$ and such that for each $s\in[a,b]$
we have that $\mathbf{E}(Z_{s})=0$, $\mathbf{E}(Z_{s}^{2})=1$ and
$\mathbf{E}(|Z_{s}|^{3})\leq\Gamma$, for a certain constant $\Gamma>0$.

Then,

$\lim_{n\to\infty}\sup_{s\in[a,b]}\int_{-\pi\sqrt{n}/h(s)}^{\pi\sqrt{n}/h(s)}\Big{|}\mathbf{E}(e^{\imath\theta
Z_{s}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

######  Remark 3.4.

The hypothesis $\sup_{s\in[a,b]}\mathbf{E}(|Z_{s}|^{3})<\infty$ may be
replaced by the assumption that
$\sup_{s\in[a,b]}\mathbf{E}(|Z_{s}|^{2+\delta})<\infty$, for some $\delta>0$,
or even by assuming that the family $(|Z_{s}|^{2})_{s\in[a,b]}$ is uniformly
integrable.

###### Proof.

First, we pay attention to the range of integration
$|\theta|\leq\sqrt{n}/(4\Gamma)$.

Theorem E gives us that

$\Big{|}\mathbf{E}\big{(}e^{\imath\theta Z_{s}/\sqrt{n}}\big{)}^{n}\Big{|}\leq
e^{-\theta^{2}/3},\quad\mbox{if $s\in[a,b]$ and
$|\theta|\leq\dfrac{\sqrt{n}}{4\Gamma}$}\,,$

while Theorem D gives us that

$\Big{|}\mathbf{E}\big{(}e^{\imath\theta
Z_{s}/\sqrt{n}}\big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\leq
C\frac{\Gamma}{\sqrt{n}}\,,\quad\mbox{if $s\in[a,b]$ and
$|\theta|\leq\dfrac{\sqrt{n}}{\Gamma}$}\,.$

Therefore, for any sequence $(s_{n})_{n\geq 1}$ drawn from $[a,b]$ we deduce,
from dominated convergence, that

$\lim_{n\to\infty}\int_{-\sqrt{n}/(4\Gamma)}^{\sqrt{n}/(4\Gamma)}\Big{|}\mathbf{E}(e^{\imath\theta
Z_{s_{n}}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,,$

and, therefore, that

$(\star)\quad\lim_{n\to\infty}\sup_{s\in[a,b]}\int_{-\sqrt{n}/(4\Gamma)}^{\sqrt{n}/(4\Gamma)}\Big{|}\mathbf{E}(e^{\imath\theta
Z_{s}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

Fix $0<J<H$ such that $J\leq h(s)\leq H$, for each $s\in[a,b]$. Take $\tau>0$
such that $\tau<1/(4\Gamma)$ and $\tau<\pi/H$.

Since, for every $s\in[a,b]$, we have that
$\mathbf{E}(|Z_{s}|)\leq\mathbf{E}(|Z_{s}|^{2})^{1/2}=1$, the family
$(Z_{s})_{s\in[a,b]}$ is bounded. Thus, Lemma 3.2 shows that there exists
$\omega_{\tau}<1$ such that

$|\mathbf{E}(e^{\imath\theta Z_{s}})|\leq\omega_{\tau},\quad\mbox{if
$s\in[a,b]$ and $\tau\leq|\theta|\leq\pi/h(s)$}\,,$

and so such that

$|\mathbf{E}(e^{\imath\theta
Z_{s}/\sqrt{n}})^{n}|\leq\omega_{\tau}^{n},\quad\mbox{if $s\in[a,b]$ and
$\tau\sqrt{n}\leq|\theta|\leq\pi\sqrt{n}/h(s)$}\,.$

Therefore, for each $s\in[a,b]$ we have that

$\int\limits_{\tau\sqrt{n}\leq|\theta|\leq\pi\sqrt{n}/h(s)}\Big{|}\mathbf{E}(e^{\imath\theta
Z_{s}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta\leq\omega_{\tau}^{n}\pi\sqrt{n}/{J}+\int\limits_{|\theta|\geq\tau\sqrt{n}}e^{-\theta^{2}/2}d\theta\,.$

Consequently,

$(\star\star)\quad\lim_{n\to\infty}\sup_{s\in[a,b]}\int\limits_{\tau\sqrt{n}\leq|\theta|\leq\pi\sqrt{n}/h(s)}\Big{|}\mathbf{E}(e^{\imath\theta
Z_{s}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

Combining $(\star)$ and $(\star\star)$ we obtain the result since
$\tau<1/(4\Gamma)$.∎

##  4\. Coefficients of large powers

We now turn to the study of the asymptotic behavior of coefficients of large
powers of functions $\psi$ in the class $\mathcal{K}$. For any
$\psi\in\mathcal{K}$, we study the coefficient

$\textsc{coeff}_{[k]}(\psi^{n}(z)),\quad\mbox{as $n\to\infty$}\,,$

under a variety of assumptions upon the joint behaviour of the index $k$ and
of the power $n$. At a later stage, we will consider the asymptotic behavior
of the coefficients of $h(z)\psi^{n}(z)$ where $h(z)\in\mathcal{K}$.

######  Remark 4.1.

The results which follow about large powers actually over the general
situation of $\psi$ with nonnegative coefficients, and not just
$\psi\in\mathcal{K}$.

If $\psi$ has nonnegative coefficients and it is not in $\mathcal{K}$, then
one of the two following possibilities occurs: 1) $\psi$ is a constant or a
monomial like $\psi(z)=bz^{m}$, for some $b\neq 0$ and integer $m\geq 1$,
which are trivial situations as coefficients and large powers are concerned,
2) $\psi(0)=0$ and $\psi$ has at least two nonnegative coefficients, in which
case, $\psi\in\mathcal{K}^{s}$ and for some integer $l\geq 1$ we have that
$\varphi(z)\triangleq\psi(z)/z^{l}\in\mathcal{K}$, and thus
$\textsc{coeff}_{[k]}(\psi^{n})=\textsc{coeff}_{[k-nl]}(\varphi^{n})$, for
$k\geq nl$.

Henceforth, $\psi(z)=\sum_{j=0}^{\infty}b_{j}z^{j}$ will denote a power series
in $\mathcal{K}$ with radius of convergence $R>0$, and from now on we let
$(Y_{t})_{t\in[0,R)}$ denote its Khinchin family.

We reserve $k$ to denote index of coefficient and $n$ to denote power of
$\psi$. In all asymptotic discussions below the power $n$ tends to $\infty$
(large powers), while the index $k$ could tend to $\infty$ in such a way that
either $k\asymp n$ (including the case when $k/n$ tends to a finite nonzero
limit) or $k=o(n)$ (including the possibility that $k$ could remain fixed) or
$n=o(k)$ (which would require the power series $\psi$ to be uniformly
Gaussian).

Some of these results are presented, for instance, in Gardy [11] in an elegant
and systematic manner, using saddle point approximation methods. Our approach
here is different: we will combine local central limit theorems for lattice
random variables, with some Hayman’s formulae which express the coefficients
of $\psi^{n}$ in terms of the characteristic function of the normalized
Khinchin family $\breve{Y_{t}}$.

If $\psi$ is the generating function of a combinatorial class, so that $b_{n}$
is the number of elements of size $n$ in the class, then
$\textsc{coeff}_{[k]}(\psi^{n}(z))$ is the number of lists of length $n$ and
total size $k$ formed with objects from the class.

If $\psi$ is the probability generating function of a random variable $Z$ so
that $\sum_{n=0}^{\infty}b_{n}=1$ and $Z_{1},Z_{2},\ldots$ are independent
copies of $Z$, then

$\textsc{coeff}_{[k]}(\psi^{n}(z))=\mathbf{P}(Z_{1}+\dots+Z_{n}=k)\,,$

###  4.1. Auxiliary function $\phi$

Recall the notation of Section 2.1.5:

$Q_{f}=\gcd\\{n\geq 1:a_{n}\neq 0\\}=\lim_{N\to\infty}\gcd\\{1\leq n\leq
N:a_{n}\neq 0\\}$

for any power series $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ in $\mathcal{K}$.

If $Q_{\psi}>1$, then $\psi(z)=\phi(z^{Q_{\psi}})$ for a certain auxiliary
power series $\phi\in\mathcal{K}$ which has radius of convergence
$R^{Q_{\psi}}$. Observe that $Q_{\phi}=1$ and that

$\textsc{coeff}_{[kQ_{\psi}]}(\psi^{n}(z))=\textsc{coeff}_{[k]}(\phi^{n}(z^{Q_{\psi}})),$

while for $q$ not a multiple of $Q_{\psi}$, we have that

$\textsc{coeff}_{[q]}(\psi^{n}(z))=0\,.$

Denote by $(Z_{t})_{t\in(0,R^{Q})}$ the Khinchin family associated to this
auxiliary function $\phi$, then, see Section 2.1.5,

$\displaystyle Y_{t}\stackrel{{\scriptstyle d}}{{=}}Q_{\psi}\cdot
Z_{t^{Q_{\psi}}},\quad\text{ for any }t\in(0,R).$

The mean and variance functions of $\psi$ and $\phi$ are related by

$m_{\psi}(t)=Q_{\psi}\cdot
m_{\phi}(t^{Q_{\psi}})\quad\mbox{and}\quad\sigma^{2}_{\psi}(t)=Q_{\psi}^{2}\cdot\sigma^{2}_{\phi}(t^{Q_{\psi}})\,.$

For each $t\in(0,R)$, the variable $Y_{t}$ is a lattice random variable with
gauge $Q_{\psi}$. Likewise, the normalized variable $\breve{Y}_{t}$ is a
lattice random variable with gauge $h(t)=Q_{\psi}/\sigma(t)$.

In subsequent analysis, we shall obtain asymptotic formulae for
$\textsc{coeff}_{[k]}(\psi(z)^{n})$ first under the assumption that
$Q_{\psi}=1$, and then in the general case when $Q_{\psi}\geq 1$, by
considering the auxiliary power series $\phi$, with $Q_{\phi}=1$, and
translating the results back to $\psi$.

###  4.2. Hayman’s identities and large powers

We may express the coefficients $\textsc{coeff}_{[k]}(\psi(z)^{n})$ in terms
of the characteristic function of the normalized variables
$(\breve{Y}_{t})_{t\in[0,R)}$ by Hayman’s identity as follows.

######  Lemma 4.2.

With the notations above,

(4.1) $\displaystyle\textsc{coeff}_{[k]}(\psi(z)^{n})$
$\displaystyle=\frac{1}{2\pi}\frac{\psi^{n}(t)}{t^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(t)}\int\limits_{|\theta|\leq\pi\sigma(t)\sqrt{n}}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{t}/\sqrt{n}}\big{)}^{n}e^{\imath(nm(t)-k)\theta/(\sigma(t)\sqrt{n})}d\theta\,,$

for any index $k\geq 1$, power $n\geq 1$ and for all $t\in(0,R)$.

###### Proof.

For $t\in(0,R)$, Cauchy’s integral formula gives that

$\displaystyle\textsc{coeff}_{[k]}(\psi(z)^{n})$
$\displaystyle=\frac{1}{2\pi\imath}\int_{|z|=t}\frac{\psi(z)^{n}}{z^{k+1}}dz=\frac{1}{2\pi}\frac{\psi(t)^{n}}{t^{k}}\int_{|\theta|<\pi}\frac{\psi(te^{\imath\theta})^{n}}{\psi(t)^{n}}e^{-\imath\theta
k}d\theta$
$\displaystyle=\frac{1}{2\pi}\frac{\psi(t)^{n}}{t^{k}}\int_{|\theta|<\pi}\mathbf{E}(e^{\imath\theta
Y_{t}})^{n}e^{-\imath\theta k}d\theta$
$\displaystyle=\frac{1}{2\pi}\frac{\psi^{n}(t)}{t^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(t)}\int_{|\theta|<\pi\sigma(t)\sqrt{n}}\mathbf{E}(e^{\imath\theta\breve{Y}_{t}/\sqrt{n}})^{n}e^{\imath(nm(t)-k)\theta/(\sigma(t)\sqrt{n})}d\theta.$

∎

Observe that the integral expression of (4.1) is greatly simplified if the
radius $t$ is such that $(\star)\ \ n\,m(t)=k$. Whenever possible we will
select and use that value of $t$ such that $(\star)$ holds exactly or at least
approximately.

In the next sections, we shall use the formula (4.1) to study the asymptotic
behaviour of $\textsc{coeff}_{[k]}(\psi^{n}(z))$, as $n\to\infty$, while
$k\asymp n$, $k/n\to 0$ or $k/n\to\infty$.

A minor variation of the proof of Lemma 4.2 above gives that if $H(z)$ is a
function holomorphic in $\mathbb{D}(0,R)$, then, with the notations above, we
have that

(4.2) $\displaystyle\textsc{coeff}_{[k]}(H(z)\psi(z)^{n})$
$\displaystyle=\frac{1}{2\pi}\frac{\psi^{n}(t)}{t^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(t)}\int\limits_{|\theta|\leq\pi\sigma(t)\sqrt{n}}H\big{(}te^{\imath\theta/(\sigma(t)\sqrt{n})}\big{)}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{t}/\sqrt{n}}\big{)}^{n}e^{\imath(nm(t)-k)\theta/(\sigma(t)\sqrt{n})}d\theta\,,$

for any index $k\geq 1$, power $n\geq 1$ and $t\in(0,R)$.

If $M_{\psi}<\infty$ (and $R<\infty$), then $\psi$ extends to be continuous in
$\mbox{\rm cl}(\mathbb{D}(0,R))$ and the Khinchin family $(Y_{t})_{t\in[0,R)}$
extends to the closed interval by adding $Y_{R}$ given by

$\mathbf{P}(Y_{R}=n)=b_{n}R^{n}/\psi(R)\,,\quad\mbox{for $n\geq 0$}\,.$

See Section 2.1.2 and Section 2.2 and the notations therein. In this case, we
have the following formula for the coefficients of $\psi$.

######  Lemma 4.3.

With the notations above, if $R<\infty$ and $M_{\psi}=m_{\psi}(R)<\infty$ and
$\sigma_{\psi}^{2}(R)<\infty$, then

(4.3) $\displaystyle\textsc{coeff}_{[k]}(\psi(z)^{n})$
$\displaystyle=\frac{1}{2\pi}\frac{\psi^{n}(R)}{R^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(R)}\int\limits_{|\theta|\leq\pi\sigma(R)\sqrt{n}}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{R}/\sqrt{n}}\big{)}^{n}e^{\imath(nm(R)-k)\theta/(\sigma(R)\sqrt{n})}d\theta\,,$

for any index $k\geq 1$ and power $n\geq 1$.

##  5\. Index $k$ and power $n$ are comparable

In this section we discuss the asymptotic behaviour of
$\textsc{coeff}_{[k]}(\psi^{n}(z))$ when the index $k$ and the power $n$ are
such that $k\asymp n$, as $n\to\infty$.

Along this section we assume that, for certain $A,B$, fixed, $0<A<B$, the
index $k$ and the power $n$ are such that $A\leq k/n\leq B$.

To deal with this case, we assume additionally that $M_{\psi}>B$. This being
the case, we denote $a=m^{-1}(A)$ and $b=m^{-1}(B)$.

