Moment Generating Function of Bernoulli Distribution

Theorem

Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$ for some $0 \le p \le 1$.

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

$\map {M_X} t = q + p e^t$

where $q = 1 - p$.

Proof

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

$\map \Pr {X = n} = \begin{cases}

q & : n = 0 \\ p & : n = 1 \\ 0 & : n \notin \set {0, 1} \\ \end{cases}$

From the definition of a moment generating function:

$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^1 \map \Pr {X = n} e^{t n}$

So:

\(\ds \map {M_X} t\)

\(=\)

\(\ds \map \Pr {X = 0} e^0 + \map \Pr {X = 1} e^t\)

\(\ds \)

\(=\)

\(\ds q + p e^t\)

Exponential of Zero

$\blacksquare$