Boundedness of Variance Functions of Natural Exponential Families with Unbounded Support

Abstract

The variance function (VF) is central to natural exponential family (NEF) theory. Prompted by an online query about whether, beyond the classical normal NEF, other real-line NEFs with bounded VFs exist, we establish three complementary sets of sufficient conditions that yield many such families. One set imposes a polynomial-growth bound on the NEF’s generating measure, ensuring rapid tail decay and a uniformly bounded VF. A second set relies on the Legendre duality, requiring a uniform positive lower bound on the second derivative of the generating function, which likewise ensures a bounded VF. The third set starts from the standard normal distribution and constructs an explicit sequence of NEFs whose Laplace transforms and VFs remain bounded. Collectively, these results reveal a remarkably broad class of NEFs whose Laplace transforms are not expressible in elementary form (apart from those stemming from the standard normal case), yet can be handled easily using modern symbolic and numerical software. Worked examples show that NEFs with bounded VFs are far more varied than previously recognized, offering practical alternatives to the normal and other classical models for real-data analysis across many fields.

1. Introduction

The motivation for this study comes from an online query asking whether natural exponential families (NEFs) with unbounded support can have bounded variance functions (VFs) beyond the classical normal NEF. In response, we establish explicit sufficient conditions for the existence of such NEFs. These conditions reveal a broad class of previously undocumented non-elementary NEFs whose Laplace transforms (LTs) or VFs cannot be expressed in elementary (algebraic) form. Thanks to modern mathematical software, such as Maple and Mathematica, as well as open-source libraries in Python and R, this obstacle is no longer a practical limitation; symbolic or numerical evaluation of the relevant transforms has become routine. The normal NEF serves as a benchmark within the subclass studied in this context. Our partial solution deepens understanding of the interplay between support structure and variance behavior in NEFs, thereby addressing a gap in the current literature. Although our focus is theoretical, we expect the subclasses introduced in this paper to assist in modeling real-world data across diverse fields.

The VF is central to the theory of natural exponential families (NEFs). Although the normal NEF is the only known example with a constant VF, it is natural to ask whether other NEFs—particularly those with unbounded support on $\mathbb{R}$—can also have bounded VFs. This issue is important both theoretically and practically, as bounded VFs often lead to more stable modeling.

This paper establishes three complementary results—Theorem 1 and Propositions 1 and 2—that provide sufficient conditions under which an NEF supported on the entire real line can still have a bounded VF. Theorem 1 imposes a polynomial-growth constraint on the cumulant generating function, ensuring sufficient tail decay of the density. Proposition 1 relies on the Legendre duality: a uniform positive lower bound on the second derivative of the generating function guarantees global convexity and, consequently, a bounded VF. Proposition 2 builds on the standard normal distribution and constructs an explicit sequence of NEFs whose VFs remain bounded; for this sequence, both the LTs and the VFs can be written in closed form. Together, these results introduce a broad new class of NEF models, greatly expanding the range of distributions available for real-data analysis and providing strong alternatives to traditional families.

This paper is structured as follows. Section 2 provides an overview of the essential background on NEFs. Section 3 presents Theorem 1, along with Propositions 1 and 2, and explains how these results are related. Section 4 provides several illustrative examples, while Section 5 offers concluding remarks.

2. Preliminaries on NEFs

The preliminaries are mainly drawn from [1,2]. We provide only the brief background necessary for the developments that follow. Let $\mu$ be a non-Dirac positive Radon measure on $\mathbb{R}$, S the support of $\mu$, and C the convex hull of S. LT of $\mu$ is a mapping $L:$$\mathbb{R}\to (0,\infty ]$ defined by

$$L\left(\theta \right)={\int }_{\mathbb{R}}{e}^{\theta x}\mu \left(dx\right).$$

Let D denotes the effective domain of $\mu$, i.e., $D=\left\{\theta \in \mathbb{R}:L\left(\theta \right)<\infty \right\}$ and assume that $\Theta =int$$D\ne \varphi .$ Let $k\left(\theta \right)=lnL\left(\theta \right)$ be the cumulant transform of $L.$ Then the NEF F generated by $\mu$ is defined by the family of probabilities

$$F=F\left(\mu \right)=\left\{P(\theta ,\mu \left(dx\right))=\exp \left\{\theta x-k\left(\theta \right)\right\}\mu \left(dx\right),\theta \in \Theta \right\}.$$

(1)

The measure $\mu$ is called a generating measure of the NEF. It is not unique, as any other measure of the form ${\mu }^{*}\left(dx\right)=e^{ax+b}\mu \left(dx\right),$ with real constants a and b, generates the same NEF (c.f., [2]). The function $k=\log L$ defined on $\Theta$ is real analytic on there, and its successive derivatives are the successive cumulants of F. In particular, $k^{\prime }$ and $k^{″}$, are the mean and variance of F.

As we consider only cases where $\mu$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}$, we henceforth restrict our attention to generating measures of the form $\mu \left(dx\right)=f\left(x\right)dx$, where f is the respective Radon–Nikodym derivative.

The following are some notions and properties related to NEFs that will be used in the sequel.

Regular NEFs. If D is an open interval, F is called regular. Particularly, if $D=\mathbb{R},{\mathbb{R}}^{+},$ or ${\mathbb{R}}^{-},$F is regular.

Mean value parameterization: The cumulant transform k is strictly convex and real analytic on $\Theta$ and

$$k^{\prime }\left(\theta \right)={\int }_{\mathbb{R}}x\exp \left\{\theta x-k\left(\theta \right)\right\}f\left(x\right)dx$$

is the mean function of F. The open interval $M=k^{\prime }(\Theta )$ is called the mean domain of F. Since the map $\theta ⟼k^{\prime }\left(\theta \right)$ is one-to-one, its inverse function $\psi :M⟶\Theta$ is well defined. Hence, the map $m⟼P(m,F)=P\left(\psi \right(m),\mu )$ is one-to-one from M onto F and is called the mean value parameterization of $F.$

Variance Function (VF). The variance corresponding to the NEF (1) is

$$V\left(m\right)=1/{\psi }^{\prime }\left(m\right)=k^{″}\left(\theta \right).$$

(2)

The map $m⟼V\left(m\right)$ from M into ${\mathbb{R}}^{+}$ is called the variance function (VF) of F. In fact, a VF of an NEF F is a pair $(V,$ $M)$. It uniquely determines an NEF within the class of NEFs. For instance, the VF $({\sigma }^2,\mathbb{R})$ with fixed $\sigma >0$ characterizes the normal NEF within the class of NEFs.