Assume first that $Q_{\psi}=1$; a restriction to be lifted shortly in Section
5.1. The family of random variables $\breve{Y}_{t}$, where $t\in[a,b]$, is a
continuous family of lattice random variables with gauge function
$h(t)=1/\sigma(t)$; see Section 3.4. In particular,
$\max\limits_{t\in[a,b]}\mathbf{E}(|\breve{Y}_{t}|^{3})<+\infty$.

For each $n\geq 1$, take $\tau_{n}$ defined by $m(\tau_{n})=k/n$. This choice
is possible because $M_{\psi}>B$. For each $n\geq 1$, we have
$\tau_{n}\in[a,b]$.

Taking $t=\tau_{n}$ in Hayman’s identity (4.1), the integral term in there
simplifies to

$I_{n}\triangleq\int\limits_{|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}}\big{)}^{n}d\theta\,.$

Since $\sigma(\tau_{n})=1/h(\tau_{n})$, Theorem 3.3 gives that

(5.1)
$\lim_{n\to\infty}\int\limits_{|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\Big{|}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}}\big{)}^{n}-e^{-\theta^{2}/2}\Big{|}d\theta=0\,.$

In particular, since $\min_{t\in[a,b]}\sigma(t)>0$, we conclude that
$\lim_{n\to\infty}I_{n}=\sqrt{2\pi}$, and, thus, that

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{\sqrt{2\pi}}\frac{\psi^{n}(\tau_{n})}{\tau_{n}^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(\tau_{n})}\,,\quad\mbox{as
$n\to\infty$}.$

###  5.1. General $Q_{\psi}$

We now lift the assumption that $Q_{\psi}=1$. To simplify notation, write
$Q=Q_{\psi}\geq 1$. Consider the auxiliary function $\phi$, so that
$\psi(z)=\phi(z^{Q})$.

Recall that $m_{\psi}(t)=Qm_{\phi}(t^{Q})$, let $A^{\prime}=A/Q$ and
$B^{\prime}=B/Q$ and observe that $M_{\phi}=M_{\psi}/Q>B/Q=B^{\prime}$.

Define $\tau_{n}^{\prime}$ by $m_{\phi}(\tau_{n}^{\prime})=k^{\prime}/n$, for
$k^{\prime}$ such that $A^{\prime}<k^{\prime}/n<B^{\prime}$.

Since $Q_{\phi}=1$, we have that

$(\natural)\quad\textsc{coeff}_{[k^{\prime}]}(\phi^{n}(z))\sim\frac{1}{\sqrt{2\pi}}\frac{\phi(\tau_{n}^{\prime})^{n}}{(\tau_{n}^{\prime})^{k^{\prime}}}\frac{1}{\sqrt{n}}\frac{1}{\sigma_{\phi}(\tau_{n}^{\prime})}\,,\quad\mbox{as
$n\to\infty$}\,.$

Let $\tau_{n}=(\tau_{n}^{\prime})^{1/Q}$. Take $k=k^{\prime}Q$, observe that
$A<\dfrac{k}{n}<B$ and that

$m_{\psi}(\tau_{n})=Q\,m_{\phi}(\tau_{n}^{\prime})=\frac{k^{\prime}Q}{n}=\frac{k}{n}\,,$

and

$\sigma_{\psi}(\tau_{n})=Q\,\sigma_{\phi}(\tau_{n}^{\prime})\,.$

Since
$\textsc{coeff}_{[k]}(\psi^{n})=\textsc{coeff}_{[k^{\prime}]}(\phi^{n})$,
translating $(\natural)$ into terms of $\psi$, we get that

$\displaystyle\textsc{coeff}_{k}(\psi^{n}(z))$
$\displaystyle\sim\frac{Q_{\psi}}{\sqrt{2\pi}}\frac{\psi(\tau_{n})^{n}}{\tau_{n}^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma_{\psi}(\tau_{n})}\,,$
$\displaystyle\quad\mbox{as $n\to\infty$ while $A<k/n<B$ and $k$ is a multiple
of $Q$}\,.$

######  Theorem 5.1.

For $0<A<B<M_{\psi}$, we have that

$\textsc{coeff}_{k}(\psi^{n}(z))\sim\frac{Q_{\psi}}{\sqrt{2\pi}}\frac{\psi(\tau_{n})^{n}}{\tau_{n}^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma_{\psi}(\tau_{n})},$

as $n\to\infty$, while $A\leq{k}/{n}\leq B$ and $k$ is a multiple of
$Q_{\psi}$, and where $\tau_{n}$ is given uniquely by
$m_{\psi}(\tau_{n})={k}/{n}$.

###  5.2. With further information on $k/n$

We assume now that for some $L$ such that $0<L<+\infty$ and $L\leq M_{\psi}$,
we have that $k/n\to L$, as $n\to\infty$. With that information, we may obtain
sharper asymptotic formula for $\textsc{coeff}_{[k]}(\psi(z)^{n})$.

Assume that $k/n\to L$, where $0<L<+\infty$ and $L\leq M_{\psi}$, and, in
fact, in such a way that for some $\omega\in\mathbb{R}$

$\frac{k}{n}=L+\omega\frac{1}{\sqrt{n}}+o\Big{(}\frac{1}{\sqrt{n}}\Big{)}\,,\quad\mbox{as
$n\to\infty$}\,.$

or, equivalently, that

(5.2) $\lim_{n\to\infty}\frac{nL-k}{\sqrt{n}}=-\omega\,.$

$\bullet$ Suppose that $L<M_{\psi}$. Choose $\tau\in(0,R)$ such that
$m(\tau)=L$. Observe that now we have a fixed value of $\tau$, and no
$\tau_{n}$ varying with $n$.

Assume first that $Q_{\psi}=1$. If we choose $t=\tau$ in formula (4.1) and
invoke the Local Central Limit Theorem F we readily find that

$\displaystyle\textsc{coeff}_{[k]}(\psi(z)^{n})$
$\displaystyle\sim\frac{1}{2\pi}\frac{\psi^{n}(\tau)}{\tau^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(\tau)}\int_{\mathbb{R}}e^{\imath\omega\theta/\sigma(\tau)}e^{-\theta^{2}/2}\,d\theta$
$\displaystyle=\frac{e^{-\omega^{2}/(2\sigma^{2}(\tau))}}{\sqrt{2\pi}\,\sigma(\tau)}\,\frac{1}{\sqrt{n}}\frac{\psi^{n}(\tau)}{\tau^{k}}\,,\,\mbox{as
$n\to\infty$}.$

An argument much like the one in Section 5.1 allows us to deduce the general
case $Q_{\psi}\geq 1$ from the case $Q_{\psi}=1$.

######  Theorem 5.2.

If $L<M_{\psi}$ and

$\lim_{n\to\infty}\frac{nL-k}{\sqrt{n}}=-\omega\,,$

then

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim
Q_{\psi}\,\frac{e^{-\omega^{2}/(2\sigma^{2}(\tau))}}{\sqrt{2\pi}\,\sigma(\tau)}\,\frac{1}{\sqrt{n}}\frac{\psi^{n}(\tau)}{\tau^{k}}\,,\,\mbox{as
$n\to\infty$}.$

where $\tau$ is given by $m_{\psi}(\tau)=L$ and $k$ is a multiple of
$Q_{\psi}$.

The case where $k=n{-}1$ and thus $L=1$ (and $\omega=0$) shall be used in the
discussion of the Otter-Meir-Moon Theorem, Theorem G, on an asymptotic formula
for the coefficients of the solutions of Lagrange’s equation.

$\bullet$ Suppose $L=M_{\psi}$ and $R<+\infty$.  Observe that
$M_{\psi}<\infty$.

Now there is no $\tau\in(0,R)$ such that $m_{\psi}(\tau)=L$. But $\psi$
extends to be continuous in $\mbox{\rm cl}(\mathbb{D}(0,R))$ and the family
$(Y_{t})_{t\in[0,R)}$ may be completed with a variable $Y_{R}$ which has
$m_{\psi}(R)=M_{\psi}=L$, see Section 2.1.3. If further
$\sigma^{2}_{\psi}(R)=\mathbf{V}(Y_{R})<\infty$, which occurs if
$\sum_{n=0}^{\infty}n^{2}b_{n}R^{n}<\infty$, then we may use Lemma 4.3 instead
of formula (4.1).

By the same argument above, we then have

######  Theorem 5.3.

If $M_{\psi}=L$ and $R<+\infty$ and
$\sum_{n=0}^{\infty}n^{2}b_{n}R^{n}<\infty$. If

$\lim_{n\to\infty}\frac{nL-k}{\sqrt{n}}=-\omega\,,$

then

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim
Q_{\psi}\,\frac{e^{-\omega^{2}/(2\sigma^{2}(R))}}{\sqrt{2\pi}\,\sigma(R)}\frac{1}{\sqrt{n}}\frac{\psi^{n}(R)}{R^{k}}\,,$

where $k$ is a multiple of $Q_{\psi}$.

######  Remark 5.4.

In the remaining case, when $L=M_{f}$ and $R=+\infty$, then since
$M_{f}<\leavevmode\nobreak\ +\infty$, we have that $f$ is a polynomial of
degree $\text{deg}(f)=L=M_{f}$. In particular $L$ is an integer, and in the
extreme case when $k=Ln$, we have that
$\textsc{coeff}_{[k]}(\psi^{n})=b_{L}^{n}$, where $b_{L}$ is the $L$-th
coefficient $\psi(z)$.

####  5.2.1. Binomial coefficients.

As an illustration, we apply next Theorem 5.2 to binomial coefficients. We
take $\psi(z)=1+z$, which belongs to $\mathcal{K}$. In this case, we have

$m_{\psi}(t)=\frac{t}{1+t}\,,\quad\mbox{and}\quad\sigma_{\psi}^{2}(t)=\frac{t}{(1+t)^{2}}\,,\quad\mbox{for
$t\in(0,+\infty)$}\,.$

In particular, $M_{\psi}=1$.

Observe that

$\textsc{coeff}_{[k]}(\psi(z)^{n})=\binom{n}{k}\,,\quad\mbox{for $n\geq 1$ and
$k\geq 1$}\,.$

$\bullet$ Let $p\in(0,1)$. For each $n\geq 1/p$, let $k=\lfloor pn\rfloor$. We
have

$0\leq p-\frac{k}{n}\leq\frac{1}{n}\,.$

Thus we may apply Theorem 5.2 with $L=p$ and $\omega=0$, to deduce

$\binom{n}{\lfloor pn\rfloor}\sim\frac{1}{\sqrt{2\pi
np(1-p)}}\frac{1}{(1-p)^{n-k}p^{k}}\,,\quad\mbox{as $n\to\infty$}\,.$

If $p=1/2$, we have

$\binom{n}{\lfloor
n/2\rfloor}\sim\frac{2^{n{+}1}}{\sqrt{2\pi}\sqrt{n}}\,,\quad\mbox{as
$n\to\infty$}\,.$

$\bullet$ Let $p\in(0,1)$ and $\lambda\in\mathbb{R}$. For $n\geq N$, let
$k=\lfloor pn+\lambda\sqrt{n}\rfloor$, where $N$ is chosen so that
$pn+\lambda\sqrt{n}\geq 1$, for $n\geq N$. Then

$\frac{k}{n}=p+\frac{\lambda}{\sqrt{n}}+O\Big{(}\frac{1}{n}\Big{)}\,,\quad\mbox{as
$n\to\infty$}\,,$

and Theorem 5.2 with $L=p$ and $\omega=\lambda$, gives us that

$\binom{n}{\lfloor pn+\lambda\sqrt{n}\rfloor}\sim\frac{1}{\sqrt{2\pi
np(1-p)}}\frac{1}{(1-p)^{n-k}p^{k}}\,e^{-\lambda^{2}/(2p(1-p))}\,,\quad\mbox{as
$n\to\infty$}\,.$

For $p=1/2$ and $\lambda\in\mathbb{R}$ fixed, we have that

$\binom{n}{\lfloor
n/2+\lambda\sqrt{n}\rfloor}\sim\frac{2^{n{+}1}}{\sqrt{2\pi}\sqrt{n}}\,e^{-2\lambda^{2}}\,,\quad\mbox{as
$n\to\infty$}\,.$

####  5.2.2. Back to Local central limit theorem

Let us assume that $\psi$ is the probability generating function of a certain
random variable $X$ (with values in $\\{0,1,2\,\ldots\\}$) which has mean
$\mu$ and standard deviation $s$.

Assume further that $\psi$ has radius of convergence $R>1$ and that
$Q_{\psi}=1$.

We have $M_{\psi}=\lim_{t\uparrow R}m_{\psi}(t)>m_{\psi}(1)=\mu$. For
$\tau=1\in(0,R)$, we have $m_{\psi}(\tau)=\mu$ and $\sigma_{\psi}(\tau)=s$.

Let $Y$ be the random variable $Y=(X-\mu)/s$. This variable $Y$ is a lattice
random variable with gauge $h=1/s$, since $Q_{\psi}=1$. The variable $Y$ has
$\mathbf{E}(Y)=0$ and $\mathbf{E}(Y^{2})=1$.

For each $n\geq 1$, let $S_{n}$ denote the random variable

$S_{n}=\frac{Y_{1}+\dots+Y_{n}}{\sqrt{n}}=\frac{X_{1}+\dots+X_{n}}{s\sqrt{n}}-\Big{(}\frac{\mu}{s}\Big{)}\sqrt{n}\,,$

where $X_{1},X_{2},\ldots$ are independent copies of $X$ and
$Y_{j}=(X_{j}-\mu)/s$, for $j\geq 1$.

Let $\mathcal{V}_{n}$ denote the collection of possible values of $S_{n}$:

$\mathcal{V}_{n}=\Big{\\{}v(k)=k\frac{1}{s\sqrt{n}}-\Big{(}\frac{\mu}{s}\Big{)}\sqrt{n}:\,k\in\mathbb{Z}\Big{\\}}\,.$

For $x\in\mathcal{V}_{n}$, let $k_{x}=s\sqrt{n}x+\mu n$, so that $v(k_{x})=x$.
We have

$\frac{k_{x}}{n}=\mu+\frac{s}{\sqrt{n}}x\,.$

Moreover, with the notations of the hypothesis of Theorem 5.2, we have that
$L=\mu$ and $\omega=sx$. Now we use that with $\tau=1$, we have that
$m_{\psi}(1)=\mu(=L)$, $\psi(1)=1$, $\sigma_{\psi}(1)=s$ and
$\omega^{2}/\sigma^{2}_{\psi}(1)=x^{2}$, and appealing to Theorem 5.2 we
obtain that

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{\sqrt{2\pi}}\frac{1}{s\sqrt{n}}e^{-x^{2}/2}\,,\quad\mbox{as
$n\to\infty$}\,.$

Since

$\mathbf{P}(S_{n}=x)=\mathbf{P}(X_{1}+\dots+X_{n}=k_{x})=\textsc{coeff}_{[k_{x}]}(\psi(z)^{n})\,,$

we deduce that

$\lim_{n\to\infty}s\sqrt{n}\,\mathbf{P}(S_{n}=x)=\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}\,,\quad\mbox{as
$n\to\infty$}\,,$

which is “consistent” with the local Central Limit Theorem as stated in
Corollary A.

##  6\. Index $k$ is little ‘o’ of $n$

Now we take care of the case where $k/n\to 0$ as $k,n\to\infty$.

For the discussion of this case and throughout this section we assume that
$\psi^{\prime}(0)\neq 0$. This hypothesis implies that $Q_{\psi}=1$.

