# Duality for real and multivariate exponential families

Gérard Letac  Institut de Mathématiques de Toulouse, 118 route de Narbonne
31062 Toulouse, France.

###### Abstract

Consider a measure $\mu$ on $\mathbb{R}^{n}$ generating a natural exponential
family $F(\mu)$ with variance function $V_{F(\mu)}(m)$ and Laplace transform

$\exp(\ell_{\mu}(s))=\int_{\mathbb{R}^{n}}\exp(-\langle s,x\rangle)\mu(dx).$

A dual measure $\mu^{*}$ satisfies
$-\ell^{\prime}_{\mu^{*}}(-\ell^{\prime}_{\mu}(s))=s.$ Such a dual measure
does not always exist. One important property is
$\ell^{\prime\prime}_{\mu^{*}}(m)=(V_{F(\mu)}(m))^{-1},$ leading to the notion
of duality among exponential families (or rather among the extended notion of
T exponential families $T\hskip-2.0ptF$ obtained by considering all
translations of a given exponential family $F$).

###### keywords:

Dilogarithm distribution, Landau distribution, large deviations, quadratic and
cubic real exponential families, Tweedie scale, Wishart distributions.

###### MSC:

[2020] Primary 62H05 , Secondary 60E10

††journal: Journal of Multivariate Analysis

##  1 Introduction

One can be surprized by the explicit formulas that one gets from large
deviations in one dimension. Consider iid random variables
$X_{1},\ldots,X_{n},\ldots$ and the empirical mean
$\overline{X}_{n}=(X_{1}+\cdots+X_{n})/n.$ Then the Cramér theorem [6] says
that the limit

$\Pr(\overline{X}_{n}>m)^{1/n}\to_{n\to\infty}\alpha(m)$

does exist for $m>\mathbb{E}(X_{1}).$ For instance, the symmetric Bernoulli
case $\Pr(X_{n}=\pm 1)=\frac{1}{2}\,$ leads, for $0<m<1$, to

$\alpha(m)=(1+m)^{-1-m}(1-m)^{-1+m}.$ (1)

For the bilateral exponential distribution $X_{n}\sim\frac{1}{2}\,e^{-|x|}dx$
we get, for $m>0$,

$\alpha(m)=\frac{1}{2}\,e^{1-\sqrt{1+m^{2}}}(1+\sqrt{1+m^{2}}).$

What strange formulas! Other ones can be found in [17], Problem 410. The
present paper is going to interpret in certain cases the function $m\mapsto
1/\alpha(m)$ as the Laplace transform of a certain dual measure $\mu^{*}$
which can be deduced explicitly from the distribution $\mu$ of $X_{n}.$ Since
the Cramér theorem can be seen as a result about one dimensional exponential
families, we develop the idea in this framework, using the large box of
examples obtained from the theory of variance functions of exponential
families initiated by the article of Carl Morris [21] in 1982. An even simpler
example of duality is provided by the Tweedie families (Barlev and Enis [1],
Jørgensen [9] and Tweedie [22]): the variance function $Am^{p}$ with $p>1$ has
dual $Bm^{q}$ with $\frac{1}{p}+\frac{1}{q}=1$ for suitable pairs $(A,B)$ (see
(8)). For instance, the Inverse Gaussian family $Am^{3}$ has dual $Bm^{3/2}$ .

The Poisson distribution, with the one of the simplest variance functions
$V_{F}(m)=m$, leads to the study of exponential family with variance function
$e^{m}$ generated up to a translation by the unsymmetric stable law with
Laplace transform $e^{-s}s^{s}$, also called Landau distribution. This gives
tools for describing duals of other familiar exponential families. The cases
of the normal and gamma families are very simple, being self dual, but other
familiar cases like the negative binomial, the Bernoulli distribution and the
cubic families are tougher. Finally, we consider another family, which is the
dilogarithm family with variance function $e^{m}-1$ as well as the one with
variance function $\sinh m.$ Like the normal and gamma families, they have the
remarkable property to be self dual.

The definition of dual measures makes sense also in $\mathbb{R}^{n}$ while the
probabilistic interpretation in terms of large deviations is lost. However we
consider several cases in $\mathbb{R}^{n}$: the multinomial distribution, the
Wishart ones and other quadratic families as classified by Casalis ([4]).

We proceed as follows: the notion of duality leads us unfortunately to change
a bit the tradition about the exponential families. Indeed we will use
$e^{-sx}$ instead of $e^{\theta x}$ in order to obtain later more readable
formulas. This is explained in Section 2, together with the description of the
classical objects attached to an exponential family.

In the preceding lines we have been vague about duality. Section 3 gives
proper definitions, explaining what we call a dual measure $\mu^{*}$ of $\mu$
and showing that some measures have no dual. We explain also what a T
exponential family $T\hskip-2.0ptF$ is. It is nothing but an exponential
family $F$ plus all its translations. Indeed, talking about the dual $F^{*}$
of an exponential family $F$ does not exactly make sense, while the dual
$T\hskip-2.0ptF^{*}$ of $T\hskip-2.0ptF$ does. Section 3 gives also the link
with large deviations.

Section 4 concentrates on the $T\hskip-2.0ptF$ when the variance function
$V_{F}(m)$ is $e^{m}$ and some parent distributions. It also give details on
what we call Lévy measures of types 0, 1 and 2. Of course, large parts of this
material are well known from probabilists and statisticians: exponential
families, variance functions, Lévy measures, Landau distribution. It was
necessary to expose them again for commodity of reading. This section contains
crucial calculations for the sequel in Proposition 7.

Section 5 applies the results of Section 4 to the description of the duals of
the Morris and the cubic families, with the surprizing fact that they exist
all with the only exception of the hyperbolic family with variance function
$m^{2}+1$.

Section 6 describes the self dual dilogarithm distribution $\mu$ on the set
$\mathbb{N}=\\{0,1,\ldots\\}$ of integers defined by

$\sum_{n=0}^{\infty}\mu(n)z^{n}=\exp\left(\sum_{k=1}^{\infty}\frac{1}{k^{2}}(z^{k}-1)\right)$

which, for $m>0$, generates the exponential family with variance function
$e^{m}-1$. Since the consideration of this exponential family and of a set of
parent distributions is not done in the literature, we develop some of their
properties, somewhat deviating from the study of duality. For instance, if $N$
is the standard Gaussian distribution, then the variance of the exponential
family generated by the convolution $N*\mu$ is $e^{m}+1.$

Section 7 considers the $\mathbb{R}^{n}$ case: The multinomial distribution
has a very explicit dual expressed in terms of the Landau distribution. The
Wishart distribution is self dual as the one dimensional gamma distribution.
We prove some negative results, like the fact that the multivariate negative
binomial law has no dual.

Section 8 discusses open problems.

##  2 Laplace and bilateral Laplace transforms

At first the exponential families and Laplace transforms are considered. A
certain tradition among statisticians (see Morris [21]) as opposed to
physicists, and may be to probabilists, defines the Laplace transform of a
positive measure $\mu$ on $\mathbb{R}$ and $\mathbb{R}^{n}$ as follows:

$L_{\mu}(\theta)=\int_{\mathbb{R}^{n}}e^{\langle\theta,x\rangle}\mu(dx).$ (2)

From the Hölder inequality the set
$D(\mu)=\\{\theta\,;\,L_{\mu}(\theta)<\infty\\}$ is a convex set, and the
function $k_{\mu}=\log L_{\mu}$ is convex on $D(\mu)$. Actually $k_{\mu}$ is
strictly convex outside of the particular case where $\mu$ is concentrated on
one point in the case of $\mathbb{R}$, or on an affine hyperplane in the case
of $\mathbb{R}^{n}$. To avoid trivialities one introduces the interior
$\Theta(\mu)$ of $D(\mu).$ One calls $\mathcal{M}(\mathbb{R}^{n})$ the set of
$\mu$ which are not concentrated on an affine hyperplane and such that
$\Theta(\mu)$ is not empty. Such a $\mu$ generates a set of probabilities

$P(\theta,\mu)(dx)=e^{\langle\theta,x\rangle-k_{\mu}(\theta)}\mu(dx)$

and $F=F(\mu)=\\{P(\theta,\mu)\,;\,\theta\in\Theta(\mu)\\}$ is called the
natural exponential family generated by $\mu.$ Note that $\mu$ is not
necessarily bounded: simple examples on $\mathbb{R}$ like
$\mu(dx)=1_{(0,\infty)}(x)dx$ or $\sum_{n=0}^{\infty}\delta_{n}$ generate the
important families of exponential distributions or geometric discrete laws.
Omiting ’natural’, we will say always exponential family for short. Objects
linked to $F$ are the mean $m$ of $P(\theta,\mu)$, the domain of the means
$M_{F}$ and the inverse function $\psi_{\mu}.$ They are defined by

$m=k^{\prime}_{\mu}(\theta)=\int_{\mathbb{R}^{n}}xP(\theta,\mu)(dx),\qquad
M_{F}=k^{\prime}_{\mu}(\Theta(\mu)),\qquad\theta=\psi_{\mu}(m).$

Note that since $k_{\mu}$ is strictly convex, then $k^{\prime}_{\mu}$ is
injective on $\Theta(\mu)$, the map $\psi_{\mu}$ from $M_{F}$ onto
$\Theta(\mu)$ is well defined and $M_{F}$ is a connected open set.

If $C_{F}$ is the closed convex set generated by the support of $\mu$ clearly
$M_{F}$ is contained in $C_{F}.$ We say that $F$ or $\mu$ are steep if $M_{F}$
is equal to the interior of $C_{F}.$ Most of the classical exponential
families are steep, but not always (see for instance the Tweedie scale below
for $p<0$). In $\mathbb{R}$ the set $M_{F}$ is an interval, but in
$\mathbb{R}^{n}$ there are non steep examples such that $M_{F}$ is non-convex
([16], p. 35).

Finally, the last important object about the exponential family $F$ is its
variance function $V_{F}$ defined on $M_{F}$ by

$V_{F}(m)=\frac{1}{\psi^{\prime}_{\mu}(m)}=k_{\mu}^{\prime\prime}(\psi_{\mu}(m))=\int_{\mathbb{R}^{n}}(x-m)\otimes(x-m)P(\psi_{\mu}(m),\mu)(dx),$

which characterizes $F$.

Now bilateral Laplace transforms are considered. In the particular case of
dimension one, an older tradition associates the name of Laplace to integrals
$\int_{0}^{\infty}e^{-sx}\mu(dx)$ which are conveniently defined for $s>0$ in
many circumstances. In the present paper we need to consider what the
physicists call the bilateral Laplace transform

$B_{\mu}(s)=\int_{\mathbb{R}^{n}}e^{-\langle s,x\rangle}\mu(dx).$ (3)

Dealing with this slight change of notation $B_{\mu}(s)=L_{\mu}(-s)$ will much
simplify the description of duality between two natural exponential families.
In the sequel, we say ’Laplace transform’ for short and by abus de langage
instead of the longer term ’bilateral Laplace transform’.

Because we will deal with these bilateral Laplace transforms, we have to
modify the description of the classical objects associated to $F=F(\mu)$ with

$\displaystyle S(\mu)$ $\displaystyle\hskip-8.0pt=$
$\displaystyle\hskip-8.0pt-\Theta(\mu),\qquad\ell_{\mu}(s)=k_{\mu}(-s),\qquad
m=-\ell^{\prime}_{\mu}(s),$ (4) $\displaystyle s=\varphi_{\mu}(m)$
$\displaystyle\hskip-8.0pt=$
$\displaystyle\hskip-8.0pt-\psi_{\mu}(m),\qquad\ell^{\prime\prime}_{\mu}(s)=k^{\prime\prime}_{\mu}(-s),\qquad
V_{F}(m)=-(\varphi_{\mu}^{\prime}(m))^{-1}.$ (5)

Let us insist on the fact that $\ell_{\mu}$ is convex and thus $s\mapsto
m=-\ell^{\prime}_{\mu}(s)$ is decreasing when $n=1$. For any $n$ the inverse
function $m\mapsto s=\varphi_{\mu}(m)$ from $M_{F}$ onto $S(\mu)$ is well
defined. Thus we have several ways of coding a member of an exponential
family:

$P(\theta,\mu),\qquad P(-s,\mu),\qquad P(\psi_{\mu}(m),\mu),\qquad
P(-\varphi_{\mu}(m),\mu).$

Finally, in one dimension, to find a generating measure $\mu$ of the
exponential family $F$ from the knowledge of $V_{F},$ we proceed as follows:

$ds=-\varphi^{\prime}_{\mu}(m)dm=\frac{dm}{V_{F}(m)}\Rightarrow-s=\int\frac{dm}{V_{F}(m)}.$

The choice of the integration constants will change $\mu$ and is arbitrary. In
practical cases, we choose these constants in order to get the simplest form
of $\mu$. Therefore this choice may depend on aesthetic considerations. Here
is an important example in dimension one:

######  Example 1 (the Tweedie scale).

This term describes a set of exponential families with variance functions of
the form $V_{F}(m)=\frac{m^{p}}{\lambda^{p-1}}$ on $M_{F}=(0,\infty)$ where
$\lambda>0$ and $p\in\mathbb{R}\setminus[0,1]$ (see [1, 9, 22]). We are going
to describe their densities or Laplace transforms. The limiting cases $p=0$
and $p=1$ correspond respectively to the Gaussian exponential family with
fixed variance $\lambda$ where $M_{F}=\mathbb{R}$ and to the Poisson family
and they need a special treatment.

  1. 1.

The stable subordinator case $p>2.$ For simplification we introduce
$\alpha=\frac{p-2}{p-1}\in(0,1).$ We write

$ds=-\frac{\lambda^{p-1}dm}{m^{p}}\Rightarrow
s=\frac{1}{p-1}\left(\frac{\lambda}{m}\right)^{p-1}\Rightarrow m=\frac{\lambda
s^{-1/(p-1)}}{(p-1)^{1/(p-1)}}=-\ell^{\prime}_{\mu}(s).$

Here $S(\mu)=(0,\infty).$ We finally obtain
$\ell_{\mu}(s)=-\lambda\frac{(p-1)^{\alpha}}{p-2}s^{\alpha}.$ This family is
generated by a stable law of parameter $\alpha$ with Lévy measure concentrated
on $(0,\infty)$ of Type 1 (see Section 4.3 for the definition of a Lévy
measure of an infinitely divisible probability and its type).

  2. 2.

The gamma case $p=2.$ Similarly we obtain $S(\mu)=(0,\infty),$
$\ell^{\prime}_{\mu}(s)=-\lambda/s$ and $B_{\mu}(s)=1/s^{\lambda}.$ This
family is generated by the measure $\mu(dx)$ with density
$x^{\lambda-1}/\Gamma(\lambda)$ and is the family of gamma distributions with
shape parameter $\lambda$ .

