Moment Generating Function of Exponential Distribution

Theorem
Let $X \sim \operatorname {Exp} \paren \beta$ for some $\beta \in \R_{> 0}$.

Then the moment generating function $M_X$ of $X$ is given by:


 * $\displaystyle \map {M_X} t = \frac 1 {1 - \beta t}$

for $t < \dfrac 1 \beta$, and is undefined otherwise.

Proof
From the definition of the Exponential distribution, $X$ has probability density function:


 * $\displaystyle \map {f_X} x = \frac 1 \beta e^{-\frac x \beta}$

From the definition of a moment generating function:


 * $\displaystyle \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$

Then:

Note that if $t > \dfrac 1 \beta$, then $\displaystyle e^{x \paren {-\frac 1 \beta + t} } \to \infty$ as $x \to \infty$ by Exponential Tends to Zero and Infinity, so the integral diverges in this case.

If $t = \dfrac 1 \beta$ then the integrand is identically $1$, so the integral similarly diverges in this case.

If $t < \dfrac 1 \beta$, then $\displaystyle e^{x \paren {-\frac 1 \beta + t} } \to 0$ as $x \to \infty$ from Exponential Tends to Zero and Infinity, so the integral converges in this case.

Therefore, the function is only well defined for $t < \dfrac 1 \beta$.

Proceeding: