# Weak convergence implies convergence in mean within GGC

Hasanjan Sayit
${}^{2}\text{Xi'Jiao Liverpool University, Suzhou, China}$

(May 7, 2024)

###### Abstract

We prove that weak convergence within generalized gamma convolution (GGC)
distributions implies convergence in the mean value. We use this fact to show
the robustness of the expected utility maximizing optimal portfolio under
exponential utility function when return vectors are modelled by hyperbolic
distributions.

Keywords: GGC; Weak convergence; Expected utility; Mean-variance mixture
models.

JEL Classification: G11

##  1 Introduction

The paper [14] gives closed form expressions for the expected utility
maximizing optimal portfolios when the returns vector of risky assets follow
hyperbolic distributions. The market model considered in this paper contains
$d+1$ assets with one risk-free asset with interest rate $r_{f}$ and $d$ risky
assets with return vector given by

$X\overset{d}{=}\mu+\gamma Z+\sqrt{Z}AN,$ (1)

where $\mu\in\mathbb{R}^{d}$ is location parameter, $\gamma\in\mathbb{R}^{d}$
controls the skewness, $Z\sim G$ is a non-negative mixing random variable,
$A\in\mathbb{R}^{d\times d}$ is a symmetric and positive definite $d\times d$
matrix of real numbers, and $N\sim N(0,I)$ is a $d-$dimensional Gaussian
random vector with identity co-variance matrix $I$ in
$\mathbb{R}^{d}\times\mathbb{R}^{d}$ and $N$ is independent from the mixing
distribution $Z$.

The mixing distribution $Z$ in the model (1) can be any non-negative random
variable. If $Z$ is a non-negative random variable with finitely many values
then $X$ is called a mixture of Normal random variables (or vectors). If $Z$
follows generalized inverse Gaussian (GIG) distribution with the density
function

$f_{GIG}(x;\lambda,a,b)=(\frac{b}{a})^{\lambda}\frac{1}{K_{\lambda}(ab)}x^{\lambda-1}e^{-\frac{1}{2}(a^{2}x^{-1}+b^{2}x)}1_{(0,+\infty)}(x),$
(2)

where $K_{\lambda}(x)$ denotes the modified Bessel function of third kind with
index $\lambda$, then the model (1) becomes generalized hyperbolic (GH)
distributions. We refer to [8] (Section 1.2, Chapter 1) for further details,
especially for the allowed range of parameters $a,b,\lambda$, of this class of
distributions.

The class of GIG distributions belong to the class of generalized gamma
convolution (GGC) random variables, see Proposition 1.23 of [8] and
Proposition 1.5 of [9] for this. A positive random variable $Z$ is a GGC, if
its Laplace transform takes the following form

$\mathcal{L}_{Z}(s)=:Ee^{-sZ}=e^{-\tau s-\int_{0}^{\infty}log(1+s/x)\nu(dx)},$
(3)

for some $\tau\geq 0$ called the drift coefficient and some positive measure
$U$ called the Thorin measure that satisfy

$\int_{0}^{1}|log(x)|\nu(dx)<\infty,\;\;\;\int_{1}^{\infty}\frac{1}{x}\nu(dx)<\infty.$
(4)

The class GGC of distributions are infinitely divisible and self-decomposable,
see [7] and [12] for these. A random variable $Z$ is a GGC if and only if its
Laplace transform $\mathcal{L}_{Z}(s)$ is a hyperbolically completely monotone
(HCM) function.

The HCM property for a function $f:(0,+\infty)\rightarrow(0,+\infty)$ means
that the function

$f(s_{1}s_{2})f(s_{1}/s_{2})$

is completely monotone (CM) as a function of $s=s_{1}+s_{2}$ for every
$s_{1}>0$. A positive random variable $Z$ is called HCM if its probability
density function $f$ is a HCM. The paper [3] proved that the class of HCM
distributions belong to GGC. The class HCM is a proper subset of GGC. All the
positive $\alpha-$stable random variables, $S_{\alpha},\alpha\in(0,1),$ belong
to GGC and a positive $\alpha-$stable random variable is in HCM if and only if
$\alpha\leq 1/2$, see [6]. The class HCM is closed under multiplication and
division of independent random variables. Also for any $Z\in HCM$ one has
$Z^{p}\in HCM$ for every real number $p$ with absolute value $|p|\geq 1$.
These facts show that $S_{\alpha}^{p}$ is in GGC for any $\alpha\in(0,1/2)$
and $|p|\geq 1$. Other examples of GGC random variables include log-normal
models and Pareto distributions.

The motivation of this paper is to show the robustness of the utility
maximizing optimal portfolio under exponential utility function when return
vectors are modelled by distributions of the form (1). For more details of the
expected utility maximization problem we refer to [14]. To give a short
review, in this paper the wealth that corresponds to portfolio weight $x$ on
the risky assets is given by

$\begin{split}W(x)=&W_{0}[1+(1-x^{T}1)r_{f}+x^{T}X]\\\
=&W_{0}(1+r_{f})+W_{0}[x^{T}(X-\textbf{1}r_{f})]\end{split}$ (5)

and the investor’s problem is

$\max_{x\in\mathbb{R}^{d}}\;EU(W(x)).\\\ $ (6)

According to Proposition 2.17 of the paper [14], a regular solution for the
utility maximizing portfolio is give by

$x^{\star}=\frac{1}{aW_{0}}\Big{[}\Sigma^{-1}\gamma-
q_{min}\Sigma^{-1}(\mu-\textbf{1}r_{f})\Big{]},$ (7)

for some

$q_{min}\in\arg min_{\theta\in(-\hat{\theta},\hat{\theta})}Q(\theta),$ (8)

where $\hat{\theta}=:\sqrt{\frac{\mathcal{A}-2\hat{s}}{\mathcal{C}}}$ and
$\hat{s}$ is the IN of the mixing distribution $Z$, see the Definition 2.4 of
[14] for this. Here $\mathcal{A},\mathcal{C},\mathcal{B},$ are given as in
equation (32) of [14] (here we assume that the model (1) is such that
$\mu-\textbf{1}r_{f}\neq 0$ as in Remark 2.2 of [14] and this guarantees
$\mathcal{C}\neq 0$). The function $Q(\theta)$ is given as

$Q(\theta)=e^{\mathcal{C}\theta}\mathcal{L}_{Z}\Big{[}\frac{1}{2}\mathcal{A}-\frac{\theta^{2}}{2}\mathcal{C}\Big{]}.$
(9)

The optimal portfolio (7) is the utility maximizing optimal portfolio under
the exponential utility function $U(x)=-e^{-ax},a>0$, and the notation $W_{0}$
denotes the initial wealth of the investor.

The closed form (7) of the expected utility maximizing optimal portfolio is
obtained without introducing any assumption on the mixing distribution $Z$ in
the paper [14]. The mixing distribution $Z$ can be a not integrable random
variable and we only need to know the IN of $Z$ to be able to write down the
utility maximizing optimal portfolio under exponential utility.

While the formula (7) is convenient in practical applications, one needs to
address the issue of robustness of this optimal portfolio with respect to
model parameters in the model (1). In practice the parameters of the NMVM
models are estimated based on Expectation-Maximization algorithm (EM) or other
statistical procedures and such estimation procedures usually give some
errors. For this reason it is important to study the robustness of the optimal
portfolio (7) with respect to model parameters $\mu,\gamma,A,Z$ in (1).

To further clarify this point, assume the model (1) is the true model for the
return vector of risky assets and assume EM-algorithm or other statistical
estimation procedures lead into a different model

$X_{e}\overset{d}{=}\mu_{e}+\gamma_{e}Z_{e}+\sqrt{Z_{e}}A_{e}N_{n},$ (10)

instead of (1). Now, assume the differences of $\mu$ with $\mu_{e}$, $\gamma$
with $\gamma_{e}$, $A$ with $A_{e}$, and $Z$ with $Z_{e}$ are _small_ in some
norms. Then we would like to show that the optimal portfolio $x^{\star}_{e}$
based on the model (10) is close to the optimal portfolio $x^{\star}$ obtained
based on the model (1) in the Euclidean norm.

To illustrate this robustness issue of the optimal portfolio in (7) with an
example, lets consider the case of GH models $X\sim
GH_{n}(\lambda,\alpha,\beta,\delta,\mu,\Sigma)$. Assume the true parameters
are $\lambda,\alpha,\beta,\delta,\mu,\Sigma$ and however EM-algorithym or
other statistical procedures lead into a different model $X_{e}\sim
GH_{n}(\lambda_{e},\alpha_{e},\beta_{e},\delta_{e},\mu_{e},\Sigma_{e})$ with
estimated parameters
$\lambda_{e},\alpha_{e},\beta_{e},\delta_{e},\mu_{e},\Sigma_{e}$. Our main
concern is to examine if the optimal portfolio $x^{\star}$ in (7) for the
model $X$ in (1) is close to the corresponding optimal portfolio
$x^{\star}_{e}$ for the model $X_{e}$ in (10) if the parameters
$\lambda_{e},\alpha_{e},\beta_{e},\delta_{e},\mu_{e},\Sigma_{e}$ are close to
the true parameter set $\lambda,\alpha,\beta,\delta,\mu,\Sigma$ in some norms.

Out of these five parameters $\mu$ and $\beta$ are vectors in $\mathbb{R}^{n}$
and the Euclidean norm can be used to measure the distance for these
parameters. The parameter $\Sigma$ is an $n\times n$ matrix and the Hilbert-
Schmit norm for matrices can be used as a measure for distance for this
parameter. The other parameters show up in the mixing distribution GIG in the
form $\lambda,\delta,\sqrt{\alpha^{2}-\beta^{T}\Sigma\beta}$. We don’t
introduce a measure for closedness for each parameter
$\lambda,\delta,\sqrt{\alpha^{2}-\beta^{T}\Sigma\beta}$ of the mixing
distribution. But we need to use a distance to measure closedness of the laws
of two different mixing distributions $Z_{1}$ and $Z_{2}$ in the nodel (1). We
have many options for this and here we review few of them. The Fortet-Mourier
distance between laws of random variables is

$d_{FM}(Z_{1},Z_{2})=\sup_{|h|_{\infty}\leq 1,|h^{\prime}|_{\infty}\leq
1}|E(h(Z_{1}))-Eh(Z_{2})|,$ (11)

where $h$ represents continuous functions, $|\cdot|_{\infty}$ is sup norm
within the class of continuous functions, and $h^{\prime}$ is first order
derivative of $h$. It is well known that this distance metrize the convergence
in law, i.e., $d_{FM}(Z_{n},Z)\rightarrow 0$ if and only if
$d_{FM}(Z_{n},Z)\rightarrow 0$ as $n\rightarrow\infty$. Another distance is
Kolmogorov’s distance

$\displaystyle d_{Kol}(Z_{1},Z_{2})=\sup_{x\in\mathbb{R}}|F_{1}(X)-F_{2}(X)|,$

where $F_{1},F_{2}$ are commutative distribution functions of $Z_{1}$ and
$Z_{2}$ respectively. This is a stronger distance than the FM distance in the
sense that $d_{Kol}(Z_{n},Z)\rightarrow 0$ implies $d_{FM}(Z_{n},Z)\rightarrow
0$. Another popular distance which is stronger than the Kolmogorov distance is
the total variation distance

$d_{TV}(Z_{1},Z_{2})=sup_{B\in\mathcal{B}}|P_{1}(B)-P_{2}(B)|,$ (12)

where $\mathcal{B}$ is the sigma algebra of Bore sets and $P_{1}$ and $P_{2}$
are distribution functions of $Z_{1}$ and $Z_{2}$ respectively. By the
Scheffe’s theorem we have
$d_{TV}(Z_{1},Z_{1})=\frac{1}{2}\int_{0}^{+\infty}|f_{1}(x)-f_{2}(x)|dx,$
where $f_{1}$ and $f_{2}$ are probability density functions of $Z_{1}$ and
$Z_{2}$. All these distances can be used to measure the closedness of the
mixing distributions. However if the mixing distributions are from the class
$GGC$ then convergence in law is equivalent to convergence in total variation
norm as Lemma 3.1 below shows.