We will use formula (4.1) with an appropriate choice of $t$. To precise the
limit value of the integral term as $n\to\infty$, we shall appeal to the
following lemma, akin to Lemma 2 in [20].

######  Lemma 6.1.

Suppose that $\psi\in\mathcal{K}$ and $\psi^{\prime}(0)\neq 0$, then for each
$\theta_{0}\in(0,\pi)$, there exists $c>0$ and $r>0$ such that

$|\mathbf{E}(e^{i\theta\breve{Y}_{t}})|=\frac{|\psi(te^{\imath\theta})|}{\psi(t)}\leq
e^{-ct}\,,\quad\mbox{if $t\leq r$ and $\theta_{0}\leq|\theta|\leq\pi$}\,.$

###### Proof.

Without loss of generality we assume that $b_{0}=1$. We have

$|\psi(te^{\imath\theta})|^{2}=1+2b_{1}t\cos\theta+O(t^{2})\,,\quad\text{ as
}\quad t\downarrow 0,$

and

$|\psi(t)|^{2}=1+2b_{1}t+O(t^{2})\,,\quad\text{ as }\quad t\downarrow 0.$

Therefore,

$|\mathbf{E}(e^{i\theta\breve{Y}_{t}})|^{2}=\frac{|\psi(te^{\imath\theta})|^{2}}{\psi(t)^{2}}=1+2b_{1}t\,(\cos\theta-1)+O(t^{2})\,,\quad\text{
as }\quad t\downarrow 0.$

For $\theta_{0}\leq|\theta|\leq\pi$, we have $\cos\theta\leq 1-\delta$, for
certain $\delta>0$, whihc may depend upon $\theta_{0}$. Since, by hypothesis,
$b_{1}>0$, we then have, for certain $r\in(0,R)$, that

$|\mathbf{E}(e^{i\theta\breve{Y}_{t}})|^{2}=\frac{|\psi(te^{\imath\theta})|^{2}}{\psi(t)^{2}}\leq
1-2b_{1}t\delta+O(t^{2})\,,\quad\mbox{for all $t\leq r$}\,,$

Therefore, for $t\leq r$ we have that

$|\mathbf{E}(e^{i\theta\breve{Y}_{t}})|^{2}=\frac{|\psi(te^{\imath\theta})|^{2}}{\psi(t)^{2}}\leq
1-b_{1}t\delta\leq e^{-b_{1}\delta t}\,,$

as claimed.∎

For $n$ large enough, we have that $k/n<M_{\psi}$ and thus we may define
uniquely $\tau_{n}\in(0,R)$ such that $m(\tau_{n})=k/n$. Observe that
$\tau_{n}\to 0$, as $n\to\infty$.

We will use here the results from Section 2.3 about the moments of the
Khinchin family $(Y_{t})_{t\in[0,R)}$ when $t$ is close to $0$. Since
$b_{1}>0$, we have that

$(\natural)\quad\lim_{t\downarrow
0}t^{1/2}\,\mathbf{E}(|\breve{Y}_{t}|^{3})=\sqrt{\frac{b_{0}}{b_{1}}}\,.$

As $m(t)=\dfrac{b_{1}}{b_{0}}t+O(t^{2})$, we have that

$\tau_{n}\sim\dfrac{b_{0}}{b_{1}}\frac{k}{n}\,,\quad\mbox{as $n\to\infty$}\,.$

Besides,

$\sigma^{2}(\tau_{n})\sim\frac{k}{n}\,,\quad\mbox{as $n\to\infty$}\,.$

For $t=\tau_{n}$ the expression $(\natural)$ above tells us that

$(\flat)\qquad\frac{\mathbf{E}(|\breve{Y}_{\tau_{n}}|^{3})}{\sqrt{n}}\sim\sqrt{\frac{1}{{k}}}\,,\quad\mbox{as
$k\to\infty$}\,.$

If we choose $t=\tau_{n}$ in formula (4.1), the integral term simplifies to

$I_{n}\triangleq\int\limits_{|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}}\big{)}^{n}d\theta\,.$

Let us see that the integral $I_{n}$ converges to $\sqrt{2\pi}$ as
$n\to\infty$.

Theorem D and $(\flat)$ give that

$\lim_{n\to\infty}\mathbf{E}\big{(}e^{\imath\breve{Y}_{\tau_{n}}\theta/\sqrt{n}}\big{)}^{n}=e^{-\theta^{2}/2},\quad\mbox{for
all $\theta\in\mathbb{R}$}\,.$

Theorem E and $(\flat)$ give that there exists $N\geq 1$ such that if $n\geq
N$, then

$\big{|}\mathbf{E}\big{(}e^{\imath\breve{Y}_{\tau_{n}}\theta/\sqrt{n}}\big{)}^{n}\big{|}\leq
e^{-\theta^{2}/3}\,,\quad\mbox{for all $|\theta|\leq\sqrt{k}/5$}\,.$

By dominated convergence we have

$\lim_{n\to\infty}\int\limits_{|\theta|\leq\sqrt{k}/5}\big{|}\mathbf{E}\big{(}e^{\imath\breve{Y}_{\tau_{n}}\theta/\sqrt{n}}\big{)}^{n}-e^{-\theta^{2}/2}\big{|}\,d\theta=0\,.$

We apply lemma 6.1 with $\theta_{0}=1/10$ to ascertain that there exists
$N\geq 1$ and $C>0$ such that for all $n\geq N$ we have that

$|\mathbf{E}(e^{i\theta
Y_{\tau_{n}}})|=\frac{|\psi(\tau_{n}e^{\imath\theta})|}{\psi(\tau_{n})}\leq
e^{-c\tau_{n}}\,,\quad\mbox{for $\frac{1}{10}\leq|\theta|\leq\pi$}\,,$

and, therefore, that

$\displaystyle|\mathbf{E}(e^{i\theta
Y_{\tau_{n}}/(\sigma(\tau_{n})\sqrt{n})})|^{n}=\Big{|}\frac{\psi(\tau_{n}e^{\imath\theta/(\sigma(\tau_{n})\sqrt{n})})}{\psi(\tau_{n})}\Big{|}^{n}$
$\displaystyle\leq e^{-cn\tau_{n}}\,,$ $\displaystyle\quad\mbox{for
$\frac{1}{10}\sigma(\tau_{n})\sqrt{n}\leq|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}$}\,.$

Since for $n$ big enough we have that
$\frac{1}{10}\sigma(\tau_{n})\sqrt{n}\leq\sqrt{k}/5$, we may bound

$\int\limits_{\sqrt{k}/5\leq|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\big{|}\mathbf{E}(e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\big{|}d\theta\leq
e^{-cn\tau_{n}}(8\pi/3)\sqrt{k}+\int\limits_{|\theta|\geq\sqrt{k}/3}e^{-\theta^{2}/2}\,d\theta\,,$

for $n$ large enough.

The first summand of this bound converges to $0$, since
$n\tau_{n}\sim(a_{0}/a_{1})k$ and $k\to\infty$, while the second converges to
0 since $k\to\infty$. We conclude that

$\lim_{n\to\infty}\int\limits_{\sqrt{k}\leq|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\big{|}\mathbf{E}(e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\big{|}d\theta=0\,,$

and, therefore, that

(6.1)
$\lim_{n\to\infty}\int\limits_{|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\big{|}\mathbf{E}(e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\big{|}d\theta=0\,.$

In particular

$\lim_{n\to\infty}I_{n}=\lim_{n\to\infty}\int\limits_{|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\mathbf{E}(e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}})^{n}\,d\theta=\sqrt{2\pi}\,.$

In sum,

######  Theorem 6.2.

If $\psi^{\prime}(0)\neq 0$, then

$\displaystyle\textsc{coeff}_{[k]}(\psi(z)^{n})$
$\displaystyle\sim\frac{1}{\sqrt{2\pi}}\frac{\psi^{n}(\tau_{n})}{\tau_{n}^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(\tau_{n})}$
$\displaystyle\sim\frac{1}{\sqrt{2\pi}}\frac{\psi^{n}(\tau_{n})}{\tau_{n}^{k}}\frac{1}{\sqrt{k}}\,,\quad\mbox{as
$n\to\infty$}\,,$

where $k/n\to 0$ and $\tau_{n}$ is given by $m(\tau_{n})=k/n$ (for $n$ large
enough).

###  6.1. With further information on $k/n$

If some further information on how fast $k/n$ tends to 0 is available, then we
may express the asymptotic formula (6.2) directly in terms of $k$ and $n$ and
not on $\tau_{n}$.

We maintain the assumption that $\psi^{\prime}(0)>0$.

The function $m_{\psi}$ is holomorphic and injective near 0. Its inverse
$m_{\psi}^{-1}$ is actually the solution of Lagrange’s equation with data
$z/m_{\psi}(z)=\psi(z)/\psi^{\prime}(z)$. Besides, $\ln\psi$ is holomorphic
near 0, since as $\psi\in\mathcal{K}$, we have $\psi(0)>0$. Thus Lagrange’s
inversion formula gives the expansion

$\ln\psi\big{(}m_{\psi}^{-1}\big{)}(z)=\ln
b_{0}+\sum_{j=1}^{\infty}B_{j}z^{j},$

where

(6.2)
$B_{j}=\frac{1}{j}\textsc{coeff}_{[j-1]}\bigg{(}\Big{(}\frac{\psi^{\prime}}{\psi}\Big{)}\Big{(}\frac{\psi}{\psi^{\prime}}\Big{)}^{j}\bigg{)}=\frac{1}{j}\textsc{coeff}_{[j-1]}\bigg{(}\Big{(}\frac{\psi}{\psi^{\prime}}\Big{)}^{j-1}\bigg{)}\,,\quad\mbox{for
$j\geq 1$}\,.$

For each $j\geq 1$, the coefficient $B_{j}$ depends only on
$b_{0},b_{1},\ldots,b_{j}$.

Likewise,

$\ln\frac{m_{\psi}^{-1}(z)}{z}=\ln\Big{(}\frac{\psi}{\psi^{\prime}}\Big{)}\big{(}m_{\psi}^{-1}\big{)}(z)=\ln\frac{b_{0}}{b_{1}}+\sum_{j=1}^{\infty}C_{j}z^{j},$

where

(6.3)
$C_{j}=\frac{1}{j}\textsc{coeff}_{[j]}\Big{(}\frac{\psi}{\psi^{\prime}}\Big{)}^{j}=\frac{j{+}1}{j}B_{j{+}1}\,,\quad\mbox{for
$j\geq 1$}\,.$

Now, with the notations of the general discussion of the case $k/n\to 0$, we
have that

$\ln\psi(\tau_{n})^{n}=n\ln\psi(\tau_{n})=n\ln\psi\Big{(}m_{\psi}^{-1}\Big{(}\frac{k}{n}\Big{)}\Big{)}=n\ln
b_{0}+\sum_{j=1}^{\infty}B_{j}(k^{j}/n^{j{-}1})\,,$

and also that,

$\ln\tau_{n}^{k}=k\ln\tau_{n}=k\ln
m_{\psi}^{-1}\Big{(}\frac{k}{n}\Big{)}=k\ln\frac{b_{0}}{b_{1}}+k\ln\frac{k}{n}+\sum_{j=1}^{\infty}C_{j}\frac{k^{j{+}1}}{n^{j}}.$

Using formula (6.3), and that $B_{1}=1$, we may write

(6.4) $\displaystyle\ln\psi(\tau_{n})^{n}-\ln\tau_{n}^{k}=(n-k)\ln b_{0}+k\ln
b_{1}+k\ln\frac{n}{k}+k-\sum_{j=2}^{\infty}\frac{B_{j}}{j{-}1}\frac{k^{j}}{n^{j{-}1}}.$

Let $\lambda=\limsup_{n\to\infty}{\ln k}/{\ln n}$ and let $J$ the smallest
integer such that $J>\lambda/(1-\lambda)$; if $\lambda=1$, we set $J=+\infty$.
Thus

$\sum_{j=J+1}^{\infty}\frac{B_{j}}{j{-}1}\frac{k^{j}}{n^{j{-}1}}=nO\Big{(}\frac{k}{n}\Big{)}^{J{+}1}=o(1)\,,\quad\mbox{as
$n\to\infty$}\,,$

and then

$\frac{\psi(\tau_{n})^{n}}{\tau_{n}^{k}}\sim
b_{0}^{n{-}k}b_{1}^{k}\frac{n^{k}\,e^{k}}{k^{k}}\exp\Big{(}-\sum_{j=2}^{J}\frac{B_{j}}{j{-}1}\frac{k^{j}}{n^{j{-}1}}\Big{)}\,,\quad\mbox{as
$n\to\infty$}\,.$

Therefore

######  Theorem 6.3.

If $\psi^{\prime}(0)\neq 0$, and $\lambda$ and $J$ as above, then

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{\sqrt{2\pi}}b_{0}^{n{-}k}b_{1}^{k}\frac{n^{k}\,e^{k}}{k^{k}\sqrt{k}}\exp\Big{(}-\sum_{j=2}^{J}\frac{B_{j}}{j{-}1}\frac{k^{j}}{n^{j{-}1}}\Big{)}\,,\quad\mbox{as
$n\to\infty$}\,.$

If $\lambda<1/2$, then $J=1$, and then

(6.5) $\textsc{coeff}_{[k]}(\psi(z)^{n})\sim
b_{0}^{n{-}k}b_{1}^{k}\frac{n^{k}}{k!}\sim\textsc{coeff}_{[k]}((b_{0}+b_{1}z)^{n})\,,\quad\mbox{as
$n\to\infty$}\,.$

For the particular case where, $k=\lfloor\sqrt{n}\rfloor$, (with
$\lambda=1/2$, and $J=2$), we have that

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim
b_{0}^{n{-}k}b_{1}^{k}\frac{n^{k}}{k!}e^{-B_{2}}\,,\quad\mbox{as
$n\to\infty$}\,.$

where $B_{2}=(1/2)-(b_{2}b_{0})/b_{1}^{2}$.

###  6.2. Particular case: $k$ fixed, while $n\to\infty$

In this particular case, where $k$ remains fixed, while $n\to\infty$, the
analysis of the asymptotic behaviour of $\textsc{coeff}_{[k]}(\psi(z)^{n})$
just requires the multinomial theorem.