Steep NEFs. An NEF $F=F\left(\mu \right)$ is called steep iff $M=int$C. Also, F is steep if its effective domain D is an open interval (see [3], Theorem 8.2), particularly, F is steep if $D=\mathbb{R},{\mathbb{R}}^{+},$ or ${\mathbb{R}}^{-}.$ Steep NEFs ensure, with probability one, the existence of the MLE for m as the stationary point of the first derivative of the likelihood function.

3. Sufficient Conditions on the Boundedness of VFs

Finding necessary and sufficient conditions for this problem is extremely challenging. Even identifying necessary conditions that closely approach sufficiency is highly non-trivial. Undoubtedly, achieving such results would constitute a significant theoretical advance, but we leave this task for future research.

Nevertheless, sufficient conditions can be established, and this paper provides them in Theorem 1 and Propositions 1 and 2. Theorem 1 imposes a polynomial-growth bound on the generating measure of the NEF, ensuring rapid tail decay and a uniformly bounded VF. Proposition 1 is based on Legendre duality and requires a uniform positive lower bound on the second derivative of the generating measure, which likewise ensures a bounded VF. Proposition 2 begins with the standard normal distribution and constructs an explicit sequence of NEFs whose LTs and VFs remain bounded.

Most examples stemming from Theorem 1 or Proposition 1 lack closed-form expressions for their LTs or VFs, but modern mathematical software can handle them efficiently in practice. By contrast, every example associated with Proposition 2 admits explicit LTs and VFs.

Collectively, these results uncover an exceptionally broad class of NEFs. To our knowledge, none of the NEFs produced by these three results have appeared previously in the literature, thereby providing a rich collection of practical models that rival the normal and other classical distributions for real-data analysis. In this paper, we do not address applications of these families to real data; this will be pursued in future research.

Theorem 1.

Let $g:\mathbb{R}\to [0,\infty )$ be a continuous function satisfying the following tail growth conditions:

$$g\left(x\right)\ge \left\{\begin{array}{cc}{C}_{+}{x}^p,\hfill & x\ge R,\hfill \\ C_{-}{\left|x\right|}^q,\hfill & x\le -R,\hfill \end{array}\right.,$$

for some constants $C_{+},C_{-,}R>0$ and $p,q>1.$ Define

$$f\left(x\right)=e^{-g\left(x\right)},x\in \mathbb{R},L\left(\theta \right)={\int }_{\mathbb{R}}{e}^{\theta x-g\left(x\right)}dx,andk\left(\theta \right)=\log L\left(\theta \right),$$

and let F be the NEF generated by $f.$ Then,

1. $f\in L^1\left(\mathbb{R}\right)$, implying that the NEF $F$ generated by f is supported on $\mathbb{R}.$

2. The LT $L\left(\theta \right)={\int }_{\mathbb{R}}{e}^{\theta x-g\left(x\right)}dx$ is finite for all $\theta \in \mathbb{R}$, i.e., the respective canonical parameter space is $\Theta =\mathbb{R}$. Thus, F is steep and $\Theta =M=\mathbb{R}$.

3. For $p,q\ge 2$, the VF $V\left(\theta \right)=k^{″}\left(\theta \right)$ of F is uniformly bounded by 1. (For $1<p<2,$ the VF is not uniformly bounded as it grows without limit as θ or m tends to ∞ or $-\infty$).

Proof. 

1. Let $\mathit{R}>\mathit{0}$ be sufficiently large. Split the integral $f\left(x\right)={\int }_{\mathbb{R}}{e}^{-g\left(x\right)}dx$ into three intervals of integration: $(-\infty ,-R),[-R,R],$ and $(R,\infty ).$ Since f is continuous on $\mathbb{R}$, then by the extreme value theorem, $f\left(x\right)={\int }_{-R}^R{e}^{-g\left(x\right)}dx$ is bounded. On $[R,\infty )$, $e^{-g\left(x\right)}\le e^{-C_{+}{x}^p}$, which integrates for $p>1$. The left tail is similar to $q>1$. Thus, $A={\int }_{\mathbb{R}}{e}^{-g\left(x\right)}dx<\infty$. Hence, $f\in L^1\left(\mathbb{R}\right).$

2. For this part, we use Young’s inequality ([4], p. 27), which states the following: Let $r,s>1$ with $\frac{1}{r}+\frac{1}{s}=1$. Then, for all real $a,b$,

$$\left|ab\right|\le {\displaystyle \frac{{\left|a\right|}^r}{r}}+{\displaystyle \frac{{\left|b\right|}^s}{s}}.$$

For this part, we split the implementation of Young’s inequality into the two parts of the tails:

(a) On the right tail ($x\ge R$, $\theta \ge 0$) we take $r=p$ and $s=p^{*}=p/(p-1)$. Set

$$a=C_{+}^{1/p}x,b=\theta C_{+}^{-1/p}.$$

Then $ab=\theta x$ and

$$\theta x\le {\displaystyle \frac{C_{+}{x}^p}{p}}+{\displaystyle \frac{{\theta }^{p^{*}}}{p^{*}{C}_{+}^{p^{*}/p}}}.$$

(b) On the left tail ($x\le -R$, $\theta \le 0$) we take $r=q$, $s=q^{*}=q/(q-1)$ and

$$a=C_{-}^{1/q}\left|x\right|,b=\left|\theta \right|C_{-}^{-1/q},$$

giving

$$\left|\theta x\right|\le {\displaystyle \frac{C_{-}{\left|x\right|}^q}{q}}+{\displaystyle \frac{{\left|\theta \right|}^{q^{*}}}{q^{*}{C}_{-}^{q^{*}/q}}}.$$

Fix $\theta \ge 0$ and use the bounds derived from Young’s inequality to obtain

$$\theta x-g\left(x\right)\le -\left(1-{\displaystyle \frac{1}{p}}\right)C_{+}{x}^p+{\displaystyle \frac{{\theta }^{p^{*}}}{p^{*}{C}_{+}^{p^{*}/p}}}(x\ge R).$$

Exponentiating and integrating yield

$${\int }_R^{\infty }{e}^{\theta x-g\left(x\right)}dx\le \exp \left\{\frac{{\theta }^{p^{*}}}{p^{*}{C}_{+}^{p^{*}/p}}\right\}{\int }_R^{\infty }\exp \left\{-\left(1-\frac{1}{p}\right)C_{+}{x}^p\right\}dx.$$

The integral converges because $p>1$. The left tail for $\theta \ge 0$ already decays exponentially; the compact part $[-R,R]$ is finite. For $\theta <0,$ a symmetric bound with q gives convergence. Hence $L\left(\theta \right)<\infty$ for all $\theta$. Thus, $\Theta =\mathbb{R},$F is regular and therefore steep with $M=\mathbb{R}$.

