Can a distribution with finite mean and infinite variance have a moment generating function? What about a distribution with finite mean and finite variance but infinite higher moments?

  • 4
    $\begingroup$ Hint: If the mgf exists in an interval around zero, say $(-t_0,t_0)$ for some $t_0 > 0$, then consider the Taylor expansion of $e^x$ and monotonicity of the integral to discover the solution. :) $\endgroup$
    – cardinal
    Jul 20, 2012 at 17:06
  • 2
    $\begingroup$ Ignoring issues of convergence (thinking about the mgf as a formal power series only), what could the mgf possibly be if any moment failed to exist? $\endgroup$
    – whuber
    Jul 24, 2012 at 21:54
  • $\begingroup$ Cardinal can you please give us some references about the propositions you provided? $\endgroup$
    – user67516
    Jan 27, 2015 at 0:34

1 Answer 1


This question provides a nice opportunity to collect some facts on moment-generating functions (mgf).

In the answer below, we do the following:

  1. Show that if the mgf is finite for at least one (strictly) positive value and one negative value, then all positive moments of $X$ are finite (including nonintegral moments).
  2. Prove that the condition in the first item above is equivalent to the distribution of $X$ having exponentially bounded tails. In other words, the tails of $X$ fall off at least as fast as those of an exponential random variable $Z$ (up to a constant).
  3. Provide a quick note on the characterization of the distribution by its mgf provided it satisfies the condition in item 1.
  4. Explore some examples and counterexamples to aid our intuition and, particularly, to show that we should not read undue importance into the lack of finiteness of the mgf.

This answer is quite long, for which I apologize in advance. If this would be better placed, e.g., as a blog post or somewhere else, please feel free to provide such feedback in the comments.

What does the mgf say about the moments?

The mgf of a random variable $X \sim F$ is defined as $m(t) = \mathbb E e^{tX}$. Note that $m(t)$ always exists since it is the integral of a nonnegative measurable function. However, if may not be finite. If it is finite (in the right places), then for all $p > 0$ (not necessarily an integer), the absolute moments $\mathbb E |X|^p < \infty$ (and, so, also $\mathbb E X^p$ is finite). This is the topic of the next proposition.

Proposition: If there exists $\newcommand{\tn}{t_{n}}\newcommand{\tp}{t_{p}}\tn < 0$ and $\tp > 0$ such that $m(\tn) < \infty$ and $m(\tp) < \infty$, then the moments of all orders of $X$ exist and are finite.

Before diving into a proof, here are two useful lemmas.

Lemma 1: Suppose such $\tn$ and $\tp$ exist. Then for any $t_0 \in [\tn,\tp]$, $m(t_0) < \infty$.
Proof. This follows from convexity of $e^x$ and monotonicity of the integral. For any such $t_0$, there exists $\theta \in [0,1]$ such that $t_0 = \theta \tn + (1-\theta) \tp$. But, then $$e^{t_0 X} = e^{\theta \tn X + (1-\theta) \tp X} \leq \theta e^{\tn X} + (1-\theta) e^{\tp X} \>.$$ Hence, by monotonicity of the integral, $\mathbb E e^{t_0 X} \leq \theta \mathbb E e^{\tn X} + (1-\theta) \mathbb E e^{\tp X} < \infty$.

So, if the mgf is finite at any two distinct points, it is finite for all values in the interval in between those points.

Lemma 2 (Nesting of $L_p$ spaces): For $0 \leq q \leq p$, if $\mathbb E |X|^p < \infty$, then $\mathbb E |X|^q < \infty$.
Proof: Two approaches are given in this answer and associated comments.

This gives us enough to continue with the proof of the proposition.

Proof of the proposition. If $\tn < 0$ and $\tp > 0$ exist as stated in the proposition, then taking $t_0 = \min(-\tn,\tp) > 0$, we know by the first lemma that $m(-t_0) < \infty$ and $m(t_0) < \infty$. But, $$ e^{-t_0 X} + e^{t_0 X} = 2 \sum_{n=0}^\infty \frac{t_0^{2n} X^{2n}}{(2n)!} \>, $$ and the right-hand side is composed of nonnegative terms, so, in particular, for any fixed $k$ $$ e^{-t_0 X} + e^{t_0 X} \geq 2 t_0^{2k} X^{2k}/(2k)! \>. $$ Now, by assumption $\mathbb E e^{-t_0 X} + \mathbb E e^{t_0 X} < \infty$. Monotonicity of the integral yields $\mathbb E X^{2k} < \infty$. Hence, all even moments of $X$ are finite. Lemma 2 immediately allows us to "fill in the gaps" and conclude that all moments must be finite.


The upshot regarding the question at hand is that if any of the moments of $X$ are infinite or do not exist, we can immediately conclude that the mgf is not finite in an open interval containing the origin. (This is just the contrapositive statement of the proposition.)

Thus, the proposition above provides the "right" condition in order to say something about the moments of $X$ based on its mgf.

Exponentially bounded tails and the mgf

Proposition: The mgf $m(t)$ is finite in an open interval $(\tn,\tp)$ containing the origin if and only if the tails of $F$ are exponentially bounded, i.e., $\mathbb P( |X| > x) \leq C e^{-t_0 x}$ for some $C > 0$ and $t_0 > 0$.