  3. 3.

The Poisson -gamma case $1<p<2.$ For simplification denote
$\beta=\frac{2-p}{p-1}\in(0,\infty).$ A computation analogous to the case
$p>2$ leads to

$\ell_{\mu}(s)=\lambda\frac{(p-1)^{-\beta}}{2-p}\frac{1}{s^{\beta}},\qquad\ell^{\prime}_{\mu}(s)=-\lambda\frac{(p-1)^{-\frac{1}{p-1}}}{2-p}\frac{1}{s^{\frac{1}{p-1}}}$
(6)

and to $S(\mu)=(0,\infty).$ This implies that the corresponding exponential
family is the set of laws of $X_{1}+\cdots+X_{N(t)}$ where
$X_{1},\ldots,X_{n},\ldots$ are iid gamma distributed with shape parameter
$\beta$ and where $N(t)$ is an independent Poisson distribution with mean
$t=\lambda\frac{(p-1)^{-\beta}}{2-p}.$

  4. 4.

The non steep stable case $p<0.$ For simplicity denote $q=-p>0$ and
$\gamma=\frac{q+2}{q+1}\in(1,2).$ Here $S(\mu)=(-\infty,0)$ and we obtain

$\ell_{\mu}(s)=\lambda\frac{(q+1)^{\gamma}}{q+2}(-s)^{\gamma}.$

The members of this family are stable laws with parameter $\gamma$ with Lévy
measure concentrated on $(-\infty,0).$ The support of such a stable law is
$\mathbb{R}$ but $M_{F}=(0,\infty):$ this is an example of a non steep family
and of an infinitely divisible distribution of Type 2 (see Section 4 and
(21)).

##  3 Duality

###  3.1 Duality between measures

Given $\mu\in\mathcal{M}(\mathbb{R}^{n})$ the function
$s\mapsto-\ell^{\prime}_{\mu}(s)=m$ maps $S(\mu)$ onto the domain of the means
$M_{F}$. Its inverse $m\mapsto s=\varphi_{\mu}(m)$ exists and maps $M_{F}$
onto $S(\mu).$ Suppose that there exists
$\mu^{*}\in\mathcal{M}(\mathbb{R}^{n})$ such that

$-\ell^{\prime}_{\mu^{*}}(-\ell_{\mu}(s))=s.$ (7)

If (7) holds we say that $\mu^{*}$ is a dual measure of $\mu.$ Note that we
say ’a dual measure’ since $\mu^{*}$ is unique only up to a multiplicative
constant. However the associated exponential family $F(\mu^{*})$ will not
change if $\mu^{*}$ is replaced by $C\mu^{*}.$ Observe also that if $\mu^{*}$
exists and if $\mu$ is steep then $\mu$ is also a dual measure of $\mu^{*}.$
In general, if $\mu^{*}$ is bounded it is natural to choose the multiplicative
constant such that $\mu^{*}$ is a probability.

######  Proposition 1.

Let $\mu\in\mathcal{M}(\mathbb{R}^{n})$ and suppose that there exists a dual
measure $\mu^{*}.$ Then
$\ell^{\prime\prime}_{\mu^{*}}(m)=(V_{F(\mu)}(m))^{-1}$. Furthermore, if $\mu$
is steep $\ell^{\prime\prime}_{\mu}(m)=(V_{F(\mu^{*})}(m))^{-1}$.

###### Proof:.

Since $s=\varphi_{\mu}(-\ell^{\prime}_{\mu}(s))$ we get by definition that

$-\varphi^{\prime}_{\mu}(-\ell^{\prime}_{\mu}(s))=-(-\ell^{\prime\prime}_{\mu}(s))^{-1}=(V_{F(\mu)}(-\ell^{\prime}_{\mu}(s)))^{-1},\
\ \ell^{\prime\prime}_{\mu^{*}}(m)=(V_{F(\mu)}(m))^{-1}.$

The second formula is obtained by symmetry. ∎

In the particular case of dimension one, we will have numerous examples of
dual measures given below. Let us give the simplest now, based on the Tweedie
scale. Essentially, duality exchanges cases 1 and 3.

######  Proposition 2.

Let $1<p<2$ and $q$ defined by $\frac{1}{p}+\frac{1}{q}=1.$ Let $\mu$, defined
by (6) generating the exponential family with variance function
$V_{F}(m)=\frac{m^{p}}{\lambda^{p-1}}.$ Then there exists a dual measure
$\mu^{*}$ and it generates the exponential family with variance function
$V_{F(\mu^{*})}(s)=(\frac{p}{q})^{q}\frac{1}{\lambda}s^{q}.$

###### Proof:.

Using formula (6) we get
$\ell_{\mu}^{\prime\prime}(s)=\lambda(p-1)^{-q}s^{-q}$. Since $p-1=p/q$
Proposition 1 gives the result. ∎

######  Remark 1.

For having a more symmetric form we can choose $\lambda$ such that

$V_{F(\mu)}(m)=\frac{R^{p}}{q^{p}}m^{p},\ \
V_{F(\mu^{*})}(s)=\frac{R^{-q}}{p^{q}}s^{q}.$ (8)

Dual measures do not always exist. Consider, for instance, Case 4 of the
Tweedie scale above generated by a stable distribution $\mu$ with parameter
$\gamma\in(1,2)$ with Lévy measure concentrated on $(-\infty,0)$ and domain of
the means $M_{F(\mu)}=(0,\infty)$. Therefore if $\mu^{*}$ was existing we
would have for positive constants $C,C_{1},C_{2}:$

$\displaystyle\ell_{\mu}(s)$ $\displaystyle\hskip-8.0pt=$
$\displaystyle\hskip-8.0ptC(-s)^{\gamma},\qquad\ell^{\prime}_{\mu}(s)=-C\gamma(-s)^{\gamma-1}=-m,$
$\displaystyle\ell^{\prime}_{\mu^{*}}(m)$ $\displaystyle\hskip-8.0pt=$
$\displaystyle\hskip-8.0ptC_{1}m^{\frac{1}{\gamma-1}},\qquad\ell_{\mu^{*}}(m)=C_{2}m^{\frac{\gamma}{\gamma-1}}.$

This would imply that $\mu^{*}$ would be a stable law with parameter
$\gamma/(\gamma-1)>2$, which is impossible.

Moreover, it is well known that a probability $P\in\mathcal{M}(\mathbb{R})$ is
infinitely divisible if and only if there exists
$\nu\in\mathcal{M}(\mathbb{R})$ such that for $s\in S(P)$ we have

$\ell^{\prime\prime}_{P}(s)=\int_{\mathbb{R}}e^{-sx}\nu(dx).$ (9)

Extension of the definition of infinite divisibility to measures
$\mu\in\mathcal{M}(\mathbb{R})$ (even the unbounded ones) is easy and is done
in Letac [16] and the same remark about $\ell^{\prime\prime}_{\mu}(s)$ holds.
Then we have the following simple proposition:

######  Proposition 3.

Let $\mu\in\mathcal{M}(\mathbb{R})$ and assume that a dual measure $\mu^{*}$
does exist. Then $\mu^{*}$ is infinitely divisible if and only if
$m\mapsto\frac{1}{V_{F(\mu)}(m)}$ is the Laplace transform of some measure
$\nu\in\mathcal{M}(\mathbb{R}).$

###### Proof:.

This an immediate consequence of Proposition 1 and of (20). ∎

######  Example 2 (Bernoulli distribution on $\pm 1$).

Proposition 3 gives a powerful tool to check quickly whether a dual measure
exists. Consider, for instance, the Bernoulli measure on $\pm 1$ with mean
$m\in(-1,1)$:

$\mu=\frac{1}{2}\,((1-m)\delta_{-1}+(1+m)\delta_{1}).$

Therefore the variance of $X\sim\mu$ is
$V_{F(\mu)}=\mathbb{E}(X^{2})-\mathbb{E}(X)^{2}=1-m^{2}.$ To prove that
$\mu^{*}$ exists we observe that

$\frac{1}{1-m^{2}}=\frac{1}{2}\,\left(\frac{1}{1-m}+\frac{1}{1+m}\right)=\frac{1}{2}\,\int_{\mathbb{R}}e^{-my-|y|}dy$

and we apply Proposition 3 to the bilateral exponential law
$\nu(dy)=\frac{1}{2}\,e^{-|y|}dy.$ One can prove that
$B_{\mu^{*}}(m)=(1+m)^{1+m}(1-m)^{1-m}$ on $S(\mu^{*})=(-1,1)$. We will
describe $\mu^{*}$ later on in Section 4, the second formula in (15) and
Section 5.

###  3.2 The role of linear transformations and the Jørgensen set

Consider the image $\mu_{a}$ of $\mu$ by the isomorphism $x\mapsto a(x)$ of
$\mathbb{R}^{n}$ into itself. Here $a$ is an invertible matrix of order $n$.
If $\mu$ has a dual $\mu^{*}$ then $(\mu^{*})_{a^{-1}}$ is a dual of
$\mu_{a}.$ Let us give the details of a tedious calculation:

$\displaystyle B_{\mu_{a}}(s)=\int_{\mathbb{R}^{n}}e^{-\langle
s,a(x)\rangle}=\int_{\mathbb{R}^{n}}e^{-\langle
a^{T}s,x)\rangle}\mu(dx),\qquad\ell_{\mu_{a}}(s)=\ell_{\mu}(a^{T}s),\qquad
m=-\ell^{\prime}_{\mu_{a}}(s)=-a^{T}\ell^{\prime}_{\mu}(a^{T}s),$
$\displaystyle(a^{-1})^{T}m=-\ell^{\prime}_{\mu}(a^{T}s),\qquad
a^{T}s=\ell^{\prime}_{\mu^{*}}((a^{-1})^{T}m),\qquad
s=(a^{-1})^{T}\ell^{\prime}_{\mu^{*}}((a^{-1})^{T}m),\ell_{(\mu_{a})^{*}}(m)=\ell_{\mu^{*}}((a^{-1})^{T}m),$
$\displaystyle
B_{(\mu_{a})^{*}}(m)=B_{\mu^{*}}((a^{-1})^{T}m)=\int_{\mathbb{R}^{n}}e^{-\langle
a^{-1})^{T}m,y\rangle}\mu^{*}(dy)=\int_{\mathbb{R}^{n}}e^{-\langle
m,a^{-1}y\rangle}\mu^{*}(dy)=B_{\mu_{a^{-1}}}(m).$

Recall also that the image of $F=F(\mu)$ by the isomorphism $a$ satisfies

$V_{F(\mu_{a})}(m)=a^{T}V_{F}(a^{-1}m)a.$

If $\mu\in\mathcal{M}(\mathbb{R}^{n})$ then the Jørgensen set $\Lambda(\mu)$
is the set of $\lambda>0$ such that there exists
$\mu_{\lambda}\in\mathcal{M}(\mathbb{R}^{n})$ such that
$S(\mu)=S(\mu_{\lambda})$ and $(B_{\mu})^{\lambda}=B_{\mu_{\lambda}}.$ Denote
$F_{\lambda}=F(\mu_{\lambda}).$ Recall that the set
$(F_{\lambda})_{\lambda\in\Lambda(\mu)}$ is called by Jørgensen an exponential
dispersion model. It is not correct to think that if $\mu^{*}$ is a dual
measure of $\mu$ then $\Lambda(\mu)=\Lambda(\mu^{*}).$ We will see in Section
5 that a dual measure $\mu^{*}$ of the Bernoulli distribution $\mu$ exists and
is infinitely divisible. Therefore

$\Lambda(\mu^{*})=(0,\infty)\neq\Lambda(\mu)=\\{1,2,\ldots\\}.$

However if $\lambda$ is both in $\Lambda(\mu)$ and $\Lambda(\mu^{*})$, then a
dual measure of $\mu_{\lambda}$ satisfies

$B_{(\mu_{\lambda})^{*}}(m)=B_{\mu^{*}}(\frac{m}{\lambda}).$ (10)

###  3.3 Duality and the change of generating measure of an exponential family

Let us clarify now what happens to $\mu^{*}$ and to $F(\mu^{*})$ when we
replace $\mu(dx)$ by $\mu_{1}(dx)=e^{\langle s_{0},x\rangle}\mu(dx).$ This is
important, since $F(\mu)=F(\mu_{1}).$

######  Proposition 4.

Let $\mu\in\mathcal{M}(\mathbb{R}^{n})$ and suppose that there exists a dual
measure $\mu^{*}.$ Let $\mu_{1}(dx)=e^{\langle s_{0},x}\rangle\mu(dx).$ Then
$\mu_{1}^{*}=\mu^{*}*\delta_{s_{0}}$ is a dual measure of $\mu_{1}.$ In
particular, the elements of $F(\mu_{1}^{*})$ are the translation by $s_{0}$ of
the elements of $F(\mu^{*}).$ Symmetrically if $\mu_{2}=\mu*\delta_{m_{0}}$ is
a translation of $\mu$ by $m_{0}$ then $\mu_{2}^{*}(dx)=e^{\langle
m_{0},y\rangle}\mu^{*}(dy)$ is a dual measure of $\mu_{2}.$ In particular
$F(\mu_{2}^{*})=F(\mu^{*}).$

We skip the proof, which is an immediate application of the definitions. Let
us consider the simple example of the gamma exponential family $F$ with shape
parameter $\lambda,$ which is Case 2 in the Tweedie scale example in Section
2. The domain of the means is $M_{F}=(0,\infty)$ and its variance function is
$V_{F}(m)=m^{2}/\lambda.$ Applying Proposition 1 we write

$\displaystyle\ell^{\prime\prime}_{\mu^{*}}=\frac{\lambda}{m^{2}},\qquad\ell^{\prime}_{\mu^{*}}=-\frac{\lambda}{m}+s_{0},\qquad\ell_{\mu^{*}}=-\lambda\log
m+s_{0}m+s_{1},\qquad
B_{\mu^{*}}(m)=\frac{e^{s_{0}m+s_{1}}}{m^{\lambda}}=e^{s_{1}}\int_{0}^{\infty}e^{-(y-s_{0})m}\frac{y^{\lambda-1}}{\Gamma(\lambda)}dy.$

If $s_{0}=0$ then $F(\mu^{*})$ is the same gamma family. If $s_{0}\neq 0$ this
is a translation of this last family. Suppose that we have started from a
translated gamma family with $M_{F}=(m_{0},\infty)$ and
$V_{F}(m)=\lambda/(m-m_{0})^{2}.$ For this case we get the same families
$F(\mu^{*})$: they do not depend on the particular translation by $m_{0}.$

###  3.4 Duality for T–exponential families

The preceding discussion shows that while the dual of a measure is well
defined up to a multiplicative constant, this is not the case for an
exponential family $F,$ since changing the generating measure $\mu$ of $F$
into some $\mu_{1}$ implies that $F(\mu)=F(\mu_{1})$ by definition, but
possibly changes $F(\mu^{*})$ into a translate $F(\mu_{1}^{*})$. For this
reason we coin the definition of a T–exponential family:

######  Definition 1.

Given an exponential family $F$ the associated T–exponential family
$T\hskip-2.0ptF$ is the union of all the translations of $F$. If $F=F(\mu)$
then

$T\hskip-2.0ptF=\\{P(-s,\mu)*\delta_{m_{0}}\ ;\ s\in S(\mu),\
m_{0}\in\mathbb{R}^{n}\\}.$

Such a $\mu\in\mathcal{M}(\mathbb{R}^{n})$ is called a generating measure of
$T\hskip-2.0ptF$.