The above discussions motivates us to measure closedness of the mixing
distributions by the weakest distance $d_{FM}$ that metrize the weak
convergence. We will focus our discussions on robustness issue for optimal
portfolios that are regular only. The reason is if an optimal portfolio
$x^{\star}$ is irregular, then there exists a sequence of portfolios
$x_{n}^{\star}$ with $|x_{n}^{\star}-x^{\star}|\rightarrow 0$ such that
$EU(W(x_{n}^{\star}))=-\infty$ while $|EU(W(x^{\star}))|<\infty$. This means
that a slight mis-specification $\tilde{x}^{\star}$ of the optimal portfolio
$x^{\star}$, which can result in from mis-specification of the model
parameters in (1), can lead to an expected utility that equals to negative
infinity while the _true_ optimal portfolio $x^{\star}$ has finite expected
utility. This makes the discussion of the robustness issue at irregular
solutions meaningless.

In this paper we use the following notations. For any vectors
$x=(x_{1},x_{2},\cdots,x_{d})^{T}$ and $y=(y_{1},y_{2},\cdots,y_{d})^{T}$ in
$\mathbb{R}^{d}$, where the superscript $T$ stands for the transpose of a
vector, $<x,y>=x^{T}y=\sum_{i=1}^{d}x_{i}y_{i}$ denotes the scalar product of
the vectors $x$ and $y$, and $|x|=\sqrt{\sum_{i=1}^{d}x_{i}^{2}}$ denotes the
Euclidean norm of the vector $x$. For any matrix $A$ we use
$|A|_{HS}=\sqrt{\sum_{i=1,j=1}^{d}|A_{ij}|^{2}}$ to denote the Hilbert-Schmidt
norm of a matrix. We use the notation $\overset{w}{\rightarrow}$ to denote
weak convergence of random variables. We sometimes use the short hand notation
$X\sim N(\mu+\gamma z,z\Sigma)\circ G$ for (1). $\mathbb{R}$ denotes the set
of real numbers and $\mathbb{R}_{+}=[0,+\infty)$ denotes the set of non-
negative real-numbers. Following the same notations of [9], $\mathcal{J}$
denotes the family of infinitely divisible random variables on
$\mathbb{R}_{+}$, $\mathcal{S}$ denotes the set of self-decomposable random
variables on $\mathbb{R}_{+}$, and $\mathcal{G}$ denotes the class of
generalized gamma convolutions (GGCs) on $\mathbb{R}_{+}$ that will be
introduced later. The Laplace transformation of any distribution $G$ is
denoted by $\mathcal{L}_{G}(s)=\int e^{-sy}G(dy)$.

The paper is organized as follows. In Section 2 below we show that weak
convergence within GGC implies convergence of the mean values for the
integrable GGC random variables. In Section 3 we use this fact to show the
robustness of the optimal portfolios for the problem (6).

##  2 Weak convergence within GGC

Our discussions about robustness of the optimal portfolio will be focused on
models of the form (1) with mixing distribution $Z$ that belong to the GGC
class of random variables. The class of GGC models include the class of GIG
models as discussed in Proposition 1.23 of [8]. The class GGC of distributions
appeared in Thorin’s work and it includes popular models like log-normal
distributions, Pareto distributions, and all positive stable random variables.
They are closed in weak limits, addition of independent random variables, and
multiplication of independent random variables, see [2] for further details.

Before we start our discussions, we first write down the definition of GGC
random variables, see [2, 5] for more details.

######  Definition 2.1.

A GGC is a probability distribution on $[0,\infty)$ with Laplace
transformation of the form $\phi(s)=exp\\{-\tau
s-\int_{0}^{\infty}log(1+\frac{s}{t})\nu(dt)\\}$, where $\nu(dt)$ is a
nonnegative measure on $(0,\infty)$ and it satisfies (4), and $\tau$ is a non-
negative number which is called left extremity of the GGC random variable. The
pair $(\tau,\nu)$ is called generator of a GGC random variable. When $\tau=0$,
we call the associated random variable a GGC without a drift.

######  Remark 2.2.

Let $Z$ be a GGC with generating pair $(\tau,\nu)$. Define a GGC random
variable $\bar{Z}$ with LT given by
$\bar{\phi}(s)=e^{-\int_{0}^{\infty}log(1+\frac{s}{t})\nu(dt)}$. Then
$Z\overset{d}{=}\bar{Z}+\tau$ or equivalently $Z-\tau\overset{d}{=}\bar{Z}$.

Before we discuss the robustness problem that is stated above, we first
collect few properties of the GGC distributions in the following Lemma.

######  Lemma 2.3.

Let $Z$ be a nondegenerate GGC random variable with generating pair
$(\tau,\nu)$. Let $\hat{s}$ be the IN of $Z$. Then $\hat{s}$ is a finite
number and the measure $\nu$ satisfies $\nu([0,-\hat{s}])=0$. We have
$\mathcal{L}_{Z}(s)=Ee^{-sZ}=e^{-\tau
s-\int_{-\hat{s}}^{\infty}log(1+\frac{s}{z})\nu(dz)}$ and there exists a
deterministic strictly positive and decreasing (almost surely) function $h(s)$
on $[-\hat{s},+\infty)$ such that
$Z-\tau\overset{d}{=}\int_{-\hat{s}}^{+\infty}h(s)d\gamma_{s}$, where
$\gamma_{s}$ is a standard gamma subordinator with Lévy measure
$\frac{e^{-x}}{x}dx,x>0$.

###### Proof.

Due to Remark 2.2, it is sufficient to consider GGC random variables with zero
drift. Therefore below we assume $Z$ is a GGC with generating pair $(0,\nu)$.
Recall the Laplace transformation
$\mathcal{L}_{Z}(s)=Ee^{-sZ}=e^{-\int_{0}^{\infty}log(1+\frac{s}{z})\nu(dz)}$
of $Z$ from the definition 1.0 of [9] at page 3.51. By the definition of
$s_{0}$, the integral $\int_{0}^{\infty}log(1+\frac{s}{z})\nu(dz)$ is finite
for all $s>\hat{s}$. This means that one should have $1+\frac{s}{z}>0$ for all
$s>\hat{s}$. But this is true only if $z>-\hat{s}$. This implies that the
Thorin measure $\nu$ needs to assign zero measure to $[0,-\hat{s}]$. We can’t
have $\hat{s}=-\infty$ as this would imply $\nu([0,+\infty))=0$, a
contradiction for the non-degenerancy of $Z$. From part 2. of Proposition 1.1
of [9], we have $Z=\int_{-\hat{s}}^{+\infty}h(s)d\gamma_{s}$ with
$h(s)=1/F^{-1}_{\nu}(x)$, where $F^{-1}_{\nu}$ is the right continuous inverse
of $F_{\nu}(x):=\int_{(-\hat{s},x]}\nu(dx)$ on $[-\hat{s},+\infty)$. The
function $F_{\nu}(x)$ is a finite valued increasing function and therefore its
right continuous inverse $F_{\nu}^{-1}$ is also finite valued and increasing
function. Therefore $h(s)$ is a decreasing function. Now if $h(s)=0$ on
$[-\hat{s},s^{\prime})$ for some $s^{\prime}>-\hat{s}$, then
$\nu([0,s^{\prime}])=0$ which contradicts with the definition of $\hat{s}$.
Therefore $h(s)>0$ almost surely on $[-\hat{s},+\infty)$. ∎

Next we state the following continuity theorem which is useful for our
discussions.

######  Theorem 2.4.

Let $\\{Z_{n}\\}$ be a family from GGC random variables with generating pairs
$\\{(\tau_{n},\nu_{n})\\}$. Assume $Z_{n}$ converges weakly to a distribution
$Z$. Then $Z$ is also a GGC with a generating pair $(\tau,\nu)$. We have
$\nu_{n}$ converges weakly to $\nu$,
$\tau=\lim_{M\rightarrow\infty}\lim_{n\rightarrow\infty}[\tau_{n}+\int_{M}^{\infty}\frac{1}{x}\nu_{n}(dx)]$,
and $\lim_{\delta\rightarrow 0}\lim_{n\rightarrow\infty}\int_{0}^{\delta}(\ln
t)\nu_{n}(dt)=0$.

###### Proof.

See page 35 of [4], also see [17] and Theorem 1.22 of [8] for this. ∎

######  Remark 2.5.

Before we state our next result we make few observations on the LT of a GGC
random variable first. According to Proposition 1 of [5], a function $\phi(s)$
is a LT of a GGC random variable iff $\phi(s)$ is analytic in $C/(-\infty,0]$
and without zeros and $\phi(0)=1$ and $Im[\phi^{\prime}(s)/\phi(s)]\geq 0$ for
$Im(s)>0$. This means that $\phi(s)$ is differentiable for any $s>0$ and

$\phi^{\prime}(s)/\phi(s)=-\tau-\int_{0}^{\infty}\frac{1}{t+s}\nu(dt),$ (13)

where $(\tau,\nu)$ is the generator of a GGC random variable $Z$ with LT
equals to $\phi(s)$. Now assume $EZ<\infty$, then from $\phi(s)=Ee^{-sZ}$ we
have $\phi^{\prime}(s)=-E[Ze^{-sZ}]$ for all $s>0$. By dominated convergence
theorem we have $\lim_{s\rightarrow 0+}\phi^{\prime}(s)=-EZ$. Now from (13) we
obtain

$EZ=\tau+\int_{0}^{+\infty}\frac{1}{t}\nu(dt).$ (14)

Note that here we used monotone convergence theorem for the integral in the
right-hand-side of (13) to obtain (14). Now recall that a Thorin measure
satisfies $\int_{1}^{+\infty}\frac{1}{t}\nu(dt)<\infty$ and
$\int_{0}^{1}|logt|\nu(dt)<\infty$. The relation (14) shows that when
$EZ<\infty$ it also satisfies $\int_{0}^{1}\frac{1}{t}\nu(dt)<\infty$.

From this we immediately obtain the following result.

######  Lemma 2.6.

Let $\\{Z_{n}\\}$ be a family from GGC and assume $Z_{n}$ converges weakly to
$Z$ and assume $EZ_{n}<\infty,EZ<\infty$. Let $(\tau_{n},\nu_{n})$ be the
generator of $Z_{n}$ for each $n$ and let $(\tau,\nu)$ be the generator of
$Z$. Define $g^{(n)}(\delta)=:\int_{0}^{\delta}\frac{1}{t}\nu_{n}(dt)$ for
each $n\geq 1$ and assume

$g(\delta)=:\lim_{n\rightarrow\infty}g^{(n)}(\delta)$ (15)

exists and finite for each $\delta\in[0,c]$ for some $c>0$. Then
$g(\delta)=q(\delta)=:\int_{0}^{\delta}\frac{1}{t}\nu(dt)$ on $[0,c]$ and at
the same time we have $EZ_{n}\rightarrow EZ$.

###### Proof.

By the continuity theorem 2.4, $\nu_{n}$ weakly converges to $\nu$. Therefore
we have
$g^{(n)}(b)-g^{(n)}(a)=\int_{a}^{b}\frac{1}{t}\nu_{n}(dt)\rightarrow\int_{a}^{b}\frac{1}{t}\nu(dt)=q(b)-q(a)$
for any $c\geq b\geq a>0$. Since $g(\delta)$ is point-wise limit of
$g^{(n)}(\delta)$ we have $g(b)-g(a)=q(b)-q(a)$ for any $c\geq b\geq a>0$.
Being a point-wise limit of monotone decreasing functions, $g(\delta)$ is a
decreasing function on $[0,c]$. The function $q(\delta)$ is also a decreasing
function. Therefore the limits of $g(a)$ and $q(a)$ when $a\rightarrow 0$
exists. By taking the limit when $a\rightarrow 0$ to the equation
$g(b)-g(a)=q(b)-q(a)$ we obtain $g(b)-g(0)=q(b)$. It remains now to show that
$g(0)=0$. Assume $g(0)>0$ and below we show that this leads into a
contradiction. Since $g^{(n)}(\delta)$ is a decreasing function with
$g_{n}(0)=0$ for each fixed $n$, we can find a convergent sequence
$\delta_{n}\rightarrow 0$ such that $g^{(n)}(\delta_{n})\leq\frac{g(0)}{2}$.
But $g_{n}(\delta_{n})\rightarrow g(0)$ and this is a contradiction. Therefore
we have $g(\delta)=\int_{0}^{\delta}\frac{1}{t}\nu(dt)$. Next we show that
$EZ_{n}\rightarrow EZ$. To see this observe that

$EZ_{n}=\tau_{n}+\int_{M}^{\infty}\frac{1}{t}\nu_{n}(dt)+\int_{\delta}^{M}\frac{1}{t}\nu_{n}(dt)+\int_{0}^{\delta}\frac{1}{t}\nu_{n}(dt)$
(16)

for any $0<\delta\leq c<M<\infty$. By the continuity theorem 2.4 we have
$\tau=\lim_{M\rightarrow\infty}\lim_{n\rightarrow\infty}[\tau_{n}+\int_{M}^{\infty}\frac{1}{x}\nu_{n}(dx)]$.
Also we have
$\lim_{M\rightarrow\infty}\lim_{n\rightarrow\infty}\int_{\delta}^{M}\frac{1}{t}\nu_{n}(dt)=\lim_{M\rightarrow\infty}[\lim_{n\rightarrow\infty}\int_{\delta}^{M}\frac{1}{t}\nu_{n}(dt)]=\lim_{M\rightarrow\infty}\int_{\delta}^{M}\frac{1}{t}\nu(dt)=\int_{\delta}^{\infty}\frac{1}{t}\nu(dt)$.
Now taking the limit $\lim_{M\rightarrow\infty}\lim_{n\rightarrow\infty}$ to
the both sides of (17) we obtain
$EZ_{n}\rightarrow\tau+\int_{\delta}^{\infty}\frac{1}{t}\nu(dt)+\int_{0}^{\delta}\frac{1}{t}\nu(dt)$=EZ.
This ends the proof. ∎

######  Lemma 2.7.

Let $\\{Z_{n}\\}$ be a family from GGC and assume $Z_{n}$ converges weakly to
$Z$ and assume $EZ_{n}<\infty,EZ<\infty$. Let $\hat{s}$ be the IN of $Z$. If
$\hat{s}\neq 0$, then we have $EZ_{n}\rightarrow EZ$. Equivalently, if the
Thorin measure $U$ of $Z$ satisfies $U([0,d])=0$ for some $d>0$, then we have
$EZ_{n}\rightarrow EZ$.

