What Is the MGF of a Double Exponential Random Variable?

“Double exponential” most commonly means the Laplace distribution, whose density has exponential tails on both sides of its center. The moment-generating function (MGF) depends on the distribution’s location and scale, so it helps to state the parameterization before giving the formula.

For a Laplace random variable $X$ with location $\mu$ and scale $b>0$, the MGF is

$$M_X(t)=\frac{e^{\mu t}}{1-b^2t^2},\qquad |t|<\frac{1}{b}.$$

The interval matters: outside it, the expectation defining the MGF diverges. If the distribution is instead written using rate $\lambda=1/b$, the formula for a location-zero variable is $\lambda^2/(\lambda^2-t^2)$, valid for $|t|<\lambda$.

Deriving the MGF

The Laplace density with location $\mu$ and scale $b$ is

$$f_X(x)=\frac{1}{2b}\exp\left(-\frac{|x-\mu|}{b}\right),\qquad -\infty<x<\infty.$$

First take $\mu=0$ and write $\lambda=1/b$. Then $f_X(x)=\frac{\lambda}{2}e^{-\lambda|x|}$. By definition, $M_X(t)=\mathbb{E}[e^{tX}]$. Split the integral at zero, where the expression for $|x|$ changes:

$$\begin{aligned} M_X(t) &=\frac{\lambda}{2}\left(\int_{-\infty}^{0}e^{(t+\lambda)x}\,dx+\int_{0}^{\infty}e^{(t-\lambda)x}\,dx\right)\\ &=\frac{\lambda}{2}\left(\frac{1}{t+\lambda}+\frac{1}{\lambda-t}\right)\\ &=\frac{\lambda^2}{\lambda^2-t^2}. \end{aligned}$$

The first integral converges only when $t+\lambda>0$, and the second only when $t-\lambda<0$. Together these conditions give $-\lambda<t<\lambda$. At either endpoint, one tail integral also diverges, so the MGF is not finite there.

For a nonzero location, write $X=\mu+Y$, where $Y$ is a centered Laplace variable with the same scale. Since $e^{tX}=e^{\mu t}e^{tY}$, the MGF is multiplied by $e^{\mu t}$. Replacing $\lambda$ with $1/b$ gives the stated location-scale form.

Worked example

Suppose $X$ has a centered double exponential distribution with rate $\lambda=2$. Its scale is $b=1/2$, and

$$M_X(t)=\frac{4}{4-t^2},\qquad |t|<2.$$

At $t=1$, which is inside the domain,

$$M_X(1)=\frac{4}{4-1}=\frac{4}{3}.$$

The first two derivatives provide a useful check. For a centered distribution, $M_X'(0)=\mathbb{E}[X]$ and $M_X''(0)=\mathbb{E}[X^2]$. Here the mean is $0$ and the variance is $M_X''(0)=2/\lambda^2=1/2$. More generally, a Laplace variable with scale $b$ has mean $\mu$ and variance $2b^2$.

Common parameterizations and interpretations

Convention	Density	MGF and domain
Location-scale	$\frac{1}{2b}e^{-	x-\mu
Centered, rate $\lambda$	$\frac{\lambda}{2}e^{-\lambda	x

Another common construction is the difference of two independent exponential random variables with the same rate $\lambda$. If $U,V\sim\operatorname{Exp}(\lambda)$ independently, then $U-V$ has the centered Laplace density above. Its MGF is also the product of the MGFs of $U$ and $-V$:

$$\frac{\lambda}{\lambda-t}\cdot\frac{\lambda}{\lambda+t}=\frac{\lambda^2}{\lambda^2-t^2},\qquad |t|<\lambda.$$

This agrees with the direct integration. If “double exponential” is being used for a different construction, such as a sum or a distribution with unequal tail rates, identify its density first; the MGF can differ.

Common mistakes

• Dropping the domain: A rational expression alone does not describe the full MGF. State the interval where the expectation is finite.
• Mixing scale and rate: Scale $b$ and rate $\lambda$ are reciprocals. The denominator is $1-b^2t^2$ in scale form and $\lambda^2-t^2$ in rate form.
• Forgetting the location: A shifted Laplace variable has the additional factor $e^{\mu t}$.
• Using the wrong sign: For the difference of independent exponentials, use the MGF of $-V$ with argument $-t$; this produces the factor $\lambda/(\lambda+t)$, not $\lambda/(\lambda-t)$ twice.