Note that $T\hskip-2.0ptF(\mu)=T\hskip-2.0ptF(\mu_{1})$ if and only if there
exists $s\in S(\mu),b\in\mathbb{R}^{n},m_{0}\in\mathbb{R}^{n}$ such that
$\mu_{1}=\delta_{m_{0}}*e^{-\langle s_{0},x\rangle+b}\mu(dx)$ or

$\ell_{\mu_{1}}(s)=-\langle s,m_{0}\rangle+b+\ell_{\mu}(s+s_{0}).$

We are now in position to clearly define the dual $T\hskip-2.0ptF^{*}$ of a
T–exponential family $T\hskip-2.0ptF$, thanks to the following proposition:

######  Proposition 5.

If $\mu$ is a generating measure of the T-exponential family TF in
$\mathbb{R}^{n}$ and if $\mu$ has a dual measure $\mu^{*}$ then any generating
measure $\mu_{1}$ has also a dual and

$T\hskip-2.0ptF(\mu^{*})=T\hskip-2.0ptF(\mu_{1}^{*})$

Under these circumstances we denote
$T\hskip-2.0ptF^{*}=T\hskip-2.0ptF(\mu^{*}).$

For instance we have seen that if $T\hskip-2.0ptF$ is the T–exponential family
of all the gamma distributions with shape parameter $\lambda$ augmented with
all its translations, then $T\hskip-2.0ptF=T\hskip-2.0ptF^{*}$ is self dual.
We will find a similar phenomena with the Gaussian distributions with variance
one, then with two unexpected examples: below in Section 6 with the
dilogarithm exponential family and the $\sinh m$ family, and, as a
generalization of the one dimensional gamma case, with the Wishart
distributions with fixed shape parameter.

###  3.5 Duality and large deviations.

Let $F=F(\mu)$ be a real natural exponential family and let $m_{0}<m$ two
points of $M_{F}.$ Let $X_{1},\ldots,X_{n},\ldots$ be independent random
variables with the same distribution in $F$ with mean $m_{0}$. The theorem of
large deviations, due to Cramér [6], says that for

$h(m_{0},m)=\int_{m_{0}}^{m}\frac{m-t}{V_{F}(t)}dt$

we have

$\left(\Pr(\frac{1}{n}(X_{1}+\cdots+X_{n})>m)\right)^{1/n}\overset{n\to\infty}{\longrightarrow}e^{-h(m_{0},m)}.$
(11)

In the next proposition, we link $h(m_{0},m)$ with $\mu^{*}$ as the rest of
the Taylor expansion of $m\mapsto\ell_{\mu^{*}}(m)$ around $m_{0}$. It shows
that $m\mapsto\exp h(m_{0},m)$ defines a member of $T\hskip-2.0ptF^{*}$ when
the distribution $P$ of $X_{n}$ is in $T\hskip-2.0ptF$. One can consider such
a proposition as a kind of probabilistic interpretation of duality in
dimension one.

######  Proposition 6.

Let $T\hskip-2.0ptF$ be a real T–exponential family such that its dual
$T\hskip-2.0ptF^{*}$ exists. Let $X_{1},\ldots,X_{n},\ldots$ be independent
random variables with the same distribution $P\in T\hskip-2.0ptF$ and mean
$m_{0}$. For $m_{0}<m$ define

$h(m_{0},m)=-\lim_{n\to\infty}\frac{1}{n}\log\Pr(\frac{1}{n}(X_{1}+\cdots+X_{n})>m).$

Suppose that the dual probability $P^{*}$ of $P$ exists. Let
$s_{0}=\ell_{P^{*}}(m_{0})$,
$c=\ell_{P^{*}}(m_{0})-m_{0}\ell^{\prime}_{P^{*}}(m_{0})$ and
$P^{*}_{m_{0}}=e^{c}P^{*}*\delta_{s_{0}}.$ Then

$B(P^{*}_{m_{0}})(m)=\int_{\mathbb{R}}e^{-my}P^{*}_{m_{0}}(dy)=e^{h(m_{0},m)}.$
(12)

###### Proof:.

From Proposition 1 we have $\ell^{\prime\prime}_{P^{*}}(m)=1/V_{F(P)}(m).$
Therefore by integration by parts

$\displaystyle
h(m_{0},m)=\int_{m_{0}}^{m}(m-t)\ell^{\prime\prime}_{P^{*}}(t)dt=(m_{0}-m)\ell^{\prime}_{P^{*}}(m)+\ell_{P^{*}}(m)-\ell_{P^{*}}(m_{0})=\ell_{P^{*}}(m)-ms_{0}+c.$

∎

###  3.6 Warnings for existence and non existence of the dual measures

To conclude Section 3, devoted to generalities on duality, we consider some
devices for proving that a function is not a Laplace transform for the case of
dimension 1. Other ones will appear in Section 7 for the $\mathbb{R}^{n}$
case.

It will be shown that the Vinogradov-Paris distribution has no dual.
Vinogradov and Paris [23] in their Theorem 6 consider the exponential family
$F$ on $(0,\infty)$ with variance function $2m^{2}/(1-m^{2})$ defined on
$M_{F}=(0,1).$ Now we show that it has no dual. Indeed if a dual measure
$\mu^{*}$ exists we have from Proposition 1

$\ell^{\prime\prime}_{\mu^{*}}(m)=\frac{1-m^{2}}{2m^{2}}\Rightarrow
B_{\mu^{*}}(m)=e^{-m^{2}/2}/\sqrt{m}.$

The function $e^{-m^{2}/2}$ is a Fourier transform, and $1/\sqrt{m}$ is a
Laplace transform. Who wins? By the theorem of maximal analyticity of Laplace
transforms (see Letac and Mora [15] and Kawata [11]) $B_{\mu^{*}}(m)$ is not
only defined on $M_{F}=(0,1)$ but also on $(0,\infty).$ However the second
derivative of $\log B_{\mu^{*}}(m)$ is negative on $m>1$ and this proves that
$B_{\mu^{*}}$ cannot be the Laplace transform of a positive measure. For
proving that some $T\hskip-2.0ptF^{*}$ does not exists, this trick can be
utilized when $F$ is not steep.

Let us apply Proposition 5 to the probability distributions

$P_{1}(dx)=\frac{1}{2}\,e^{-|x|}dx,\qquad
P_{2}(dx)=\frac{dx}{2\cosh(\frac{1}{2}\,\pi x)}$

and to a distribution $P_{3}$ described in Letac [16] generating the
exponential family with variance function $(1+m^{2})^{3/2}.$ The probabilities
$P_{1}$ and $P_{2}$ are called the bilateral exponential and the hyperbolic
distribution. They have means $m_{0}=0$, and respective Laplace transforms

$B_{1}(s)=\frac{1}{1-s^{2}},\qquad B_{2}(s)=\frac{1}{\cos s},$

which generate two exponential families with respective variance functions

$V_{1}(m)=m^{2}+1+\sqrt{m^{2}+1},\qquad V_{2}(m)=m^{2}+1.$

Performing the calculation of $h(0,m)=\ell_{\mu*}(m)$ for $P_{1},P_{2},P_{3}$,
we obtain

$\displaystyle
h_{1}(m)=2\frac{m^{2}+1-\sqrt{m^{2}+1}}{m^{2}}e^{\sqrt{m^{2}+1}-1},\qquad
h_{2}(m)=\frac{1}{\sqrt{m^{2}+1}}e^{m\arctan m},\qquad
h_{3}(m)=e^{\sqrt{m^{2}+1}-1}.$ (13)

For seeing that $P_{1}^{*}$ , $P_{2}^{*}$ and $P_{3}^{*}$ do not exist we use
Mathematica for computing

$\displaystyle h_{1}(m)=1+\frac{3m^{2}}{4}+o(m^{5}),\qquad
h_{2}(m)=1+\frac{m^{2}}{2}+\frac{m^{4}}{24}+\frac{m^{6}}{80}-\frac{283m^{8}}{3020}+o(m^{9}),\qquad
h_{3}(m)=1+\frac{m^{2}}{2}+o(m^{5}).$

This proves that
$\int_{\mathbb{R}}x^{4}P_{1}^{*}(dx)=\int_{\mathbb{R}}x^{4}P_{3}^{*}(dx)=0,\ \
\int_{\mathbb{R}}x^{8}P_{2}^{*}(dx)<0$: none of them is possible. Thus
$P_{1}^{*}$, $P_{2}^{*}$, and $P_{3}^{*}$ do not exist. My thanks go to Lev
Klebanov for suggesting this method for proving the lack of duality of some
measures.

##  4 The variance function $e^{m}$, the Landau law $\varphi$ and the dual of
Poisson family

In this section, we restrict ourselves to the one dimensional case.

###  4.1 The Landau distribution $\varphi$.

We will study $T\hskip-2.0ptF^{*}$ when $T\hskip-2.0ptF$ is the set of Poisson
distributions augmented with its translations. In Section 5 we are going to
study the dual pairs $(T\hskip-2.0ptF,T\hskip-2.0ptF^{*})$ issued of the case
where $V_{F}(m)$ is a quadratic or a cubic polynomial. For studying their dual
we will use a particular probability $\varphi$ on $\mathbb{R}$ called the
Landau law, which can be defined by its Laplace transform defined on
$S(\varphi)=(0,\infty)$ by $B_{\varphi}(s)=e^{-s}s^{s}$.

As we are going to see, $\varphi$ exists and has $\mathbb{R}$ as support.
Sometimes the term of Landau distribution is given to the convolution
$\varphi*\delta_{-1}$, because its Laplace transform has the more elegant form
$s^{s}$. It is the law of $X-1$ when $X\sim\varphi.$ A detailed study of the
Landau distribution can be found, for instance, in Marucho et al. [20].

The law $\varphi$ generates the exponential family with variance function
$e^{m}.$ For seeing this we write using for instance (5):

$\ell_{\varphi}(s)=s\log s-s,\qquad\ell^{\prime}_{\varphi}(s)=\log
s=-m,\qquad\ell^{\prime\prime}_{\varphi}(s)=\frac{1}{s}=e^{m}.$ (14)

An important feature of $\varphi$ for our purposes is the existence of the
following dual $\varphi^{*}$:

$s=-\ell^{\prime}_{\varphi}(-\ell^{\prime}_{\varphi^{*}}(s))=\log\ell^{\prime}_{\varphi^{*}}(s),\qquad\ell^{\prime}_{\varphi^{*}}(s)=e^{-s},\qquad\ell_{\varphi^{*}}(s)=e^{-s}-1,\qquad
B_{\varphi^{*}}(s)=e^{e^{-s}-1}.$

Thus a dual of the Landau distribution is the Poisson distribution with mean
1.

###  4.2 Existence of $\varphi$ and parents

######  Proposition 7.

For $s>0$ and $R>1$ we have

$\displaystyle\hskip-28.45274pte^{-s}s^{s}=\exp\int_{0}^{\infty}(e^{-sx}-1+sx\,e^{-x})x^{-2}dx,\qquad(1+s)^{1+s}(1-s)^{1-s}=\exp\int_{\mathbb{R}}(e^{-sx}-1-sx)x^{-2}e^{-|x|}dx,$
(15)
$\displaystyle\hskip-28.45274pt\tfrac{s^{s}}{(s+1)^{s+1}}=\exp-\int_{0}^{\infty}(1-e^{-sx})(1-e^{-x})x^{-2}dx,\qquad\tfrac{s^{s}}{(s+1)^{s}}=\exp-\int_{0}^{\infty}(1-e^{-sx})(1-e^{-x}(1+x))x^{-2}dx,$
(16)
$\displaystyle\hskip-28.45274pt\tfrac{s^{s}(s+1)^{\tfrac{s+1}{R-1}}}{(Rs+1)^{\tfrac{Rs+1}{R-1}}}=R^{-\tfrac{Rs}{R-1}}\exp-\int_{0}^{\infty}(1-e^{-sx})f(x)dx,\qquad
f(x)=\tfrac{R-1+e^{-x}-Re^{-x/R}}{(R-1)x^{2}},$ (17)
$\displaystyle\hskip-28.45274pt\tfrac{1}{4}\left(\tfrac{s+2}{s+1}\right)^{s+2}=\exp-\int_{0}^{\infty}(1-e^{-sx})(x-1+e^{-x})x^{-2}dx.$
(18)

###### Proof:.

We prove first that for $a>0$ we have

$F(a)=\int_{0}^{\infty}(e^{-u}-1+ue^{-au})u^{-2}du=-1-\log a.$

For seeing this we note that $F^{\prime}(a)=-1/a$ and doing an obvious
integration by parts we observe that $F(1)=-1.$ We pass from the value of
$F(a)$ to the first statement in (15) by the change of variable $u=sx.$

For proving the second statement in (15) we apply the first one while
replacing $s$ by $1+s$ and $1-s$, thus obtaining two integrals. In the second
one we change $x$ into $-x$ and thereafter summing the two integrals yield the
second expression in (15) plus the extra term

$2+\int_{\mathbb{R}}(e^{-|x|}-1-|x|e^{-|x|})\frac{dx}{x^{2}}=0,$

which is obtained by an integration by parts.

