Proof: Moment-generating function of the exponential distribution
Theorem: Let $X$ be a random variable following an exponential distribution:
\[\label{eq:exp} X \sim \mathrm{Exp}(\lambda) \; .\]Then, the moment generating function of $X$ is
\[\label{eq:exp-mgf} M_X(t) = \frac{\lambda}{\lambda - t}\]which is well-defined for $t < \lambda$.
Proof: Suppose $X$ follows an exponential distribution with rate $\lambda$; that is, $X\sim \mathrm{Exp}(\lambda)$. Then, the probability density function is given by
\[\label{eq:exp-pdf} f_X(x) = \lambda e^{-\lambda x}\]and the moment-generating function is defined as
\[\label{eq:mgf} M_X(t) = \mathrm{E} \left[ e^{tX} \right] \; .\]Using the definition of expected value for continuous random variables, the moment-generating function of $X$ is thus:
\[\label{eq:exp-mgf-s1} \begin{split} M_X(t) &= \int_0^{\infty} e^{tx} \cdot f_X(x) dx\\ &= \int_0^{\infty} e^{tx}\cdot \lambda e^{-\lambda x} dx\\ &= \int_0^{\infty} \lambda e^{x(t-\lambda)} dx\\ &= \frac{\lambda}{t-\lambda} e^{x(t-\lambda)} \Big|_{x = 0}^{x = \infty}\\ &= \lim_{x\rightarrow \infty} \left[ \frac{\lambda}{t-\lambda} e^{x(t-\lambda)} - \frac{\lambda}{t-\lambda}\right]\\ &= \frac{\lambda}{t-\lambda} \left[ \lim_{x\rightarrow \infty} e^{x(t-\lambda)} -1 \right] \; . \end{split}\]Note that $t$ cannot be equal to $\lambda$, else $M_X(t)$ is undefined. Further, if $t > \lambda$, then $\lim_{x\rightarrow \infty} e^{x(t-\lambda)} = \infty$, which implies that $M_X(t)$ diverges for $t \geq \lambda$. So, we must restrict the domain of $M_X(t)$ to $t<\lambda$. Assuming this, we can further simplify \eqref{eq:exp-mgf-s1}:
\[\label{eq:exp-mgf-s2} \begin{split} M_X(t) &= \frac{\lambda}{t-\lambda} \left[ \lim_{x\rightarrow \infty} e^{x(t-\lambda)} -1 \right] \\ &= \frac{\lambda}{t-\lambda} \left[ 0 - 1 \right] \\ &= \frac{\lambda}{\lambda - t} \; . \end{split}\]This completes the proof of \eqref{eq:exp-mgf}.
Metadata: ID: P403 | shortcut: exp-mgf | author: tomfaulkenberry | date: 2023-04-19, 12:00.