Observe that

$\textsc{coeff}_{[k]}(\psi(z)^{n})=\textsc{coeff}_{[k]}\Big{(}\sum_{j=0}^{k}b_{j}z^{j}\Big{)}^{n}\,.$

The multinomial theorem gives then

$\textsc{coeff}_{[k]}(\psi(z)^{n})=\sum\frac{n!}{j_{0}!\dots
j_{k}!}b_{0}^{j_{0}}\dots b_{k}^{j_{k}}\,,$

where the sum extends to all $k{+}1$-tuples $(j_{0},\ldots,j_{k})$ of
nonnegative integers such that

$\displaystyle j_{0}+j_{1}+\dots+j_{k}$ $\displaystyle=n,$ $\displaystyle
j_{1}+2j_{2}+\dots+kj_{k}$ $\displaystyle=k.$

For such a tuple, we have $n-k\leq j_{0}\leq n$ and $j_{i}\leq k$, for $1\leq
i\leq k$. We write $j_{0}=n-l$, with $0\leq l\leq k$ and classify by $l$:

$\textsc{coeff}_{[k]}(\psi(z)^{n})=\sum_{l=0}^{k}\binom{n}{l}b_{0}^{n{-}l}\,\underset{\triangleq
C_{l}}{\underbrace{\sum\begin{pmatrix}j_{1}+\dots+j_{k}\\\
j_{1},\dots,j_{k}\end{pmatrix}b_{1}^{j_{1}}\dots b_{k}^{j_{k}}}}\,.$

For each integer $l,0\leq l\leq k$, the interior sum $C_{l}$ extends to all
$k$-tuples with $j_{1}+\dots+j_{k}=l$ and $j_{1}+2j_{2}+\dots+kj_{k}=k$.
Observe that $C_{l}$ does not depend upon $n$. Therefore

$\textsc{coeff}_{[k]}(\psi(z)^{n})=\sum_{l=0}^{k}\binom{n}{l}b_{0}^{n{-}l}C_{l},$

is a polynomial in $n$ whose degree $\gamma$ is given by the largest $l\leq k$
such that $C_{l}\neq 0$, and

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{\gamma!}b_{0}^{n{-}\gamma}C_{\gamma}\,n^{\gamma}\,,\quad\mbox{as
$n\to\infty$}\,.$

$\bullet$ If $b_{1}\neq 0$, then $C_{k}\neq 0$, and $\gamma=k$. In fact, the
sum $C_{k}$ contains just one summand corresponding to $j_{1}=k$ and
$j_{2}=\ldots=j_{k}=0$, and thus $C_{k}=b_{1}^{k}$. We have in this case that

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{k!}b_{0}^{n{-}k}b_{1}^{k}\,n^{k}\sim\binom{n}{k}b_{0}^{n{-}k}b_{1}^{k}\,,\quad\mbox{as
$n\to\infty$}\,.$

In other terms and asymptotically speaking, the $k$-th coefficient of
$\psi(z)^{n}$ behaves as the $k$-th coefficient of $(b_{0}+b_{1}z)^{n}$, as
$n\to\infty$. Compare with the formula (6.5), where $k\to\infty$ with $n$, but
slowly.

$\bullet$ If $b_{1}=0$, then the nonzero summands of $C_{l}$ must have
$j_{1}=0$ and they must satisfy $j_{2}+\dots+j_{k}=l$ and
$2j_{2}+\dots+kj_{k}=k$. Thus, if $C_{l}$ is not zero, then $2l\leq k$, and
so, $\gamma\leq k/2$.

Assume further that $b_{2}\neq 0$. Distinguish: $k$ even and $k$ odd.

If $k=2q$, then $C_{q}$ has only one nonzero summand corresponding to
$j_{2}=q$, and all other $j$s equal to 0. Thus $C_{q}=b_{2}^{q}$,
$\gamma=q=k/2$, and

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{(k/2)!}b_{0}^{n{-}k/2}b_{2}^{k/2}\,n^{k/2}\,,\quad\mbox{as
$n\to\infty$}\,.$

If $k=2q+1$, then $\gamma\leq q$. The only nonzero summand of $C_{q}$ has
$j_{2}=q-1,j_{3}=1$ and all other $j$s equal to 0. Thus
$C_{q}=qb_{2}^{q-1}b_{3}$.

If $b_{3}\neq 0$, then $C_{q}\neq 0$, $\gamma=q$ and

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{((k{-}3)/2){\mbox{\large{!}}}}\,b_{0}^{n-(k-1)/2}b_{2}^{(k{-}3)/2}b_{3}\,n^{(k{-}1)/2}\,,\quad\mbox{as
$n\to\infty$}\,.$

If $b_{3}=0$, then $C_{q}=0$ and actually the degree $\gamma$ is at most
$k/3$.

$\bullet$ In general, if $b_{1}=\ldots=b_{m{-}1}=0$, and $b_{m}\neq 0$, then
for each $h,0\leq h\leq m{-}1$, the coefficients of index $k\equiv h\mod(m)$,
satisfy an asymptotic formula

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\Omega_{k,h}n^{\gamma_{k,h}}\,,\quad\mbox{as
$n\to\infty$}\,,$

where $\gamma_{k,h}\leq(k-h)/m$ and $\Omega_{k,h}>0$.

For $h=0$, i.e., for those $k$ which are multiples of $m$, we have
$\gamma_{k,0}=k/m$ and, actually, that

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{(k/m)!}b_{0}^{n{-}k/m}b_{m}^{k/m}\,n^{k/m}\,,\quad\mbox{as
$n\to\infty$}\,.$

In particular, if $Q=Q_{\psi}>1$ and $k$ is a multiple of $Q$, then

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{(k/Q)!}b_{0}^{n-k/Q}b_{Q}^{k/Q}\,n^{k/Q}\,,\quad\mbox{as
$n\to\infty$}\,.$

Observe the $b_{Q}>0$. If $Q$ is not a divisor of $k$, then, of course,
$\textsc{coeff}_{[k]}(\psi(z)^{n})=0$. for every $n\geq 1$.

##  7\. Power $n$ is little ‘o’ of index $k$

To obtain an asymptotic formula in this case where $k/n\to\infty$, as
$n\to\infty$, we require more from $\psi$, namely, that $\psi$ is uniformly
Gaussian. In an appendix of this paper, we describe at some length these so
called uniformly Gaussian power series.

Since uniformly Gaussian power series are strongly Gaussian, Hayman’s
asymptotic formula (2.6) is valid for $\psi$ and thus $Q_{\psi}=1$.

The assumption of uniform Gaussianity on $\psi$ is to be compared with those
of Gardy in [11, Theorems 5 and 6], see also [11, Section 6].

The exponential generating function of set partitions, $e^{e^{z}-1}$, and the
ordinary generating function of partitions of numbers,
$\prod_{j=1}^{\infty}1/(1-z^{j})$ are examples of uniformly Gaussian power
series.

Since $\psi$ is uniformly Gaussian, we have that $M_{\psi}=+\infty$, and thus
we may define (uniquely) $\tau_{n}\in(0,R)$ as being such that
$m(\tau_{n})=k/n$. Observe that $\tau_{n}\uparrow R$, as $n\to\infty$.

Now, formula (4.1) of Lemma 4.2, with $t=\tau_{n}$, gives us that

$\textsc{coeff}_{[k]}(\psi(z)^{n})=\frac{1}{2\pi}\frac{\psi^{n}(\tau_{n})}{\tau_{n}^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(\tau_{n})}\int\limits_{|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}}\big{)}^{n}d\theta\,.$

Since $\tau_{n}\to R$, uniform Gaussianity of $\psi$ implies that

$\lim_{n\to\infty}\int\limits_{|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}}\big{)}^{n}d\theta=\sqrt{2\pi}\,,$

and we conclude that

######  Theorem 7.1.

If $\psi$ is uniformly Gaussian, then

$\textsc{coeff}_{[k]}(\psi(z)^{n})\sim\frac{1}{\sqrt{2\pi}}\frac{\psi^{n}(\tau_{n})}{\tau_{n}^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(\tau_{n})}\,,\quad\mbox{as
$n\to\infty$}\,,$

where $k/n\to\infty$ as $n\to\infty$ and $\tau_{n}$ is such that
$m_{\psi}(\tau_{n})=k/n$.

##  8\. Coefficients of $h(z)\psi(z)^{n}$

We assume here from the start that $Q_{\psi}=1$.

Let $h(z)=\sum_{j=0}^{\infty}c_{j}z^{j}$ be a power series in $\mathcal{K}$.
We assume throughout that the radius of convergence of $h$ is at least
$R_{\psi}$, the radius of convergence of $\psi$.

We treat jointly the cases in which $k\asymp n$ and $k=o(n)$, with
$k\to\infty$, by adjusting the previous arguments (whose notations we will use
liberally) in which $h\equiv 1$.

Assume the hypothesis of Theorem 5.1, when $k\asymp n$, and of Theorem 6.2,
when $k=o(n)$ and $k\to\infty$.

The $k$-th coefficient of $h(z)\psi(z)^{n}$ is given, see (4.2), by

$\frac{1}{2\pi}\frac{\psi^{n}(\tau_{n})}{\tau_{n}^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(\tau_{n})}\int\limits_{|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}}h(\tau_{n}e^{\imath\theta/(\sigma(\tau_{n})\sqrt{n})})\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}}\big{)}^{n}d\theta\,,$

where we are taking $\tau_{n}$ such that $m(\tau_{n})=k/n$.

Denote by $I_{n}$ the integral on the right hand-side of the previous
expressions.

For $0\leq s\leq r<R$ and $\phi\in\mathbb{R}$ we have

$\big{|}h(se^{\imath\phi})-h(s)\big{|}\leq\max\limits_{|z|\leq
r}|h^{\prime}(z)|\,s\,|\phi|\,.$

For $k\asymp n$, we have $\tau_{n}\leq m^{-1}(B)<R$, and for $k=o(n)$ we have
$\tau_{n}\to 0$, so, in both cases, there exists a constant $K>0$ such that

$(\nabla)\qquad\Big{|}h(\tau_{n}e^{\imath\theta/(\sigma(\tau_{n})\sqrt{n})})-h(\tau_{n})\Big{|}\leq
K\frac{|\theta|}{\sigma(\tau_{n})\sqrt{n}}\,.$

Denote by $\mathcal{D}_{n}$ the interval
$\\{\theta\in\mathbb{R}:|\theta|\leq\pi\sigma(\tau_{n})\sqrt{n}\\}$.

Using the bound $(\nabla)$ and writing

$\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{t}/\sqrt{n}}\big{)}^{n}=\Big{(}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{t}/\sqrt{n}}\big{)}^{n}-e^{-\theta^{2}/2}\Big{)}+e^{-\theta^{2}/2}\,,$

we conclude that

$\displaystyle I_{n}$
$\displaystyle=h(\tau_{n})\int\limits_{\mathcal{D}_{n}}e^{-\theta^{2}/2}d\theta+h(\tau_{n})\int\limits_{\mathcal{D}_{n}}\Big{(}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}}\big{)}^{n}-e^{-\theta^{2}/2}\Big{)}d\theta$
$\displaystyle+O\Big{(}\int\limits_{\mathcal{D}_{n}}\Big{|}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{\tau_{n}}/\sqrt{n}}\big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\frac{|\theta|}{\sigma(\tau_{n})\sqrt{n}}d\theta\Big{)}+O\Big{(}\int\limits_{\mathcal{D}_{n}}e^{-\theta^{2}/2}\frac{|\theta|}{\sigma(\tau_{n})\sqrt{n}}d\theta\Big{)}\,.$

For $\theta\in\mathcal{D}_{n}$, we have that
$\dfrac{|\theta|}{\sigma(\tau_{n})\sqrt{n}}\leq\pi$, and thus we see that the
third term in the sum tends to 0, by virtue of (5.1) (if $k\asymp n$) or (6.1)
(if $k=o(n)$).

The second term in the sum tends to 0, since $h(\tau_{n})$ is bounded and the
integral converges to 0, by (5.1) and (6.1), respectively.

The fourth term in the sum tends to 0 since
$\sigma(\tau_{n})\sqrt{n}\asymp\sqrt{k}$ and
$\int\limits_{\mathbb{R}}e^{-\theta^{2}/2}|\theta|d\theta=2$.

The first term in the sum is $h(\tau_{n})\sqrt{2\pi}+o(1)$, since
$h(\tau_{n})$ is bounded, and so

$I_{n}=h(\tau_{n})\sqrt{2\pi}+o(1)\,,\quad\mbox{as $n\to\infty$}\,,$

but, given that $h(\tau_{n})$ is bounded from below by $h(0)$ and recalling
that $h\in\mathcal{K}$, we conclude that

$I_{n}=h(\tau_{n})\sqrt{2\pi}(1+o(1))\,,\quad\mbox{as $n\to\infty$}\,.$

For the case $k\asymp n$ we obtain

$\textsc{coeff}_{[k]}(h(z)\psi(z)^{n})\sim\frac{1}{\sqrt{2\pi}}\frac{h(\tau_{n})\psi^{n}(\tau_{n})}{\tau_{n}^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(\tau_{n})}\,,\quad\mbox{as
$n\to\infty$}\,.$

For the case $k=o(n)$ and $k\to\infty$, we find that

$\textsc{coeff}_{[k]}(h(z)\psi(z)^{n})\sim\frac{h(0)}{\sqrt{2\pi}}\frac{\psi^{n}(\tau_{n})}{\tau_{n}^{k}}\frac{1}{\sqrt{k}}\,,\quad\mbox{as
$n\to\infty$}\,.$

Observe that, in this last case, $h(\tau_{n})\to h(0)$ as $n\to\infty$.

If more information is available on how $k/n$ tends to 0, as discussed in
Section 6.1, we could apply the argument and the conclusions obtained there to
refine the above asymptotic formula.

Suppose that $k/n\rightarrow L>0$, as $n\rightarrow\infty$ and that equation
(5.2) is satisfied. Assume that $L<M_{\psi}$. Then for $\tau=m^{-1}(L)$, we
deduce, using the Hayman’s formula (4.2) and arguing as in the case $h\equiv
1$, that

(8.1)
$\textsc{coeff}_{[k]}(h(z)\psi(z)^{n})\sim\frac{1}{\sqrt{2\pi}}\frac{h(\tau)\psi^{n}(\tau)}{\tau^{k}}\frac{1}{\sqrt{n}}\frac{1}{\sigma(\tau)}\,,\quad\mbox{as
$n\to\infty$}\,.$

Recall that the power series $h$ has radius of convergence $R_{\psi}$ and that
$\tau\in(0,R_{\psi})$.

To conclude, we consider the case in which $k$ is fixed and $n\to\infty$.
Denote $Q=Q_{\psi}=\\{n\geq 1:b_{n}>0\\}$. Then

$\displaystyle\textsc{coeff}_{[k]}(h(z)\psi(z)^{n})$
$\displaystyle=\sum_{j=0}^{k}c_{j}\textsc{coeff}_{[k]}(z^{j}\psi(z)^{n})$
$\displaystyle=\sum_{j=0}^{k}c_{j}\textsc{coeff}_{[k{-}j]}(\psi(z)^{n}).$

The only $(k{-}j)$-th coefficients of $\psi(z)^{n}$ which are nonzero are the
coefficients with indices $j\equiv k,\mod Q$. Appealing to the case $h\equiv
1$, we have that

$\displaystyle\textsc{coeff}_{[k]}(h(z)\psi(z)^{n})$
$\displaystyle=\sum_{\begin{subarray}{c}0\leq j\leq k\\\ j\equiv
k\,,\text{mod}Q\end{subarray}}c_{j}\textsc{coeff}_{[k{-}j]}(\psi(z)^{n})$
$\displaystyle=\sum_{\begin{subarray}{c}0\leq j\leq k\\\ j\equiv
k\,,\text{mod}Q\end{subarray}}c_{j}\frac{1}{((k{{-}}j)/Q)!}b_{0}^{n{-}(k{-}j)/Q}\,b_{Q}^{(k{-}j)/Q}\,n^{(k{-}j)/Q}\,(1+o(1))\,.$

Denoting $j_{0}=\displaystyle\min\\{0\leq j\leq k:j\equiv k,\mod Q\text{ and
}c_{j}\neq 0\\}$, we have that

$\displaystyle\textsc{coeff}_{[k]}(h(z)\psi(z)^{n})$ $\displaystyle\sim
c_{j_{0}}\frac{1}{((k{-}j_{0})/Q)!}b_{0}^{n{-}(k{-}j_{0})/Q}\,b_{Q}^{(k{-}j_{0})/Q}\,n^{(k{-}j_{0})/Q},\quad\mbox{as
$n\to\infty$}\,.$

In particular, if $h(0)=c_{0}\neq 0$ and if $Q$ is a divisor of $k$, then

$\displaystyle\textsc{coeff}_{[k]}(h(z)\psi(z)^{n})$ $\displaystyle\sim
c_{0}\frac{1}{(k/Q)!}\,b_{0}^{n{-}k/Q}\,b_{Q}^{k/Q}\,n^{k/Q},\quad\mbox{as
$n\to\infty$}\,.$

Observe that in this case, $j_{0}=0$.