###### Proof.

Let $\hat{s}_{n}$ and $\nu_{n}$ denote the IN and the Thorin measure of
$Z_{n}$ for each $n\geq 1$ respectively. Let $\nu$ denote the Thorin measure
of $Z$. From Lemma 2.16 below, we have $\hat{s}_{n}\rightarrow\hat{s}$.
Therefore there exists positive integer $n_{0}$ such that
$|\hat{s}_{n}|\geq\delta=:|\hat{s}|/2$ for all $n\geq n_{0}$. Define
$g^{(n)}(\theta)=:\int_{0}^{\theta}\frac{1}{t}\nu_{n}(dt)$ and
$g(\theta)=:\int_{0}^{\theta}\frac{1}{t}\nu(dt)$ for any $\theta\geq 0$. By
Lemma 2.3, when $\theta<\delta$ we have $\nu_{n}([0,\theta])=0$ for all $n\geq
n_{0}$ and $\nu([0,\theta])=0$. This shows that
$g^{(n)}(\theta)=0,\;g(\theta)=0$ for all $\theta<\delta$. Therefore the
condition (15) is trivially satisfied with $c=\frac{\delta}{2}$. From these
analysis observe that $U([0,d])=0$ for some $d>0$ if and only if the IN
$\hat{s}$ of $Z$ satisfies $\hat{s}\neq 0$. This completes the proof. ∎

######  Lemma 2.8.

Let $\\{Z_{n}\\}$ be a family from GGC and assume $Z_{n}$ converges weakly to
$Z$ with $EZ<\infty$. Let $(\tau_{n},\nu_{n})$ be the generator of $Z_{n}$ for
each $n$ and let $(\tau,\nu)$ be the generator of $Z$. Assume $\\{Z_{n}\\}$
has a sub-sequence $\\{Z_{n_{k}}\\}$ such that $EZ_{n_{k}}<\infty$ for all
$k\geq 1$ and the sequence $\\{EZ_{n_{k}}\\}_{k\geq 1}$ converges to a finite
number, then $EZ_{n_{k}}\rightarrow EZ$. Especially, if, in addition, $EZ_{n}$
is a bounded sequence then we have $EZ_{n}\rightarrow EZ$.

###### Proof.

Since any sub-sequence of the weakly convergent sequence converges weakly, it
is sufficient to prove if $EZ_{n}<\infty$ for all $n$ and if $EZ_{n}$
converges to a finite number then $EZ_{n}\rightarrow EZ$. For this it is
sufficient to show $g^{(n)}(\delta)=:\int_{0}^{\delta}\frac{1}{t}\nu_{n}(dt)$
converges point-wise to a finite valued function $g(\delta)$ on $[0,c]$ for
some $c>0$ by the above Lemma 2.6. Fix any number $c>0$ and observe that

$EZ_{n}=\tau_{n}+\int_{M}^{\infty}\frac{1}{t}\nu_{n}(dt)+\int_{\delta}^{M}\frac{1}{t}\nu_{n}(dt)+\int_{0}^{\delta}\frac{1}{t}\nu_{n}(dt)$
(17)

for any $0<\delta\leq c<M<\infty$. By the continuity theorem the limit
$\lim_{M\rightarrow\infty}\lim_{n\rightarrow}[\tau_{n}+\int_{M}^{\infty}\frac{1}{t}\nu_{n}(dt)]$
exists. Also the limit
$\lim_{M\rightarrow\infty}\lim_{n\rightarrow\infty}\int_{\delta}^{M}\frac{1}{t}\nu_{n}(dt)=\lim_{M\rightarrow\infty}[\lim_{n\rightarrow\infty}\int_{\delta}^{M}\frac{1}{t}\nu_{n}(dt)]=\lim_{M\rightarrow\infty}\int_{\delta}^{M}\frac{1}{t}\nu(dt)=\int_{\delta}^{\infty}\frac{1}{t}\nu(dt)$
exists for each fixed $0<\delta<c$. By the assumption the sequence $EZ_{n}$
converges to a finite limit. Therefore from (17) the limit of
$\lim_{n\rightarrow\infty}\int_{0}^{\delta}\frac{1}{t}\nu_{n}(dt)$ exists for
each $0<\delta\leq c$ and it is finite valued. This completes the proof. To
see the second claim of the Lemma it is sufficient to show any convergent sub-
sequence of $\\{EZ_{n}\\}$ converges to the same number $EZ$. Let $EZ_{n_{k}}$
be a convergent sub-sequence of $\\{EZ_{n}\\}$. Since by assumption
$\\{EZ_{n}\\}$ is a bounded sequence, $EZ_{n_{k}}$ converges to a finite
number. Therefore from the claim in the first half of the Lemma we have
$EZ_{n_{k}}\rightarrow EZ$. This completes the proof for the Lemma. ∎

The above Lemmas 2.6 and 2.8 give some sufficient conditions for the
convergence of the mean when a sequence of GGC converges weakly to a GGC.
Below we use these Lemmas to show that actually weak convergence in the family
of GGC random variables imply convergence of the corresponding mean values.
Namely we will show that the condition (15) in Lemma 2.6 and the boundedness
assumption of the expected values in Lemma 2.8 can be dropped. Before we prove
this result we need some preparations.

A Gamma distribution $\xi\sim G(\alpha,\beta)$ has density function
$g(x)=x^{\alpha-1}e^{-x/\beta}/(\beta^{\alpha}\Gamma(\alpha))$. A right-
shifted Gamma distribution is given by $\eta=\xi+\tau$, where $\xi\sim
G(\alpha,\beta)$ and $\tau\geq 0$ (see section 1.5 at page 28 of [8] for more
details). We use the notation $G(\alpha,\beta,\tau):=Law(Y)$ to denote the
right-shifted gamma distributions. In our discussions we need to consider
independent sequences $G(\alpha_{i},\beta_{i},\tau_{i})$ of right-shifted
Gamma distributions. We write
$G(\alpha_{i},\beta_{i},\tau_{i})=\xi_{i}+\tau_{i}$, where
$\\{\xi_{i}\\}_{1\leq i\leq n}$ is then a sequence of independent gamma random
variables with probability density functions

$f_{i}(x_{i})=f_{i}(x_{i};\alpha_{i},\beta_{i})=x^{\alpha_{i}-1}e^{-x_{i}/\beta_{i}}/[\beta_{i}^{\alpha_{i}}\Gamma(\alpha_{i})],\;\alpha_{i}>0,\;\beta_{i}>0,\;x_{i}>0.$
(18)

We have $\xi_{i}\sim\beta_{i}\bar{\xi}_{i}$ where $\bar{\xi}_{i}\sim
f_{i}(x_{i};\alpha_{i},1)$ and $\\{\bar{\xi}_{i}\\}_{1\leq i\leq n}$ are
mutually independent. With these we have
$\bar{Z}=:\bar{\xi}_{1}+\bar{\xi}_{2}+\cdots+\bar{\xi}_{n}\sim f(x;\alpha,1)$,
where $\alpha=\sum_{i=1}^{n}\alpha_{i}$. The distribution of
$Z=:\xi_{1}+\xi_{2}+\cdots+\xi_{n}$ is not known in closed form. Let $g(x)$
denote the probability density function of $Z$. Denote $\beta_{m}=:\min_{1\leq
i\leq n}\beta_{i}$ and for notational simplicity, without loosing any
generality, we can assume that $\beta_{m}=\beta_{1}$. Then, when not all of
$\\{\beta_{i}\\}$ are equal to each other, the paper [10] in its equation
(2.12) gives the following bound for $g$

$g(x)\leq[C/(\beta_{m}^{\rho}\Gamma(\rho))]x^{\rho-1}e^{-x(1-v)/\beta_{m}},$
(19)

where

$C=\prod_{i=1}^{n}(\beta_{m}/\beta_{i})^{\alpha_{i}},\;\rho=\sum_{i=1}^{n}\alpha_{i},\;v=\max_{2\leq
i\leq n}(1-\beta_{m}/\beta_{i}).$ (20)

######  Remark 2.9.

Since for two independent gamma random variables $\xi_{1}\sim
f(x;\alpha_{1},\beta_{1})$ and $\xi_{2}\sim f(x;\alpha_{2},\beta_{2})$ with
$\beta=:\beta_{1}=\beta_{2}$, we have $\xi_{1}+\xi_{2}\sim
f(x;\alpha_{1}+\alpha_{2},\beta)$, in (19) we can assumed that
$\beta_{1}=\beta_{m}<\min_{2\leq i\leq n}\beta_{i}$. Also we observe that the
numbers $C$ and $\nu$ in (19) are bounded numbers.

In our next result, we will consider sequences

$\bar{Z}_{n}=\prod_{i=1}^{k_{n}}\ast
G(\alpha_{i}^{(n)},\beta_{i}^{(n)},\tau_{i}^{(n)}),\;\;n\geq 1,$ (21)

of finite convolutions of right-shifted gamma distributions, where
$k_{n},n\geq 1,$ are positive integers. If we denote
$\tau^{(n)}=\sum_{i}^{k_{n}}\tau_{i}^{(n)}$, then we have

$\bar{Z}_{n}\sim\tau^{(n)}+\xi^{(n)}_{1}+\xi_{2}^{(n)}+\cdots+\xi_{k_{n}}^{(n)},\;\;\xi_{i}^{(n)}\sim
f(x;\alpha^{(n)}_{i},\beta^{(n)}_{i}),\;1\leq i\leq k_{n}.$ (22)

We denote

$Z_{n}=:\xi^{(n)}_{1}+\xi_{2}^{(n)}+\cdots+\xi_{k_{n}}^{(n)},n\geq 1$ (23)

and

$C^{(n)}=:\prod_{i=1}^{k_{n}}(\beta_{m}^{(n)}/\beta_{i}^{(n)})^{\alpha_{i}^{(n)}},\;\rho^{(n)}=:\sum_{i=1}^{k_{n}}\alpha_{i}^{(n)},\;v^{(n)}=\max_{2\leq
i\leq k_{n}}(1-\beta_{m}^{(n)}/\beta_{i}^{(n)}),$ (24)

where $\beta_{m}^{(n)}=\min_{1\leq i\leq
k_{n}}\beta^{(n)}_{i}=\beta_{1}^{(n)}$ (again $\beta_{1}^{(n)}$ is assumed to
be minimum of $\\{\beta_{i}^{(n)}\\}_{1\leq i\leq k_{n}}$). Let $g_{n}(x)$
denote the probability density function of $Z_{n}$ for each $n\geq 1$. Then
assuming that not all of $\beta_{i}^{(n)},1\leq i\leq k_{n}$ are equal to each
other, we have

$g_{n}(x)\leq[C^{(n)}(\beta_{m}^{(n)})^{-\rho^{(n)}}/\Gamma(\rho^{(n)})]x^{\rho^{(n)}-1}e^{-x(1-v^{(n)})/\beta_{m}^{(n)}}.$
(25)

Next we prove the following Lemma.

