What Is the Moment-Generating Function of an Exponential Distribution?

Let $X$ have an exponential distribution with rate $\lambda>0$, so its density is $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$. Its moment-generating function (MGF) is

$$M_X(t)=\frac{\lambda}{\lambda-t},\qquad t<\lambda.$$

The condition on $t$ matters: the defining expectation is finite for every $t<\lambda$, including negative values, and diverges for $t\ge\lambda$. In particular, it is finite on an interval around zero, as required for the usual use of an MGF.

Deriving the MGF

By definition, $M_X(t)=\mathbb{E}[e^{tX}]$. Integrate against the exponential density:

$$\begin{aligned} M_X(t) &=\int_0^\infty e^{tx}\lambda e^{-\lambda x}\,dx\\ &=\lambda\int_0^\infty e^{-(\lambda-t)x}\,dx. \end{aligned}$$

This improper integral converges when $\lambda-t>0$. Under that condition, its value is $1/(\lambda-t)$, so

$$M_X(t)=\frac{\lambda}{\lambda-t}.$$

If $t=\lambda$, the integrand in the last integral is the constant $\lambda$, so the integral diverges. If $t>\lambda$, the integrand grows exponentially and the integral also diverges. Thus the formula is not a finite MGF at or above $t=\lambda$.

Rate and scale forms

Some books parameterize an exponential distribution by its scale $\theta=1/\lambda$ instead of its rate. The density is then $f_X(x)=\frac{1}{\theta}e^{-x/\theta}$, and the MGF can be written as

$$M_X(t)=\frac{1}{1-\theta t},\qquad t<\frac{1}{\theta}.$$

These are the same formula in different parameters. Check which convention is being used before substituting a number: the rate is the reciprocal of the mean, while the scale is the mean itself.

Using the MGF to find moments

The $n$th raw moment is the $n$th derivative of the MGF evaluated at zero, provided the MGF exists around zero. For the rate form,

$$M_X'(t)=\frac{\lambda}{(\lambda-t)^2},\qquad M_X''(t)=\frac{2\lambda}{(\lambda-t)^3}.$$

Therefore,

$$\mathbb{E}[X]=M_X'(0)=\frac{1}{\lambda},\qquad \mathbb{E}[X^2]=M_X''(0)=\frac{2}{\lambda^2}.$$

Since $\operatorname{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2$, the variance is $1/\lambda^2$. More generally, differentiating repeatedly gives

$$\mathbb{E}[X^n]=\frac{n!}{\lambda^n}$$

for nonnegative integers $n$.

Worked example: rate 2

Suppose $X\sim\operatorname{Exponential}(2)$, using the rate convention. Then

$$M_X(t)=\frac{2}{2-t},\qquad t<2.$$

The first two derivatives are $M_X'(t)=2/(2-t)^2$ and $M_X''(t)=4/(2-t)^3$. Evaluating at zero gives

$$\mathbb{E}[X]=\frac{2}{2^2}=\frac12, \qquad \mathbb{E}[X^2]=\frac{4}{2^3}=\frac12.$$

So $\operatorname{Var}(X)=\frac12-(\frac12)^2=\frac14$. These values also match the rate-based formulas $1/\lambda$ and $1/\lambda^2$.

MGF of a sum of independent exponentials

If $X_1,\ldots,X_n$ are independent, the MGF of their sum is the product of their MGFs:

$$M_{X_1+\cdots+X_n}(t)=\prod_{i=1}^n M_{X_i}(t).$$

For example, if two independent variables both have rate $\lambda$, then

$$M_{X_1+X_2}(t)=\left(\frac{\lambda}{\lambda-t}\right)^2,\qquad t<\lambda.$$

This is the MGF of a gamma distribution with shape $2$ and rate $\lambda$. The restriction $t<\lambda$ is shared by the product because both individual MGFs must be finite.

Common mistakes

• Dropping the domain: $\lambda/(\lambda-t)$ is valid as an MGF only for $t<\lambda$. At $t=\lambda$ the defining integral diverges.
• Mixing up rate and scale: With rate $\lambda$, the mean is $1/\lambda$. With scale $\theta$, the mean is $\theta$.
• Using the MGF formula outside its domain: The algebraic expression may still produce a number for some $t$, but that does not make it the expectation $\mathbb{E}[e^{tX}]$ when the integral diverges.
• Confusing raw and central moments: $M_X''(0)$ gives $\mathbb{E}[X^2]$, not the variance directly. Subtract the square of the mean to obtain the variance.