##  9\. Coefficients of solutions of Lagrange’s equation

Next we shall apply the asymptotic results about large powers of Section 4 to
obtain asymptotic formulae for the coefficients of solutions of Lagrange’s
equation when the data $\psi$ is in $\mathcal{K}$.

###  9.1. Lagrange’s equation

We start with a function $\psi(z)$ holomorphic in a neighborhood of $z=0$,
with radius of convergence $R>0$ and such that $\psi(0)\neq 0$. Consider
Lagrange’s equation:

$({\dagger})\qquad g(w)=w\psi(g(w))\,.$

The solution $g(w)$ of Lagrange’s equation with data $\psi(z)$ is a
holomorphic function $g(w)=\sum_{n=1}^{\infty}A_{n}w^{n}$ which satisfies
$({\dagger})$ in a neighborhood of $w=0$.

Assume that the data $\psi(z)$ of Lagrange’s equation has power series
expansion

$\psi(z)=\sum_{n=0}^{\infty}b_{n}z^{n},\quad\mbox{for all
$z\in\mathbb{D}(0,R)$}.$

The holomorphic function $g(w)$, solution of Lagrange’s equation with data
$\psi(z)$, is unique. In fact, the coefficients $A_{n}$ of the Taylor
expansion of $g(w)$ around $w=0$ are given by Lagrange’s inversion formula:

$A_{n}=\frac{1}{n}\textsc{coeff}_{[n{-}1]}(\psi(z)^{n})\,,\quad\mbox{for all
$n\geq 1$}\,.$

For $n=0$, we have $A_{0}=g(0)=0$.

This formula is exact for each coefficient $A_{n}$ of $g$ such that $n\geq 1$.

In a more general setting, consider the coefficients of $H(g(z))$, where $g$
is the solution of Lagrange’s equation with data $\psi$ and $H$ is a
holomorphic function with nonnegative coefficients around $z=0$, then

(9.1)
$\displaystyle\textsc{coeff}_{[n]}(H(g(z)))=\frac{1}{n}\textsc{coeff}_{[n{-}1]}(H^{\prime}(z)\psi(z)^{n}),\quad\mbox{for
all $n\geq 1$}\,,$

and $\textsc{coeff}_{[0]}(H(g(z)))=H(0)$.

A particular instance of this formula, that will be of interest later, is that
given by $H(z)=z^{q}$ for an integer $q\geq 1$. In this case we have

(9.2)
$\displaystyle\textsc{coeff}_{[n]}(g(z)^{q})=\frac{q}{n}\textsc{coeff}_{[n{-}q]}(\psi(z)^{n}),\quad\mbox{for
all $n\geq 1$}\,,$

and $\textsc{coeff}_{[0]}(g(z)^{q})=H(0)=0.$

If $Q_{\psi}>1$, then the only nonzero coefficients $b_{n}$ are those where
the index $n$ is a multiple of $Q_{\psi}$.Therefore, for the solution $g(w)$
of Lagrange’s equation with data $\psi$, the only nonzero coefficients $A_{n}$
are those where the index $n$ verifies that $n-1$ is a multiple of $Q_{\psi}$.

If $\psi(0)=0$, then the only solution of Lagrange’s equation is $g\equiv 0$,
but, recall, nonetheless that $\psi\in\mathcal{K}$ implies that $\psi(0)>0$.

###  9.2. The Otter-Meir-Moon theorem and some extensions

The Otter-Meir-Moon Theorem, Theorem G below, comes from [24, Theorem 4] and
[19, Theorem 3.1]. It gives an asymptotic formula for the coefficients $A_{n}$
of $g$, when $n{-}1$ is a multiple of $Q_{\psi}$ and $n\to\infty$, using
minimal (but crucial) information about the function $\psi$.

We distinguish three cases, according as $M_{\psi}$ is $>1,=1$ or $<1$.

$\blacklozenge$ $M_{\psi}>1$. This is the original assumption of both Otter
and Meir-Moon; it means that there is a unique $\tau\in(0,R)$ such that
$m_{\psi}(\tau)=1$.

######  Theorem G (Otter-Meir-Moon theorem).

Let $\psi(z)$ be a power series in $\mathcal{K}$ with radius of convergence
$R>0$ and let $\psi(z)=\sum_{n=0}^{\infty}b_{n}z^{n}$ be its power series
expansion.

Assume that $M_{\psi}>1$ and let $\tau\in(0,R)$ be given by
$m_{\psi}(\tau)=1$.

Then the coefficients $A_{n}$ of the solution $g(w)$ of Lagrange’s equation
verify that

  * •

if $n\not\equiv 1\mod{Q_{\psi}}$, then $A_{n}=0$,

  * •

for the indices $n$ such that $n\equiv 1\mod{Q_{\psi}}$ we have the asymptotic
formula

(9.3)
$A_{n}\sim\frac{Q_{\psi}}{\sqrt{2\pi}}\,\frac{\tau}{\sigma_{\psi}(\tau)}\frac{1}{n^{3/2}}\,\Big{(}\frac{\psi(\tau)}{\tau}\Big{)}^{n}\,,\quad\mbox{as
$n\to\infty$}\,.$

Theorem G of Otter and Meir-Moon follows readily from Theorem 5.2 with $L=1$
and $\omega=0$, since $k=n-1$.

Given that

$\sigma^{2}_{\psi}(t)=m_{\psi}(t)(1-m_{\psi}(t))+t^{2}\frac{\psi^{\prime\prime}(t)}{\psi(t)}\,,\quad\mbox{for
all $t\in[0,R)$}\,,$

we have

(9.4)
$\displaystyle\frac{\psi^{\prime\prime}(\tau)}{\psi(\tau)}=\frac{\sigma^{2}_{\psi}(\tau)}{\tau^{2}}\,.$

We thus may rewrite the conclusion of Theorem G (in Laplace’s method or saddle
point approximation style) as

$A_{n}\sim
Q_{\psi}\,\sqrt{\frac{\psi(\tau)}{2\pi\psi^{\prime\prime}(\tau)}}\,\frac{1}{n^{3/2}}\Big{(}\frac{\psi(\tau)}{\tau}\Big{)}^{n}\,,\quad\mbox{when
$n{-}1$ is a multiple of $Q_{\psi}$ and $n\to\infty$}\,.$

$\blacklozenge$ $M_{\psi}=1$ (and $R<+\infty$). For the case $M_{\psi}=1$,
there is an Otter-Meir-Moon like theorem under a certain non-degeneracy
condition on $\psi$. Recall the notations of Section 2.1.2.

######  Theorem 9.1.

Let $\psi(z)$ be a power series in $\mathcal{K}$ with radius of convergence
$R>0$ and let $\psi(z)=\sum_{n=0}^{\infty}b_{n}z^{n}$ be its power series
expansion.

Assume that $M_{\psi}=1$ and that $R<\infty$. And besides that
$\sum_{n=0}^{\infty}n^{2}b_{n}R^{n}<+\infty$ or equivalently
$\psi^{\prime\prime}(R)<+\infty$.

Then the coefficients $A_{n}$ of the solution $g(w)$ of Lagrange’s equation
verify that

  * •

if $n\not\equiv 1\mod{Q_{\psi}}$, then $A_{n}=0$,

  * •

for the indices $n$ such that $n\equiv 1\mod{Q_{\psi}}$ we have the asymptotic
formula

(9.5)
$A_{n}\sim\frac{Q_{\psi}}{\sqrt{2\pi}}\,\frac{R}{\sigma_{\psi}(R)}\frac{1}{n^{3/2}}\,\Big{(}\frac{\psi(R)}{R}\Big{)}^{n}\,,\quad\mbox{as
$n\to\infty$}\,.$

The power series $\psi$ extends to be continuous in $\mbox{\rm
cl}(\mathbb{D}(0,R))$ and the Khinchin family $(Y_{t})_{t\in[0,R)}$, extends
(continuously in distribution) to include a variable $Y_{R}$ with mean 1 and
variance $\sigma_{\psi}^{2}(R)=\lim_{t\uparrow R}\sigma_{\psi}^{2}(t)$ which
is finite because by hypothesis $\sum_{n=0}^{\infty}n^{2}b_{n}R^{n}<+\infty$.
See Section 2.1.2.

Theorem 9.1 follows readily from Theorem 5.3 with $L=1$ and $\omega=0$, since
$k=n-1$.

This limit case of the Otter-Meir-Moon Theorem appears as Remark 3.7 in [9,
Chapter 3]; the proof suggested there, of a different nature than the one
above, comes from the Appendix of [17]. See also [22, Section 2.3.1]. In there
you may find the same asymptotic result under the assumption that $\psi$ is a
probability generating function satisfying $\psi^{(4)}(R)=\lim_{t\uparrow
R}\psi^{(4)}(t)<\infty$.

Observe that, in particular, it follows from Theorem G that when $M_{\psi}>1$
and with $m_{\psi}(\tau)=1$ the radius of convergence of the solution $g$ is
$\tau/\psi(\tau)$. When $M_{\psi}=1$ (and $R<+\infty$), the radius of
convergence of $g$ is $R/\psi(R)$.

If $M_{\psi}=1$ and $R=+\infty$, then $\psi$ is a polynomial of degree 1. Thus
$\psi(z)=a+bz$, with both $a,b>0$. In this case the solution $g(z)$ of
Lagrange equation with data $\psi$ is simply

$g(z)=\frac{az}{1-bz}=\sum_{n=1}^{\infty}(ab^{n{-}1})z^{n}\,.$

and thus

$A_{n}=ab^{n{-}1}\,,\quad\mbox{para $n\geq 1$}\,.$

$\blacklozenge$ $M_{\psi}<1$. In this case, we necessarily have that
$R<\infty$, since $R=\infty$ and $M_{\psi}<1$ would imply that $\psi$ is a
constant, which is not allowed. See Lemma A.

We assume as above that $\sum_{n=0}^{\infty}n^{2}b_{n}R^{n}<\infty$. As in the
preceding case $M_{\psi}=1$ above, the power series $\psi$ is continuous in
$\mbox{\rm cl}(\mathbb{D}(0,R))$ and the Khinchin family extends to include
continuously a random variable $Y_{R}$ with mean $M_{\psi}<1$ and variance
$\lim_{t\uparrow R}\sigma_{\psi}^{2}(t)=\sigma_{\psi}^{2}(R)<\infty$. Assume
for simplicity that $Q_{\psi}=1$

In this case, we do not obtain a proper asymptotic formula, just that

(9.6) $\lim_{n\to\infty}A_{n}\frac{R^{n-1}}{\psi^{n}(R)}n^{3/2}=0\,.$

For $A_{n}$ we have that

$A_{n}=\frac{1}{n}\textsc{coeff}_{[n-1]}(\psi^{n})=\frac{1}{2\pi}\frac{\psi^{n}(R)}{R^{n-1}}\frac{1}{n^{3/2}}\frac{1}{\sigma_{\psi}(R)}\,I_{n}$

where

$I_{n}\triangleq\int\limits_{|\theta|\leq\pi\sigma(R)\sqrt{n}}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{R}/\sqrt{n}}\big{)}^{n}e^{\imath\theta\alpha(n)}d\theta\,,\quad\mbox{
for $n\geq 1$}\,,$

and

$\alpha(n)=\frac{nM_{\psi}-(n-1)}{\sigma(R)\sqrt{n}}\,,\quad\mbox{ for $n\geq
1$}\,.$

We shall now verify that $I_{n}$ tends towards 0 as $n\to\infty$. We split
$I_{n}$ as

$\displaystyle I_{n}$
$\displaystyle=\int\limits_{|\theta|\leq\pi\sigma(R)\sqrt{n}}\Big{(}\mathbf{E}\big{(}e^{\imath\theta\breve{Y}_{R}/\sqrt{n}}\big{)}^{n}-e^{-\theta^{2}/2}\Big{)}e^{\imath\theta\alpha(n)}d\theta$
$\displaystyle+\int_{\mathbb{R}}e^{-\theta^{2}/2}e^{\imath\theta\alpha(n)}d\theta$
$\displaystyle-\int\limits_{|\theta|>\pi\sigma(R)\sqrt{n}}e^{-\theta^{2}/2}e^{\imath\theta\alpha(n)}d\theta\,.$

The third summand obviously tends to 0, while the first summand tends to 0 on
account of Theorem F, the integral form of the Local Central Limit Theorem.
The second summand may be written as

$\int_{\mathbb{R}}e^{-\theta^{2}/2}e^{\imath\theta\alpha(n)}d\theta=e^{-\alpha(n)^{2}/2}$

to observe that since $M_{\psi}<1$, we have that
$\lim_{n\to\infty}\alpha(n)=-\infty$, and thus that this second summand too
converges to 0, as $n\to\infty$.

Minami in [22, Section 2.3.2] obtains a faster rate of convergence (higher
power of $n$) in (9.6) under the stronger assumption that $\lim_{t\uparrow
R}\psi^{(k)}(t)<\infty$ for some $k\geq 3$.

###  9.3. Coefficients of powers of solutions of Lagrange’s equation

We assume $Q_{\psi}=1$. We are interested now, see Meir-Moon [20], in
asymptotic formulas for the coefficients of powers of the solution of
Lagrange’s equation.

For $q\geq 1$ and $n\geq 1$, the $n$-th coefficient of $g(w)^{q}$, which we
will denote by $B_{n,q}$, is given by the exact formula

$B_{n,q}=\frac{q}{n}\textsc{coeff}_{[n-q]}(\psi(z)^{n})\,.$

$\bullet$ For $q\geq 1$ fixed, Theorem 5.2 gives, under the assumption
$M_{\psi}>1$ and with $\tau$ given by $m_{\psi}(\tau)=1$, that

$B_{n,q}\sim\frac{q}{\sqrt{2\pi}}\frac{\tau^{q}}{\sigma_{\psi}(\tau)}\frac{1}{n^{3/2}}\Big{(}\frac{\psi(\tau)}{\tau}\Big{)}^{n}\,,\quad\mbox{as
$n\to\infty$ }.$

$\bullet$ If $q$ and $n$ are related by $q=\alpha n+\beta\sqrt{n}+o(\sqrt{n})$
as $n\to\infty$, where $\alpha\in[0,1)$ and $\beta\in\mathbb{R}$, then for
$\tau_{\alpha}$ given by $m(\tau_{\alpha})=1-\alpha$, we obtain, analogously
to (8.1), that

$B_{n,q}\sim
e^{-\beta^{2}/(2\sigma_{\psi}^{2}(\tau_{\alpha}))}\frac{q}{\sqrt{2\pi}}\frac{\tau_{\alpha}^{q}}{\sigma_{\psi}(\tau_{\alpha})}\frac{1}{n^{3/2}}\Big{(}\frac{\psi(\tau_{\alpha})}{\tau_{\alpha}}\Big{)}^{n}\,,\quad\mbox{as
$n\to\infty$}\,.$

Observe that

$\sigma_{\psi}(\tau_{\alpha})=\alpha(1-\alpha)+\tau^{2}_{\alpha}\frac{\psi^{\prime\prime}(\tau_{\alpha})}{\psi(\tau_{\alpha})}\,.$

####  9.3.1. Coefficients of functions of solutions of Lagrange’s equation

In a more general setting, let $H$ be a power series with nonnegative
coefficients.