3. For $p\ge 2$, $q\ge 2,$ we shall find a quadratic bound on the cumulant transform $k.$ For this, we first show that

$$L\left(\theta \right)\le C{e}^{{\theta }^2/2},\mathrm{for}\mathrm{some}\mathrm{constant}C.$$

(3)

Let $R>0$ and split

$$L\left(\theta \right)={\int }_{-\infty }^{-R}{e}^{\theta x}f\left(x\right)dx+{\int }_{-R}^R{e}^{\theta x}f\left(x\right)dx+{\int }_R^{\infty }{e}^{\theta x}f\left(x\right)dx.$$

(4)

Consider the central interval $[-R,R]$ and use the elementary inequality $2ab\le a^2+b^2$ with $a=\left|\theta \right|$ and $b=R$, then

$$\left|\theta \right|R\le {\displaystyle \frac{{\left|\theta \right|}^2}{2}}+{\displaystyle \frac{R^2}{2}}\mathrm{or}{e}^{\left|\theta \right|R}\le e^{{\displaystyle \frac{R^2}{2}}}{e}^{{\displaystyle \frac{{\theta }^2}{2}}}.$$

Insert this into the central integral in (4), then

$${\int }_{-R}^R{e}^{\theta x}f\left(x\right)dx\le C_c{e}^{\left|\theta \right|R}\le C_c{e}^{{\displaystyle \frac{R^2}{2}}}{e}^{{\displaystyle \frac{{\theta }^2}{2}}}\doteq C_{cent}{e}^{{\displaystyle \frac{{\theta }^2}{2}}}.$$

(5)

For the right and left tails, there exist constants A and B such that

$$x\ge R\Rightarrow f\left(x\right)\le A{e}^{-x^p},x\le -R\Rightarrow f\left(x\right)\le B{e}^{-{\left|x\right|}^q}.$$

Similarly, for $a=\left|\theta \right|$ and $b=x,$ we get

$$\theta x\le {\displaystyle \frac{{\theta }^2}{2}}+{\displaystyle \frac{x^2}{2}},\mathrm{or}{e}^{\theta x}\le e^{{\theta }^2/2}{e}^{x^2/2}\mathrm{for}\mathrm{all}\theta ,x.$$

Thus, for the right tail, we have

$${\int }_R^{\infty }{e}^{\theta x}f\left(x\right)dx\le e^{{\theta }^2/2}A{\int }_R^{\infty }\exp \left(\frac{x^2}{2}-x^p\right)dx\le e^{{\theta }^2/2}A{C}_p,$$

where $C_p={\int }_R^{\infty }\exp \left(\frac{x^2}{2}-x^p\right)dx<\infty$. The left tail is similar with a constant $C_q$. Let

$$C\doteq C_{cent}+A{C}_p+B{C}_q$$

then (3) holds and thus $k\left(\theta \right)=\log L\left(\theta \right)\le C_0+{\theta }^2/2$, where $C_0=\log C.$

We now show that the VF is bounded. Consider the second finite difference

$${\Delta }_h^{2}k\left(\theta \right)=k(\theta +h)-2k\left(\theta \right)+k(\theta -h).$$

The strict convexity of k yields a lower bound: ${\Delta }_h^{2}k\left(\theta \right)\ge 0$. An upper bound of ${\Delta }_h^{2}k\left(\theta \right)$ gives

$$\begin{array}{cc}\hfill {\Delta }_h^{2}k\left(\theta \right)& \le \left[C_0+{\displaystyle \frac{{(\theta +h)}^2}{2}}\right]-2\left[C_0+{\displaystyle \frac{{\theta }^2}{2}}\right]+\left[C_0+{\displaystyle \frac{{(\theta -h)}^2}{2}}\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[{(\theta +h)}^2-2{\theta }^2+{(\theta -h)}^2\right]=h^2.\hfill \end{array}$$

Hence, $0<{\Delta }_h^{2}k\left(\theta \right)\le h^2.$ Because $k\left(\theta \right)$ is real analytic, $k^{″}\left(\theta \right)$ exists everywhere and

$$k^{″}\left(\theta \right)=\underset{h\to 0}{\lim }{\displaystyle \frac{{\Delta }_h^{2}k\left(\theta \right)}{h^2}}=\underset{h\to 0}{\lim }{\displaystyle \frac{h^2}{h^2}}=1.$$

This implies that $V\left(\theta \right)=k^{″}\left(\theta \right)\le 1.$ □

Remark 1.

By Theorem 1 and all of the propositions below, $f\in L^1\left(\mathbb{R}\right)$ is not necessarily a probability density function. To make it such, we need to divide it by $c_g$, where $c_g={\int }_{\mathbb{R}}f\left(x\right)dx$ is the normalizing constant, which may depend on the various parameters involved. In most of the following examples, these normalizing constants can be computed numerically or expressed in terms of certain transcendental functions.

In the following proposition, the tail condition required by Theorem 1 is satisfied. However, we prefer to present it separately as Proposition 1, since it assumes that $g^{″}\left(x\right)$ is bounded below by some constant $\delta >0$, which provides a stronger and more self-contained control over the variance function. The proof, based on Legendre duality, is both elegant and innovative, and we believe that its presentation is valuable in its own right. Moreover, the condition on $g^{″}$ is relatively easy to verify in concrete examples and facilitates the construction of NEFs that satisfy it. This property of $g^{″}$ will be illustrated in Examples 3–5 below.

