📂Probability Distribution

Moment Generating Function of the Laplace Distribution

Formula

$X \sim$ When $\operatorname{Laplace}(\mu, b)$, the moment-generating function of $X$ is as follows.

$$m(t) = \dfrac{1}{1 - b^{2}t^{2}} e^{\mu t} \qquad \text{for } |t| < \dfrac{1}{b}$$

Proof

By the definition of the moment-generating function,

$$\begin{align*} E(e^{tX}) &= \int\limits_{-\infty}^{\infty} e^{tx} f(x) dx \\ &= \int\limits_{-\infty}^{\infty} e^{tx} \dfrac{1}{2b} e^{-|x - \mu|/b} dx \\ &= \dfrac{a}{2}e^{\mu t} \int\limits_{-\infty}^{\infty} e^{ty} e^{-a|y|} dx \qquad (y \equiv x - \mu, \quad a \equiv 1/b) \\ &= \dfrac{a}{2}e^{\mu t} \left( \int\limits_{-\infty}^{0} e^{(t+a)y} dx + \int\limits_{0}^{\infty} e^{(t-a)y} dx \right) \\ &= \dfrac{a}{2}e^{\mu t} \left( \left[ \dfrac{1}{t+a}e^{(t+a)y} \right]_{-\infty}^{0} + \left[ \dfrac{1}{t-a}e^{(t-a)y} \right]_{0}^{\infty} \right) \\ \end{align*}$$

Here, the first integral converges when $t + a \lt 0$, and the second integral converges when $t - a \gt 0$. Therefore, it is integrable when $|t| < a$.

$$\begin{align*} E(e^{tX}) &= \dfrac{a}{2}e^{\mu t} \left( \left[ \dfrac{1}{t+a}e^{(t+a)y} \right]_{-\infty}^{0} + \left[ \dfrac{1}{t-a}e^{(t-a)y} \right]_{0}^{\infty} \right) \\ &= \dfrac{a}{2}e^{\mu t} \left( \dfrac{1}{t+a} - \dfrac{1}{t-a} \right) & \text{for } |t| < a \\ &= \dfrac{a}{2}e^{\mu t} \dfrac{-2a}{t^{2} - a^{2}} & \text{for } |t| < a \\ &= \dfrac{a^{2}}{a^{2} - t^{2}}e^{\mu t} & \text{for } |t| < a \\ &= \dfrac{1}{b^{2}}\dfrac{1}{1/b^{2} - t^{2}}e^{\mu t} & \text{for } |t| < \dfrac{1}{b} \\ &= \dfrac{1}{1 - b^{2}t^{2}}e^{\mu t} & \text{for } |t| < \dfrac{1}{b} \\ \end{align*}$$

■