For proving the first relation in (16) we write patiently

$\displaystyle\hskip-28.45274pt\log\frac{s^{s}}{(s+1)^{s+1}}=s+\int_{0}^{\infty}(e^{-sx}-1+sx\,e^{-x})x^{-2}dx-(s+1)-\int_{0}^{\infty}(e^{-sx}e^{-x}-1+sx\,e^{-x}+xe^{-x})x^{-2}dx$
$\displaystyle=-1+\int_{0}^{\infty}(e^{-sx}(1-e^{-x})-xe^{-x})x^{-2}dx=-1+\int_{0}^{\infty}(e^{-sx}-\tfrac{xe^{-x}}{1-e^{-x}})(1-e^{-x})x^{-2}dx$
$\displaystyle=-1+\int_{0}^{\infty}(e^{-sx}-1)(1-e^{-x})x^{-2}dx+\int_{0}^{\infty}(1-\tfrac{xe^{-x}}{1-e^{-sx}})(1-e^{-x})x^{-2}dx=-\int_{0}^{\infty}(1-e^{sx})x^{-1}(1-e^{-x})x^{-1}dx.$

The last line uses the fact observed above that $F(1)=-1.$ For proving the
second statement in (16) we use the Frullani integral

$\log(1+s)=\int_{0}^{\infty}(1-e^{-sx})e^{-x}x^{-1}dx$ (19)

that we add to $\log\frac{s^{s}}{(s+1)^{s+1}}$ for obtaining the desired
result.

The proof of (17) follows the same lines: we start from the log of (15) and
replace $s$ by $s+1$ and $s+\frac{1}{R}$. We watch the coefficient $f(x)$ of
$e^{-sx}$ defined by (17) and we rearrange the remainder terms. The integral

$\int_{0}^{\infty}f(x)dx=\frac{\log R}{R-1}$

is computed by an integration by parts followed by the application of (19).

For proving (18) we combine (19) and the first statement in (16) applied to
$s+1$ instead of $s$:

$\displaystyle\tfrac{1}{4}\left(\tfrac{(s+2)}{s+1}\right)^{s+2}$
$\displaystyle=$
$\displaystyle\tfrac{1}{4}\left(\tfrac{(s+2)^{s+2}}{(s+1)^{s+1}}\right)\tfrac{1}{s+1}=\tfrac{1}{4}\exp\int_{0}^{\infty}\\{(e^{-sx}-1)e^{-x}x^{-1}+(1-e^{-sx}e^{-x})(1-e^{-x})x^{-2}\\}dx$
$\displaystyle=$
$\displaystyle\tfrac{1}{4}\exp\Big{\\{}\int_{0}^{\infty}(1-e^{-sx})(x-1+e^{-x})x^{-2}dx+\int_{0}^{\infty}\left(\tfrac{1-e^{-x}}{x}\right)^{2}dx\Big{\\}}.$

To conclude we have by an integration by parts and (19) that
$\int_{0}^{\infty}\left(\frac{1-e^{-x}}{x}\right)^{2}dx=\log 4$. This ends the
proof of (18). ∎

###  4.3 Lévy measures.

Recall that we concentrate on the one dimensional case in this section.
Extension to the $\mathbb{R}^{n}$ case is possible, but we will have no use of
it in this paper. Recall also that if a probability $P$ is in
$\mathcal{M}(\mathbb{R})$ then $P$ is infinitely divisible if there exists a
positive measure $\nu(dx)$ on $\mathbb{R}\setminus\\{0\\}$ such that

$\int_{\mathbb{R}\setminus\\{0\\}}\min(1,x^{2})\nu(dx)<\infty,$

and there exist two numbers $a\in\mathbb{R}$ and $\sigma\geq 0$ such that for
all $s\in S(P)$ we have

$\ell_{P}(s)=as+\frac{1}{2}\,\sigma^{2}s^{2}+\int_{\mathbb{R}\setminus\\{0\\}}(e^{-sx}-1+s\tau(x))\nu(dx).$
(20)

Here $\tau$ is a bounded function such that $\tau(x)/x\overset{x\to
0}{\longrightarrow}1$. Feller [7] chooses $\tau(x)=\sin x$, the Russian school
chooses $\tau(x)=x/(1+x^{2}).$ For getting simple formulas here we have chosen
$\tau(x)=xe^{-|x|}.$ Changing $\tau$ may change $a$ but neither $\sigma$ nor
the measure $\nu$, which is called the Lévy measure of $P$. The number
$\sigma$ is called the Gaussian part. If $\nu$ is bounded, we say that $P$ or
$\nu$ is Type 0. If $\nu$ is unbounded, but
$\int_{\mathbb{R}\setminus\\{0\\}}\min(1,|x|)\nu(dx)<\infty,$ we say that $P$
is Type 1. For Types 0 and 1, the representation of the Laplace transform
$B_{P}(s)$ does not need $\tau$ and we can write

$\ell_{P}(s)=as+\frac{1}{2}\,\sigma^{2}s^{2}-\int_{\mathbb{R}\setminus\\{0\\}}(1-e^{-sx})\nu(dx).$

In other cases we say that $P$ is of Type 2 and $\tau$ is necessary. The
convex hull of the support of such a probability of Type 2 is always
$\mathbb{R}$.

######  Example 3.

In Section 2 we have mentioned in the Tweedie scale the unsymmetric stable
distribution with parameter $\gamma\in(1,2)$ defined by its Laplace transform

$e^{s^{\gamma}}=\exp\left(as+\int_{0}^{\infty}(e^{-sx}-1+sxe^{-x})x^{-\gamma-1}\frac{dx}{\gamma(\gamma-1)\Gamma(2-\gamma)}\right).$
(21)

This formula can be guessed by differentiating $s^{\gamma}$ twice, and can be
checked by two integrations by parts, yielding also the exact value of the
constant $a.$ The Lévy measure is of Type 2 and equals

$\nu(dx)=1_{(0,\infty)}(x)x^{-\gamma-1}\frac{dx}{\gamma(\gamma-1)\Gamma(2-\gamma)}.$

From Proposition 7, the first formula in (15), we see that the Landau
distribution $\varphi$ exists, is infinitely divisible without Gaussian part
and has $\nu(dx)=1_{(0,\infty)}(x)\frac{dx}{x^{2}}$ for Lévy measure. It is of
Type 2.

From Proposition 7, the second formula in (15), we see that there exists an
infinitely divisible distribution $P$ such that
$B_{P}(s)=(1+s)^{1+s}(1-s)^{1-s}$ with $S(P)=\mathbb{R}$ and with Lévy measure

$\nu(dx)=\frac{e^{-|x|}}{x^{2}}dx.$

Its type is 2. We will see in Section 6 that $P$ is the dual of the symmetric
binomial distribution $\frac{1}{2}\,(\delta_{-1}+\delta_{1})$ as expected from
(1) of the introduction.

From Proposition 7, the first formula in (16), we see that there exists an
infinitely divisible distribution $P$ such that $B_{P}(s)=s^{s}/(s+1)^{s+1}$
with $S(P)=(0,\infty)$ and with Lévy measure

$\nu(dx)=1_{(0,\infty)}(x)\frac{1-e^{-x}}{x^{2}}dx.$

Its type is 1. We will see later that $P$ is the dual of the negative binomial
distribution.

From Proposition 7, the second formula in (16), we see that there exists an
infinitely divisible distribution $P$ such that $B_{P}(s)=s^{s}/(s+1)^{s}$
with $S(P)=(0,\infty)$ and with Lévy measure

$\nu(dx)=1_{(0,\infty)}(x)\frac{1-e^{-x}(1+x)}{x^{2}}dx.$

Its type is 0. We will see later that $P$ is the dual of an Abel distribution
sometimes called generalized Poisson distribution, generating an exponential
family on non-negative integers with cubic variance function $m(1+m)^{2}.$

From Proposition 7, the first formula in (17), we see that there exists an
infinitely divisible distribution $P$ such that
$B_{P}(s)=\frac{s^{s}(s+1)^{\frac{s+1}{R-1}}}{(Rs+1)^{\frac{Rs+1}{R-1}}}$ with
$S(P)=(0,\infty)$ and with Lévy measure

$\nu(dx)=1_{(0,\infty)}(x)f(x)dx$

where the positive function $f$ is defined by (17). Its type is 0, since the
integral of $f$ is finite. We will see later that $P$ is the dual of a Takács
distribution, generating an exponential family on non negative integers with
cubic variance function $m(1+m)(1+Rm)$.

From Proposition 7, the second formula in (17),we see that there exists an
infinitely divisible distribution $P$ such that
$B_{P}(s)=(\frac{s+2}{s+1})^{s+2}$ with $S(P)=(0,\infty)$ and with Lévy
measure

$\nu(dx)=1_{(0,\infty)}(x)\frac{x-1+e^{-x}}{x^{2}}dx.$

Its type is 0. We will see later that $P$ is the dual of a Kendall-Ressel
distribution, generating an exponential family with cubic variance function
$m^{2}(1+m)$.

##  5 A dictionary of dual pairs for quadratic and cubic families in
$\mathbb{R}$

We split the section in two parts: duals of Morris families, duals of cubic
families. The case of the Tweedie scale has been already ruled out in Section
3.1 and in particular, by Proposition 2. Most of the considered distributions
are infinitely divisible, which is saying that the corresponding Jørgensen set
is $(0,\infty).$ Formulas for passing from $\mu$ to $\mu_{\lambda}$ and from
$\mu^{*}$ to $(\mu_{\lambda})^{*}$ have been given above in Section 3.2. For
this reason, except in the binomial case, we give simplified versions,
ignoring the Jørgensen parameter $\lambda$. For instance, looking at the dual
of the negative binomial family with variance function
$m+\frac{m^{2}}{\lambda}$, we deal only with $m+m^{2}$ in order to have more
readable formulas for the duals.

The presentation of each pair $(T\hskip-2.0ptF,T\hskip-2.0ptF^{*})$ is done by
selecting a $\mu$ and a $\nu$ which may be only a translation of $\mu^{*}$.
These measures are generating $T\hskip-2.0ptF$ and $T\hskip-2.0ptF^{*}$
respectively, namely $F(\mu)\subset T\hskip-2.0ptF,\ F(\nu)\subset
T\hskip-2.0ptF^{*}.$ We give one of the two variance functions $V_{F(\mu)}(m)$
and $V_{F(\nu)}(m)$ as well one of the two Laplace transforms $B_{\mu}(s)$ and
$B_{\nu}(s)$. In most of the cubic cases, the densities on $\mathbb{R}$, or
the weights in the discrete cases, are not available and either the Laplace
transform or the variance function is computable explicitly: hence question
marks may appear in the description of the dual.

Logic would have imposed to write $V_{F(\nu)}(s)$ and $B_{\nu}(m)$
systematically. However we do not stick always to this rule. There are two
reasons: variance functions of exponential families have become familiar
objects during the last forty years from Morris [21], with many examples, and
we are used to write $m$ not $s$ for the mean, exactly like we are reluctant
to call $c$ the radius of a circle and $r$ its center. The second reason is
after all that $T\hskip-2.0ptF^{**}=T_{F}$. Which is the first,
$T\hskip-2.0ptF$ or $T\hskip-2.0ptF^{*}?$

###  5.1 Duals of Morris families

  1. 1.

The Poisson case

$(m,\exp(e^{-s});e^{m},e^{-s}s^{s}).$

The Poisson exponential family is generated by
$\mu=\sum_{n=0}^{\infty}\frac{\delta_{n}}{n!}$ with Laplace transform
$\exp(e^{-s})$. We have seen in Section 3.1 that $\ell_{\varphi}(m)=m\log m-m$
and therefore $\ell^{\prime}_{\varphi}(m)=\log m=-s$, $m=e^{-s}$
$\ell^{\prime}_{\varphi^{*}}(s)=-e^{-s}$ and, ignoring the integration
constant, $\ell_{\varphi^{*}}(s)=e^{-s}.$ Therefore $\mu$ is a dual of the
Landau distribution $\varphi.$

  2. 2.

The Bernoulli case

$(m-m^{2},1+e^{-s};4(\cosh\frac{m}{2})^{2},s^{s}(1-s)^{1-s}).$

The Bernoulli distribution is generated by $\mu=\delta_{0}+\delta_{1}$ leading
to

$\ell(\mu)=\log(1+e^{-s}),\ \ell^{\prime}(\mu)(s)=-\frac{1}{1+e^{s}}=-m,$ (22)

where $m$ is in the domain of the means $(0,1)$ of the Bernoulli distribution.
Therefore $\ell^{\prime}_{mu^{*}}(m)=-s=-\log(1-m)+\log m$ and, ignoring the
integration constant:

$\ell_{\mu^{*}}(m)=(1-m)\log(1-m)+m\log m,\ B_{\mu^{*}}(m)=m^{m}(1-m)^{1-m}.$

One can describe $\mu^{*}$ by introducing the law $\tilde{\varphi}$ of $X$
when $-X$ has the Landau distribution $\varphi$, thus with Laplace transform
$e^{m}(-m)^{-m}$ defined for $m<0.$ Now we introduce the measure
$\varphi_{1}(dx)=e^{x-1}\tilde{\varphi}(dx)$ with Laplace transform, defined
for $m<1$,

$B_{\varphi_{1}}(m)=\int_{\mathbb{R}}e^{-mx}\varphi_{1}(dx)=e\int_{\mathbb{R}}e^{-mx+x}\tilde{\varphi}(dx)=e^{m}(1-m)^{1-m}.$

Note, by assuming $m=0$ in $B_{\varphi_{1}}(m)$, $\varphi_{1}$ is a
probability. Finally from the observation of their Laplace transforms we get a
dual probability of $\mu$ as the convolution of $\varphi$ and $\varphi_{1}.$
We obtain

$\mu^{*}=\varphi*\varphi_{1},\qquad
B_{\varphi*\varphi_{1}}(m)=e^{-m}m^{m}e^{m}(1-m)^{1-m}=m^{m}(1-m)^{1-m},$

since the product of two Laplace transforms is well defined on the
intersection of their existence domain (here $(-\infty,1)$ and $(0,\infty)$.
For computing the variance function we use Proposition 1 and (24):

$\frac{1}{V_{mu^{*}}(m)}=\ell^{\prime\prime}_{\mu}(m)=\frac{d}{dm}\frac{-1}{1+e^{m}}=\frac{4}{\cosh^{2}\frac{m}{2}}.$

In Proposition 7 and the second statement in (15) we have considered the case
with the symmetric binomial distribution
$\mu=\frac{1}{2}(\delta_{-1}+\delta_{1})$ which is an affine transformation of
the ordinary Bernoulli distribution. We have seen that its dual exists and has
Laplace transform $(1-s)^{1-s}(1+s)^{1+s}$, and is infinitely divisible with
Lévy measure of Type 2. Since $\ell_{\mu}(s)=\log\cosh s$ we get the variance
function with Proposition 1:

$V(m)=\frac{1}{\ell^{\prime\prime}_{\mu}(m)}=\cosh^{2}m.$

  3. 3.

The binomial case

$(m-\frac{m^{2}}{N},(1+e^{-s})^{N};\frac{4}{N}(\cosh\frac{m}{2})^{2},s^{s}(N-s)^{N-s}).$
(23)

For describing the dual of the binomial distribution from the study of the
Bernoulli case, enough is to use (10).

