How to Find the Expected Value of an Exponential Function

The phrase “expected value of an exponential function” most often means finding $\mathbb{E}[e^X]$ for a random variable $X$. The answer depends on the distribution of $X$, and the expectation may be infinite.

The general method is to average $e^x$ over the probability distribution of $X$. This article works out the common Beta-distribution case and explains how the answer differs when $X$ itself has an exponential distribution.

The general expectation formula

For a discrete random variable with possible values $x_i$ and probabilities $p_i$,

For a continuous random variable with density $f_X(x)$,

provided the integral is finite. If $X$ is supported only on an interval, integrate over that interval instead. More generally, the moment-generating function (MGF) is $M_X(t)=\mathbb{E}[e^{tX}]$, so the quantity in question is $M_X(1)$ when that value exists.

If X has a Beta distribution

Suppose $X\sim\operatorname{Beta}(a,b)$, with $a,b>0$. Its density on $0<x<1$ is

where $B(a,b)$ is the beta function. Direct substitution gives

Here ${}_1F_1$ is the confluent hypergeometric function. Therefore,

This result can also be seen from the power series for the exponential. The $k$th raw moment of a Beta random variable is $\mathbb{E}[X^k]=(a)_k/(a+b)_k$, where $(q)_k$ is the rising factorial. Thus

Because $0\leq X\leq1$, this expectation is finite for every real $t$.

Worked example: X ~ Beta(2,3)

For $X\sim\operatorname{Beta}(2,3)$, the density is $12x(1-x)^2$ on $0<x<1$. The expectation is

The result is plausible: $X$ lies between 0 and 1, so $e^X$ lies between 1 and $e$. In fact, $\mathbb{E}[X]=2/5$, and the convexity of $e^x$ implies $\mathbb{E}[e^X]\geq e^{\mathbb{E}[X]}\approx1.4918$, consistent with the computed value.

If X is exponentially distributed

A different interpretation is that $X$ itself has an exponential distribution. Use the rate convention $X\sim\operatorname{Exp}(\lambda)$, whose density is $\lambda e^{-\lambda x}$ for $x\geq0$. Then

This integral converges only if $\lambda>1$. In that case,

For $\lambda\leq1$, the expectation is infinite. More generally, $\mathbb{E}[e^{tX}]=\lambda/(\lambda-t)$ for $t<\lambda$. If the exponential distribution is parameterized by its mean (scale) $\theta=1/\lambda$, the condition for $\mathbb{E}[e^X]$ to be finite is $\theta<1$, and the value is $1/(1-\theta)$.

Common mistakes

• Confusing $e^X$ with an exponential random variable. $e^X$ is a function of $X$; an exponential random variable instead has an exponential probability density.
• Using only the mean of $X$. In general, $\mathbb{E}[e^X]\ne e^{\mathbb{E}[X]}$. For example, the Beta example gives about $1.5225$, while $e^{\mathbb{E}[X]}$ is about $1.4918$.
• Ignoring whether the expectation is finite. For an exponential random variable, the integral diverges when the rate is at most 1.
• Mixing up rate and scale. The rate is $\lambda$ in $\lambda e^{-\lambda x}$; the scale is $1/\lambda$. Check which convention is being used before applying a formula.