Assume further that the radius of convergence of $H$ is at least the radius of
convergence $R$ of $\psi$ and also that $M_{\psi}>1$. Let $\tau$ be such that
$m_{\psi}(\tau)=1$.

It follows from formulas (9.1) and (8.1) that the coefficients of $H(g(z))$,
where $g(z)$ is the solution of Lagrange’s equation with data $\psi$, satisfy:

$\textsc{coeff}_{[n]}(H(g(z)))\sim\frac{1}{\sqrt{2\pi}}\frac{H^{\prime}(\tau)\tau}{\sigma_{\psi}(\tau)}\frac{1}{n^{3/2}}\Big{(}\frac{\psi(\tau)}{\tau}\Big{)}^{n}\,,\quad\mbox{as
$n\to\infty$}\,.$

##  10\. Probability generating functions

Let $\psi$ be the probability generating function of a random variable $X$.

We assume for convenience that $\psi(0)\neq 0$ and that $\psi^{\prime}(0)\neq
0$, and further that $Q_{\psi}=1$, that $M_{\psi}=+\infty$ and that the radius
of convergence of $\psi$ is $R>1$. This last assumption implies, in
particular, that the variable $X$ has finite moments of all orders.

Let $F$ be the fulcrum of $\psi$, see Section 2.1.6, given by
$F(z)=\ln\psi(e^{z})$, in a region containing $[0,R_{\psi})$. Recall that in
terms of then fulcrum $F$ we have that $m_{\psi}(t)=F^{\prime}(s)$ and
$\sigma^{2}_{\psi}(t)=F^{\prime\prime}(s)$, where $s$ and $t$ are related by
$e^{s}=t$.

The function $e^{F(z)}=\psi(e^{z})=\sum_{n=0}^{\infty}b_{n}e^{nz}$, which is
holomorphic in a neighborhood of $z=0$, is the moment generating function of
$X$.

If $X_{1},X_{2},\ldots$ are independent copies of $X$, then

$\textsc{coeff}_{[k]}(\psi(z)^{n})=\mathbf{P}\left(\frac{X_{1}+\dots+X_{n}}{n}=\frac{k}{n}\right)\,.$

Let $m_{\psi}(\tau_{n})=k/n$ and let $s_{n}$ be given by $e^{s_{n}}=\tau_{n}$.
Observe that $\psi(\tau_{n})=e^{F(s_{n})}$ and that
$\sigma^{2}_{\psi}(\tau_{n})=F^{\prime\prime}(s_{n})$.

In terms of the fulcrum $F$ (or the $\ln$ of the moment generating function),
we have that if $k\asymp n$, see Theorem 5.1 or $k/n\to 0$, see Theorem 6.2,
that

$\displaystyle\mathbf{P}\left(\frac{X_{1}+\dots+X_{n}}{n}=\frac{k}{n}\right)$
$\displaystyle\sim\frac{1}{\sqrt{2\pi
n}\,\sigma_{\psi}(\tau_{n})}\frac{\psi(\tau_{n})^{n}}{\tau_{n}^{k}}$
$\displaystyle=\frac{1}{\sqrt{2\pi
n\,F^{\prime\prime}(s_{n})}}e^{n[F(s_{n})-s_{n}(k/n)]}\,,\quad\mbox{as
$n\to\infty$}\,.$

If $\psi$ is uniformly Gaussian, the asymptotic formula above holds also if
$k/n\to\infty$.

###  10.1. Lagrangian distributions

We now introduce the so called Lagrangian probability distributions. We refer
to [28] for a neat presentation of the basic theory of these probability
distributions. See also [5], for a comprehensive treatment, and [25].

We start with a probability generating function $\phi$ of a variable $Y$
taking values in $\\{0,1,\ldots\\}$. We assume that $\phi$ is in
$\mathcal{K}$.

This variable $Y$ generates a cascade or Galton-Watson random process starting
(initial stage) with a single individual. The variable $Y$ gives the random
number of immediate descendants, the offspring, of each individual in every
generation.

The random number of individuals in generation $n$ is denoted by $G_{n}$; the
generation $0$ is the initial stage. We have $G_{0}\equiv 1$ and

$(\star)\quad G_{n+1}=\sum_{j=1}^{G_{n}}Y_{n,j}\,,\quad\mbox{for $n\geq
0$}\,,$

where the $Y_{n,j}$ are independent copies of $Y$.

Let us denote $\phi^{\prime}(1)\triangleq\lim_{t\uparrow
1}\phi^{\prime}(t)=\mathbf{E}(Y)$. Observe that
$\phi^{\prime}(1)=m_{\phi}(1)$.

The total progeny $Z$ of the single individual of the generation 0 (including
itself) is the random variable $Z=\sum_{n=0}^{\infty}G_{n}$. This $Z$ could
take the value $\infty$, but it is a proper random variable (i.e.,
$\mathbf{P}(Z<\infty)=1$) if and only if $\psi^{\prime}(1)\leq 1$.

Assume thus that $\psi^{\prime}(1)\leq 1$. The probability generating function
$g$ of the total progeny $Z$ is actually the solution of Lagrange’s equation
with data $\phi$.

Furthermore, let $f$ be a power series with nonnegative coefficients which is
the probability generating function of a random variable $X$ taking values in
$\\{0,1,\ldots\\}$.

We enhance the process by allowing the size of the initial stage (generation
$0$) to be randomly chosen following the distribution of $X$. Thus $G_{0}=X$
and the evolution is determined by the recurrence $(\star)$. The total progeny
$Z$ including the individuals of the initial stage has probability generating
function $f\circ g$. This size of progeny $Z$ is said to follow a Lagrangian
distribution $\mathcal{L}(\phi,f)$; $\phi,f$ are called the generators of
$\mathcal{L}(\phi,f)$.

If $f(z)\equiv z$, then $X\equiv 1$, and there is (deterministically) a single
individual in the initial generation.

We have, see (9.1), that

$\mathbf{P}(Z=n)=\textsc{coeff}_{[n]}(f(g(z)))=\frac{1}{n}\textsc{coeff}_{[n{-}1]}(f^{\prime}(z)\phi^{n}(z))\,,\quad\mbox{for
$n\geq 1$}\,,$

and $\mathbf{P}(Z=0)=f(0)$.

###  10.2. Lagrangian distributions and Khinchin families

We will have two ingredients: a Khinchin family of offspring probability
distributions and a Khinchin family of probability distribution for the
initial distribution.

(a) Let $\psi(z)=\sum_{n=0}^{\infty}b_{n}z^{n}$ be a power series in
$\mathcal{K}$ with radius of convergence $R$. We assume from the outset that
$Q_{\psi}=1$. We let $(Y_{t})_{t\in[0,R)}$

Assume that either $M_{\psi}>1$ and then we let $\tau\in(0,R)$ be such that
$m_{\psi}(\tau)=1$ or $M_{\psi}=1$ and $R<\infty$ and $\lim_{t\uparrow
R}\sigma_{\psi}(t)=\sigma_{\psi}(R)<\infty$ and then we let $\tau=R$.

For each $t\in(0,\tau]$, we let $\psi_{t}$ denote the power series
$\psi_{t}(z)=\psi(tz)/\psi(t)$. This $\psi_{t}$ is the probability generating
function of $Y_{t}$, and besides, since $t\leq\tau$, we have that

$(\flat)\quad\psi_{t}^{\prime}(1)=m_{\psi}(t)\leq 1\,.$

For each $t\in(0,\tau]$, we let $g_{t}$ be the solution of Lagrange’s equation
with data $\psi_{t}$:

(10.1) $\displaystyle g_{t}(z)=z\psi_{t}(g_{t}(z))\,.$

Because of $(\flat)$ and the discussion above in Section 10.1, $g_{t}$ is the
probability generating function of the random distribution of the total
progeny of a single individual with offspring distribution $Y_{t}$.

We let $g$ denote the solution of Lagrange’s equation with data $\psi$, then
we may write each $g_{t}$ in terms of $g$ as follows

######  Lemma 10.1.

With the notations above

$g_{t}(z)=\frac{1}{t}g\big{(}(t/\psi(t))z\big{)}\,,\quad\mbox{for
$t\leq\tau$}\,.$

###### Proof.

This is a consequence of the uniqueness of solution of Lagrange’s equation.
Let $\widetilde{g}_{t}(z)=\frac{1}{t}g\big{(}(t/\psi(t))z\big{)}$. Now,

$\displaystyle z\psi_{t}(\widetilde{g}_{t}(z))$
$\displaystyle=\frac{z}{\psi(t)}\psi(t\,\widetilde{g}_{t}(z))=\frac{z}{\psi(t)}\psi(g\big{(}(t/\psi(t))z\big{)}$
$\displaystyle=\frac{1}{t}\frac{zt}{\psi(t)}\psi(g\big{(}(t/\psi(t))z\big{)}=\frac{1}{t}g\big{(}(t/\psi(t))z\big{)}$
$\displaystyle=\widetilde{g}_{t}(z)\,.$

Uniqueness gives that $\widetilde{g}_{t}(z)=g_{t}(z)$, as claimed.∎

######  Remark 10.2.

The $g_{t}$ are not the probability generating functions of a Khinchin family,
but if we change parameters and substitute $t$ by $g(u)$ and let

$\widetilde{g}_{u}(z)=g_{g(u)}(z)$

we have that $\widetilde{g}_{u}(z)=g(uz)/g(u)$. The power series $g$ is not in
$\mathcal{K}$ but $g(z)/z$ is in $\mathcal{K}$, since
$g^{\prime}(0)=\psi(0)>0$. If $(W_{u})$ is the Khinchin family of $g(z)/z$
then $\widetilde{g}_{u}$ is the probability generating functions of $W_{u}+1$.

(b) Next, we let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be a nonconstant power
series with nonnegative coefficients and radius of convergence $S>0$. We do
not require $f$ to be in $\mathcal{K}$; in fact $f(z)=z^{m}$, for integer
$m\geq 1$, is particularly relevant.

For $s\in(0,S)$ we denote by $f_{s}$ the power series

$f_{s}(z)=\frac{f(sz)}{f(s)}$

which is the probability generating function of a random variable $X_{s}$,
say.

######  Remark 10.3.

If the power series $f$ were not in $\mathcal{K}$, then since it is not a
constant, there should exist at least one index $m\geq 1$ so that $a_{m}\neq
0$. Thus only two cases could occur. In the first case, $f(z)=a_{m}z^{m}$, for
some integer $m\geq 1$. And in the second; there is $\phi\in\mathcal{K}$ and
an integer $l\geq 0$ so that $f(z)=z^{l}\phi(z)$.

In the first case $(X_{s})$ is not a Khinchin family, but $X_{s}\equiv m$, for
all $s\in(0,S)$, and therefore there are exactly $m\geq 1$ nodes in the first
generation; this is the deterministic case $f_{s}(z)=z^{m}$.

In the second case $(X_{s})_{s\in[0,S)}$ is a shifted Khinchin family. If
$f(0)>0$, then $(X_{s})_{s\in[0,S)}$ is a proper Khinchin family.

Now, for $s\in(0,S)$ and $t\in(0,\tau]$, the composition $f_{s}(g_{t}(z))$ is
the probability generating function of the total progeny $Z_{s,t}$ of a
Galton-Watson process with initial distribution $f_{s}$ and offspring
distribution $g_{t}$. The progeny $Z_{s,t}$ has Lagrange distribution
$\mathcal{L}(\psi_{t},f_{s})$.

For radius $u>0$ such that $us<S$ and $ut\leq\tau$ and appealing to (9.1) we
may write

$\displaystyle\mathbf{P}(Z_{s,t}=n)$
$\displaystyle=\textsc{coeff}_{[n]}\big{(}f_{s}(g_{t}(z))\big{)}=\frac{1}{n}\textsc{coeff}_{[n{-}1]}\big{(}f_{s}^{\prime}(z)\psi_{t}^{n}(z)\big{)}$
$\displaystyle=\frac{s}{f(s)}\frac{\psi(tu)^{n}}{\psi(t)^{n}}\frac{1}{u^{n{-}1}}\frac{1}{n}\,\frac{1}{2\pi}\int\limits_{|\theta|\leq\pi}f^{\prime}(sue^{\imath\theta})\frac{\psi(tue^{\imath\theta})^{n}}{\psi(tu)^{n}}e^{-\imath(n{-}1)\theta}d\theta\,.$

If the parameter $s$ is further restricted to $s\tau<tS$, we may take
$u=\tau/t$ in the expression above and write

$\displaystyle\mathbf{P}(Z_{s,t}=n)$
$\displaystyle=\frac{1}{2\pi}\frac{s}{f(s)}\frac{\psi(\tau)^{n}}{\psi(t)^{n}}\left(\frac{t}{\tau}\right)^{n{-}1}\frac{1}{n}\int\limits_{|\theta|\leq\pi}f^{\prime}\left(\frac{s\tau}{t}e^{\imath\theta}\right)\mathbf{E}(e^{\imath
Y_{\tau}\theta})^{n}e^{-\imath(n{-}1)\theta}d\theta$
$\displaystyle=\frac{1}{2\pi}\frac{s}{f(s)}\frac{\psi(\tau)^{n}}{\psi(t)^{n}}\Big{(}\frac{t}{\tau}\Big{)}^{n{-}1}\frac{1}{n^{3/2}}\frac{1}{\sigma_{\psi}(\tau)}$
$\displaystyle\int\limits_{|\theta|\leq\pi\sigma_{\psi}(\tau)\sqrt{n}}f^{\prime}\left(\frac{s\tau}{t}e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}\right)\mathbf{E}(e^{\imath\breve{Y}_{\tau}\theta/\sqrt{n}})^{n}e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}d\theta$

For $\theta$ fixed, we have that

$\lim_{n\to\infty}f^{\prime}\left(\frac{s\tau}{t}e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}\right)e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}=f^{\prime}\left(\frac{s\tau}{t}\right)$

and

$\Big{|}f^{\prime}\left(\frac{s\tau}{t}e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}\right)e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}\Big{|}\leq
f^{\prime}\left(\frac{s\tau}{t}\right)\,.$

And thus taking into account the integral form of the local central limit
theorem, Theorem F, we deduce that

$\lim_{n\to\infty}\int\limits_{|\theta|\leq\pi\sigma_{\psi}(\tau)\sqrt{n}}f^{\prime}\left(\frac{s\tau}{t}e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}\right)\mathbf{E}(e^{\imath\breve{Y}_{\tau}\theta/\sqrt{n}})^{n}e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}d\theta=\sqrt{2\pi}f^{\prime}\left(\frac{s\tau}{t}\right)$

and, consequently, that

(10.2)
$\mathbf{P}(Z_{s,t}=n)\sim\frac{1}{\sqrt{2\pi}}\frac{s}{f(s)}\frac{\psi(\tau)^{n}}{\psi(t)^{n}}\Big{(}\frac{t}{\tau}\Big{)}^{n{-}1}\frac{1}{n^{3/2}}\frac{1}{\sigma_{\psi}(\tau)}f^{\prime}\left(\frac{s\tau}{t}\right)\,,\quad\mbox{as
$n\to\infty$}\,,$

as long as $s\tau<tS$, which amounts to no restriction if $S=+\infty$.