  4. 4.

The negative binomial case

$(m+m^{2},(1-e^{-s})^{-1};4(\sinh\frac{m}{2})^{2},s^{s}(1+s)^{-1-s}).$

The negative binomial is generated by $\mu=\sum_{n=0}^{\infty}\delta_{n}$
leading to

$\ell_{\mu}(s)=-\log(1-e^{-s}),\
\ell^{\prime}_{\mu}(s)=-\frac{1}{e^{s}-1}=-m,$ (24)

where $m$ is in the domain of the means $(0,\infty)$. Therefore
$\ell^{\prime}_{\mu^{*}}(m)=-s=-\log(1+m)+\log m$ and, ignoring the
integration constant

$\ell_{\mu^{*}}(m)=-(1+m)\log(1+m)+m\log m,\
B_{\mu^{*}}(m)=m^{m}(1+m)^{-1-m}.$

Proposition 7, the first relation in (16), has shown the existence of the
probability $\mu^{*}$. It is an infinitely divisible distribution of Type 1.
Since $\ell^{\prime}_{\mu}(s)=-\frac{1}{e^{s}-1}$, the variance function of
$F(\mu^{*})$ is easily obtained:

$V_{F(\mu^{*}}(m)=\frac{1}{\ell^{\prime\prime}_{\mu}(m)}=4(\sinh\frac{m}{2})^{2}.$

  5. 5.

The gamma case

$(\frac{m^{2}}{\lambda},s^{-\lambda};\frac{m^{2}}{\lambda},s^{-\lambda}).$

This self dual has been detailed in Section 3 as a particular case of the
Tweedie scale.

  6. 6.

The Gaussian case

$(\sigma^{2},\exp(\frac{1}{2}\,\sigma^{2}s^{2});\sigma^{-2},\exp(\frac{1}{2}\,\sigma^{-2}s^{2})).$

This case is very simple since
$\ell_{N(0,\sigma^{2})}(s)=\frac{1}{2}\,\sigma^{2}s^{2},\
\ell^{\prime}_{N(0,\sigma^{2})}(s)=\sigma^{2}s$ and therefore
$\ell^{\prime}_{N^{*}(0,\sigma^{2})}(s)=\sigma^{-2}s.$

  7. 7.

The hyperbolic case We have seen in Section 3.6 with equation (13) that these
laws have no dual.

###  5.2 Duals of cubic families

  1. 1.

The Abel case

$(m(1+m)^{2},?;?,s^{s}/(s+1)^{s}).$

  2. 2.

The Takacs case

$(m(1+m)(1+Rm),?;?,\frac{s^{s}(s+1)^{\frac{s+1}{R-1}}}{(Rs+1)^{\frac{Rs+1}{R-1}}}.$

  3. 3.

The Kendall Ressel case

$(m^{2}(1+m),?;?,(\frac{s+1}{s})^{s+1}).$

The above given three cases are similar. The densities of these three cubic
families are explicit (see Letac and Mora [15]) but not the Laplace
transforms. Similarly the variance function of the dual is not computable
(except if we accept a description by a sum of a series obtained by the
Lagrange formula). However using Proposition 1 and the relation
$\ell^{\prime\prime}_{\mu*}(m)=1/V_{F}(m)$ we can catch the Laplace transform
of the dual. Therefore we apply this principle to the three functions

$\frac{1}{m(1+m)^{2}},\qquad\frac{1}{m(1+m)(1+Rm)},\qquad\frac{1}{m^{2}(1+m)}.$

Fortunately these relations can be integrated twice leading to the three
Laplace transforms appearing in the second expression of (16), (17) and (18),
respectively. For seeing that these three duals do exist, the hard work has
been done in Proposition 7, where it is shown that these three duals are
infinitely divisible, as commented on in Section 4.3. About (18), observe that
the Laplace transform $(\frac{s+1}{s})^{s+1}$ of an unbounded measure $\mu$ is
changed by the substitution $s\mapsto s+1$ into the Laplace transform
$(\frac{s+2}{s+1})^{s+2}$, then converted into the Laplace transform
$\frac{1}{4}(\frac{s+2}{s+1})^{s+2}$ of a probability of $F(\mu)$:

  1. 1.

The Inverse Gaussian case

$(\frac{m^{3}}{\lambda},e^{-\lambda\sqrt{2s}};2^{3/2}\lambda
m^{3/2},\exp{\lambda^{2}/4s}).$

This is an application of Proposition 2 with $q=3$.

  2. 2.

The strict arcsine family case

$(m(1+m^{2}),?;?,(\frac{s}{\sqrt{1+s^{2}}})^{s}e^{\arctan s}).$

For computing the Laplace transform of $\mu^{*}$ we use Proposition 1,

$\ell^{\prime\prime}_{\mu^{*}}(s)=\frac{1}{s(1+s^{2})}=\frac{1}{s}-\frac{s}{1+s^{2}},\qquad\ell^{\prime}_{\mu^{*}}(s)=\log
s-\frac{1}{2}\,\log(1+s^{2}),$

leading to $B_{\mu^{*}}(s)=(\frac{s}{\sqrt{1+s^{2}}})^{s}e^{\arctan s}.$ For
seeing that the dual exists we note that

$\int_{0}^{\infty}e^{-sx}(1-\cos
x)dx=\frac{1}{s}-\frac{s}{1+s^{2}}\ell_{\mu^{*}}(s)=\int_{0}^{\infty}(e^{-sx}-1)\frac{1-\cos
x}{x^{2}}dx.$

Therefore $\mu^{*}$ exists, is infinitely divisible with Lévy measure
$\frac{1-\cos x}{x^{2}}1_{(0,\infty)}(x)dx$ and is of Type 0.

  3. 3.

The large arcsine family case $(m(1+2m\cos a+m^{2}),?;?,B_{\mu^{*}}(s))$ with
$0<a<\pi/2.$ The Laplace transform of the dual Proposition 1 is obtained as
follows:

$\ell^{\prime\prime}_{\mu^{*}}(s)=\frac{1}{s(1+2s\cos
a+s^{2})}=\frac{1}{s}-\frac{1}{\sin^{2}a}\frac{s+2\cos a}{\left(\frac{s+\cos
a}{\sin a}\right)^{2}+1}.$

Observing that $\ell^{\prime\prime}_{\mu^{*}}$ is the Laplace transform of a
positive measure is not quite obvious. Introducing $u=\frac{s+\cos a}{\sin a}$
leads to $s=u\sin a-\cos a$ and to

$\displaystyle\frac{1}{\sin^{2}a}\frac{s+2\cos a}{\left(\frac{s+\cos a}{\sin
a}\right)^{2}+1}$ $\displaystyle=$
$\displaystyle\frac{1}{\sin^{2}a}\frac{u\sin a+\cos
a}{u^{2}+1}=\frac{1}{\sin^{2}a}\int_{0}^{\infty}e^{-ut}\sin(a+t)dt$
$\displaystyle=$
$\displaystyle\frac{1}{\sin^{2}a}\int_{0}^{\infty}e^{-t\frac{s+\cos a}{\sin
a}}\sin(a+t)dt=\frac{1}{\sin a}\int_{0}^{\infty}e^{-sx}e^{-x\cos
a}\sin(a+x\sin a)dx.$

We now show that for $x>0$ and $0\leq a\leq\pi/2$ we have

$\frac{1}{\sin a}e^{-x\cos a}\sin(a+x\sin a)\leq 1$

or, equivalently, that $f(x)=\sin x\ e^{x\cos a}-\sin(a+x\sin a)\geq 0$. For
seeing this we use the two inequalities $e^{c}\geq 1+c$ and $r\geq\sin r$
respectively applied to $c=x\cos a$ and $r=x\sin a$ and we get

$\displaystyle f(x)\geq\sin a(1+x\cos a)-\sin a\cos r-\sin r\cos a=\sin
a(1-\cos r)+\cos a(r-\sin r)\geq 0.$

As a consequence, denoting

$g(x)=1-\frac{1}{\sin a}e^{-x\cos a}\sin(a+x\sin a)\overset{x\to
0}{\sim}\frac{x^{2}}{2}$

we can claim that $\mu^{*}$ exists, is infinitely divisible with Lévy measure
$\frac{g(x)}{x^{2}}1_{(0,\infty)}(x)dx$ and is Type 0. One can note that the
limit values $a=\pi/2$ and $a=0$ yield the Abel and the strict arcsine cases.
The computation of the function $B_{\mu^{*}}(s)$ is a painful exercise of
calculus. We get indeed

$\displaystyle\ell_{\mu^{*}}(s)$ $\displaystyle=$ $\displaystyle s\log
s-\tfrac{1}{2}(s+\cos a)\log(s+\cos a)+(-\tfrac{1}{2}+(\tfrac{\pi}{2}-a)\cos
a)s-\tfrac{1}{2}(\mathrm{cotan}\,a)\log(1+2s\cos a+s^{2})$
$\displaystyle+(\mathrm{cotan}\,a)\left(\frac{s+\cos a}{\sin
a}\right)\arctan\left(\frac{s+\cos a}{\sin a}\right)+\tfrac{1}{2}\cos
a\log\cos a+(\mathrm{cotan}\,a)^{2}(\tfrac{\pi}{2}-a).$

##  6 The variance function $e^{m}-1$ and the dilogarithm law

In this section we still continue our study of duality in the one dimensional
case. We introduce the dilogarithm law which generates the exponential family
with variance function $e^{m}-1.$ This gives the opportunity to introduce
parent distributions $\mu_{r},$ $\sigma,$ $\sigma_{r},$ $\eta$ and $\alpha$
with interesting properties, and exponential families with unexpected variance
functions, like $e^{m}+1$ and $\sinh m.$ We will see that $\mu$ and $\alpha$
are self dual.

###  6.1 The generating probability $\mu$

From the Bar Lev criteria described in Letac and Mora [15] (Corollary 3.3 and
Proposition 4.4), $V(m)=e^{m}-1$ defined on $(0,\infty)$ is the variance
function of an exponential family $F$ concentrated on the non-negative
integers. Let us compute a particular generating probability
$\mu=\sum_{n=0}^{\infty}\mu(n)\delta_{n}$ of $F$. We use

$ds=-\frac{dm}{V(m)}=-\frac{dm}{e^{m}-1}=-\frac{e^{-m}dm}{1-e^{-m}}=-d\log(1-e^{-m}).$

Therefore $S(\mu)=(0,\infty)$ and

$e^{-s}=1-e^{-m}\Rightarrow\ell^{\prime}_{\mu}(s)=-m=\log(1-e^{-s})\Rightarrow\ell^{\prime}_{\mu}(s)=-\sum_{n=1}^{\infty}\frac{e^{-sn}}{n}.$
(25)

Recall that

$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}.$

Continuing the calculation and choosing properly the integration constant such
that $k_{\mu}(0)=0$ we have

$\ell_{\mu}(s)=-\frac{\pi^{2}}{6}+\sum_{n=1}^{\infty}\frac{e^{-ns}}{n^{2}}\Rightarrow
B_{\mu}(s)=\sum_{n=0}^{\infty}\mu(n)e^{-ns}=\exp\left(-\frac{\pi^{2}}{6}+\sum_{n=1}^{\infty}\frac{e^{-ns}}{n^{2}}\right).$

As a consequence

$\sum_{n=0}^{\infty}\mu(n)z^{n}=\exp\left(\sum_{n=1}^{\infty}\frac{z^{n}-1}{n^{2}}\right).$
(26)

This generating function is linked to the classical special function
$Li_{2}(z)$, called dilogarithm, defined on the unit disk $\\{z\in\mathbb{C}\
;|z|\leq 1\\}$ by $Li_{2}(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{2}},$ for
which a vast literature exists from Euler: let us quote Lewin [19], Zagier
[24] and Kirilov [12]. The measure $\mu$ is a probability as we can see by
putting $z=1$ in (26). It is an infinitely divisible distribution with Lévy
mesure $\sum_{n=1}^{\infty}\frac{\delta_{n}}{n^{2}}$. For this reason we call
$\mu$ the dilogarithm law in the sequel. Here is a surprising property that
$\mu$ shares with the normal and the gamma laws, namely the self duality.

######  Proposition 8.

The dilogarithm distribution $\mu$ is dual of itself.

###### Proof:.

The result is immediate from (25) which implies the symmetric relation
$e^{-s}+e^{-m}=1$. ∎

Consider $Y$ and $Y^{\prime}$ which are independent with the same distribution
$\mu$. We denote by $\sigma$ the distribution of $Y-Y^{\prime}$ in
$\mathbb{Z}$. Consider also an element $P(-s,\mu))$ of the exponential family
$F(\mu)$. Since $S(\mu)=(0,\infty)$ we rather write $r=e^{-s}\in(0,1)$ and we
consider the probability $\mu_{r}=P(-s,\mu)$ which satisfies

$\sum_{n=0}^{\infty}\mu_{r}(n)z^{n}=e^{Li_{2}(rz)-Li_{2}(r)}.$ (27)

Similarly, we introduce $Y_{r}$ and $Y_{r}^{\prime}$ independent with the same
distribution $\mu_{r}$ and denote by $\sigma_{r}$ the distribution of
$Y_{r}-Y_{r}^{\prime}$ in $\mathbb{Z}$. In Sections 6.2 and 6.3 we study the
four probabilities $\mu,\sigma,\mu_{r},\sigma_{r}.$

###  6.2  Some properties of $\mu$ and $\sigma$

For having a probabilistic interpretation of the probability $\mu$ defined by
(26), consider the probability $\nu$ on positive integers defined by

$\nu=\frac{6}{\pi^{2}}\sum_{n=1}^{\infty}\frac{\delta_{n}}{n^{2}}$

Suppose that $X_{1},\ldots,X_{n},\ldots$ are iid with distribution $\nu$ and
consider an independent Poisson random variable $N$ with mean
$\lambda=\pi^{2}/6.$ Then the distribution of $X_{1}+\cdots+X_{N}$ is $\mu$ by
a classical calculation using conditioning by $N.$ Of course, this shows that
$\mu$ is infinitely divisible of type 0.

######  Proposition 9.

If $Y$ and $Y^{\prime}$ are independent with the same distribution $\mu$ then
the characteristic function of the random variable $Y-Y^{\prime}\sim\sigma$ on
$\mathbb{Z}$ is the function with period $2\pi$ given for $0\leq t\leq 2\pi$
by $\mathbb{E}(e^{it(Y-Y^{\prime})})=e^{-\frac{1}{2}t(2\pi-t)}$.