######  Lemma 2.10.

Let $\bar{Z}_{n}$ be a sequence of finite convolutions of right-shifted gamma
distributions. If $\bar{Z}_{n}$ converges weakly to a non-degenerate random
variable $Z$ with $EZ<\infty$, we have $E\bar{Z}_{n}\rightarrow EZ$.

###### Proof.

Let $(\tau,U)$ denote the generating pair for $Z$. We assume that the sequence
$\bar{Z}_{n}$ is given by (21). We denote by $U^{(n)}$ the Thorin measure of
$\bar{Z}_{n}$ for each $n\geq 1$. To prove the claim in the Lemma it is
sufficient, by Lemma 2.8, to prove that $\\{E\bar{Z}_{n}\\}$ is a bounded
sequence. Below we divide the proof into two parts a) and b).

(a) For arbitrarily fixed positive numbers $0<\delta<K<+\infty$, we define

$\begin{split}&I^{(n)}_{0,\delta}=\\{i:\frac{1}{\beta_{i}^{(n)}}\leq\delta,\;1\leq
i\leq k_{n}\\},\\\
&I^{(n)}_{\delta,K}=\\{i:K\geq\frac{1}{\beta_{i}^{(n)}}>\delta,\;1\leq i\leq
k_{n}\\},\\\ &I^{(n)}_{K,+\infty}=\\{i:\frac{1}{\beta_{i}^{(n)}}>K,\;1\leq
i\leq k_{n}\\},\end{split}$ (26)

and we write
$\bar{Z}_{n}=\tau^{(n)}+\tilde{Z}_{1}^{(n)}+\tilde{Z}_{2}^{(n)}+\tilde{Z}_{3}^{(n)}$,
where $\tilde{Z}_{1}^{(n)}=\sum_{i\in
I_{0,\delta}^{(n)}}\xi_{i}^{(n)},\tilde{Z}_{2}^{(n)}=\sum_{i\in
I_{\delta,K}^{(n)}}\xi_{i}^{(n)},$ and $\tilde{Z}_{3}^{(n)}=\sum_{i\in
I_{K,+\infty}^{(n)}}\xi_{i}^{(n)}$. We write the Laplace transformation of $Z$
as $\mathcal{L}_{Z}(s)=\mathcal{L}_{1}\mathcal{L}_{2}\mathcal{L}_{3}$ with

$\mathcal{L}_{1}=e^{\int_{0}^{\delta}log(\frac{1}{1+s/t})U(dt)},\mathcal{L}_{2}=e^{\int_{\delta}^{K}log(\frac{1}{1+s/t})U(dt)},\mathcal{L}_{3}=e^{\tau+\int_{K}^{+\infty}log(\frac{1}{1+s/t})U(dt)},$
(27)

and denote by $\tilde{Z}_{1}$ the GGC random variable that corresponds to
$\mathcal{L}_{1}$, and by $\tilde{Z}_{2}$ the GGC random variable that
corresponds to $\mathcal{L}_{2}$, and similarly by $\tilde{Z}_{3}$ the GGC
that corresponds to $\mathcal{L}_{3}$ (see for a similar idea in the proof of
Theorem 3.1 at page 130 of [1]). The Laplace distribution of $Z$ is the
multiplication of three Laplace transformations and therefore $Z$ is equal in
distribution to the independent sum of $\tilde{Z}_{1},\tilde{Z}_{2}$, and
$\tilde{Z}_{3}$, i.e., $Z\sim\tilde{Z}_{1}+\tilde{Z}_{2}+\tilde{Z}_{3}$. We
also have that $\tilde{Z}_{1}^{(n)}$ converges weakly to $\tilde{Z}_{1}$,
$\tilde{Z}_{2}^{(n)}$ converges weakly to $\tilde{Z}_{2}$, and
$\tau_{n}+\tilde{Z}_{3}^{(n)}$ converges weakly to $\tilde{Z}_{3}$. To see
this, observe that the Laplace distribution of $\tilde{Z}_{1}^{(n)}$ does not
contribute to $\mathcal{L}_{2}$ and $\mathcal{L}_{3}$ in the limit due to the
restriction in the set $I_{0,\delta}^{(n)}$ above. Therefore its Laplace
distribution need to converge $\mathcal{L}_{1}$, which in turn implies
$\tilde{Z}_{1}^{(n)}$ converges weakly to $\tilde{Z}_{1}$. Similar arguments
hold for the other two.

To show $\\{E\bar{Z}_{n}\\}$ is a bounded sequence, it is sufficient to show
all of $E\tilde{Z}_{1}^{(n)},E\tilde{Z}_{2}^{(n)},$ and
$\tau_{n}+E\tilde{Z}_{3}^{(n)}$ are bounded sequences. The rest of the proof
is devoted to show this claim. First observe that

$E(\tau_{n}+\tilde{Z}_{3}^{(n)})=\tau_{n}+\int_{K}^{+\infty}\frac{1}{t}U^{(n)}(dt).$
(28)

By the continuity Theorem 2.4 we have

$\lim_{K\rightarrow+\infty}\lim_{n+\infty}[\tau_{n}+\int_{K}^{+\infty}\frac{1}{t}U^{(n)}(dt)]=\tau<\infty.$

Therefore for a fixed number $\epsilon>0$ there exists positive integer
$n_{0}$ and positive number $K_{0}$ such that
$\tau_{n}+\int_{K_{0}}^{+\infty}\frac{1}{t}U^{(n)}(dt)$ lies in the interval
$[\tau-\epsilon,\tau+\epsilon]$ when $n\geq n_{0}$. For the simplicity of
notations we denote $K_{0}$ by $K$ and the $\\{\tilde{Z}_{3}^{(n)}\\}$ (these
are defined to be $\tilde{Z}_{3}^{(n)}=\sum_{i\in
I_{K_{0},+\infty}^{(n)}}\xi_{i}^{(n)}$ in fact and we use $K$ instead of
$K_{0}$ for notational simlicity) are defined as above. We conclude that
$E(\tau_{n}+Z_{3}^{(n)})$ is a bounded sequence of $n$. Also we have
$EZ_{2}^{(n)}=\int_{\delta}^{K}\frac{1}{t}U^{(n)}(dt)$ and as $Z_{n}$
converges weakly to $Z$, the measure $U^{(n)}$ converges weakly to $U$ by the
continuity Theorem 2.4. Therefore
$EZ_{2}^{(n)}=\int_{\delta}^{K}\frac{1}{t}U^{(n)}(dt)\rightarrow\int_{\delta}^{K}\frac{1}{t}U(dt)<\infty$.
From this we conclude that $EZ_{2}^{(n)}$ is also a bounded sequence of $n$.
Next we show that $EZ_{1}^{(n)}$ is a bounded sequence of $n$ also and we do
this in part b) below.

b) Recall that $\tilde{Z}_{1}^{(n)}=\sum_{i\in
I_{0,\delta}^{(n)}}\xi_{i}^{(n)}$. We have
$E\tilde{Z}^{n}_{1}=\int_{0}^{\delta}\frac{1}{t}U^{(n)}(dt)$. Here we can’t
apply the same idea that we have used for $\\{Z_{2}^{(n)}\\}$ and
$\\{Z_{3}^{(n)}\\}$ in part a) above as we don’t know if
$\int_{0}^{\delta}\frac{1}{t}U^{(n)}(dt)\rightarrow\int_{0}^{\delta}\frac{1}{t}U(dt)$
holds true. However we have an upper bound as in (19) for the density
functions of finite gamma convolutions and we use this fact to show that
$EZ_{1}^{(n)}$ is a bounded sequence. First observe that
$\int_{0}^{\delta}U^{(n)}(dt)=\sum_{i\in
I_{0,\delta}^{(n)}}\alpha_{i}^{(n)}\rightarrow\int_{0}^{\delta}U(dt)<\infty$
as $U^{(n)}$ weakly converges to $U$. Therefore $\rho_{1}^{(n)}=:\sum_{i\in
I_{0,\delta}^{(n)}}\alpha_{i}^{(n)}$ is a bounded sequence of $n$. We have
$EZ_{1}^{(n)}=\sum_{i\in I_{0,\delta}^{(n)}}\alpha_{i}^{(n)}\beta_{i}^{(n)}.$
Now if $\\{\beta_{i}^{(n)},i\in I^{(n)}_{0,\delta}\\}_{n\geq 1}$ is a
uniformly bounded family of $n$, then clearly we have $EZ_{1}^{(n)}$ is a
bounded sequence of $n$. Therefore we exclude this case from our discussions
below and we assume that $\\{\beta_{i}^{(n)},n\geq 1,i\in
I^{(n)}_{0,\delta}\\}$ is an unbounded family of $n$.

Below for the sake of notations we assume that the family $\\{Z_{1}^{(n)}\\}$
is given by (23) and we denote this family, with abuse of notations, by
$\\{Z_{n}\\}$. So we have $Z_{n}$ converges weakly to $\tilde{Z}_{1}$ in part
a) above. The corresponding probability density functions have the upper
bounds as in (25). We use the same notations $\rho^{(n)},\beta_{m}^{(n)},$ as
in (24). We also define $\beta_{M}^{(n)}=\max_{1\leq i\leq
k_{n}}\beta_{i}^{(n)}$. We divide the family $\\{Z_{n}\\}_{n\geq 1}$ (namely
the family $\\{Z_{1}^{(n)}\\}$) into two disjoint sets $\\{Z_{n}\\}_{n\geq
1}=\\{Z_{n}^{\prime}\\}_{n\geq 1}\cup\\{Z_{n}^{{}^{\prime\prime}}\\}_{n\geq
1}$, where each member of $\\{Z_{n}^{\prime}\\}$ is finite gamma convolutions
with not all $\\{\beta^{\prime}_{i}\\}$ are equal to each other and the family
$\\{Z_{n}^{{}^{\prime\prime}}\\}$ is such that each member of it is finite
gamma convolutions with equal $\beta^{{}^{\prime\prime}}_{i}$. Both of
$Z_{n}^{{}^{\prime}}$ and $Z_{n}^{{}^{\prime\prime}}$ (being sub-sequences of
$Z_{n}$, more precisely of $Z_{1}^{(n)}$) converge weakly to $\tilde{Z}_{1}$.

First consider the sequence $Z_{n}^{{}^{\prime\prime}}$. It is clear that, in
fact, each $Z_{n}^{{}^{\prime\prime}}$ are single gamma random variables with
$Z_{n}^{{}^{\prime\prime}}\sim
G(\rho{{}^{\prime\prime}}_{n},\beta^{{}^{\prime\prime}}_{n})$ for some
$\rho_{n}^{{}^{\prime\prime}}$ and $\beta_{n}^{{}^{\prime\prime}}$. Clearly
$\rho_{n}^{{}^{\prime\prime}}$ is a sub-sequence of $\rho_{1}^{(n)}$ defined
in the above paragraph. As such $\rho_{n}^{{}^{\prime\prime}}$ is a bounded
sequence. Now assume $\rho_{n}^{{}^{\prime\prime}}\rightarrow 0$. Let
$U_{n}^{{}^{\prime\prime}}$ denote the Thorin measure of
$Z_{n}^{{}^{\prime\prime}}$ for each $n\geq 1$. Then since
$Z_{n}^{{}^{\prime\prime}}$ weakly converges to $\tilde{Z}_{1}$, we have
$\int_{0}^{\delta}U(dt)=\lim_{n\rightarrow+\infty}\int_{0}^{\delta}U_{n}^{{}^{\prime\prime}}=\lim_{n\rightarrow+\infty}\rho_{n}^{{}^{\prime\prime}}=0$,
which shows $U([0,\delta])=0$. But if this is true then we have
$E\bar{Z}_{n}\rightarrow EZ$ already by Lemma 2.7. So we assume
$\inf_{n}\rho_{n}^{{}^{\prime\prime}}>0$ below. Now if
$EZ_{n}^{{}^{\prime\prime}}$ is an unbounded sequence then we should have
$\beta_{n}^{{}^{\prime\prime}}\rightarrow+\infty$ as
$EZ_{n}^{{}^{\prime\prime}}=\rho_{n}^{{}^{\prime\prime}}\beta_{n}^{{}^{\prime\prime}}$
and $\rho_{n}^{{}^{\prime\prime}}$ is a bounded sequence from below and above
as explained above. Then the sequence of probability density functions
$f_{n}^{{}^{\prime\prime}}(x)=x^{\rho_{n}^{{}^{\prime\prime}}-1}e^{-x/\beta_{n}^{{}^{\prime\prime}}}/[(\beta_{n}^{{}^{\prime\prime}})^{\rho_{n}^{{}^{\prime\prime}}}\Gamma(\rho_{n}^{{}^{\prime\prime}})]$
of $Z_{n}^{{}^{\prime\prime}}$ converges to zero almost surely. But at the
same time $f_{n}^{{}^{\prime\prime}}(x)$ should converge to the density
function of $\tilde{Z}_{1}$ by Theorem 4.1.5 of [4] implying that the
probability density function of $\tilde{Z}_{1}$ is zero, a contradiction.
Therefore $EZ_{n}{{}^{\prime\prime}}$ is a bounded sequence.