Proof. We'll deal with the right tail separately. The left tail is handled completely analogously.

$(\Rightarrow)$ Suppose $m(t_0) < \infty$ for some $t_0 > 0$. Then, the right tail of $F$ is exponentially bounded; in other words, there exists $C > 0$ and $b > 0$ such that $$ \mathbb P(X > x) \leq C e^{-b x} \>. $$ To see this, note that for any $t > 0$, by Markov's inequality, $$ \mathbb P(X > x) = \mathbb P(e^{tX} > e^{tx}) \leq e^{-tx} \mathbb E e^{t X} = m(t) e^{-t x} \>. $$ Take $C = m(t_0)$ and $b = t_0$ to complete this direction of the proof.

$(\Leftarrow)$ Suppose there exists $C >0$ and $t_0 > 0$ such that $\mathbb P(X > x) \leq C e^{-t_0 x}$. Then, for $t > 0$, $$ \mathbb E e^{t X} = \int_0^\infty \mathbb P( e^{t X} > y)\,\mathrm dy \leq 1 + \int_1^\infty \mathbb P( e^{t X} > y)\,\mathrm dy \leq 1 + \int_1^\infty C y^{-t_0/t} \, \mathrm dy \>, $$ where the first equality follows from a standard fact about the expectation of nonnegative random variables. Choose any $t$ such that $0 < t < t_0$; then, the integral on the right-hand side is finite.

This completes the proof.

A note on uniqueness of a distribution given its mgf

If the mgf is finite in an open interval containing zero, then the associated distribution is characterized by its moments, i.e., it is the only distribution with the moments $\mu_n = \mathbb E X^n$. A standard proof is short once one has at hand some (relatively straightforward) facts about characteristic functions. Details can be found in most modern probability texts (e.g., Billingsley or Durrett). A couple related matters are discussed in this answer.

Examples and counterexamples

(a) Lognormal distribution: $X$ is lognormal if $X = e^Y$ for some normal random variable $Y$. So $X \geq 0$ with probability one. Because $e^{-x} \leq 1$ for all $x \geq 0$, this immediately tells us that $m(t) = \mathbb E e^{t X} \leq 1$ for all $t < 0$. So, the mgf is finite on the nonnegative half-line $(-\infty,0]$. (NB We've only used the nonnegativity of $X$ to establish this fact, so this is true from all nonnegative random variables.)

However, $m(t) = \infty$ for all $t > 0$. We'll take the standard lognormal as the canonical case. If $x > 0$, then $e^{x} \geq 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3$. By change of variables, we have $$ \mathbb E e^{t X} = (2\pi)^{-1/2} \int_{-\infty}^\infty e^{t e^u - u^2/2} \,\mathrm d u \>. $$ For $t > 0$ and large enough $u$, we have $t e^u - u^2/2 \geq t+tu$ by the bounds given above. But, $$ \int_{K}^\infty e^{t + tu} \,\mathrm du = \infty $$ for any $K$, and so the mgf is infinite for all $t > 0$.

On the other hand, all moments of the lognormal distribution are finite. So, the existence of the mgf in an interval about zero is not necessary for the conclusion of the above proposition.

(b) Symmetrized lognormal: We can get an even more extreme case by "symmetrizing" the lognormal distribution. Consider the density $f(x)$ for $x \in \mathbb R$ such that $$ f(x) = \frac{1}{2\sqrt{2\pi}|x|} e^{-\frac{1}{2} (\log |x|)^2} \>. $$ It is not hard to see in light of the previous example that the mgf is finite only for $t = 0$. Yet, the even moments are exactly the same as those of the lognormal and the odd moments are all zero! So, the mgf exists nowhere (except at the origin where it always exists) and yet we can guarantee finite moments of all orders.

(c) Cauchy distribution: This distribution also has an mgf which is infinite for all $t \neq 0$, but no absolute moments $\mathbb E|X|^p$ are finite for $p \geq 1$. The result for the mgf follows for $t > 0$ since $e^x \geq x^3 / 6$ for $x > 0$ and so $$ \mathbb E e^{tX} \geq \int_1^\infty \frac{t^3 x^3}{6\pi(1+x^2)} \,\mathrm dx \geq \frac{t^3}{12\pi} \int_1^\infty x \,\mathrm dx = \infty \>. $$ The proof for $t < 0$ is analogous. (Perhaps somewhat less well known is that the moments for $0 < p < 1$ do exist for the Cauchy. See this answer.)

(d) Half-Cauchy distribution: If $X$ is (standard) Cauchy, call $Y = |X|$ a half-Cauchy random variable. Then, it is easy to see from the previous example that $\mathbb E Y^p = \infty$ for all $p \geq 1$; yet, $\mathbb E e^{tY}$ is finite for $t \in (-\infty,0]$.

  • 9
    $\begingroup$ Thanks for posting this - this is surprisingly easy to understand, considering how technical it is - well done. $\endgroup$
    – Macro
    Jul 24, 2012 at 23:05
  • $\begingroup$ Do you know any results about the mgf in Hilbert space? $\endgroup$
    – trying
    Nov 4, 2019 at 10:51

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