Moment Generating Function of Exponential Distribution

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Theorem

Let $X$ be a continuous random variable with an exponential distribution with parameter $\beta$ 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:

\(\ds \map {M_X} t\) \(=\) \(\ds \frac 1 \beta \int_0^\infty e^{x \paren {-\frac 1 \beta + t} } \rd x\) Exponential of Sum
\(\ds \) \(=\) \(\ds \frac 1 {\beta \paren {-\frac 1 \beta + t} } \sqbrk {e^{x \paren {-\frac 1 \beta + t} } }_0^\infty\) Primitive of Exponential Function

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:

\(\ds \frac 1 {\beta \paren {-\frac 1 \beta + t} } \sqbrk {e^{x \paren {-\frac 1 \beta + t} } }_0^\infty\) \(=\) \(\ds \frac 1 {\beta \paren {-\frac 1 \beta + t} } \paren {0 - 1}\) Exponential Tends to Zero and Infinity, Exponential of Zero
\(\ds \) \(=\) \(\ds \frac 1 {\beta \paren {\frac 1 \beta - t} }\)
\(\ds \) \(=\) \(\ds \frac 1 {1 - \beta t}\)

$\blacksquare$