Now it remains to show that $EZ_{n}^{{}^{\prime}}$ is a bounded sequence.
Below, with another abuse of notation, we denote the family
$\\{Z_{n}^{\prime}\\}$ by $\\{Z_{n}\\}$ and show that $EZ_{n}$ is a bounded
sequence. The probability density function of $Z_{n}$ is denoted by $g_{n}(x)$
for each $n$. We have

$g_{n}(x)\leq[C^{(n)}(\beta_{m}^{(n)})^{-\rho^{(n)}}/\Gamma(\rho^{(n)})]x^{\rho^{(n)}-1}e^{-x(1-v^{(n)})/\beta_{m}^{(n)}}.$
(29)

In (29), the family $\\{\rho^{(n)}\\}_{n\geq 1}$ is uniformly bounded as
explained above. Since the Thorin measure of $\tilde{Z}_{1}^{(n)}$ has support
in the interval $[0,\delta]$, we have $\frac{1}{\beta_{m}^{(n)}}\leq\delta$
for all $n$ and this gives a lower bound $\beta_{m}^{(n)}\geq\frac{1}{\delta}$
for the family $\\{\beta_{m}^{(n)}\\}_{n\geq 1}$. Also we can assume that the
family $\\{\beta_{m}^{(n)}\\}_{n\geq 1}$ is uniformly bounded. If not, since
the Thorin measure $U^{(n)}|_{[0,\delta]}$ (the restriction of the Thorin
measure of $\bar{Z}_{n}$ to $[0,\delta]$) has support in
$[\frac{1}{\beta_{M}^{(n)}},\frac{1}{\beta_{m}^{(n)}}]$, we have
$U^{(n)}([0,\delta])\rightarrow 0$ as $n\rightarrow 0$. This implies that
$U([0,\delta])=0$ and we are reduced to the case of Lemma 2.7 and so we have
$E\bar{Z}_{n}\rightarrow EZ$ trivially. Therefore we assume
$\\{\beta_{m}^{(n)}\\}_{n\geq 1}$ is uniformly bounded. Also, as explained
above, we can assume that $\beta_{M}^{(n)}$ is an unbounded sequence (this
corresponds to the Thorin measure $U$ has support in any close neighborhoods
of $0$) as if it is bounded sequence then the Thorin measure $U$ satisfies
$U([0,\inf_{n}{(1/\beta_{M}^{(n)})}])=0$ with
$\inf_{n}{(1/\beta_{M}^{(n)})}>0$ and again we are reduced to the case of
Lemma 2.7.

Now we look at $v^{(n)}=\max_{2\leq i\leq
k_{n}}(1-\beta_{m}^{(n)}/\beta_{i}^{(n)})$ in (29). As explained above we can
assume that $\\{\beta_{m}^{(n)}\\}$ is a bounded family and
$\\{\beta_{M}^{(n)}\\}$ is an unbounded family. Therefore there exists $n_{0}$
such that for all $n\geq n_{0}$ we have $v^{(n)}\leq\frac{1}{2}$. Therefore
when $n\geq n_{0}$ we have
$(1-v^{(n)})/\beta_{m}^{(n)}\geq\beta_{m}^{(n)}\geq\frac{1}{2B}$, where $B$ is
any fixed upper bound for $\\{\beta_{m}^{(n)}\\}$. Also observe that

$\ln C^{(n)}=\beta_{m}^{(n)}\\{\sum_{i\in
I_{0,\delta}^{(n)}}\alpha_{i}^{(n)}\\}-\sum_{i\in
I_{0,\delta}^{(n)}}\alpha_{i}^{(n)}\ln\beta_{i}^{(n)}.$

Since $\sum_{i\in
I_{0,\delta}^{(n)}}\alpha_{i}^{(n)}\ln\beta_{i}^{(n)}=\int_{0}^{\delta}(\ln
t)U_{n}(dt)\rightarrow 0$ by the continuity Theorem 2.4, we can conclude that
$\\{C^{(n)}\\}_{n\geq 1}$ is a bounded sequence. Therefore for all $n\geq
n_{0}$ we have

$g_{n}(x)\leq\bar{C}x^{\bar{\rho}-1}e^{-\bar{B}x},$ (30)

where $\bar{B}=\frac{1}{2B}$, $\bar{\rho}$ is any fixed upper bound for
$\rho^{(n)}$, and $\bar{C}$ is any fixed upper bound for the family
$\\{C^{(n)}(\beta_{m}^{(n)})^{-\rho^{(n)}}/\Gamma(\rho^{(n)})\\}_{n\geq 1}$.
Next, we show that (30) implies that $EZ_{n}$ is a bounded sequence. To see
this, recall that $Z_{n}$ converges weakly to $\tilde{Z}_{1}$ and
$E\tilde{Z}_{1}<\infty$. Let $g(x)$ denote the probability density function of
$\tilde{Z}_{1}$. By Theorem 4.1.5 of [4], the sequence $g_{n}(x)$ converges
point-wise to $g(x)$. We have the following

$|EZ_{n}-E\tilde{Z}_{1}|\leq\int_{0}^{+\infty}x|g_{n}(x)-g(x)|dx=J_{1}^{(n)}(L)+J_{2}^{(n)}(L),$
(31)

where $J_{1}^{(n)}(L)=:\int_{0}^{L}x|g_{n}(x)-g(x)|dx$ and
$J_{2}^{(n)}(x)=:\int_{L}^{+\infty}x|g_{n}(x)-g(x)|dx$ and $L$ can be any
positive number but we require $L>1$ . If we can show that both
$\\{J_{1}^{(n)}(L)\\}_{n\geq 1}$ and $\\{J_{2}^{(n)}(L)\\}_{n\geq 1}$ are
bounded family then from (31) we can conclude that $\\{EZ_{n}\\}$ is a bounded
family. For each fixed $L>1$, we have $J_{1}^{(n)}(L)$ converges to zero by
the dominated convergence Theorem. For $J_{2}(L)$ we have

$J_{2}(L)\leq\int_{L}^{+\infty}xg_{n}(x)dx+\int_{L}^{+\infty}xg(x)dx.$

We have $\int_{L}^{+\infty}xg(x)dx\leq E\tilde{Z}_{1}\leq EZ<\infty$ for all
$L>1$. For $\int_{L}^{+\infty}xg_{n}(x)dx$ we have

$\int_{L}^{+\infty}xg_{n}(x)dx\leq\bar{C}\int_{L}^{+\infty}x^{\bar{\rho}}e^{-\bar{B}x}dx,$

due to (30) whenever $n\geq n_{0}$. By applying integration by parts multiple
times we can obtain $\int_{L}^{+\infty}x^{\bar{\rho}}e^{-\bar{B}x}dx\leq
Q(L)e^{-\tilde{B}L}$, where $Q(L)$ is a polynomial of $L$ (here we need to use
the requirement $L>1$). But $Q(L)e^{-\tilde{B}L}\rightarrow 0$ as
$L\rightarrow+\infty$ for any polynomial $Q(L)$ of $L$. Therefore for a given
finite number $M>0$, we have a positive number $L_{0}>1$ such that
$Q(L)e^{-\bar{B}L}\leq M$ for all $L\geq L_{0}$. Then $J_{2}^{(n)}(L_{0})\leq
EZ+M$ for all $n\geq n_{0}$. We also have $J_{1}^{(n)}(L_{0})\rightarrow 0$ as
explained above. Therefore there exists a positive integer $n_{0}^{\prime}$
such that $J_{1}^{(n)}(L_{0})\leq M$ for all $n\geq n_{0}^{\prime}$. Then for
all $n\geq\max\\{n_{0},n_{0}^{\prime}\\}$ we have

$|EZ_{n}-E\tilde{Z}_{1}|\leq 2M+EZ.$

This shows that $\\{EZ_{n}\\}$ is a bounded sequence as
$E\tilde{Z}_{1}<\infty$. This completes the proof. ∎

######  Remark 2.11.

The above Lemma 2.16 shows that if the members of the sequence $\\{Z_{n}\\}$
are finite convolutions of right shifted gamma random variables, then the weak
convergence of $Z_{n}$ to a non-degenerate random variable $Z$ with
$EZ<\infty$ implies the convergence of the mean value, i.e, $EZ_{n}\rightarrow
EZ$. Here, unlike Lemma 2.8, we don’t have to require the boundedness of the
sequence $EZ_{n}$. As the proof of this Lemma shows the properties of the
finite gamma convolutions play important role for the proof of this Lemma. We
wish to show a similar result for the general class of GGC random variables.
For this, we need to use a result from the paper [11]: let $Z_{n}$ be a
sequence of random variables with the Laplace transformations
$\mathcal{L}_{Z_{n}}(s)$ and $Z$ be a random variable with Laplace
transformation $\mathcal{L}_{Z}(s)$. Then, according to [11], the sequence
$\mathcal{L}_{Z_{n}}(s)$ converges to $\mathcal{L}_{Z}(s)$ point-wise on some
interval $(c,d)$ if and only if $Z_{n}$ converges weakly to $Z$ and
$\sup_{n}\mathcal{L}_{Z_{n}}(s)<\infty$ for all $s\in(c,d)$. Here $(c,d)$ can
be any open interval. Also we need to use the fact that if a sequence of
monotone continuous functions $g_{n}(x)$ converge point-wise to a continuous
function $g(x)$ on a compact interval $[c,d]$, then the convergence is uniform
on $[c,d]$.

######  Theorem 2.12.

Let $Z_{n}$ be a sequence of non-degenerate random variables from GGC with
$EZ_{n}<\infty$ for each $n\geq 1$. Assume $Z_{n}$ weakly converges to a non-
degenerate random variable $Z$ with $EZ<\infty$. Then we have
$EZ_{n}\rightarrow EZ$.