(c) As an illustration, consider the case where $\psi(z)=e^{z}$ and
$f(z)=z^{j}$, for some integer $j\geq 1$.

In this case, $R=S=\infty$, $m_{\psi}(t)=t$ and $\sigma_{\psi}^{2}(t)=t$. Also
$M_{\psi}=\infty$ and $\tau=1$.

For $t\leq 1=\tau$, we have that $\psi_{t}(z)=e^{t(z-1)}$ and for $s<\infty$,
we have that $f_{s}(z)=z^{j}$. Observe that for any $s$, $f_{s}$ is the
probability generating function of the constant $j$.

For $0<t\leq 1$ and $0<s<\infty$, the variable $Z_{s,t}$ is the total progeny
of a Galton-Watson process, where the initial generation consists of exactly
$j$ individuals and the offspring of each individual is given by a Poisson
variable of parameter $t$. This distribution, $\mathcal{L}(e^{t(z-1)},z^{j})$,
is the Borel-Tanner distribution with parameters $t$ and $j$. The case $j=1$
is the Borel distribution. See [5], [26] and [28], and also the original
sources [2], [15] and [29].

Using (10.2), we deduce that

$\mathbf{P}(Z_{s,t}=n)\sim\frac{j}{\sqrt{2\pi}}\frac{1}{n^{3/2}}{t^{n-j}}e^{n(1-t)}\,,\quad\mbox{as
$n\to\infty$}\,.$

In fact, for the Borel-Tanner distribution with parameters $t$ and $j$ we have
the exact formula

(10.3)
$\mathbf{P}(Z_{s,t}=n)=\frac{j}{n}\frac{e^{-tn}(tn)^{n{-}j}}{(n-j)!}\,,\quad\mbox{for
$n\geq j$}\,.$

The asymptotic formula above follows then from Stirling’s formula.

(d) Consider now the case $\psi(z)=e^{z}$ and $f(z)=e^{z}$, so that
$R=S=\infty$, $m_{\psi}(t)=t$ and $\sigma_{\psi}^{2}(t)=t$ and, also,
$M_{\psi}=\infty$ and $\tau=1$.

For $t\leq 1=\tau$, we have that $\psi_{t}(z)=e^{t(z-1)}$ and for $s<\infty$,
we have that $f_{s}(z)=e^{s(z-1)}$.

For $0<t\leq 1$ and $0<s<\infty$, the variable $Z_{s,t}$ is the total progeny
of a Galton-Watson process, where the size of the initial generation is drawn
from a Poisson distribution of parameter $s$ individuals and the offspring of
each individual is given by a Poisson variable of parameter $t$. This
distribution, in Lagrangian distribution notation, is
$\mathcal{L}(e^{t(z-1)},e^{s(z-1)})$.

From (10.2), we have that

$\mathbf{P}(Z_{s,t}=n)\sim\frac{1}{\sqrt{2\pi}}e^{s/t-s}st^{n{-}1}e^{n(1-t)}\frac{1}{n^{3/2}}\,,\quad\mbox{as
$n\to\infty$}\,.$

Conditioning on the size of the initial generation and using (10.3), we deduce
for $\mathcal{L}(e^{t(z-1)},e^{s(z-1)})$ that

$\mathbf{P}(Z_{s,t}=n)=\frac{1}{n!}e^{-tn-s}(tn+s)^{n{-}1}s\,,\quad\mbox{para
$n\geq 1$}\,.$

See [28]. The asymptotic formula above follows then from Stirling’s formula.

(e) Back to Galton-Watson. Assume that $\psi$ is a probability generating
function and that $f(z)=z$. We assume that $\psi^{\prime}(1)\leq 1$, so that
the total progeny $Z$ of the Galton-Watson process starting with a single
individual and with offspring distribution $\psi$ is a (proper) random
variable.

Assume that $M_{\psi}>1$ or $M_{\psi}=1$ with $R<\infty$ and
$\sigma_{\psi}(R)<\infty$. In the first case we take $\tau\in(0,R)$ such that
$m_{\psi}(\tau)=1$. Since $\psi^{\prime}(1)\leq 1$, we have that
$m_{\psi}(1)=1$, and since $m_{\psi}$ is increasing we see that $1\leq\tau<R$.
In the second case we take $\tau=R$; observe that $R\geq 1$.

With $t=1$ (so that $\psi_{1}(z)\equiv\psi(z)$) and $s=1$ (an immaterial
choice since $f(z)=z$), we have

$\mathbf{P}(Z=n)\sim\frac{1}{\sqrt{2\pi}}\frac{\tau}{\sigma_{\psi}(\tau)}\Big{(}\frac{\psi(\tau)}{\tau}\Big{)}^{n}\frac{1}{n^{3/2}}\,,\quad\mbox{as
$n\to\infty$}\,.$

This is Theorem G applied to the solution of Lagrange’s equation with data
$\psi$. Recall that
$\sigma_{\psi}^{2}(\tau)={\tau^{2}\psi^{\prime\prime}(\tau)}/{\psi(\tau)}$,
see formula (9.4).

####  10.2.1. Limit cases

By limit cases we mean $t=\tau$ and $s\to S$, with $S<\infty$. See [23] for
related results.

For $t=\tau$ and $s<S$ we have the exact formula

$(\flat)\qquad\begin{aligned}
\mathbf{P}(Z_{s,\tau}=n)&=\frac{1}{2\pi}\frac{s}{f(s)}\frac{1}{n^{3/2}}\frac{1}{\sigma_{\psi}(\tau)}\\\
&\int\limits_{|\theta|\leq\pi\sigma_{\psi}(\tau)\sqrt{n}}f^{\prime}\big{(}se^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}\big{)}\mathbf{E}(e^{\imath\breve{Y}_{\tau}\theta/\sqrt{n}})^{n}e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}d\theta\end{aligned}$

and the asymptotic formula.

$\mathbf{P}(Z_{s,\tau}=n)\sim\frac{1}{\sqrt{2\pi}}m_{f}(s)\frac{1}{\sigma_{\psi}(\tau)}\frac{1}{n^{3/2}}\,,\quad\mbox{as
$n\to\infty$}\,,$

where with a slight abuse of notation we have written
${sf^{\prime}(s)}/{f(s)}=m_{f}(s)$, see Remark 10.3.

To consider $s\to S$, we first rewrite $(\flat)$ as

$(\flat\flat)\qquad\begin{aligned}
\mathbf{P}(Z_{s,\tau}=n)&=\frac{1}{2\pi}\frac{1}{n^{3/2}}{m_{f}(s)}\frac{1}{\sigma_{\psi}(\tau)}\\\
&\int\limits_{|\theta|\leq\pi\sigma_{\psi}(\tau)\sqrt{n}}\dfrac{f^{\prime}\big{(}se^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}\big{)}}{f^{\prime}(s)}\mathbf{E}(e^{\imath\breve{Y}_{\tau}\theta/\sqrt{n}})^{n}e^{\imath\theta/(\sigma_{\psi}(\tau)\sqrt{n})}d\theta\end{aligned}$

If $\sum_{n=0}^{\infty}na_{n}S^{n}<\infty$, then $f^{\prime}$ extends
continuously to $\mbox{\rm cl}(\mathbb{D}(0,S))$, and by appealing to Theorem
F, we readily see that

$\lim_{\begin{subarray}{c}s\uparrow S;\\\
n\to\infty\end{subarray}}\frac{n^{3/2}}{m_{f}(s)}\mathbf{P}(Z_{\tau,s}=n)=\frac{1}{\sigma_{\psi}(\tau)\sqrt{2\pi}}\,.$

More generally, if $s_{n}<S$ and $n$ are such that

$(\sharp)\quad\lim_{n\to\infty}\frac{f^{\prime}(s_{n}e^{\imath\phi/\sqrt{n}})}{f^{\prime}(s_{n})}=1\,,\quad\mbox{for
each $\phi\in\mathbb{R}$}\,,$

then

$\lim_{n\to\infty}\frac{n^{3/2}}{m_{f}(s_{n})}\mathbf{P}(Z_{\tau,s_{n}}=n)=\frac{1}{\sigma_{\psi}(\tau)\sqrt{2\pi}}\,.$

For instance for $f(z)=1/(1-z)$, condition $(\sharp)$ is satisfied if $s_{n}$
and $n$ are related so that $\lim_{n\to\infty}(1-s_{n})\sqrt{n}=\infty$.

## References

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##  11\.  Appendix. Uniformly Gaussian Khinchin families

Let $f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}$ be a power series in $\mathcal{K}$
with radius of convergence $R>0$ and let $(X_{t})_{t\in[0,R)}$ be its Khinchin
family.

We have encountered two integral convergence results: the integral central
limit Theorem F and the notion of strongly gaussian power series, definition
2.

$\bullet$ Assume that $Q_{f}=\gcd\\{n\geq 1:b_{n}>0\\}=1$. For each fixed
$t\in(0,R)$, the normalized variable $\breve{X}_{t}$ is a lattice random
variable with gauge function $1/\sigma(t)$, since $Q_{f}=1$. Because of
Theorem F we then have that

$({\dagger})\qquad\lim_{n\to\infty}\int\limits_{|\theta|\leq\pi\sigma(t)\sqrt{n}}\Big{|}\mathbf{E}\Big{(}e^{\imath\theta\breve{X}_{t}/\sqrt{n}}\Big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

$\bullet$ If the Khinchin family $(X_{t})_{t\in[0,R)}$ is strongly Gaussian,
then we have for each fixed $n\geq 1$, that

$({\ddagger})\qquad\lim_{t\uparrow
R}\int\limits_{|\theta|\leq\pi\sigma(t)\sqrt{n}}\Big{|}\mathbf{E}\Big{(}e^{\imath\theta\breve{X}_{t}/\sqrt{n}}\Big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

This fact follows from the bound

$\displaystyle\int\limits_{|\theta|\leq\pi\sigma(t)\sqrt{n}}\Big{|}\mathbf{E}\Big{(}e^{\imath\theta\breve{X}_{t}/\sqrt{n}}\Big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta$
$\displaystyle=\sqrt{n}\,\int\limits_{|\varphi|\leq\pi\sigma(t)}\Big{|}\mathbf{E}\Big{(}e^{\imath\varphi\breve{X}_{t}}\Big{)}^{n}-e^{-\varphi^{2}n/2}\Big{|}\,d\varphi$
$\displaystyle\leq
n^{3/2}\int\limits_{|\varphi|\leq\pi\sigma(t)}\Big{|}\mathbf{E}\Big{(}e^{\imath\varphi\breve{X}_{t}}\Big{)}-e^{-\varphi^{2}/2}\Big{|}\,d\varphi\,,$

where, after the change of variables $\theta=\varphi\sqrt{n}$, we have used
that for complex numbers $z,w$ such that $|z|,|w|\leq 1$ we have that
$|z^{n}-w^{n}|\leq n|z-w|$.

Thus, for a Gaussian power series the integral

$I_{n}(t)=\int\limits_{|\theta|\leq\pi\sigma(t)\sqrt{n}}\Big{|}\mathbf{E}\Big{(}e^{\imath\theta\breve{X}_{t}/\sqrt{n}}\Big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta$

converges to 0 as $n\to\infty$ with $t$ fixed and as $t\uparrow R$ with $n$
fixed.

Power series $f$ in $\mathcal{K}$ with Khinchin family $(X_{t})_{t\in[0,R)}$
are power series for which $({\dagger})$ and $({\ddagger})$ hold
simultaneously in the sense that the involved integrals converges to 0, as
$n\to\infty$ or $t\uparrow R$.

######  Definition 4.

A power series $f$ and its Khinchin family $(X_{t})_{t\in[0,R)}$ are called
uniformly Gaussian if the following two conditions are satisfied:

$a)\quad\lim_{t\uparrow R}\sigma(t)=\infty\quad\mbox{and}\quad
b)\quad\lim_{[n\to\infty\,\vee\,t\uparrow
R]}\int\limits_{|\theta|\leq\pi\sigma(t)\sqrt{n}}\Big{|}\mathbf{E}\Big{(}e^{\imath\theta\breve{X}_{t}/\sqrt{n}}\Big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=0\,.$

By $[n{\to}\infty\,\vee\,t{\uparrow}R]$ we mean that $1\leq n\to\infty$ or
(inclusive) $0\leq t_{0}\leq t\uparrow R$. The restriction $t>t_{0}$ is there
to exclude the possibility of $t\to 0$ and $n\to\infty$, simultaneously.

By fixing $n=1$ and letting $t\uparrow R$, we observe that uniformly Gaussian
power series are strongly Gaussian. In particular, if $f$ is a uniformly
Gaussian power series then

$M_{f}=\lim_{t\uparrow R}m(t)=\infty\,.$

Moreover, the coefficients $b_{n}$ of $f$ satisfy Hayman’s asymptotic formula
and in particular $Q_{f}=1$.

Let us verify that the exponential $f(z)=e^{z}$ is uniformly Gaussian. We have
$\sigma(t)=\sqrt{t}$, for $t>0$ and so $\sigma(t)\sqrt{n}=\sigma(nt)$, for
$t>0$ and $n\geq 1$. For the characteristic function of the normalized
variable $\breve{X}_{t}$ we have that

$\mathbf{E}\big{(}e^{\imath\theta\breve{X}_{t}}\big{)}=\exp\big{(}t\,\big{(}e^{\imath\theta/\sqrt{t}}-1-\imath\theta/\sqrt{t}\big{)}\Big{)}\,,$

and thus that

$\mathbf{E}\big{(}e^{\imath\theta\breve{X}_{t}/\sqrt{n}}\big{)}^{n}=\mathbf{E}\big{(}e^{\imath\theta\breve{X}_{nt}}\big{)}\,.$

Therefore,

$\int\limits_{|\theta|\leq\pi\sigma(t)\sqrt{n}}\Big{|}\mathbf{E}\Big{(}e^{\imath\theta\breve{X}_{t}/\sqrt{n}}\Big{)}^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta=\int\limits_{|\theta|\leq\pi\sigma(nt)}\Big{|}\mathbf{E}\Big{(}e^{\imath\theta\breve{X}_{nt}}\Big{)}-e^{-\theta^{2}/2}\Big{|}\,d\theta\,.$

Since $e^{z}$ is strongly Gaussian, this integral tends to 0 as
$(nt)\to\infty$, and thus as $[n{\to}\infty\,\vee\,t{\uparrow}+\infty]$.
Observe that in this case $R=+\infty$.

The notion of uniformly Hayman power series which we are about to introduce
generalizes that of Hayman power series, see Section 2.7. We will verify
shortly that uniformly Hayman power series are uniformly Gaussian, much like
power series in the Hayman class are strongly gaussian. Theorem 11.2 provides
us with an ample class of power series which are uniformly Hayman.

######  Definition 5.

Let $f(z)$ be a power series in $\mathcal{K}$ with radius of convergence $R>0$
and let $(X_{t})_{t\in[0,R)}$ be its Khinchin family.