The Legendre duality presents the cumulant transform $k=\log L$ and the convex generating function g as Legendre conjugates (see [5], Chapter 26; [3]). Concretely,

$$g\left(x\right)=\underset{\theta \in \mathbb{R}}{\sup }\{\theta x-k\left(\theta \right)\},k\left(\theta \right)=\underset{x\in \mathbb{R}}{\sup }\{\theta x-g\left(x\right)\},$$

and, for each $\theta \in \mathbb{R}$, the maximizer $x_{\theta }$ of the second expression satisfies $g^{\prime }\left(x_{\theta }\right)$ and $k\left(\theta \right)=\theta x_{\theta }-g\left(x_{\theta }\right).$

Proposition 1.

Let $g:\mathbb{R}\to [0,\infty )$ be a twice continuously differentiable and convex function satisfying the following condition: There exists a constant $\delta >0$ such that $g^{″}\left(x\right)\ge \delta$ for all $x\in \mathbb{R}$. Define $f\left(x\right)=e^{-g\left(x\right)}$ and F the NEF generated by f. Then,

1. $f\in L^1\left(\mathbb{R}\right)$, i.e., the support of F is $\mathbb{R}$.

2. The LT $L\left(\theta \right)$ of f is finite for all $\theta \in \mathbb{R}$, and thus F is regular and steep with $\Theta =M=\mathbb{R}$.

3.

$$E_{\theta }\left[X\right]=x_{\theta },Va{r}_{\theta }\left(X\right)={\displaystyle \frac{1}{g^{″}\left(x_{\theta }\right)}}\le {\displaystyle \frac{1}{\delta }}.$$

where $x_{\theta }$ is the unique minimizer of $h_{\theta }\left(x\right)=g\left(x\right)-\theta x.$

4. g satisfies Theorem 1.

Proof. 

1. Since $g^{″}\left(x\right)\ge \delta >0$, we can integrate this inequality:

$$g^{\prime }\left(x\right)=g^{\prime }\left(0\right)+{\int }_0^x{g}^{″}\left(t\right)dt\ge g^{\prime }\left(0\right)+\delta x.$$

Hence, $g^{\prime }\left(x\right)\to \infty$ as $x\to \infty$, and $g^{\prime }\left(x\right)\to -\infty$ as $x\to -\infty$.

Now integrate again:

$$g\left(x\right)=g\left(0\right)+{\int }_0^x{g}^{\prime }\left(t\right)dt\to \infty \mathrm{as}\left|x\right|\to \infty .$$

So for large $\left|x\right|$, $g\left(x\right)\ge {\displaystyle \frac{\delta }{2}}{x}^2-C$ for some constant C. Therefore,

$$f\left(x\right)=e^{-g\left(x\right)}\le C{e}^{-\delta x^2/2},$$

and this is integrable over $\mathbb{R}$. Thus, $f\in L^1\left(\mathbb{R}\right)$.

2. For showing $\Theta =\mathbb{R}$, we analyze $L\left(\theta \right)={\int }_{\mathbb{R}}{e}^{\theta x-g\left(x\right)}dx.$ From above, $g\left(x\right)\ge {\displaystyle \frac{\delta }{2}}{x}^2-C$, so,

$$e^{\theta x-g\left(x\right)}\le e^{\theta x-\delta x^2/2+C}=e^C·e^{\theta x-\delta x^2/2}.$$

This is integrable over $\mathbb{R}$ for all $\theta \in \mathbb{R}$ since the exponent is quadratic in x, and the quadratic term dominates. Therefore, $L\left(\theta \right)<\infty$ for all $\theta$, and the canonical parameter space $\Theta =\mathbb{R}=M$.

3. We prove this part by using the Legendre duality. Let $h_{\theta }\left(x\right):=g\left(x\right)-\theta x$. The unique minimizer $x_{\theta }$ satisfies $g^{\prime }\left(x_{\theta }\right)=\theta$, due to the strict convexity of g. Since g is convex and differentiable, the Legendre transform of g is

$$k\left(\theta \right)=\underset{x\in \mathbb{R}}{\sup }\left\{\theta x-g\left(x\right)\right\},\mathrm{attained}\mathrm{at}x=x_{\theta }\mathrm{where}{g}^{\prime }\left(x_{\theta }\right)=\theta .$$

So,

$$k\left(\theta \right)=\theta x_{\theta }-g\left(x_{\theta }\right),k^{\prime }\left(\theta \right)=x_{\theta }.$$

Hence,

$$E_{\theta }\left(X\right)=k^{\prime }\left(\theta \right)={\displaystyle \frac{d}{d\theta }}[\theta x_{\theta }-g\left(x_{\theta }\right)]=x_{\theta }-\theta {\displaystyle \frac{d{x}_{\theta }}{d\theta }}-g^{\prime }\left(x_{\theta }\right){\displaystyle \frac{d{x}_{\theta }}{d\theta }}=x_{\theta },\theta \in \mathbb{R}.$$

and

$$V_{\theta }\left(X\right)=k^{″}\left(\theta \right)={\displaystyle \frac{d{x}_{\theta }}{d\theta }}.$$

Differentiate both sides of $g^{\prime }\left(x_{\theta }\right)=\theta$ using the chain rule:

$$g^{″}\left(x_{\theta }\right)·{\displaystyle \frac{d{x}_{\theta }}{d\theta }}=1\Rightarrow {\displaystyle \frac{d{x}_{\theta }}{d\theta }}={\displaystyle \frac{1}{g^{″}\left(x_{\theta }\right)}}.$$

Thus,

$$Va{r}_{\theta }\left(X\right)={\displaystyle \frac{1}{g^{″}\left(x_{\theta }\right)}}\le {\displaystyle \frac{1}{\delta }},\sin \mathrm{ce}{g}^{″}\left(x_{\theta }\right)\ge \delta .$$

4. We show that there exist constants $C_1,C_2>0$ and $p_1,p_2>1$ such that

$$g\left(x\right)\ge C_1{x}^{p_1}\mathrm{for}\mathrm{large}x>0,g\left(x\right)\ge C_2{\left|x\right|}^{p_2}\mathrm{for}\mathrm{large}x<0.$$