###### Proof.

Let $\\{Z^{(n)}_{k}\\}_{k\geq 1}$ denote a sequence of finite convolutions of
right-shifted gamma distributions that converges weakly to $Z_{n}$ for each
$n$. Let $g_{k}^{(n)}(s)=:\mathcal{L}_{Z^{(n)}_{k}}(s)$ denote the Laplace
transformations of $Z_{k}^{(n)}$ respectively for all $k\geq 1,n\geq 1$. Let
$g^{(n)}(s)=:\mathcal{L}_{Z_{n}}(s)$ denote the Laplace transformation of
$Z_{n}$ for each $n\geq 1$. Let $g(s)=:\mathcal{L}_{Z}(s)$ denote the Laplace
transormation of the limit random variable $Z$. For each fixed $n$, we have
$g_{k}^{(n)}(s)$ converges point-wise to $g^{(n)}(s)$ on the compact interval
$[0,1]$ due to weak convergence of $Z_{k}^{(n)}$ to $Z^{(n)}$. Since all of
$g_{k}^{(n)}(s),g^{(n)}(s)$ are decreasing functions on $[0,1]$, the
convergence is uniform. Also, by following a similar argument, we have
$g^{(n)}(s)$ converges uniformly to $g(s)$ on $[0,1]$. Now, from above Lemma
2.16, we have $EZ_{k}^{(n)}\rightarrow EZ_{n}$ for each fixed $n\geq 1$.
Therefore for each fixed $n\geq 1$, we can pick a finite gamma convolution
$Z_{k_{n}}^{(n)}$ with the property

$|EZ_{n}-EZ_{k_{n}}^{n}|\leq\frac{1}{n},\;\;\sup_{s\in[0,1]}|\mathcal{L}_{Z_{n}}(s)-\mathcal{L}_{Z_{k_{n}}^{(n)}}(s)|\leq\frac{1}{n},$
(32)

at the same time. We now show that the sequence $Z_{k_{n}}^{(n)}$ converges
weakly to $Z$. Let $\epsilon>0$ be an arbitrary small number. Since $Z_{n}$
weakly converges to $Z$, there exists a positive integer $n_{0}$ such that
$\sup_{s\in[0,1]}|\mathcal{L}_{Z}(s)-\mathcal{L}_{Z_{n}}(s)|\leq\epsilon$ and
$\frac{1}{n}<\epsilon$ at the same time for all $n\geq n_{0}$. Then for all
$n\geq n_{0}$ we have

$\begin{split}\sup_{s\in[0,1]}|\mathcal{L}_{Z}(s)-\mathcal{L}_{Z_{k_{n}}^{(n)}}(s)|\leq&\sup_{s\in[0,1]}|\mathcal{L}_{Z}(s)-\mathcal{L}_{Z_{n}}(s)|+\sup_{s\in[0,1]}|\mathcal{L}_{Z_{n}}(s)-\mathcal{L}_{Z_{k_{n}}^{(n)}}(s)|,\\\
\leq&2\epsilon.\end{split}$ (33)

Since $\epsilon>0$ is an arbitrary small number, we conclude that
$\mathcal{L}_{Z_{k_{n}}^{(n)}}(s)$ converges to $\mathcal{L}_{Z}(s)$ point-
wise on $(0,1)$. Then by Remark 2.11, the sequence of finite gamma
convolutions $Z_{k_{n}}^{(n)}$ converges weakly to $Z$. Now if we have
$EZ_{n}\rightarrow+\infty$, then due to (32) we have
$EZ_{k_{n}}^{(n)}\rightarrow+\infty$ also. But this contradicts with Lemma
2.16 above. Therefore $\\{EZ_{n}\\}$ is a bounded sequence. Then
$EZ_{n}\rightarrow EZ$ follows from Lemma 2.8. ∎

######  Lemma 2.13.

Let $\\{Z_{n}\\}$ be a family from GGC with corresponding generating pairs
$\\{(\tau_{n},\nu_{n})\\}$. Assume $Z_{n}$ converges weakly to $Z$ with
generating pair $(\tau,\nu)$. Let
$Z_{n}-\tau_{n}\overset{d}{=}\int_{0}^{\infty}h_{n}(s)d\gamma_{s}$,
$Z-\tau\overset{d}{=}\int_{0}^{\infty}h(s)d\gamma_{s}$ denote Wiener-Gamma
representations with unique increasing functions $h_{n}(s),h(s)$. Then
$h_{n}(s)\rightarrow h(s)$ pointwise.

###### Proof.

By the continuity Theorem 2.4, $\nu_{n}$ weakly converges to $\nu$. Denote
$F_{\nu_{n}}(x)=\int_{(0,x]}\nu_{n}(dy)$ and $F_{\nu}(x)=\int_{(0,x]}\nu(dy)$
and denote by $F_{\nu_{n}}^{-1},F_{\nu}^{-1}$ their respective right-
continuous inverses. From part 2 of Proposition 1.1 of [9] we have
$h_{n}(s)=1/F_{\nu_{n}}^{-1}$ and $h(s)=1/F_{\nu}^{-1}$. Now from Theorem 2.4,
weak convergence implies $F_{\nu_{n}}\rightarrow F_{\nu}$ pointwise and this
in turn implies $F_{\nu_{n}}^{-1}\rightarrow F_{\nu}^{-1}$. Therefore we have
$h_{n}(s)\rightarrow h(s)$ pointwise. ∎

Recall that our goal is to discuss the robustness problem of the optimal
portfolio in (7). The optimal portfolio in this theorem involves the Laplace
transformation of the mixing distribution. Therefore, we first need to study
the properties of the Laplace transformation of the GGC random variables.
Especially, we would like to study the relation of weak convergence with the
convergence of the corresponding Laplace transformations within the class of
GGC distributions. First recall the classical result that a sequence $Z_{n}$
converges to $Z$ in distribution, i.e., $Z_{n}\overset{w}{\rightarrow}Z$, if
and only if $Ef(Z_{n})\rightarrow Ef(Z)$ for any bounded and continuous
function $f$. This result clearly does not imply that the Laplace
transformations $\mathcal{L}_{Z_{n}}(s)$ of the random variables $Z_{n}$
converge to the Laplace transformation $\mathcal{L}_{Z}(s)$ of the random
variable $Z$ under the condition that $Z_{n}\overset{w}{\rightarrow}Z$ as the
functions $f(x)=e^{-sx}$ are not bounded functions when $s<0$. In our setting
all the random variables in GGC are non-negative and therefore
$\mathcal{L}_{Z_{n}}(s)\rightarrow\mathcal{L}_{Z}(s)$ for all $s\geq 0$ as
long as $Z_{n}\overset{w}{\rightarrow}Z$. But when $s<0$ such result does not
immediately follow as then the function $e^{-sx}$ is no longer bounded on
$(0,+\infty)$ when $s<0$ as mentioned above. But interestingly we will show
that $Z_{n}\overset{w}{\rightarrow}Z$ implies
$\mathcal{L}_{Z_{n}}(s)\rightarrow\mathcal{L}_{Z}(s)$ whenever
$\mathcal{L}_{Z}(s)<\infty$ as long as $Z_{n},Z$ are within the class GGC of
distributions. First, in the next simple Lemma we state some useful facts on
the GGC random variables.

######  Lemma 2.14.

Let $Z_{n},Z$ be a family from GGC with respective generators
$(\tau_{n},\nu_{n})$ and $(\tau,\nu)$. Assume $EZ_{n}<\infty,EZ<\infty$. Let
$\hat{s}_{n},\hat{s}$ denote their corresponding IN. Let $b_{n}>0$ be any
sequence of real numbers with $b_{n}\rightarrow b>0$, $b_{n}<|\hat{s}_{n}|$
for all $n\geq 1$, and $b<|\hat{s}|$. Then if $Z_{n}\overset{w}{\rightarrow}Z$
we have $Ee^{b_{n}Z_{n}}\rightarrow Ee^{bZ}$.

###### Proof.

Since $b_{n}\rightarrow b$ we have $b_{n}Z_{n}\overset{w}{\rightarrow}bZ$.
Since $e^{x}$ is a continuous function we have
$\eta_{n}=:e^{b_{n}Z_{n}}\overset{w}{\rightarrow}\eta=:e^{bZ}$. By the
definitions of $b,\;b_{n}$ we have $Ee^{b_{n}Z_{n}}<\infty,Ee^{bZ}<\infty$ for
all $n$. Exponentials of GGC are again GGC (see [5] for this). So
$\eta_{n},\eta$ are GGC. The claim now follows from Theorem 2.12. ∎

######  Remark 2.15.

From the above Lemma 2.14, it can be seen that if a real number $s>0$
satisfies $s<|\hat{s}_{n}|$ for all $n\geq 1$ and $s<|\hat{s}|$ then
$Ee^{sZ_{n}}\rightarrow Ee^{sZ}$ whenever $Z_{n},Z$ satisfy the hypothesis of
the Lemma 2.14.

######  Lemma 2.16.

Assume all of $Z_{n},Z$ are nondegenerate GGC random variables with
corresponding generating pairs $(\tau_{n},\nu_{n}),(\tau,\nu)$. Let
$\hat{s}_{n}$ and $\hat{s}$ denote their corresponding IN respectively. If
$Z_{n}\overset{w}{\rightarrow}Z$, then we have
$\hat{s}_{n}\rightarrow\hat{s}$.

###### Proof.

Due to Remark 2.2, we can assume $\tau_{n}=0,\tau=0$. First we show
$\\{|\hat{s}_{n}|\\}$ is a bounded sequence. Assume $\\{|\hat{s}_{n}|\\}$ has
an unbounded sub-sequence. We show that this leads into a contradiction.
Without loss of any generality we assume $|\hat{s}_{n}|\rightarrow\infty$.
From Lemma 2.3 we have
$Z_{n}=\int_{|\hat{s}_{n}|}^{+\infty}h_{n}(s)d\gamma_{s}=\int_{0}^{+\infty}h_{n}(s)1_{[|\hat{s}_{n}|\;\;\infty)}(s)d\gamma_{s}$
and
$Z=\int_{|\hat{s}|}^{+\infty}h(s)d\gamma_{s}=\int_{0}^{+\infty}h(s)1_{[|\hat{s}|,\infty)]}(s)d\gamma_{s}$
for some decreasing deterministic functions $h^{(n)}(s)$ and $h(s)$. Since
$Z_{n}$ converges weakly to $Z$, by lemma 2.13 we should have
$h_{n}(s)1_{[|\hat{s}_{n}|\;\;\infty)}(s)\rightarrow
h(s)1_{[|\hat{s}|,\infty)]}(s)$ point-wise almost surely. By Lemma 2.3 we have
$h(s)>0$ on $[|\hat{s}|,\infty)$. But this is not possible if
$|\hat{s}_{n}|\rightarrow\infty$ while $|\hat{s}|<\infty$. Therefore
$\\{|\hat{s}_{n}|\\}$ is a bounded sequence. Next we show
$\hat{s}_{n}\rightarrow\hat{s}$. For this it is sufficient to show that any
convergent sub-sequence of $\\{\hat{s}_{n}\\}$ converges to $\hat{s}$. To show
this, without loss of any generality, we assume that $\hat{s}_{n}\rightarrow
s^{\prime}$ and show that $\hat{s}=s^{\prime}$. We first assume
$|s^{\prime}|<|\hat{s}|$ and find a contradiction. From the definitions of the
numbers $|\hat{s}_{n}|$ (recall $Ee^{|\hat{s}_{n}|Z_{n}}=+\infty$) we can
claim the existence of real numbers $\delta_{n}>0$ with
$\lim_{n\rightarrow\infty}\delta_{n}=0$ such that
$Ee^{(|\hat{s}_{n}|-\delta_{n})Z_{n}}\rightarrow+\infty$. We let
$b_{n}=|\hat{s}_{n}|-\delta_{n},n\geq 1,$ and $b=|s^{\prime}|$. Then these
numbers $\\{b_{n},b\\}$ satisfy the conditions of Lemma 2.14. Therefore we
should have $sup_{n}Ee^{b_{n}Z_{n}}<\infty$ and this contradicts with
$Ee^{(|\hat{s}_{n}|-\delta_{n})Z_{n}}\rightarrow+\infty$. Now we assume
$|s^{\prime}|>|\hat{s}|$ and find a contradiction. By Lemma 2.3 we have
$Z_{n}=\int_{|\hat{s}_{n}|}^{\infty}h_{n}(s)d\gamma_{s}=\int_{0}^{\infty}h_{n}(s)1_{[|\hat{s}_{n}|,\infty)]}(s)d\gamma_{s}$,
$Z=\int_{|\hat{s}|}^{\infty}h(s)d\gamma_{s}=\int_{0}^{\infty}h(s)1_{[|\hat{s}|,\infty)]}(s)d\gamma_{s}$
and $h(s)>0$ on $[|\hat{s}|,\infty)$. Since $Z_{n}$ converges weakly to $Z$,
by Lemma 2.13 we have $h_{n}(s)1_{[|\hat{s}_{n}|,\infty)]}(s)$ converges
pointwise to $h(s)1_{[|\hat{s}|,\infty)]}(s)$. But this is not possible if
$|\hat{s}_{n}|\rightarrow|s^{\prime}|>|\hat{s}|$ while $h(s)>0$ on
$[|\hat{s}|,\infty)$.

∎

######  Lemma 2.17.

Assume all of $Z_{n},Z$ are nondegenerate GGC random variables. Let
$(\tau_{n},\nu_{n})$ and $(\tau,\nu)$ be their respective generators. Assume
$EZ_{n}<\infty,\;EZ<\infty$. Then if $Z_{n}\overset{w}{\rightarrow}Z$, we have

$\mathcal{L}_{Z_{n}}(s)\rightarrow\mathcal{L}_{Z}(s),$ (34)

whenever $\mathcal{L}_{Z}(s)<\infty$.