###### Proof:.

Since
$\mathbb{E}(e^{itY})=\exp\left(\sum_{n=1}^{\infty}\frac{e^{int}-1}{n^{2}}\right)$
we have

$\mathbb{E}(e^{it(Y-Y^{\prime})})=\exp\left(2\sum_{n=1}^{\infty}\frac{\cos
nt-1}{n^{2}}\right)=\exp\left(-\frac{1}{2}t(2\pi-t)\right),$

and due to elementary calculation of Fourier series we have, for $0\leq t\leq
2\pi$, $\sum_{n=1}^{\infty}\frac{\cos
nt}{n^{2}}=\frac{\pi^{2}}{6}-\frac{1}{4}t(2\pi-t)$. ∎

######  Remark 2.

1) Since $S(\mu)=(0,\infty)$, then $\sigma$ has no Laplace transform and
cannot generate an exponential family. 2) The calculation of
$\Pr(Y=Y^{\prime})$ is not elementary. More specifically by the change of
variable $s=t-\pi$ and Mathematica:

$Pr(Y=Y^{\prime})=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-\frac{1}{2}t(2\pi-t)}dt=\frac{1}{\pi}\int_{0}^{\pi}e^{\frac{1}{2}\,s^{2}-\frac{1}{2}\,\pi^{2}}ds=0.11751.$

By a similar calculation for $n\in\mathbb{Z}$

$Pr(Y-Y^{\prime}=n)=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-\frac{1}{2}t(2\pi-t)+int}dt=\frac{(-1)^{n}}{\pi}\int_{0}^{\pi}e^{\frac{1}{2}\,s^{2}-\frac{1}{2}\,\pi^{2}}\cos(ns)ds.$

The fact that $\mathbb{E}(e^{it(Y-Y^{\prime})})=e^{\frac{t^{2}}{2}-\pi t}$ in
$[0,2\pi]$ calls for a link with the Gaussian distribution, as it is shown in
the next proposition.

######  Proposition 10.

If $Z\sim N(0,1)$ and $Y,Y^{\prime}\sim\mu$ are independent, then the
logarithm of the characteristic function of the continuous random variable
$W=Z+Y-Y^{\prime}$ is made of piecewise affine functions and is concave. More
specifically, if $t-2\pi k\in[0,2\pi)$ with $k\in\mathbb{Z}$ then

$\log\mathbb{E}(e^{itW})=2k(k+1)\pi^{2}-(2k+1)\pi t$

and the density $f(x)$ of $W$ is, using the notation
$e_{k}(x)=e^{-2k^{2}\pi^{2}-2i\pi x}$, equals

$f(x)=\frac{1}{2\pi}\sum_{k\in\mathbb{Z}}\frac{e_{k}(x)-e_{k+1}(x)}{(2k+1)\pi+ix}.$
(28)

###### Proof:.

If $t-2\pi k\in[0,2\pi)$ with $k\in\mathbb{Z}$ we have

$\log\mathbb{E}(e^{itW})=-\frac{1}{2}t^{2}+\frac{1}{2}(t-2k\pi)^{2}-\pi(t-2k\pi)=2k(k+1)\pi^{2}-(2k+1)\pi
t.$

To compute $f(x)$ we use the inverse Fourier transform leads easily to (28):

$\displaystyle
f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-itx}\mathbb{E}(e^{itW})dt=\frac{1}{2\pi}\sum_{k\in\mathbb{Z}}\int_{2k\pi}^{2(k+1)}e^{-itx+2k(k+1)\pi^{2}-(2k+1)\pi
t}dt.$

∎

###  6.3  Some properties of $\mu_{r}$ and $\sigma_{r}$

######  Proposition 11.

Let $0<r<1$. If $Y_{r}$ and $Y_{r}^{\prime}$ are independent with the same
distribution $\mu_{r}$ then the Laplace transform of the law $\sigma_{r}$ of
$Y_{r}-Y^{\prime}_{r}$ is defined for $s$ in $S(\sigma_{r})=(\log r,\log(1/r)$
by

$\mathbb{E}(e^{-s(Y_{r}-Y^{\prime}_{r})})=\exp\left(-\int_{0}^{r}\log(1-2x\cosh\theta+x^{2})\frac{dx}{x}-2\int_{0}^{r}\log(1-x)\frac{dx}{x}\right).$

The distribution $\sigma_{r}$ generates an exponential family $F(\sigma_{r})$,
where $\mathbb{R}$ is the domain of the means and the variance function is,
with $a=\sinh\frac{m}{2}$,

$V_{F(\sigma_{r})}(m)=\frac{2}{1-r^{2}}\sqrt{r^{2}+a^{2}}(\sqrt{r^{2}+a^{2}}+\sqrt{1+a^{2}}).$

###### Proof:.

From (27) $B_{\sigma_{r}}(s)$ exists if and only if $re^{-s}<1$ and
$re^{-s}<1,$ in other terms if $|s|<-\log r.$ Since
$Li_{2}(r)=-\int_{0}^{r}\log(1-x)\frac{dx}{x}$ we get

$Li_{2}(re^{-s})=-\int_{0}^{re^{-s}}\log(1-x)\frac{dx}{x}=-\int_{0}^{r}\log(1-xe^{-s})\frac{dx}{x}$

and $B_{\sigma_{r}}(s)$ is easily deduced from this expression. ∎

For computing the variance function, we observe that

$m=k^{\prime}_{\sigma_{r}}(s)=\log(1-re^{-s})-\log(1-re^{s}).$ (29)

Equality (29) shows that $\mathbb{R}$ is the domain of the means by letting
$s\to\pm\log r.$ Also, (29) can be rewritten,

$a=\sinh\frac{m}{2}=r\sinh\left(-s+\frac{m}{2}\right)$ (30)

With the classical notations $\theta=\psi(m)$ and
$V_{F(\sigma_{r})}(m)=1/\psi^{\prime}(m)$ we can write, by differentiating
(30) with respect to $m$,

$\displaystyle\tfrac{1}{2}\sqrt{1+a^{2}}=\tfrac{1}{2}\cosh\tfrac{m}{2}=r\cosh\left(\psi(m)+\tfrac{m}{2}\right)(\psi^{\prime}(m)+\tfrac{1}{2})=r\sqrt{1+\frac{a^{2}}{r^{2}}}\left(\frac{1}{V_{F(\sigma_{r})}(m)}+\tfrac{1}{2}\right).$

This transform leads to the desired formula:

$V_{F(\sigma_{r})}(m)=\frac{2\sqrt{r^{2}+a^{2}}}{\sqrt{1+a^{2}}-\sqrt{r^{2}+a^{2}}}=\frac{2}{1-r^{2}}\sqrt{r^{2}+a^{2}}(\sqrt{1+a^{2}}+\sqrt{r^{2}+a^{2}}).$

The calculation of the dual of $\sigma_{r}$ involves elliptic integrals and
its existence is unproven.

###  6.4 The variance function $e^{m}+1$

In this section we study the exponential family $F(\eta)$ on $\mathbb{R}$ with
variance function $e^{m}+1$ and with the most surprising property that
probability $\eta$ is equal to $N(0,1)*\mu$ where $\mu$ is the dilogarithm
probability defined by (26).

Since the function $e^{m}+1$ is real analytic and positive on $\mathbb{R}$
then, if it is a variance fuction, the domain of the means of the
corresponding natural exponential family will be $\mathbb{R}.$ Now, to decide
whether it is a variance function or not, we have to compute a potential
Laplace transform in the usual way:

$ds=-\frac{dm}{e^{m}+1}=\frac{e^{-m}dm}{1+e^{-m}}=-\frac{du}{1+u}=-d\log(1+u)=-d\log(1+e^{-m}).$

Since $m>0$ we can take $S(\eta)=(0,\infty)$ and $s=\log(1+e^{-m}).$ This
leads to

$\ell^{\prime}_{\eta}(s)=-m=\log(e^{s}-1)=s+\log(1-e^{-s})=s-\sum_{n=1}^{\infty}\frac{e^{-ns}}{n}.$

By choosing properly integration constant such that $\eta$ is a probability,
we get

$\ell_{\eta}(s)=\tfrac{1}{2}s^{2}-\tfrac{\pi^{2}}{6}+\sum_{n=1}^{\infty}\frac{e^{-ns}}{n}=\ell_{N(0,1)}(s)+\ell_{\mu}(s).$

Showing the unexpected result that $\eta=N(0,1)*\mu$ proves also that $\eta$
exists and is a positive measure, and that $e^{m}+1$ is the variance function
of an exponential family. Proposition 11 given above can be reformulated in
terms of $\mu$ and $\eta$. If $X\sim\eta$ and $Y^{\prime}\sim\mu$ are
independent, then $W\sim X-Y^{\prime}$. One can remark that since $e^{m}$ and
$e^{m}\pm 1$ are variance functions, therefore using affinities and the
Jørgensen set would enable us to describe all exponential families with
variance functions $Ae^{m/\lambda}+B.$ We leave to the reader to prove that
$\eta$ has no dual.

###  6.5 The variance function $\sinh(m)$

In this section we study the exponential family $F(\alpha)$ on $\mathbb{N}$
with variance function $\sinh(m)$ and we describe the link between the
probability $\alpha=(\alpha(n))_{n=0}^{\infty}$ and the dilogarithm function
$Li_{2}(z).$ The existence of $\alpha$ is granted by the Bar-Lev theorem since
the Taylor expansion of $\sinh(m)$ has only nonnegative coefficients (see
[15], Corollary 3.3). The fact that $\alpha$ is concentrated on $\mathbb{N}$
is granted by the fact that $\sinh(m)\overset{m\to 0}{\sim}m$ (see [15],
Proposition 4.4). For computing $\alpha$ we proceed in the usual way

$\displaystyle
ds=-\frac{dm}{\sinh(m)}=-\frac{2e^{m}dm}{e^{2m}-1}=-\frac{2du}{u^{2}-1}=-\left(\frac{1}{u-1}-\frac{1}{u+1}\right)du=-d\log\frac{u-1}{u+1}=-d\log\frac{e^{m}-1}{e^{m}+1}.$

Since $m>0$ we can take $S(\alpha)=(0,\infty)$ and
$s=-\log\frac{e^{m}-1}{e^{m}+1}.$ This leads to

$\ell^{\prime}_{\alpha}(s)=-m=-\log\frac{1+e^{-s}}{1-e^{-s}}=-2\sum_{n=1}^{\infty}\frac{e^{-(2n-1)s}}{2n-1}.$
(31)

Writing $z=e^{-s}\in(0,1)$ we get

$\ell_{\alpha}(s)=C+2\sum_{n=1}^{\infty}\frac{z^{2n-1}}{(2n-1)^{2}}=C+2Li_{2}(z)-\frac{1}{2}Li_{2}(z^{2}).$

Since $Li_{2}(1)=\pi^{2}/6$, in order to have $\alpha$ of mass 1, we take
$C=-\pi^{2}/4$ and we finally obtain

$\sum_{n=0}^{\infty}\alpha(n)z^{n}=e^{-\frac{\pi^{2}}{4}+2Li_{2}(z)-\frac{1}{2}Li_{2}(z^{2})}.$
(32)

Of course, if

$\beta(dx)=\frac{8}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}\delta_{2n-1},$

if $N$ is Poisson distributed with mean $\pi^{2}/4$, and if
$X_{1},\ldots,X_{n},\ldots$ are iid with distribution $\beta$ and independent
of $N$, we have $X_{1}+\cdots+X_{N}\sim\alpha$.

######  Proposition 12.

The probability $\alpha$ is dual of itself.

###### Proof:.

From (31) we obtain $e^{-m}+e^{-s}+e^{-s-m}=1.$ The symmetry between $m$ and
$s$ implies the result. ∎

The last proposition links $\alpha$ and $\mu$. The proof follows from (32).

######  Proposition 13.

If $Y\sim\mu$ and $X\sim\alpha^{*2}$ are independent, then $X+2Y\sim\mu^{*4}$.

##  7 Examples of duality in $\mathbb{R}^{n}$

The classification of exponential families in $\mathbb{R}^{n}$ has been done
in the literature with the same guidelines as for $\mathbb{R}$: from the
simplest variance functions like the Tweedie scale, or like the quadratic ones
of Morris [21] to more complicated ones like the cubic families (see [15]) or
the Babel class (see [16]). For $\mathbb{R}^{n},$ several choices of
classification through a definition of simple variances have been carried out:

  1. 1.

The first one has been to extend the gamma family with shape parameter $p>0$,
whose variance function is $m^{2}/p$, to homogeneous quadratic variance
functions in $\mathbb{R}^{n}.$ More specifically one had to find which
variance matrices $V_{F}(m)$ of order $n$ are made of homogeneous quadratic
polynomials with respect to $m=(m_{1},\ldots,m_{n}).$ The answer has been
given by Casalis [3]: out of trivial cases, the only exponential families with
homogeneous quadratic variance functions are the Wishart ones, on symmetric,
Hermitian, quaternionic Hermitian matrices, and analogous objects on the
Lorentz cones and on the exceptional Albert algebra.

  2. 2.

The second choice has been to consider the so called simple quadratic families
in $\mathbb{R}^{n}.$ While a quadratic family has variance function of the
form $V_{F}(m)=A(m)+B(m)+C$, where $m\mapsto A(m)$ is quadratic homogeneous,
$m\mapsto B(m)$ is linear and $C$ is a constant symmetric matrix of order $n$,
such a family is said to be simple if $A(m)$ has rank one, and therefore of
the form $\frac{1}{\lambda}m\otimes m$; recall that if $E$ is a Euclidean
space and $a,b$ are in $E$ then $a\otimes b$ is the endomorphism of $E$
defined by

$x\mapsto(a\otimes b)(x)=a\langle b,x\rangle.$

Here again their classification has been done by Casalis [4], while the
classification of general quadratic families is an open problem.

  3. 3.

The third block is the simple cubic exponential families, obtained from the
simple quadratic ones by a Möbius transformation of their variance function.
They have been classified by Hassaïri [8].

  4. 4.

The last choice is the so called diagonal families in $\mathbb{R}^{n}$ defined
by the fact that the diagonal part of $V_{F}(m)$ has the form
$(V_{1}(m_{1}),\ldots,V_{n}(m_{n})).$ They have been classified in a paper
with six coauthors: see Bar-Lev et al. [2].