For this, fix any point $x_0\in \mathbb{R}$. We can integrate this inequality because $g^{″}\left(x\right)\ge \delta >0$. By integrating from $x_0$ to $x>x_0$, we get

$$g^{\prime }\left(x\right)=g^{\prime }\left(x_0\right)+{\int }_{x_0}^x{g}^{″}\left(t\right)dt\ge g^{\prime }\left(x_0\right)+\delta (x-x_0).$$

Now, integrating $g^{\prime }\left(t\right)$ from $x_0$ to x, then

$$g\left(x\right)=g\left(x_0\right)+{\int }_{x_0}^x{g}^{\prime }\left(t\right)dt.$$

Using the lower bound for $g^{\prime }\left(t\right)$ implies

$$g\left(x\right)\ge g\left(x_0\right)+{\int }_{x_0}^x\left[g^{\prime }\left(x_0\right)+\delta (t-x_0)\right]dt\ge g\left(x_0\right)+g^{\prime }\left(x_0\right)(x-x_0)+{\displaystyle \frac{\delta }{2}}{(x-x_0)}^2.$$

Hence, as $x\to \infty$, this inequality shows

$$g\left(x\right)\ge {\displaystyle \frac{\delta }{2}}{x}^2-C,$$

for some constant $C>0$ (since $x_0$, $g\left(x_0\right)$, and $g^{\prime }\left(x_0\right)$ are fixed). So, for sufficiently large x,

$$g\left(x\right)\ge C_1{x}^2$$

for some $C_1>0$. A symmetric argument applies for $x\to -\infty$, using integration from x to $x_0$. Therefore, $g\left(x\right)$ satisfies the tail lower bounds required by Theorem 1, with $p_1=p_2=2$. □

The following proposition is based on the standard normal distribution and generates many examples that, in some cases, satisfy Theorem 1 while violating Proposition 1. The LTs and VFs for these examples can be expressed explicitly, and the corresponding NEFs may have broad practical applicability.

Proposition 2.

Let

$$f_n\left(x\right)={\displaystyle \frac{1}{\sqrt{2\pi }}}{x}^{2n}{e}^{-x^2/2},x\in \mathbb{R},n\in \mathbb{N},$$

(6)

and $F_n$ be the NEF generated by $f_n$. Denote by $L_n$ and $V_n$, respectively, the LT and VF corresponding to $F_n$. Then,

1. The support and the canonical parameter space of $F_n$ is $\mathbb{R}$.

2. The LT of $f_n$ has the form

$$L_n\left(\theta \right)={\int }_{\mathbb{R}}{x}^{2n}{e}^{\theta x-x^2/2}{\displaystyle \frac{dx}{\sqrt{2\pi }}}=e^{{\theta }^2/2}{P}_n\left(\theta \right),\theta \in \mathbb{R},$$

where

$$P_n\left(\theta \right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{2k}}\right)(2k-1)!!{\theta }^{2(n-k)}.$$

(7)

Here, $k!!$ is the double factorial, and $P_n$ is an even polynomial of degree $2n$ whose all coefficients are strictly positive.

3. Fix $n\in \mathbb{N},$ then the VF $V_n$ satisfies

$$V_n\left(\theta \right)<V_n\left(0\right)=2n+1,\forall \theta \ne 0,$$

(8)

4. $f_n\left(x\right),n\in \mathbb{N},$ does satisfy the premises of Theorem 1 but not of Proposition 1.

5. The LT and VF of $f_n$ in (6) are given, respectively, by

$$L_n\left(\theta \right)=e^{{\theta }^2/2}{P}_n\left(\theta \right),$$

and

$$V_n\left(\theta \right)={\displaystyle \frac{d^2}{d{\theta }^2}}\log L_n\left(\theta \right)=1+{\displaystyle \frac{P_n^{″}\left(\theta \right)}{P_n\left(\theta \right)}}-{\left({\displaystyle \frac{P_n^{\prime }\left(\theta \right)}{P_n\left(\theta \right)}}\right)}^2.$$

where $P_n$ is given by (7). The special cases $n=1,2,3$ are given, respectively, by (a) $n=1,$

$$P_1\left(\theta \right)={\theta }^2+1,L_1\left(\theta \right)=e^{{\theta }^2/2}({\theta }^2+1),$$

$$V_1\left(\theta \right)={\displaystyle \frac{{\theta }^4+6{\theta }^2+3}{{({\theta }^2+1)}^2}},V_1\left(0\right)=3.$$

(9)

(b) $n=2,$

$$P_2\left(\theta \right)={\theta }^4+6{\theta }^2+3,L_2\left(\theta \right)=e^{{\theta }^2/2}({\theta }^4+6{\theta }^2+3),$$

$$V_2\left(\theta \right)={\displaystyle \frac{{\theta }^8+8{\theta }^6+30{\theta }^4+45}{{\theta }^8+12{\theta }^6+42{\theta }^4+36{\theta }^2+9}},V_2\left(0\right)=5.$$

(10)

(c) $n=3$

$$P_3\left(\theta \right)={\theta }^6+15{\theta }^4+45{\theta }^2+15,L_3\left(\theta \right)=e^{{\theta }^2/2}({\theta }^6+15{\theta }^4+45{\theta }^2+15),$$

$$V_3\left(\theta \right)={\displaystyle \frac{{\theta }^{12}+24{\theta }^{10}+225{\theta }^8+840{\theta }^6+1575{\theta }^4+1575}{{\theta }^{12}+30{\theta }^{10}+315{\theta }^8+1380{\theta }^6+2475{\theta }^4+1350{\theta }^2+225}},V_3\left(0\right)=7.$$

(11)

Proof. 

1. This part is trivial

2. Clearly,

$$L_n\left(\theta \right)=E\left[X^{2n}{e}^{\theta X}\right]=e^{{\theta }^2/2}{P}_n\left(\theta \right),$$

where

$$P_n\left(\theta \right)=E\left[{(X+\theta )}^{2n}\right].$$

To get the explicit form of $P_n\left(\theta \right),$ write

$${(X+\theta )}^{2n}=\sum _{k=0}^{2n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)X^k{\theta }^{2n-k},X\sim N(0,1).$$

Taking expectations

$$P_n\left(\theta \right)=\sum _{k=0}^{2n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)E\left[X^k\right]{\theta }^{2n-k}.$$

Only even powers contribute due to the symmetry of $X\sim N(0,1)$, and thus,

$$E\left[X^{2k}\right]=(2k-1)!!,$$

so that

$$P_n\left(\theta \right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{2k}}\right)(2k-1)!!{\theta }^{2(n-k)},$$

is an even polynomial of degree $2n$ with strictly positive coefficients. Note, however, that $P_n=2^{-n}{H}_{2n}$, where $H_m$ denotes the Hermite polynomial.

