Moment Generating Function of Geometric Distribution

Theorem
Let $X \sim \operatorname {G_0} \paren p$ for some $0 < p \le 1$.

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


 * $\displaystyle \map {M_X} t = \frac p {1 - \paren {1 - p} e^t}$

for $t < -\map \ln {1 - p}$, and is undefined otherwise.

Proof
From the definition of the Geometric distribution, $X$ has probability mass function:


 * $\displaystyle \map \Pr {X = k} = \paren {1 - p}^k p$

From the definition of a moment generating function:


 * $\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}$

So:

By Sum of Infinite Geometric Progression, for this sum to be convergent we must have:


 * $\displaystyle \size {\paren {1 - p} e^t} < 1$

In the case $p = 1$, this demand is satisfied immediately regardless of $t$.

Otherwise, as both $e^t$ and $1 - p$ are positive:


 * $\displaystyle e^t < \frac 1 {1 - p}$

So, by Logarithm of Power:


 * $\displaystyle t < -\ln \paren {1 - p}$

is the range of $t$ for which the expectation is well-defined.

Now applying Sum of Infinite Geometric Progression, we have for this range of $t$:


 * $\displaystyle \map {M_X} t = p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p} e^t}^k = \frac p {1 - \paren {1 - p} e^t}$