###  7.1 The Wishart families are self dual

In the Euclidean space $V$ of real symmetric matrices of order $n$ with scalar
product $\langle x,y\rangle=\mathrm{trace}\,(xy)$ consider the cone
$V_{+}\subset V$ of positive definite matrices and the cone
$\overline{V_{+}}\subset V$ of semipositive definite matrices. If

$p\in\Lambda=\left\\{\frac{1}{2}\,,1,\frac{3}{2},\ldots,\frac{n-1}{2}\right\\}\cup\left(\frac{n-1}{2},\infty\right),$

the family of the Wishart distributions of shape parameter $p$ is generated by
the unbounded measure $\mu_{p}$ on $\overline{V_{+}}$ which can be defined on
$s\in V_{+}$ by its Laplace transform

$\int_{\overline{V_{+}}}e^{-\langle s,x\rangle}\mu_{p}(dx)=\frac{1}{(\det
s)^{p}}.$

As a consequence $\ell_{\mu_{p}}(s)=-p\det s,\
\ell^{\prime}_{\mu_{p}}(s)=-ps^{-1}$ since the gradient of $s\mapsto\log\det
s$ is $s^{-1}.$ Clearly
$-\ell_{\mu_{p}}(-\ell_{\mu_{p}}(s))=p(ps^{-1})^{-1}=s$ and this shows that
$\mu_{p}$ is a dual of itself.

We are not going to give details for the Wishart distributions defined on the
other symmetric cones: after introducing the necessary definitions, the simple
calculation above remains the same. One can consult Casalis and Letac [5], for
instance.

###  7.2 The dual of the multinomial distribution

Let $(e_{1},e_{2},\ldots,e_{n})$ be the canonical basis of $\mathbb{R}^{n}$
and let us call Bernoulli distribution any law concentrated on the $n+1$
points $(0,e_{1},e_{2},\ldots,e_{n}),$ namely of the form

$p_{0}\delta_{0}+p_{1}\delta_{e_{1}}+p_{2}\delta_{e_{2}}+\cdots+p_{n}\delta_{e_{n}},$

where $p_{i}>0$ for $i\in\\{0,\ldots,n\\}$ and $p_{0}+p_{1}+\cdots+p_{n}=1.$
The set of all these Bernoulli distributions is an exponential family
generated by the measure $\mu=\delta_{0}+\delta_{e_{1}}+\cdots+\delta_{e_{n}}$
whose Laplace transform is

$\Delta=B_{\mu}(s_{1},\ldots,s_{n})=1+e^{-s_{1}}+\cdots+e^{-s_{n}}$

Therefore

$\ell_{\mu}(s)=\log\Delta,\qquad m_{1}=-\frac{\partial}{\partial
s_{1}}\ell_{\mu}(s)=\frac{e^{-s_{1}}}{\Delta},\ldots,\
m_{n}=-\frac{\partial}{\partial
s_{n}}\ell_{\mu}(s)=\frac{e^{-s_{n}}}{\Delta}.$

For deciding whether $\mu^{*}$ exists or not we compute $\ell_{\mu^{*}}(m)$
such that $-\ell^{\prime}_{\mu^{*}}(-\ell^{\prime}_{\mu}(s)=s.$ Thus we have
to compute $(s_{1},\ldots,s_{n})$ with respect to $(m_{1},\ldots,m_{n})$ as
follows:

$-m_{1}\Delta=e^{-s_{1}},\ \ldots,-m_{n}\Delta=e^{-s_{n}},\qquad
1+(m_{1}+\cdots+m_{n})\Delta=\Delta,\qquad\Delta=\frac{1}{1-m_{1}-\cdots-
m_{n}}.$

Therefore

$s_{1}=-\log m_{1}+\log(1-m_{1}-\cdots-m_{n}),\ldots,s_{n}=-\log
m_{n}+\log(1-m_{1}-\cdots-m_{n}).$

As a consequence, if $\mu^{*}$ does exist, it must satisfy

$\ell^{\prime}_{\mu^{*}}(m)=\left(\log m_{1}-\log(1-m_{1}-\cdots-
m_{n}),\ldots,\log m_{n}-\log(1-m_{1}-\cdots-m_{n})\right).$

It is easy to see that up to an additive constant we have

$\displaystyle\ell_{\mu^{*}}(m)$ $\displaystyle=$ $\displaystyle(m_{1}\log
m_{1}+\cdots+m_{n}\log m_{n}+(1-m_{1}-\cdots-m_{n})\log(1-m_{1}-\cdots-
m_{n}),$ $\displaystyle B_{\mu^{*}}(m)$ $\displaystyle=$ $\displaystyle
m_{1}^{m_{1}}\ldots m_{n}^{m_{n}}(1-m_{1}-\cdots-m_{n})^{1-m_{1}-\cdots-
m_{n}}$

Recall that the Landau distribution $\varphi$ is defined by
$\int_{\mathbb{R}}e^{-sx}\varphi(dx)=e^{-s}s^{s}.$ In order to describe the
above $\mu^{*}$ let us coin a simple lemma:

######  Lemma 1.

Let $f(s)=f(s_{1},\cdots,s_{n})=a_{1}s_{1}+\cdots+a_{n}s_{n}+b=\langle
a,s\rangle+b$ be a non-constant affine form of $\mathbb{R}^{n}$. Then
$e^{-f(s)}f(s)^{f(s)}$, defined on the half plane $H=\\{s;f(s)>0\\}$, is the
Laplace transform of a positive measure $\mu(dx)$ on $\mathbb{R}^{n}$,
concentrated on the line $\mathbb{R}a$, defined as the image of the measure
$\varphi_{1}(dt)=e^{-bt}\varphi(dt)$ by the map $t\mapsto x=at.$ This measure
is bounded if $b\geq 0$, of mass $e^{b}b^{b}$ when $b>0$, and mass 1 if $b=0.$

###### Proof:.

By definition we have

$\int_{\mathbb{R}^{n}}e^{-\langle
s,x\rangle}\mu(dx)=\int_{\mathbb{R}}e^{-t\langle
s,a\rangle}\varphi_{1}(dt)=\int_{\mathbb{R}}e^{-t\langle s,a\rangle-
bt}\varphi(dt)=\int_{\mathbb{R}}e^{-tf(s)}\varphi(dt).$

The remainder is plain. ∎

With this lemma we can describe $\mu^{*}(dx)$ which satisfies

$\int_{\mathbb{R}^{n}}e^{-s_{1}x_{1}-\cdots-
s_{n}x_{n}}\mu^{*}(dx)=s_{1}^{s_{1}}\ldots s_{n}^{s_{n}}(1-s_{1}-\cdots-
s_{n})^{1-s_{1}-\cdots-s_{n}}$

on the tetrahedron $\\{s;s_{1},\ldots,s_{n},1-s_{1}-\cdots-s_{n}>0\\}$ which
is the domain of the means of the exponential family of the Bernoulli
distribution. From the lemma, applied to the $n+1$ affine forms
$f_{0}(s)=-s_{1}-\cdots-s_{n}+1$, $f_{1}(s)=s_{1},\ \ldots\ f_{n}(s)=s_{n}$,
the measure $\mu^{*}$ is the convolution of $n+1$ probabilities respectively
concentrated on the lines $\mathbb{R}e_{1},\ldots,Re_{n}$ and
$\mathbb{R}(-e_{1}-\cdots-e_{n}).$ For $f_{0}$ the measure described in the
lemma is bounded and has mass $e^{-1}$ and has to be normalized to become a
probability:

$e^{1-f_{0}}f_{0}^{f_{0}}e^{-f_{1}}f_{1}^{f_{1}}\ldots,e^{-f_{n}}f_{n}^{f_{n}}=(1-s_{1}-\cdots-
s_{n})^{1-s_{1}-\cdots-s_{2}}s_{1}^{s_{1}}\ldots s_{n}^{s_{n}}.$

It is interesting to compute the variance function of $F(\mu^{*}):$

######  Proposition 14.

Let $\Delta=1+e^{-s_{1}}+\cdots+e^{-s_{n}}$,
$D=\mathrm{diag}(e^{-s_{1}},\ldots,e^{-s_{n}})$ and $J_{n}$ the $(n,n)$ matrix
with all entries equal to one. Let $\mu^{*}$ be a dual of
$\mu=\delta_{0}+\delta_{e_{1}}+\cdots+\delta_{e_{n}}.$ Then the variance
function of $F(\mu^{*})$ is

$V_{F(\mu^{*})}(s)=\Delta(D^{-1}+J_{n})=(1+e^{-s_{1}}+\cdots+e^{-s_{n}})\left[\begin{array}[]{ccc}e^{s_{1}}+1&\ldots&1\\\
\vdots&\ddots&\vdots\\\ 1&\ldots&e^{s_{n}}+1\end{array}\right].$

###### Proof:.

Let $v=v(s)=(e^{-s_{1}},\ldots,e^{-s_{n}})^{T}$. Note that $\Delta^{\prime}=v$
and that $v^{\prime}=-D.$ With this notation we have, since
$\ell_{\mu}=\log\Delta$,

$\ell^{\prime}_{\mu}(s)=-\frac{1}{\Delta}v,\qquad\ell^{\prime\prime}_{\mu}(s)=\frac{1}{\Delta}D-\frac{1}{\Delta^{2}}v\otimes
v,\qquad V_{F(\mu*)}(s)=[\ell^{\prime\prime}_{\mu}(s)]^{-1}.$

For computing the inverse matrix we use the following identity: if
$a\in\mathbb{R}^{n}$ such that $\|a\|^{2}=a_{1}^{2}+\cdots+a_{n}^{2}$ is
different from 1, then

$(I_{n}-a\otimes a)^{-1}=I_{n}+\frac{a\otimes a}{1-\|a\|^{2}}.$

We are going to apply this expresion to

$a=\Delta^{-1/2}D^{-1/2}v=\Delta^{-1/2}(e^{-s_{1}/2},\ldots,e^{-s_{n}/2})^{T},$

which satisfies

$\|a\|^{2}=\frac{\Delta-1}{\Delta},\qquad
1-\|a\|^{2}=\frac{1}{\Delta},\qquad\frac{a\otimes
a}{1-\|a\|^{2}}=(D^{-1/2}v)\otimes(D^{-1/2}v),$ $\displaystyle V_{F(\mu*)}(s)$
$\displaystyle=$
$\displaystyle\left(\frac{1}{\Delta}D-\frac{1}{\Delta^{2}}v\otimes
v\right)^{-1}=\Delta D^{-1/2}(I_{n}-a\otimes a)^{-1}D^{-1/2}=\Delta
D^{-1/2}(I_{n}+\frac{a\otimes a}{1-\|a\|^{2}})D^{-1/2}$ $\displaystyle=$
$\displaystyle\Delta(D^{-1}+\Delta(D^{-1/2}a)\otimes(D^{-1/2}a))=\Delta(D^{-1}+J_{n}).$

∎

######  Remark 3.

Note that when $n=1$ the variance is
$(1+e^{-s})(1+e^{s})=4\cosh^{2}\frac{s}{2}$ as seen in (23) .

###  7.3 The other simple quadratic families.

Casalis [4] splits the set of simple quadratic families in $\mathbb{R}^{n}$ in
$2n+4$ types, extending the famous Morris classification for $n=1$ in six
families. Among them $n+1$ are trivial from our point of view, with variance
function of the form $B(m)+C$ where $m\mapsto B(m)$ is linear. In fact, with a
proper choice of a basis of $\mathbb{R}^{n}$ they are product of $k$ Poisson
families and $n-k$ one dimensional Gaussian families with
$k\in\\{0,\ldots,n\\}$ (see also [14]). Since trivially a product of duals is
dual of the product, there is nothing to add. The remaining $n+3$ families are
made with two exceptional ones, the multinomial and the hyperbolic, and a set
of $n+1$ types denoted $(\mathrm{NM-ga})_{k}$ families with
$k\in\\{0,\ldots,n\\}$ that we will describe later. In the sequel we are going
to prove that neither the hyperbolic type nor the $(\mathrm{NM-ga})_{k}$ have
a dual when $k>0$. The case $k=n$ is referred to as the multivariate negative
binomial case. The case $k=0$ has a dual if and only if $n=2$ or $n=3.$

Now it is shown that there is no dual for the multivariate negative binomial
distribution. Denote by $\mathbb{N}$ the set of nonnegative integers. Consider
the measure on $\mathbb{N}^{n}$ defined by
$\mu(dx)=(\delta_{0}-\delta_{e_{1}}-\cdots-\delta_{n})^{-1*}$ where the
inverse is taken in the sense of convolution. Its Laplace transform is

$B_{\mu}(s)=\frac{1}{1-e^{-s_{1}}-\cdots-e^{-s_{n}}}=\frac{1}{\Delta},$

defined on the set

$S(\mu)=\\{s\in\mathbb{R}^{n};\Delta>0\\}.$

We show that no dual $\mu^{*}$ of $\mu$ exists. Suppose the existence of
$\mu^{*}$. Since $\ell_{\mu}=-\log\Delta$ we have

$m=(m_{1},\ldots,m_{n})=-\ell^{\prime}_{\mu}(s)=\frac{1}{\Delta}(e^{-s_{1}},\ldots,e^{-s_{n}}).$

Since $e^{s_{i}}=m_{i}\Delta$ we get
$\Delta(m_{1}+\cdots+m_{n})=e^{-s_{1}}+\cdots+e^{-s_{n}}=1-\Delta$ leading to
$\Delta=1/(1+m_{1}+\cdots+m_{n}).$ Finally we have on the domain $m_{i}\geq
0$, for all $i\in\\{1,\ldots,n\\}$,

$\ell^{\prime}_{\mu^{*}}(m)=(-s_{1},\ldots,-s_{n})$

with $-s_{i}=\log m_{i}-\log(1+m_{1}+\cdots+m_{n})$ and therefore

$\displaystyle\ell_{\mu^{*}}(m)$ $\displaystyle=$ $\displaystyle m_{1}\log
m_{1}+\cdots+m_{n}\log
m_{n}-(1+m_{1}+\cdots+m_{n})\log(1+m_{1}+\cdots+m_{n}),$ $\displaystyle
B_{\mu^{*}}(m)$ $\displaystyle=$ $\displaystyle\frac{m_{1}^{m_{1}}\cdots
m_{n}^{m_{n}}}{(1+m_{1}+\cdots+m_{n})^{1+m_{1}+\cdots+m_{n}}}.$ (33)

For fixed $m$ and $m^{\prime}$ the matrix

$M=\left[\begin{array}[]{cc}B_{\mu^{*}}(2m)&B_{\mu^{*}}(m+m^{\prime})\\\
B_{\mu^{*}}(m+m^{\prime})&B_{\mu^{*}}(2m^{\prime})\end{array}\right]$ (34)

is semi positive definite since

$\displaystyle(a,b)M\left(\begin{array}[]{c}a\\\ b\end{array}\right)$
$\displaystyle=$ $\displaystyle
a^{2}B_{\mu^{*}}(2m)+2abB_{\mu^{*}}(m+m^{\prime})+b^{2}B_{\mu^{*}}(2m^{\prime})=\int_{\mathbb{R}^{n}}(ae^{-\langle
m,x\rangle}+be^{-\langle m^{\prime},x\rangle})^{2}\mu^{*}(dx)\geq 0.$

This implies that $\det M\geq 0.$ Let us apply this remark to the particular
case $m=(0,1,0,\ldots,0)$ and $m^{\prime}=(1,1,0,\ldots,0).$ It leads to

$\det
M=\frac{1}{3^{3}4^{7}5^{5}}(4^{8}-3^{3}5^{5})=-\frac{19339}{3^{3}4^{7}5^{5}}<0$

which is the desired contradiction.