###### Proof.

The claim (34) is true for any $s\geq 0$ as in this case $e^{-sx}$ are bounded
continuous functions on $[0,+\infty)$. Therefore we need to show it for
negative $s$. Let $\hat{s}_{n},\hat{s}$ denote the IN of $Z_{n},Z$
respectively. We exclude the case $\hat{s}=0$ as in this case
$\mathcal{L}_{Z}(s)=+\infty$ for any $s<0$. Then it is sufficient to assume
$0<s<|\hat{s}|$ and show that $Ee^{sZ_{n}}\rightarrow Ee^{sZ}$. Fix such $s$
and denote $\delta=(|\hat{s}|-s)/2$. In Lemma 2.3 we showed that
$\hat{s}_{n}\rightarrow\hat{s}$. Therefore there exists a positive integer
$n_{0}$ such that $|\hat{s}_{n}|>s+\delta$ for all $n\geq n_{0}$. Note that we
have $|\hat{s}|>s+\delta$ also. Then from Remark 2.15 we have
$Ee^{sZ_{n}}\rightarrow Ee^{sZ}$. This completes the proof. ∎

In the next Lemma we show that for non-degenerate GGC random variables with IN
number $\hat{s}$, if $\hat{s}\neq 0$ then necessarily
$\mathcal{L}_{Z}(\hat{s})=+\infty$. This fact will be used in the proof of
Proposition 3.3 below.

######  Lemma 2.18.

Let $\hat{s}$ be the IN of a non-degenerate GGC random variable $Z$. If
$\hat{s}\neq 0$, then we have $\mathcal{L}_{Z}(\hat{s})=+\infty$.

###### Proof.

First look at the case of a gamma random variable $Z\sim G(\alpha,\beta)$ with
shape parameter $\alpha$ and scale parameter $\frac{1}{\beta}$. We have
$\mathcal{L}_{Z}(s)=\frac{1}{(1+\beta s)^{\alpha}}$. In this case the IN is
$\hat{s}=-\frac{1}{\beta}$ and clearly we have
$\mathcal{L}_{Z}(-\frac{1}{\beta})=+\infty$. If $Z$ is finite gamma
convolution $Z\sim\sum_{i=1}^{n}G(\alpha_{i},\beta_{i})$, we have
$\mathcal{L}_{Z}(s)=\prod\frac{1}{(1+\beta_{i}s)^{\alpha_{i}}}$ and in this
case the IN of $Z$ is $\hat{s}=\max_{1\leq i\leq n}\\{-\frac{1}{\beta_{i}}\\}$
and one can easily check that $\mathcal{L}_{G}(\hat{s})=+\infty$ in this case
also. Now any GGC random variable $Z$ with zero drift is a weak limit of a
sequence $Z_{n}$ of finite gamma convolutions. Denote the IN of $Z$ by
$\hat{s}$ and the IN number of $Z_{n}$ by $\hat{s}_{n}$ for each $n\geq 1$. We
have $\mathcal{L}_{Z_{n}}(\hat{s}_{n})=+\infty$ for each $n\geq 1$. Therefore
there exists a non-negative sequence of deterministic numbers $\epsilon_{n}$
with $\epsilon_{n}\rightarrow 0$ such that
$\mathcal{L}_{Z_{n}}(\hat{s}_{n}+\epsilon_{n})=Ee^{(|\hat{s}_{n}|-\epsilon_{n})Z_{n}}<+\infty$
and
$\lim_{n\rightarrow\infty}\mathcal{L}_{Z_{n}}(\hat{s}_{n}+\epsilon_{n})=\lim_{n\rightarrow\infty}Ee^{(|\hat{s}_{n}|-\epsilon_{n})Z_{n}}\rightarrow+\infty$.
By Lemma 2.16, we have $|\hat{s}_{n}|-\epsilon_{n}\rightarrow|\hat{s}|$. Now
if we assume $\mathcal{L}_{Z}(\hat{s})=Ee^{|\hat{s}|Z}<+\infty$, then by Lemma
2.14 we should have $Ee^{(|\hat{s}_{n}|-\epsilon_{n})Z_{n}}\rightarrow
Ee^{|\hat{s}|Z}<\infty$, a contradiction. For a GGC random variable $G$ with
generating pair $(\tau,\nu)$, the random variable $G-\tau$ is a GGC with
generating pair $(0,\nu)$ and the IN for both $Z$ and $Z-\tau$ are equal to
each other. For any $s$ we have $e^{\tau
s}\mathcal{L}_{Z}(s)=\mathcal{L}_{Z-\tau}(s)$. From this we conclude that for
any non-degenerate GGC random variable $Z$ we have
$\mathcal{L}_{Z}(\hat{s})=+\infty$ when the IN of $Z$ is not zero. ∎

##  3 Robustness of the exponential utility maximizing portfolio

In this subsection, we address the robustness issue of the optimal portfolio
in the paper [14] as an application of our results in Section 2 above. First
we prove the following Lemma.

######  Lemma 3.1.

Assume $Z_{n},Z$ are nondegenerate random variables from the class GGC. Then
$Z_{n}\overset{w}{\rightarrow}Z$ if and only if $d_{TV}(Z_{n},Z)\rightarrow
0$. Also $Z_{n}\overset{w}{\rightarrow}Z$ implies $d_{Kol}(Z_{n},Z)\rightarrow
0$.

###### Proof.

All nondegenerate random variables in GGC have probability density functions
and they are unimodel, see part vi) of [2] for this (also see the introduction
of [18]). Then the claim follows from [15] (also see page 383 of [13]). Since
the limit distribution $Z$ has probability density function convergence in law
implies convergece in the Kolmogorov distance, a fact that can be derived by
using Dini’s second theorem (see the introduction of [13] for this). ∎

Next we define the models that are necessary for the discussion of robustness.
Let $(\mu_{n})$ and $(\gamma_{n})$ be any family of vectors in
$\mathbb{R}^{d}$. Let $(A_{n})$ be a family of symmetric and positive definite
matrices in $\mathbb{R}^{d}\times\mathbb{R}^{d}$. Let $(Z_{n})$ be a family of
non-negative random variables that are independent from $N_{n}$ (the
$d-$dimensional standard normal random variables). We define the following
models

$X_{n}=\mu_{n}+\gamma_{n}Z_{n}+\sqrt{Z_{n}}A_{n}N_{n}.$ (35)

Also we define $\Sigma_{n}=A_{n}A_{n}^{T}$ and

$\mathcal{A}_{n}=\gamma^{T}_{n}\Sigma^{-1}_{n}\gamma_{n},\;\mathcal{C}_{n}=(\mu_{n}-\textbf{1}r_{f})^{T}\Sigma^{-1}_{n}(\mu_{n}-\textbf{1}r_{f}),\;\mathcal{B}_{n}=\gamma^{T}_{n}\Sigma^{-1}_{n}(\mu_{n}-\textbf{1}r_{f}),$
(36)

for each positive integer $n$. For each $n\geq 1$, the corresponding utility
maximization problem for the model (35) is

$\max_{x\in\mathbb{R}^{d}}EU(W_{n}(x)),$ (37)

where

$\begin{split}W_{n}(x)=&W_{0}[1+(1-x^{T}1)r_{f}+x^{T}X_{n}]\\\
=&W_{0}(1+r_{f})+W_{0}[x^{T}(X_{n}-\textbf{1}r_{f})].\end{split}$

As discussed at the beginning of Section 3, we assume the model (1) is the
_true_ model and the parameters of the models (35) converge to the
corresponding parameters of the _true_ model. Namely we assume the following
holds

$\mu_{n}\rightarrow\mu,\;\gamma_{n}\rightarrow\gamma,\;A_{n}\rightarrow A.$
(38)

Denote the solution of (37) by $x_{n}^{\star}$ for each $n\geq 1$ and the
solution of (6) by $x^{\star}$. In this section, we would like to show that
$x_{n}^{\star}\rightarrow x^{\star}$ under some conditions.

We first prove the following Lemma. Since all the matrices here are symmetric
matrix we drop the transpose operator ”T” in our calculations below. We also
drop the symbol “HS” in the norm $|\cdot|_{HS}$ whenever there is no confusion
arises.

######  Lemma 3.2.

Assume the $(\mu_{n})$ and $(\gamma_{n})$ in the models (35) are convergent
sequences of real vectors with limits $\mu$ and $\gamma$ in the model (1). Let
the $(A_{n})$ in (35) be a sequence of $d\times d$ symmetric positive definite
matrices that satisfy $|A_{n}-A|_{HS}\rightarrow 0$, where $A$ is the matrix
in the model (1). Then we have

$\mathcal{A}_{n}\rightarrow\mathcal{A},\;\mathcal{C}_{n}\rightarrow\mathcal{C},\;\mathcal{B}_{n}\rightarrow\mathcal{B},$
(39)

where $\mathcal{A},\;\mathcal{B},\;\mathcal{C},$ are given as in (32) of [14].
We also have

$\Sigma_{n}^{-1}\gamma_{n}\rightarrow\Sigma^{-1}\gamma,\;\;\Sigma_{n}^{-1}(\mu_{n}-\textbf{1}r_{f})\rightarrow\Sigma^{-1}(\mu-\textbf{1}r_{f}).$
(40)

###### Proof.

Note that
$\mathcal{A}_{n}=\gamma^{T}_{n}(A_{n}^{T})^{-1}A_{n}^{-1}\gamma_{n}=(A_{n}^{-1}\gamma_{n})^{T}A_{n}^{-1}\gamma_{n}$
and also
$\mathcal{A}=\gamma^{T}A^{-1}(A^{-1})^{T}\gamma=(A^{-1}\gamma)^{T}(A^{-1}\gamma)$.
Therefore it is sufficient to show
$\gamma_{n}^{T}A_{n}^{-1}\rightarrow\gamma^{T}A^{-1}$ in Euclidean norm. Since
$A_{n}\rightarrow A$, from [16] we have $A_{n}^{-1}\rightarrow A_{n}^{-1}$
(inverse of non-singular matrix is a continuous function of the elements of
the matrix). We have

$\begin{split}|\gamma^{T}_{n}A_{n}^{-1}-\gamma^{T}A^{-1}|\leq&|\gamma^{T}_{n}(A_{n}^{-1}-A^{-1})|+|(\gamma_{n}^{T}-\gamma^{T})A^{-1}|\\\
\leq&|\gamma_{n}||A_{n}^{-1}-A^{-1}|_{HS}+|\gamma_{n}-\gamma||A^{-1}|_{HS}\\\
\leq&|\gamma_{n}-\gamma||A_{n}^{-1}-A^{-1}|_{HS}+||\gamma||A_{n}^{-1}-A^{-1}||_{HS}+|\gamma_{n}-\gamma||A^{-1}|_{HS}.\end{split}$

From this the claim follows. The other cases
$\mathcal{C}_{n}\rightarrow\mathcal{C},\mathcal{B}_{n}\rightarrow\mathcal{B}$
can be proved similarly. To show (40) it is sufficient to show
$\Sigma_{n}^{-1}\rightarrow\Sigma^{-1}$ as the remaining parts of the proof
follows from similar arguments as above. The relation
$\Sigma_{n}^{-1}\rightarrow\Sigma^{-1}$ follows from

$\begin{split}|\Sigma_{n}^{-1}-\Sigma^{-1}|=&|A_{n}^{-1}A_{n}^{-1}-A^{-1}A^{-1}|=|(A_{n}^{-1}-A^{-1})A_{n}^{-1}+A^{-1}(A_{n}^{-1}-A^{-1})|\\\
\leq&|(A_{n}^{-1}-A^{-1})A_{n}^{-1}|+|A^{-1}(A_{n}^{-1}-A^{-1})|\\\
\leq&|(A_{n}^{-1}-A^{-1})(A_{n}^{-1}-A^{-1})|+|(A_{n}^{-1}-A^{-1})A^{-1}|+|A^{-1}(A_{n}^{-1}-A^{-1})|\\\
\leq&|(A_{n}^{-1}-A^{-1})||(A_{n}^{-1}-A^{-1})|+|(A_{n}^{-1}-A^{-1})||A^{-1}|+|A^{-1}||(A_{n}^{-1}-A^{-1})|.\end{split}$

∎

Now, for each random variable $Z_{n}$ in (35), let $\mathcal{L}_{Z_{n}}(s)$
denote its Laplace transformation and let $\hat{s}_{n}$ denote its IN. We
define

$Q_{n}(\theta)=e^{\mathcal{C}_{n}\theta}\mathcal{L}_{Z_{n}}\Big{[}\frac{1}{2}\mathcal{A}_{n}-\frac{\theta^{2}}{2}\mathcal{C}_{n}\Big{]},\;\;\hat{\theta}_{n}=:\sqrt{\frac{\mathcal{A}_{n}-2\hat{s}_{n}}{\mathcal{C}_{n}}}$
(41)

for each $n\geq 1$. The functions $Q_{n}(\theta)$ are strictly convex on
$(-\hat{\theta}_{n},\hat{\theta}_{n})$ for each $n$ when $\hat{s}_{n}$ is
finite with $\mathcal{L}_{Z}(\hat{s}_{n})=+\infty$ or when
$\hat{s}_{n}=-\infty$ and it is strictly convex in
$[-\hat{\theta}_{n},\hat{\theta}_{n}]$ when $\hat{s}_{n}$ is finite and
$\mathcal{L}_{Z}(\hat{s}_{n})<\infty$ as will be proved in Lemma 4 of [14].
These properties are important for the proof of our result as we shall see.