3. To prove (8), note that

$$V_n\left(\theta \right)={\displaystyle \frac{L_n^{″}\left(\theta \right)}{L_n\left(\theta \right)}}-{\left({\displaystyle \frac{L_n^{\prime }\left(\theta \right)}{L_n\left(\theta \right)}}\right)}^2,L_n\left(\theta \right)=E\left[X^{2n}{e}^{\theta X}\right],X\sim \mathcal{N}(0,1).$$

The following steps carry out the proof of this part: (a) computing $V_n$ at 0; (b) defining the tilted measure ${\mu }_{\theta }$; (c) showing that $V_n\left(\theta \right)<R_n\left(\theta \right)$ for $\theta \ne 0$; (d) showing the symmetry of $R_n\left(\theta \right),$ where $R_n$ is defined in (12); (e) justifying the log-convexity of $M_k\left(\theta \right)$ in k; (f) showing the monotonicity of $R_n\left(\theta \right).$

(a) Evaluation at $\theta =0$: Compute

$$L_n\left(0\right)=\mathbb{E}\left[X^{2n}\right]=(2n-1)!!,L_n^{\prime }\left(0\right)=\mathbb{E}\left[X^{2n+1}\right]=0,L_n^{″}\left(0\right)=\mathbb{E}\left[X^{2n+2}\right]=(2n+1)(2n-1)!!.$$

So,

$$V_n\left(0\right)={\displaystyle \frac{L_n^{″}\left(0\right)}{L_n\left(0\right)}}=2n+1.$$

(b) Let

$${\mu }_{\theta }\left(dx\right)\doteq {\displaystyle \frac{x^{2n}{e}^{\theta x-x^2/2}}{L_n\left(\theta \right)\sqrt{2\pi }}}dx.$$

Then,

$$E_{\theta }\left[X^k\right]={\displaystyle \frac{M_{2n+k}\left(\theta \right)}{M_{2n}\left(\theta \right)}},$$

and

$$V_n\left(\theta \right)=Va{r}_{{\mu }_{\theta }}\left(X\right)={\displaystyle \frac{M_{2n+2}\left(\theta \right)}{M_{2n}\left(\theta \right)}}-{\left({\displaystyle \frac{M_{2n+1}\left(\theta \right)}{M_{2n}\left(\theta \right)}}\right)}^2.$$

(c) Define

$$R_n\left(\theta \right)\doteq {\displaystyle \frac{M_{2n+2}\left(\theta \right)}{M_{2n}\left(\theta \right)}},{\mu }_n\left(\theta \right)\doteq {\displaystyle \frac{M_{2n+1}\left(\theta \right)}{M_{2n}\left(\theta \right)}},$$

(12)

thus,

$$V_n\left(\theta \right)=R_n\left(\theta \right)-{\mu }_n{\left(\theta \right)}^2<R_n\left(\theta \right)\mathrm{for}\mathrm{all}\theta \ne 0.$$

(d) Since

$$M_k(-\theta )={(-1)}^k{M}_k\left(\theta \right),$$

it follows that

$$R_n\left(\theta \right)=R_n(-\theta ).$$

So, it suffices to prove that $R_n^{\prime }\left(\theta \right)<0$ for $\theta >0$.

(e) Let

$$M_k\left(\theta \right)=\mathbb{E}\left[X^k{e}^{\theta X}\right]=e^{{\theta }^2/2}E\left[{(X+\theta )}^k\right],$$

and define $Z=X+\theta$, then

$$M_k\left(\theta \right)=e^{{\theta }^2/2}E\left(Z^k\right),Z\sim N(0,1).$$

We now analyze the sequence $\left\{M_k\left(\theta \right)\right\}$ to show that, for each fixed $\theta ,$

$${\displaystyle \frac{M_{k+1}\left(\theta \right)}{M_k\left(\theta \right)}}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}k.$$

(13)

To this end, we establish the following: strict log-convexity of the sequence $\left\{M_k\left(\theta \right)\right\}$; convexity of $\log M_k\left(\theta \right)$ in k; monotonicity of successive ratios, followed by the application of (13). Recall that a sequence ${\left\{a_k\right\}}_{k\ge 0}$ of positive numbers is log-convex if

$$a_k^2<a_{k-1}{a}_{k+1}(k\ge 1).$$

(14)

For the normal moments, this strict inequality follows from Turán’s inequality for Hermite polynomials (c.f., [6,7,8]); hence, the sequence ${\left\{M_k\left(\theta \right)\right\}}_{k\ge 0}$ is strictly log-convex. Setting $a_k=M_k$ and taking logarithms in (14) yields

$$\log M_{k-1}-2\log M_k+\log M_{k+1}>0.$$

That is, the discrete second difference of the function $k\mapsto \log M_k\left(\theta \right)$ is positive. In discrete calculus, this is precisely the definition of strict convexity. Moreover, for any positive sequence, strict log-convexity is equivalent to the property that

$${\displaystyle \frac{a_{k+1}}{a_k}}\mathrm{increases}\mathrm{strictly}\mathrm{with}k.$$

Applying this to $M_k\left(\theta \right)$ yields (13), for every fixed $\theta$.