Next it is shown that there is no dual for the $(\mathrm{NM-ga})_{n-1}$ case.
We use the notation of [4]. This exponential family is generated by the
measure $\mu(dx_{1},\ldots,dx_{n-1},dy)$ on $\mathbb{N}^{n-1}\times(0,\infty)$
defined by

$(\delta_{0}-\delta_{e_{1}}-\cdots-\delta_{e_{n-1}})^{-1*}(dx_{1},\ldots,dx_{n-1})\frac{y^{x_{1}+\cdots+x_{n-1}}}{(x_{1}+\cdots+x_{n-1})!}1_{(0,\infty)}(y)dy.$

The exponential family $F(\mu)$ is the set of the laws of $(X,Y)$ when
$X=(X_{1},\ldots,X_{n-1})$ is multivariate negative binomial on
$\mathbb{N}^{n-1}$ and the conditional law of $Y$ knowing $X$ is gamma with
shape parameter $1+|X|=1+X_{1}+\cdots+X_{n-1}.$ Its variance function is

$V_{\mu}(m)=m\otimes m+\mathrm{diag}(m_{1},\ldots,m_{n-1},0)$

and the Laplace transform of $\mu$ is

$B_{\mu}(s)=(s_{n}-e^{-s_{1}}-\ldots-e^{s_{n-1}})^{-1}=\frac{1}{\Delta}.$

We show first the following: If $\mu^{*}$ exists then its Laplace transform is
defined on $m_{i}>0$, $i\in\\{1,\cdots,n\\}$, by

$B_{\mu^{*}}(m)=\frac{m_{1}^{m_{1}}\ldots
m_{n-1}^{m_{n-1}}}{m_{n}^{m_{1}+\cdots+m_{n-1}+1}}.$ (36)

To see this, since $\Delta^{\prime}=(e^{-s_{1}},\ldots,e^{-s_{n-1}},1)$ and
that $m=-\ell^{\prime}_{\mu}(s)=\Delta^{\prime}/\Delta$ one gets
$m_{n}=1/\Delta$ and $e^{-s_{i}}=m_{i}/m_{n}$ for $i<n$ leading to

$\ell^{\prime}_{\mu^{*}}(m)=-s=\log(m_{1}/m_{n}),\ldots,\log(m_{n-1}/m_{n}),(1+m_{1}+\cdots+m_{n-1})/m_{n}$

from which the result is obtained.

To see that actually $\mu^{*}$ does not exist we use the same method as for
the multivariate negative binomial distribution by observing that the matrix
$M$ defined by (34) has a negative determinant when choosing
$m=(0,\ldots,0,1)$ and $m^{\prime}=(1,\ldots,0,1).$

Now it will be established that there is no dual for the $(\mathrm{NM-
ga})_{k}$ case with $0<k<n-1$. This exponential family is generated by the
measure $\mu(dx_{1},\ldots,dx_{k},dy,dz)$ on
$\mathbb{N}^{k}\times(0,\infty)\times\mathbb{R}^{n-k-1}$ defined by

$(\delta_{0}-\delta_{e_{1}}-\cdots-\delta_{e_{k}})^{-1*}(dx_{1},\ldots,dx_{k})\frac{y^{x_{1}+\cdots+x_{k}}}{(x_{1}+\cdots+x_{k})!}1_{(0,\infty)}(y)dy\times
N(0,yI_{n-k-1})(dz),$

where $N(0,yI_{n-k-1})$ is the Gaussian distribution on $\mathbb{R}^{n-k+1}$
with covariance matrix $yI_{n-k-1}$. Its Laplace transform is
$B_{\mu}(s)=1/\Delta$ with

$\Delta=s_{k+1}-\frac{1}{2}\,\sum_{j=k+2}^{n}s_{j}^{2}-\sum_{i=1}^{k}e^{-s_{i}}.$

Computations, quite analogous to the previous one, yield that the Laplace
transform of $\mu^{*}$ would be

$B_{\mu^{*}}(m)=\frac{\prod_{i=1}^{k}e^{-m_{i}}m_{i}^{m_{i}}}{m_{k+1}^{1+m_{1}+\cdots+m_{k}}}\exp\left(\frac{1}{2}\,\sum_{j=k+2}^{n}\frac{m_{j}^{2}}{m_{k+1}}\right).$
(37)

If $k\neq 0$, putting $m_{k+2}=\ldots=m_{n}=0$ shows that $B_{\mu^{*}}(m)$
cannot be a Laplace transform as we have seen for (36).

Consider now the $(\mathrm{NM-ga})_{0}$ case $B_{\mu}(s)=1/\Delta$ with

$\Delta=s_{1}-\frac{1}{2}\,\sum_{j=2}^{n}s_{j}^{2}.$

This generates the exponential family containing the distribution of
$(X_{1},Z_{2}\sqrt{X_{1}},\ldots,Z_{n}\sqrt{X_{1}}),$ where
$X_{1},Z_{2},\ldots,Z_{n}$ are independent, with $Z_{i}\sim N(0,1)$ and where
$X_{1}$ is exponential with mean 1. Standard calculations show that if
$\mu^{*}$ exists then

$B_{\mu^{*}}(m)=\frac{1}{m_{1}}\exp\left(\frac{1}{2}\,\sum_{j=2}^{n}\frac{m_{j}^{2}}{m_{1}}\right).$

Denoting for simplicity $\vec{m}=(m_{2},\ldots,m_{n})$ we can write

$B_{\mu^{*}}(m_{1},\vec{m})=\int_{\mathbb{R}^{n-1}}e^{-\langle\vec{m},x\rangle-\frac{1}{2}\,m_{1}\|x\|^{2}}m_{1}^{(n-3)/2}\frac{dx}{(2\pi)^{(n-1)/2}}.$

If $n-3\leq 0$ we prove that $B_{\mu^{*}}$ is a Laplace transform as follows:

If $n=2$ we use

$\frac{1}{\sqrt{m_{1}}}=\int_{0}^{\infty}e^{-tm_{1}}\frac{dt}{\sqrt{\pi t}}$

for writing

$\displaystyle
B_{\mu^{*}}(m_{1},m_{2})=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-m_{2}x}\int_{0}^{\infty}e^{-m_{1}(t+\frac{1}{2}\,x^{2})}\frac{dt}{\sqrt{\pi
t}}dx=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-m_{2}x}\int_{x^{2}/2}^{\infty}e^{-m_{1}y}\frac{dy}{\sqrt{\pi(y-\frac{1}{2}\,x^{2})}}dx.$

This shows that $B_{\mu^{*}}$ is the Laplace transform of the density
$\frac{1}{\sqrt{2\pi}\sqrt{\pi(y-\frac{1}{2}\,x^{2})}}$ restricted to the
interior of the parabola $y=x^{2}/2.$

If $n=3$ we have

$B_{\mu^{*}}(m_{1},\vec{m})=\tfrac{1}{2\pi}\int_{\mathbb{R}^{2}}e^{-\langle\vec{m},x\rangle-\frac{1}{2}\,m_{1}\|x\|^{2}}dx.$

This is the Laplace transform of the following singular measure in
$\mathbb{R}^{3}$: it is concentrated on the cone
$y=\frac{1}{2}\,(x_{1}^{2}+x_{2}^{2})$ and it is the image of the Lebesgue
measure $\frac{dx_{1}dx_{2}}{2\pi}$ on $\mathbb{R}^{2}$ by the map
$x\mapsto(x,\frac{1}{2}\,\|x\|^{2}).$

To complete the study of these last two measures $\mu^{*}$, one can describe
the variance function of $F(\mu^{*})$ with the following matrices for $n=2$
and $n=3:$

$\displaystyle V_{F(\mu^{*})}(s_{1},s_{2})$ $\displaystyle=$
$\displaystyle\left(s_{1}-\frac{1}{2}\,s_{2}^{2}\right)\left[\begin{array}[]{cc}s_{1}+\frac{1}{2}\,s_{2}^{2}&\frac{1}{2}\,s_{2}\\\
\frac{1}{2}\,s_{2}&1\end{array}\right],$ $\displaystyle
V_{F(\mu^{*})}(s_{1},s_{2},s_{3})$ $\displaystyle=$
$\displaystyle\left(s_{1}-\frac{1}{2}\,(s_{2}^{2}+s_{3}^{2})\right)\left[\begin{array}[]{ccc}s_{1}+\frac{1}{2}\,(s_{2}^{2}+s_{3}^{2})&\frac{1}{2}\,s_{2}&\frac{1}{2}\,s_{3}\\\
\frac{1}{2}\,s_{2}&1&0\\\ \frac{1}{2}\,s_{3}&0&1\end{array}\right].$

We skip this standard computation.

For $n\geq 4$ one can suspect that $\mu^{*}$ does not exist. Enough is to
prove it for $n=4,$ since we can do $m_{5}=\ldots=m_{n}=0$ to pass from the
case $n>4$ to the case $n=4.$ For simplicity of calculations in the case $n=4$
we rather set $m_{1}=t$ and $m_{2}=s_{1},\ m_{3}=s_{2},\ m_{y}=s_{3}.$ And we
show now that

$L(\vec{s},t)=\frac{1}{t+1}\exp\left(\frac{1}{2(t+1)}\|\vec{s}\|^{2}\right)$

is not a Laplace transform on $(-1,\infty)\times\mathbb{R}^{3}.$ Suppose the
contrary. Consider a random variable $(\vec{X},Y)$ such that for $t>-1$

$A=e^{-\langle\vec{s},\vec{X}\rangle-tY},\ \mathbb{E}(A)=L(\vec{s},t).$

Denote $\vec{\mathbf{1}}=(1,1,1)$ We are going to prove that for a suitable
$(r,\vec{s},t)$ we have

$\mathbb{E}(\|\vec{X}-rY\vec{\mathbf{1}}\|^{2}A)<0$ (40)

which proves the impossibility. By taking partial derivatives of
$L(\vec{s},t)$ a calculation gives
$\mathbb{E}(\|\vec{X}-rY\vec{\mathbf{1}}\|^{2}A)=\frac{1}{(t+1)^{5}}e^{\frac{u}{t+1}}B$
with the shorter notations $u=\frac{1}{2}\,\|\vec{s}\|^{2}$ and

$B=2u(t+1)^{2}+2r\langle\vec{s},\vec{\mathbf{1}}\rangle((t+1)^{3}+(t+1)^{2}(1+u))+r^{2}(2(t+1)^{2}+4u(t+1)+u^{2})$

We now choose $\vec{s}=-2r\vec{\mathbf{1}}$. This gives $u=6r^{2}$ and
$\langle\vec{s},\vec{\mathbf{1}}\rangle=6r.$ The quantity $B$ becomes

$B=-12r((t+1)^{3}+(t+1)^{2}(1+6r^{2}))+r^{2}(14(t+1)^{2}+24(t+1)r^{2}+36r^{4}).$

For fixed $t$ we can now choose $r$ small enough such that $B<0$. Therefore
(40) is proved.

Finally it is shown that there is no dual for the multivariate hyperbolic
case. Following Casalis [4], p. 1836, the multivariate hyperbolic case is
described by the Laplace transform

$B_{\mu}(s)=(\cos s_{n}-e^{-s_{1}}-\cdots-e^{s_{n-1}})^{-1}$

defined on

$S(\mu)=\\{s\in\mathbb{R}^{n-1}\times(-\frac{1}{2}\,\pi,\frac{1}{2}\,\pi)\
;\cos s_{n}-e^{-s_{1}}-\cdots-e^{s_{n-1}}>0\\}.$

We spare to the reader the proof of the fact that if a dual $\mu^{*}$ exists
then its Laplace transform is, with the notation $S=1+m_{1}+\cdots+m_{n-1}$,

$B_{\mu^{*}}(m)=\frac{\prod_{i=1}^{n-1}m_{i}^{m_{1}}}{(S^{2}+m_{n}^{2})^{S/2}}\exp(m_{n}\arctan(m_{n}/S)).$

Note that if $m_{1}=\cdots=m_{n-1}=0$ we get $B_{\mu^{*}}(m)=h_{2}(m_{n})$
where $h_{2}$ is defined in (13). But is has been proven in Section 3 that
$h_{2}$ cannot be a Laplace transform. As a consequence the present $\mu^{*}$
does not exist.

##  8 Remarks.

Duality, with its links with the large deviations as described in Proposition
6, has not a strong probabilistic interpretation compared to the notion of
reciprocity, which was used in Letac and Mora [15]. However it widens the zoo
of variance functions, and it creates unexpected links like the pair Poisson-
Landau. The problem of deciding of the existence of a dual is unsolved in many
circumstances, and we need more tools than we described in Section 3.5. Why do
some measures have dual and why others have not? The Babel class has been
defined and described in Letac [16], as the set the variance functions
$b\Delta+(am+c)\sqrt{\Delta}$ where $\Delta$ is a polynomial of degree $\leq
2$. Its study from the duality view point has not been considered in the
present paper. The same problem arises with the elliptic variance functions
(Letac [18]) or the Seshadri class (Kokonendji [13]). The deep functional
equations of the dilogarithm described in [12, 19, 24] should lead to new
properties of the dilogarithm distribution. A complete characterization of
self duality would also be in order. The role of steepness is not well
understood: all tractable examples of a non steep $\mu$ lead to the non
existence of a dual . Does non steepness imply no dual? Dealing with
exponential families in $\mathbb{R}^{n}$ is generally hard: the simplest non
trivial diagonal family in $\mathbb{R}^{2}$ is generated by the measure
$\delta_{0}+\delta_{e_{1}}+\delta_{e_{2}}+c\delta_{e_{1}+e_{2}}$ where $c\neq
1$ (see [2] p. 895) and leads to computations of great complexity. We have not
tried yet the cubic families in $\mathbb{R}^{n}$ for $n>1$ characterized by
Hassairi [8].

## Acknowledgments

I have had some discussions with Shaul Bar-Lev, Lev Klebanov and Vladimir
Vinogradov about various parts of the paper. I thank them all.

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