Below are the optimal portfolios for the problems (37) and (6) and we write
them down here for convenience.

$\begin{split}q_{min}^{(n)}&=:\arg
min_{\theta\in(-\hat{\theta}_{n},\hat{\theta}_{n})}Q_{n}(\theta),\;\;\;x^{\star}_{n}=\frac{1}{aW_{0}}\Big{[}\Sigma^{-1}_{n}\gamma_{n}-q_{min}^{(n)}\Sigma^{-1}_{n}(\mu_{n}-\textbf{1}r_{f})\Big{]},\\\
q_{min}&=:\arg
min_{\theta\in(-\hat{\theta},\hat{\theta})}Q(\theta),\;\;\;x^{\star}=\frac{1}{aW_{0}}\Big{[}\Sigma^{-1}\gamma-
q_{min}\Sigma^{-1}(\mu-\textbf{1}r_{f})\Big{]}.\end{split}$ (42)

######  Proposition 3.3.

Consider the model (1) and assume $Z$ is a non-degenerate GGC random variable
with $EZ<\infty$. Assume the associated problem (6) with the model (1) has a
regular solution $x^{\star}$. Let (35) be a sequence of models that satisfy
(38). Assume $Z_{n}$ in (35) are non-degenerate mixing distributions from the
class GGC also and $EZ_{n}<\infty$. Assume for each $n\geq 1$, the problem
(37) has regular solution $x_{n}^{\star}$. Then if
$Z_{n}\overset{w}{\rightarrow}Z$, we have
$|x_{n}^{\star}-x^{\star}|\rightarrow 0$.

###### Proof.

Let $\hat{s}_{n}$ and $\hat{s}$ denote the IN of $Z_{n}$ and $Z$ respectively.
Let $\hat{\theta}_{n}=\sqrt{(\mathcal{A}_{n}-2\hat{s}_{n})/\mathcal{C}_{n}}$
and $\hat{\theta}=\sqrt{(\mathcal{A}-2\hat{s})/\mathcal{C}}$. The assumption
on the regularity of $x_{n}^{\star}$ and $x^{\star}$ imply all of
$(-\hat{\theta}_{n},\hat{\theta}_{n})$ and $(\hat{\theta},\hat{\theta})$ are
non-empty open intervals. By Proposition 2.17 of [14], the solutions
$x_{n}^{\star}$ and $x^{\star}$ to the problems (37) and (6) are given by (42)
respectively. We have $q^{(n)}_{min}\in(-\hat{\theta}_{n},\hat{\theta}_{n})$
and $q_{min}\in(-\hat{\theta},\hat{\theta})$ as the solutions $x_{n}^{\star}$
and $x^{\star}$ are regular by the assumption. Due to lemma 3.2, we only need
to prove $q_{min}^{(n)}\rightarrow q_{min}$.

First observe that $\\{q_{min}^{(n)}\\}$ is a bounded sequence as
$\\{\hat{s}_{n}\\}$ is a bounded sequence due to Lemma 2.16 and hence
$\hat{\theta}_{n}$ is a bounded sequence by Lemma 3.2. So to prove
$q_{min}^{(n)}\rightarrow q_{min}$ it is sufficient to prove that any
convergent sub-sequence $q_{min}^{(n_{k})}$ converges to the same number
$q_{min}$. Without loss of any generality, below we show that if
$q_{min}^{(n)}\rightarrow q$ then $q=q_{min}$. For the simplicity of notations
below we write $q_{n}=:q_{min}^{(n)}$ for all $n\geq 1$. By Lemma 4.1 of [14]
we have $q_{n}\in(-\hat{\theta}_{n},0)$ and $q_{min}\in(-\hat{\theta},0)$.
Being a limit of $q_{n}$ and also since
$\hat{\theta}_{n}\rightarrow\hat{\theta}$, we can assume
$q\in[-\hat{\theta},0]$ below. We divide the proof into two cases.

Case 1: Assume $\hat{s}\neq 0$. Then by Lemma 2.18 we have
$\mathcal{L}_{Z}(\hat{s})=+\infty$ and therefore $Q(-\hat{\theta})=+\infty$.
We first show that $Q_{n}(\theta)\rightarrow Q(\theta)$ for all
$\theta\in\bar{\Theta}=:(-\hat{\theta},-\sqrt{\frac{\mathcal{A}}{\mathcal{C}}})\cup(-\sqrt{\frac{\mathcal{A}}{\mathcal{C}}},0)$.
Denote
$\eta_{n}(\theta)=:\frac{1}{2}\mathcal{A}_{n}-\frac{\theta^{2}}{2}\mathcal{C}_{n}$
and $\eta(\theta)=:\frac{1}{2}\mathcal{A}-\frac{\theta^{2}}{2}\mathcal{C}$.
Here we singled out the point
$\tilde{\theta}=:-\sqrt{\mathcal{A}/\mathcal{C}}$ as $\eta(\tilde{\theta})=0$.
It is easy to see that if $\theta\in(-\hat{\theta}_{n},0)$ we have
$\hat{s}_{n}<\eta_{n}(\theta)<\frac{1}{2}\mathcal{A}_{n}$ and if
$\theta\in(-\hat{\theta},0)$ we have
$\hat{s}<\eta(\theta)<\frac{1}{2}\mathcal{A}$. Observe that
$\eta_{n}(\theta)\rightarrow\eta(\theta)$ for all $\theta\in\mathbb{R}$ due to
Lemma 3.2. Take any $\theta_{0}\in\bar{\Theta}$. Then either
$\eta(\theta_{0})>0$ or $\eta(\theta_{0})<0$. If $\eta(\theta_{0})>0$ then we
have $\eta_{n}(\theta_{0})>0$ for all $n$ that are sufficiently large. Since
$Z_{n}\overset{w}{\rightarrow}Z$ and the function $e^{-sx}$ is bounded
continuous function of $x\geq 0$ for each $s>0$, we have
$\mathcal{L}_{Z_{n}}(\eta_{n}(\theta_{0}))\rightarrow\mathcal{L}_{Z}(\eta(\theta_{0}))$.
This and Lemma 3.2 then implies $Q(\eta_{n}(\theta_{0}))\rightarrow
Q(\eta(\theta_{0}))$. If $\eta(\theta_{0})<0$ then we have
$\eta_{n}(\theta_{0})<0$ for all $n$ that are sufficiently large. Therefore in
this case the sequence $\eta_{n}(\theta_{0})$ and $\eta(\theta_{0})$ satisfy
the conditions of Lemma 2.14. Therefore we still have
$\mathcal{L}_{Z_{n}}(\eta_{n}(\theta_{0}))\rightarrow\mathcal{L}_{Z}(\eta(\theta_{0}))$
and thus $Q(\eta_{n}(\theta_{0}))\rightarrow Q(\eta(\theta_{0}))$.

Now if $q=-\hat{\theta}$ then since $Q_{n}(\theta)\rightarrow Q(\theta)$ for
any $\theta\in(-\hat{\theta},-\sqrt{\mathcal{A}/\mathcal{C}})$ and
$Q(\theta)\rightarrow+\infty$ when $\theta$ converges to $-\hat{\theta}$ from
the right, we conclude that $Q_{n}(q_{n})\rightarrow+\infty$. But $q_{n}$ is
the minimizing point of $Q_{n}(\theta)$ in
$(-\hat{\theta}_{n},\hat{\theta}_{n})$, a contradiction. Therefore we assume
$q\in(-\hat{\theta},0]$ and $q\neq q_{min}$ below. First consider the case
$q\neq\tilde{\theta}$. Then we have $Q_{n}(q_{n})\rightarrow Q(q)>Q(q_{min})$
(here we don’t rule out the case $q_{min}=\tilde{\theta}$). As $Q(\theta)$ is
a continuous function on $(-\hat{\theta},0)$ and
$q_{min}\in(-\hat{\theta},0)$, $q_{min}$ has a small neighborhood
$(q_{min}-\delta,q_{min}+\delta)$ with some $\delta>0$ such that for all
$\theta\in(q_{min}-\delta,q_{min}+\delta)$ we have $Q(\theta)<Q(q)$. This
contradicts with the fact that $Q_{n}(\theta)\rightarrow Q(\theta)$ for all
$\theta\in(q_{min}-\delta,q_{min}+\delta)$ except possibly
$\theta=q_{min}=\tilde{\theta}$. Now assume $q=\tilde{\theta}$. We have
$Q_{n}(q_{n})=e^{\mathcal{C}_{n}q_{n}}Ee^{(\frac{1}{2}\mathcal{C}_{n}q_{n}-\frac{1}{2}\mathcal{A}_{n})Z_{n}}<Q_{n}(\theta)$
for all $\theta\in(-\hat{\theta}_{n},\hat{\theta}_{n})$. When
$q_{n}\rightarrow q=-\frac{\mathcal{A}}{\mathcal{C}}$ we have
$\frac{1}{2}\mathcal{C}_{n}q_{n}-\frac{1}{2}\mathcal{A}_{n}\rightarrow 0$.
Therefore the random variables
$(\frac{1}{2}\mathcal{C}_{n}q_{n}-\frac{1}{2}\mathcal{A}_{n})Z_{n}$ converge
almost surely to zero. We apply Fatou’s lemma to the expression of
$Q_{n}(q_{n})$ above and obtain

$Q(q)=e^{\mathcal{C}q}\leq\liminf_{n}Q_{n}(q_{n})\leq\liminf_{n}Q_{n}(\theta)=Q(\theta),\;\;\forall\theta\in\bar{\Theta}.$

This implies $Q(q)\leq Q(q_{min})$ and this contradicts with
$Q(q_{min})<Q(q)$.

Case 2: Assume $\hat{s}=0$. In this case
$\hat{\theta}=\sqrt{\frac{\mathcal{A}}{\mathcal{C}}}$ and
$0\leq\eta(\theta)\leq\frac{1}{2}\mathcal{A}$ for
$\theta\in[-\hat{\theta},0]$. We have $\eta(\theta)>0$ for all
$\theta\in(-\hat{\theta},0]$. Therefore $Q_{n}(\theta)\rightarrow Q(\theta)$
for all $\theta\in(-\hat{\theta},0]$. If $q\in(-\hat{\theta},0]$ then we have
$Q_{n}(q_{n})\rightarrow Q(q)>Q(q_{min})$. This contradicts with the fact that
$Q_{n}(\theta)$ converges to $Q(\theta)$ in the neighbourhood of
$q_{min}\in(-\hat{\theta},0)$. Now assume $q=-\hat{\theta}$. We have
$Q_{n}(q_{n})=e^{\mathcal{C}_{n}q_{n}}Ee^{(\frac{1}{2}\mathcal{C}_{n}q_{n}-\frac{1}{2}\mathcal{A}_{n})Z_{n}}$
and when $q_{n}\rightarrow-\hat{\theta}$ we have
$\frac{1}{2}\mathcal{C}_{n}q_{n}-\frac{1}{2}\mathcal{A}_{n}\rightarrow 0$.
Therefore $(\frac{1}{2}\mathcal{C}_{n}q_{n}-\frac{1}{2}\mathcal{A}_{n})Z_{n}$
converge almost surely to zero. Again we apply Fatou’s lemma and obtain

$Q(-\hat{\theta})=e^{-\mathcal{C}\hat{\theta}}\leq\liminf_{n}Q_{n}(q_{n})\leq\liminf_{n}Q_{n}(\theta)=Q(\theta),\;\forall\theta\in(-\hat{\theta},0].$

This implies $Q(-\hat{\theta})\leq Q(q_{min})$. But
$Q(q_{min})<Q(-\hat{\theta})$ due to strict convexity of $Q$ in
$[-\hat{\theta},0]$ (note here that $\mathcal{L}_{Z}(\hat{s})<\infty$).

∎

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