(f) Differentiate $R_n\left(\theta \right)$ to obtain

$$R_n^{\prime }\left(\theta \right)={\displaystyle \frac{M_{2n+3}{M}_{2n}-M_{2n+2}{M}_{2n+1}}{M_{2n}^2}}.$$

Hence, from the final inequality in (e), it follows that

$${\displaystyle \frac{M_{2n+3}}{M_{2n+2}}}<{\displaystyle \frac{M_{2n+2}}{M_{2n+1}}}\Rightarrow M_{2n+3}{M}_{2n+1}<M_{2n+2}^2,\Rightarrow M_{2n+3}{M}_{2n}<M_{2n+2}{M}_{2n+1},\Rightarrow R_n^{\prime }\left(\theta \right)<0.$$

All the preceding steps demonstrate that

$$V_n\left(\theta \right)<R_n\left(\theta \right)<R_n\left(0\right)=2n+1,$$

and

$$V_n\left(\theta \right)<V_n\left(0\right)=2n+1\mathrm{for}\mathrm{all}\theta \ne 0.$$

4.

$$g_n\left(x\right)={\displaystyle \frac{x^2}{2}}-2n\log \left|x\right|+\mathrm{const}.$$

For $\left|x\right|\ge 1,$ we use the inequality $\log \left|x\right|\le \left|x\right|/2,$ and then note that $\frac{x^2}{2}-n\left|x\right|\ge \frac{x^2}{4}$ once $\left|x\right|\ge 2n$. Hence,

$$g_n\left(x\right)\ge {\displaystyle \frac{x^2}{2}}-2n\log \left|x\right|\ge {\displaystyle \frac{x^2}{2}}-n\left|x\right|\ge {\displaystyle \frac{x^2}{4}}\left(\left|x\right|\ge 2n\right),$$

Thus, the quadratic tail condition of Theorem 1 (with $p=q=2$) is indeed satisfied. However, $g_n$ violates Proposition 1 as

$$g_n^{\prime }\left(x\right)=x-{\displaystyle \frac{2n}{x}},x\ne 0,$$

$$g_n^{″}\left(x\right)=1+{\displaystyle \frac{2n}{x^2}},x\ne 0.$$

Since $g_n$ is not defined everywhere, we cannot even begin to verify a uniform bound for all x.

5. A straightforward calculation yields these results. □

To illustrate the functional relationship between the variance and the canonical parameter $\theta$, as well as the upper bound of the variance function, Figure 1 displays $V_n$, $=1,2,$ and 3, corresponding to Equations (9), (10), and (11), respectively.

4. Examples

None of the following examples has appeared in the literature in NEF form, likely because their Laplace transforms lack closed-form expressions (except for those related to Proposition 2). As noted in the Introduction, modern mathematical software can easily handle such cases numerically.

Collectively, these examples provide a rich class of NEFs with unbounded support and bounded variance functions, offering valuable alternatives to the normal NEF and other standard distributions in statistical modeling. We present only a small selection from a much larger set, omitting additional cases for brevity.

Example 1.

The Generalized normal distributions (GNDs). Let $g\left(x\right)={\left|x\right|}^p$ and

$$f\left(x\right)=e^{-{\left|x\right|}^p},p>1,x\in \mathbb{R},$$

(15)

where the appropriate normalizing constant making f a density is $c_g\doteq c_p=$$2p/(\Gamma (1/p\left)\right)$.

The class of generalized normal distributions (GNDs) has a long history, beginning with Russian researchers (see [9], and the references therein). A GND with $1<p<2$ provides a robust, flexible alternative to the normal distribution, making it suitable for data that exhibit sharp peaks and moderate outliers. It is especially useful in signal processing, robust regression, and sparsity-based inference. The density given in (15) is symmetric and unimodal, with heavier tails and a sharper peak than the normal distribution; these features make it leptokurtic. and well suited to settings where moderate outliers or pronounced central tendencies are expected. GNDs with $p\in (1,2)$ appear as error distributions in robust estimation ([10]), Bayesian regression ([11]), signal and audio processing ([12]), and independent component analysis and sparse coding ([13]). These examples satisfy Theorem 1 but not Proposition 1—except when $p=2,$ which corresponds to the normal model. Specifically, Theorem 1 requires $g\left(x\right)\ge C_1{x}^{p_1}$ for large $x>0$, and $g\left(x\right)\ge C_2{\left|x\right|}^{p_2}$ for large $x<0$, with constants $C_1,C_2>0$ and exponents $p_1,p_2>1$. This condition yields a bounded variance for the associated NEF. For this case,

$$g_p\left(x\right)=\left\{\begin{array}{cc}{x}^p\hfill & \mathrm{if}x\ge 0,\hfill \\ {(-x)}^p={\left|x\right|}^p\hfill & \mathrm{if}x<0,\hfill \end{array}\right.$$

which meets the first two conditions of Theorem 1 with $C_1=C_2=1$ and $p_1=p_2=p>1$. For $1<p<2$, Part 3 of Theorem 1 only hints that the VF is not uniformly bounded; indeed, it diverges as the canonical parameter $\theta ,$ or m, tends to ∞ or $-\infty$.

Uniform boundedness of V requires $p\ge 2,$ so Proposition 1 applies only to the Gaussian case $p=2.$ Specifically,

$$g^{″}\left(x\right)=\left[p(p-1)\right]{\left|x\right|}^{(p-2)},x\ne 0.$$

For $1<p<2$, this second derivative vanishes at the origin, whereas for $p>2$, it diverges there. Consequently, Proposition 1 fails whenever $p\ne 2.$ In particular,

$$g^{″}\left(0\right)=\left\{\begin{array}{cc}2,\hfill & p=2,\hfill \\ 0,\hfill & p>2,\hfill \\ \infty ,\hfill & 1<p<2.\hfill \end{array}\right.$$

(Details are omitted for brevity.)

Example 2.

The extended generalized normal distributions (EGNDs). Let

$$g\left(x\right)={\displaystyle \frac{{\left|x\right|}^p}{a}}+{\displaystyle \frac{{\left|x\right|}^q}{b}},p\ge 2,q\ge 2,a>0,b>0.$$

(16)

We refer to the NEF $F,$ generated by $f=e^{-g\left(x\right)},$ as the EGND NEF, as we intend to explore it further and apply it to real data sets. As in Example 1, all the assumptions of Theorem 1 are satisfied. Proposition 1 holds only when either p or q equals 2. For example, if $p=2$, then $g_{a,b,2}\left(x\right)={\displaystyle \frac{a+b}{ab}}{x}^2$ with $a^{*}=ab/(a+b)$, in which case,

$$L_{a,b,2}\left(\theta \right)=\sqrt{{\displaystyle \frac{\pi a^{*}}{2}}}{e}^{{\theta }^2{a}^{*}/4},V_{a,b,2}\left(\theta \right)={\displaystyle \frac{a^{*}}{2}}.$$

Example 3.