We say that $f(z)$ and $(X_{t})_{t\in[0,R)}$ are uniformly Hayman if for each
$n\geq 1$ and $t\in(0,r)$ there exists $h(n,t)\in(0,\pi)$ (called cuts) such
that the following requirements are satisfied:

(11.1) $\displaystyle\sup_{|\theta|\leq
h(n,t)\sigma(t)\sqrt{n}}\Big{|}\mathbf{E}(e^{\imath\theta\breve{X}_{t}/\sqrt{n}})^{n}e^{\theta^{2}/2}-1\Big{|}\rightarrow
0,\quad\mbox{as $[n{\to}\infty\,\vee\,t{\uparrow}R]$}\,,$ (11.2)
$\displaystyle\sqrt{n}\sigma(t)\,\sup_{h(n,t)\sigma(t)\sqrt{n}\leq|\theta|\leq\pi\sigma(t)\sqrt{n}}\big{|}\mathbf{E}(e^{\imath\theta\breve{X}_{t}/\sqrt{n}})^{n}\big{|}\rightarrow
0,\quad\mbox{as $[n{\to}\infty\,\vee\,t{\uparrow}R]$}\,,$ (11.3)
$\displaystyle\lim_{t\to R}\sigma(t)=\infty\,.$

Condition (11.2) may be written equivalently as

(11.4)
$\qquad\sqrt{n}\sigma(t)\,\sup_{h(n,t)\leq|\theta|\leq\pi}\big{|}\mathbf{E}(e^{\imath\theta
X_{t}})^{n}\big{|}\rightarrow 0,\quad\mbox{as
$[n{\to}\infty\,\vee\,t{\uparrow}R]$}\,.$

As announced,

######  Theorem 11.1.

Uniformly Hayman power series are uniformly Gaussian.

The proof below is analogous to the proof of Theorem C of Section 2.7 which
claims that Hayman power series are strongly Gaussian.

###### Proof.

Denote $\theta(n,t)=h(n,t)\sigma(t)\sqrt{n}$. First we show, that

(11.5) $\theta(n,t)\to\infty\,,\quad\mbox{as
$[n{\to}\infty\,\vee\,t{\uparrow}R]$}\,.$

Abbreviate $\widehat{\theta}=\theta(n,t)$. By (11.1), we have that

$\mathbf{E}(e^{\imath\widehat{\theta}\breve{X}_{t}/\sqrt{n}})e^{\widehat{\theta}^{2}/2}\to
1\,,\quad\mbox{as $[n{\to}\infty\,\vee\,t{\uparrow}R]$}\,,$

while, from (11.2) we obtain that

$\sqrt{n}\sigma(t)\mathbf{E}\big{(}e^{\imath\widehat{\theta}\breve{X}_{t}/\sqrt{n}}\big{)}^{n}\to
0\,,\quad\mbox{as $[n{\to}\infty\,\vee\,t{\uparrow}R]$}\,.$

From these two limits we deduce that

$\sqrt{n}\sigma(t)e^{-\widehat{\theta}^{2}/2}\to 0\,,\quad\mbox{as
$[n{\to}\infty\,\vee\,t{\uparrow}R]$}\,,$

and thus, since $\sqrt{n}\sigma(t)\to\infty$ as
$[n{\to}\infty\,\vee\,t{\uparrow}R]$ we deduce that

$\widehat{\theta}\to\infty\,,\quad\mbox{as
$[n{\to}\infty\,\vee\,t{\uparrow}R]$}\,.$

Denote by $A(n,t),B(n,t)$, respectively, the supremums in (11.1) and (11.2).

We bound

$\displaystyle\int\limits_{|\theta|\leq h(n,t)\sigma(t)\sqrt{n}}$
$\displaystyle\Big{|}\mathbf{E}(e^{\imath\theta\breve{X}_{t}/\sqrt{n}})^{n}-e^{-\theta^{2}/2}\Big{|}\,d\theta$
$\displaystyle=\int\limits_{|\theta|\leq h(n,t)\sigma(t)\sqrt{n}}$
$\displaystyle\Big{|}\mathbf{E}(e^{\imath\theta\breve{X}_{t}/\sqrt{n}})^{n}e^{\theta^{2}/2}-1\Big{|}e^{-\theta^{2}/2}\,d\theta$
$\displaystyle\leq A(n,t)\sqrt{2\pi}\,,$

and

$\displaystyle\int\limits_{h(n,t)\sigma(t)\sqrt{n}\leq|\theta|\leq\pi\sigma(t)\sqrt{n}}$
$\displaystyle\Big{|}\mathbf{E}(e^{\imath\theta\breve{X}_{t}/\sqrt{n}})-e^{-\theta^{2}/2}\Big{|}\,d\theta$
$\displaystyle\leq$ $\displaystyle
2\pi\sigma(t)\sqrt{n}B(n,t)+\int_{|\theta|\geq
h(n,t)\sigma(t)\sqrt{n}}e^{-\theta^{2}/2}\,d\theta\,.$

These two bounds and conditions (11.1) and (11.2) combined with (11.5) give
the result.

∎

###  11.1. Uniformly Hayman exponentials

Let $g(z)=\sum_{n=0}^{\infty}b_{n}z^{n}$, with $b_{n}\geq 0$, for $n\geq 0$,
and radius of convergence $R>0$. Let $f\in\mathcal{K}$ be given by $f=e^{g}$.

Exponentials of power series with positive coefficients are very relevant, in
particular, in Combinatorics since they codify (most) generating functions of
the set construction, which includes among them generating functions of
partitions of many sorts.

One of the main results of [4], see [3, Theorem 4.1] and [4, Theorem F] gives
conditions on the powers series $g$ that guarantees that $f=e^{g}$ is in the
Hayman class, and thus strongly Gaussian, and, therefore, amenable to the
Hayman asymptotic formula, (2.6).

It turns out that these same conditions on $g$ are enough for $f$ being
uniformly Hayman; this is the content of Theorem 11.2.

Denote

(11.6)
$\omega_{g}(t)\triangleq\frac{1}{6}(b_{1}t+8b_{2}t^{2}+\frac{9}{2}t^{3}g^{\prime\prime\prime}(t))\,,\quad\mbox{for
$t\in(0,R)$}\,.$

######  Theorem 11.2.

Let $g$ be a nonconstant power series with radius of convergence $R>0$ and
nonnegative coefficients.

Assume that the variance condition is satisfied

(11.7) $\lim_{t\uparrow
R}\big{(}tg^{\prime}(t)+t^{2}g^{\prime\prime}(t)\big{)}=+\infty\,.$

Assume further that there is a cut function $h(t)$ satisfying

(11.8) $\lim_{t\uparrow R}\omega_{g}(t)h(t)^{3}=0\,.$

and that there are positive functions $U,V$ defined in $(t_{0},R)$, for some
$t_{0}\in(0,R)$, where $U$ takes values in $(0,\pi]$ and $V$ in $(0,\infty)$
and are such that

(11.9) $\sup\limits_{|\varphi|\geq\omega}\big{(}\Re
g(te^{\imath\varphi})-g(t)\big{)}\leq-V(t)\,\omega^{2}\,,\quad\mbox{for
$\omega\leq U(t)$ and $t\in(t_{0},R)$}\,,$

and thus that the cut $h$ is such that

(11.10) $h(t)\leq U(t),\,\mbox{for
$t\in(t_{0},1)$}\quad\mbox{and}\quad\lim_{t\uparrow
R}\sigma(t)\,e^{-V(t)h(t)^{2}}=0\,.$

then the function $f=e^{g}$ is uniformly Hayman.

###### Proof.

Condition (11.7) is directly condition (11.3).

We shall verify the conditions on the cuts of the definition of uniformly
Hayman power series (11.1) and (11.4) with cuts $h(n,t)$ given by

$h(n,t)=h(t)\,n^{-\beta}$

where the parameter $\beta$ satisfies $1/3<\beta<1/2$.

From the discussion of Section 4 of [3], we have that

$\Big{|}\ln\mathbf{E}(e^{\imath\theta\breve{X}_{t}})+\frac{\theta^{2}}{2}\Big{|}\leq\omega_{g}(t)\frac{|\theta|^{3}}{\sigma^{3}(t)}\,,\quad\mbox{for
$t\in(0,R)$ and $\theta\in\mathbb{R}$}\,,$

and, thus, that

$\Big{|}n\ln\mathbf{E}(e^{\imath\theta\breve{X}_{t}/\sqrt{n}})+\frac{\theta^{2}}{2}\Big{|}\leq\omega_{g}(t)\frac{|\theta|^{3}}{\sigma^{3}(t)\sqrt{n}}\,,\quad\mbox{for
$t\in(0,R)$ and $\theta\in\mathbb{R}$}\,.$

For $|\theta|\leq h(n,t)\sigma(t)\sqrt{n}$ we deduce that

$\Big{|}n\ln\mathbf{E}(e^{\imath\theta\breve{X}_{t}/\sqrt{n}})+\frac{\theta^{2}}{2}\Big{|}\leq\omega_{g}(t)\,h(t)^{3}\,n^{1{-}3\beta}$

Hypothesis (11.8) on $h(t)$ and the fact that $1-3\beta<0$ gives us that

$\lim\limits_{[n{\to}\infty\vee
t{\uparrow}R]}\omega_{g}(t)\,h(t)^{3}\,n^{1{-}3\beta}=0\,,$

and, thus, that condition (11.1) is satisfied.

Since $h(n,t)\leq h(t)\leq U(t)$, condition (11.9) gives us that

$n\sup\limits_{h(n,t)\leq|\theta|\leq\pi}\big{(}\Re
g(te^{\imath\theta})-g(t)\big{)}\leq-V(t)h(t)^{2}n^{1{-}2\beta}\,,$

and so that

$\sup\limits_{h(n,t)\leq|\theta|\leq\pi}\big{|}\mathbf{E}(e^{\imath\theta
X_{t}})^{n}\Big{|}\leq\exp\Big{(}-V(t)h(t)^{2}n^{1{-}2\beta}\Big{)}\,.$

Since $\lim_{t\uparrow R}V(t)h(t)^{2}=\infty$, we have, for a certain
$t_{0}\in(0,R)$ that $V(t)h(t)^{2}\geq 1$, for $t\in(t_{0},R)$, and thus that

$V(t)h(t)^{2}n^{1{-}2\beta}\geq V(t)h(t)^{2}+n^{1{-}2\beta}\,.$

We deduce that

$\sqrt{n}\sigma(t)\,\sup\limits_{h(n,t)\leq|\theta|\leq\pi}\big{|}\mathbf{E}(e^{\imath\theta
X_{t}})^{n}\big{|}\leq\sqrt{n}e^{-n^{1{-}2\beta}}\,\sigma(t)e^{-V(t)h(t)^{2}}\,,$

and, because of hypothesis (11.10) and since $\beta<1/2$, that condition
(11.4) of the definition of uniformly Hayman is satisfied. ∎

In [4] a large number of exponentials $f=e^{g}$ where $g$ is a power series
with nonnegative coefficients which satisfy the conditions of Theorem 11.2 are
exhibited. For instance, the egf of the Bell numbers, or the generating
functions $P$ of partitions or $Q$ of partitions into distinct parts, are
actually uniformly Hayman, and also are uniformly Hayman related examples like
the egf of sets of pointed sets, the egf of sets of functions or the ogf of
plane partitions or of some colored partitions.

###  11.2. Exponential of polynomials

Let $g$ be a polynomial with nonnegative coefficients

$g(z)=\sum_{n=0}^{N}b_{n}z^{n}$

and of degree $N$, so that $b_{N}>0$.

Assume that $Q_{g}=\gcd\\{1\leq n\leq N:b_{n}>0\\}=1$. Then $f=e^{g}$ is in
the Hayman. This is a particular case of a result of Hayman [16, Theorem X].
See [3, Proposition 5.1] for a simpler proof of this particular case.

We are going to show next that $f=e^{g}$ is actually uniformly Hayman with an
argument similar to the one used to show in [3, Proposition 5.1] that
$f=e^{g}$ is in the Hayman class.

Observe first that

$\sigma^{2}(t)=tg^{\prime}(t)+t^{2}g^{\prime\prime}(t)\sim
N^{2}b_{N}t^{N},\quad\mbox{as $t\to\infty$}\,.$

Thus, the variance condition (11.3) of being uniformly Hayman is satisfied.

We have

$\omega_{g}(t)=\frac{1}{6}\big{(}b_{1}t+8b_{2}t^{2}+\frac{9}{2}t^{3}g^{\prime\prime\prime}(t)\big{)}=O(t^{N})\,,\quad\mbox{as
$t\to\infty$}\,.$

For cuts we propose $h(n,t)=h(t)n^{-\beta}=t^{-N\alpha}n^{-\beta}$, with
$\alpha,\beta$ in the interval $(1/3,1/2)$. For concreteness, we take
$\alpha=\beta=5/12$.

From the proof of Theorem 11.2 we have that

$\displaystyle\Big{|}n\ln\mathbf{E}(e^{\imath\theta\breve{X}_{t}/\sqrt{n}})+\frac{\theta^{2}}{2}\Big{|}$
$\displaystyle\leq\omega_{g}(t)t^{-5N/4}n^{-1/4}$
$\displaystyle=O\big{(}t^{-N/4}n^{-1/4}\big{)}\,,\quad\mbox{for $|\theta|\leq
h(n,t)\sigma(t)\sqrt{n}$}\,,$

and thus, we see that condition (11.1) is satisfied.

Now, the proof that $f=e^{g}$ is in the Hayman class of [3] gives
$\eta\in(0,\pi)$ and $t_{0}>0$, depending on $g$, so that

$\sup\limits_{|\theta|>\omega}\big{(}\Re g(te^{\imath\theta})-g(t)\big{)}\leq-
C_{g}\min\big{\\{}t,t^{N}\omega^{2}\big{\\}},\quad\mbox{for $\omega\geq\eta$
and $t>t_{0}$}\,,$

for some constant $C_{g}$ depending on $g$.

Thus for some $t_{1}>t_{0}$, so that $h(t)<\eta$, for $t>t_{1}$, we have that

$\displaystyle n\sup\limits_{h(n,t)\leq|\theta|\leq\pi}\big{(}\Re
g(te^{\imath\theta})-g(t)\big{)}$
$\displaystyle\leq-C\min\big{\\{}tn,t^{N}t^{-5N/6}n^{1-5/6}\big{\\}}$
$\displaystyle=-C\min\big{\\{}tn,t^{N/6}n^{1/6}\big{\\}}\,.$

With $\delta=\min\\{1,N/6\\}$, we have, for $t\geq 1$, that

$\min\\{tn,t^{N/6}n^{1/6}\\}\geq
t^{\delta}n^{1/6}\geq(1/2)(t^{\delta}+n^{1/6})\,,$

and thus, since $|\mathbf{E}(e^{\imath\theta X_{t}})|=e^{\Re
g(te^{\imath\theta})-g(t)}$, that

$\displaystyle\sqrt{n}\sigma(t)\sup\limits_{h(n,t)\leq|\theta|\leq\pi}\big{|}\mathbf{E}(e^{\imath\theta
X_{t}})^{n}\big{|}$
$\displaystyle=O\Big{(}\sqrt{n}t^{N/2}\exp\big{(}-C\min\\{tn,t^{N/6}n^{1/6}\\}\big{)}\Big{)}$
$\displaystyle=O\Big{(}\sqrt{n}\exp\big{(}-(C/2)n^{1/6}\big{)}\Big{)}\,O\Big{(}t^{N/2}\exp\big{(}-(C/2)t^{\delta}\big{)}\Big{)}\,,$

and, therefore, condition (11.2) is satisfied.