Let

$$g\left(x\right)=(1+\delta )x^2+{\sin }^2\left(x\right),\delta >0,f\left(x\right)=e^{-g\left(x\right)}.$$

This g satisfies both Theorem 1 and Proposition 1. Theorem 1 requires $g\left(x\right)\ge C_1{x}^{p_1}$ as $x\to \infty$ and $g\left(x\right)\ge C_2{\left|x\right|}^{p_2}$ as $x\to -\infty$, for some $p_1,p_2>1$ and $C_1,C_2>0$.

Here,

$$g\left(x\right)=(1+\delta )x^2+{\sin }^2\left(x\right),$$

and ${\sin }^2\left(x\right)\in [0,1]$ is bounded for all x.

As $\left|x\right|\to \infty$, we have $g\left(x\right)\sim (1+\delta )x^2,$ so for sufficiently large $\left|x\right|$,

$$g\left(x\right)\ge (1+\delta )x^2\ge C{x}^2,$$

with $C=1+\delta >0$. Hence, both Theorem 1 and Proposition 1 are satisfied, with $g^{″}\left(x\right)\ge 2\delta .$ The choice $\delta >0$ is is essential, as the function loses strict convexity when $\delta =0$.

Example 4.

Let

$$g\left(x\right)=e^x+e^{-x}.$$

First, we show that $f\left(x\right)=e^{-g\left(x\right)}=e^{-e^x-e^{-x}}$ is integrable over $\mathbb{R}$; that is,

$${\int }_{-\infty }^{\infty }f\left(x\right)dx<\infty .$$

For $A>0$, consider the left tail:

$${\int }_{-\infty }^{-A}{e}^{-e^{-x}}dx={\int }_{e^A}^{\infty }{\displaystyle \frac{e^{-u}}{u}}du<\infty ,$$

where the change of variables $u=e^{-x}$ is used. A similar change of variables handles the right tail. Hence, f is integrable.

Next, $g^{″}\left(x\right)=e^x+e^{-x}>0,\forall x\in \mathbb{R}$ with a single minimum at $x=0,$ so $g^{″}\left(x\right)\ge \delta =2,$$\forall x\in \mathbb{R}.$ Consequently, both Theorem 1 and Proposition 1 hold.

Example 5.

Let

$$g\left(x\right)=\log (1+e^x)+{\displaystyle \frac{x^2}{2}}.$$

Differentiating twice gives

$$g^{″}\left(x\right)={\displaystyle \frac{d^2}{d{x}^2}}\left(\log (1+e^x)+{\displaystyle \frac{x^2}{2}}\right)={\displaystyle \frac{e^x}{{(1+e^x)}^2}}+1=\sigma \left(x\right)(1-\sigma \left(x\right))+1,$$

where $\sigma \left(x\right)=e^x/(1+e^x)$. Since $\sigma \left(x\right)(1-\sigma (x\left)\right)\in (0,1/4)$, it follows that $g^{″}\left(x\right)\in (1,{\displaystyle \frac{5}{4}}).$ Thus, Proposition 1 is satisfied with $\delta =1.$ Theorem 1 also holds: for large positive $x,$$g\left(x\right)\ge C_1{x}^2$ with $C_1={\displaystyle \frac{1}{2}};$ and for large negative x, $g\left(x\right)\ge {\displaystyle \frac{1}{2}}{x}^2=C_2{\left|x\right|}^2$ with $C_2={\displaystyle \frac{1}{2}}.$ Hence, the required growth conditions are met, and since $g^{″}\left(x\right)\ge \delta =1$ everywhere, Proposition 1 is confirmed.

5. Concluding Remarks

NEFs supported on the entire real line and possessing bounded VFs are uncommon; historically, the normal family has been the primary example. This paper broadens that landscape by presenting three complementary sets of sufficient conditions that ensure unbounded support and bounded VFs.

• Tail-growth control (Theorem 1). If the cumulant generator g dominates two even polynomials of degree $p,q>2$, then $f=e^{-g}$ is integrable, its Laplace transform is finite on $\mathbb{R}$, and the corresponding VF is bounded by 1.
• Uniform strong convexity (Proposition 1). Whenever $g^{″}\left(x\right)\ge \epsilon >0$ for all $x\in \mathbb{R}$, Legendre duality yields $V\left(m\right)\le 1/\epsilon$ for every mean m.
• A closed-form Gaussian sequence (Proposition 2). Weighting the standard normal density by even powers of x produces a family of NEFs whose VFs satisfy $V_n\le 2n+1$ and whose Laplace transforms remain elementary.

A set of twelve fully worked examples demonstrates how each route succeeds—or fails—in practice, highlighting the logical independence of the two analytic criteria. Although most examples lack closed-form Laplace transforms, modern symbolic and numerical software make them readily applicable. These results can generate almost infinitely many NEFs of this type, satisfying at least one of the three statements in Theorem 1 or Propositions 1 and 2.

For an empirical study, it is often advantageous to embed a given g in a richer parametric family. Simple rescaling

$$g\mapsto g/a(a>0)$$

(17)

or decompositions

$$g=g_1+g_2\mapsto g_1/a+g_2/b(a,b>0)$$

(18)

preserve the hypotheses of Theorem 1 and Propositions 1 and 2, while endowing the resulting NEFs with additional shape control (see Example 2 for an illustration). These extra degrees of freedom, in conjunction with the canonical parameter, make the families more adaptable to real data and will be further explored in forthcoming applied studies.

The purpose of presenting so many examples is to underscore the virtually endless supply of exponential families supported on the real line that can be constructed—none of which were previously known. Moreover, each of these families can be extended to a multi-parameter family, as shown in Equations (17) and (18).

Looking ahead, two applied research projects are currently underway with several collaborators, aiming to apply the theoretical results of this paper to modeling real data in the various fields indicated below.

1. Model competition. The proposed NEFs, all defined on $\mathbb{R}$ with bounded VFs, offer natural competitors to the normal, Laplace, logistic, hyperbolic, and secant distributions—as well as other classical distributions—in contexts where tail behavior or regularization is essential.

2. Statistical methodology. Investigating maximum likelihood and Bayesian inference for these families—particularly those lacking closed-form transforms—promises new tools for robust regression, signal processing, and econometrics.

3. Focus. In particular, we will focus on the NEFs generated by the generalized normal distributions (Example 1) and the extended generalized normal distributions (Example 2). We will explore their probabilistic properties and apply them to real data arising from signal and audio processing, independent component analysis, and sparse coding.