Moment Generating Function of Bernoulli Distribution

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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:

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

So:

\(\displaystyle \map {M_X} t\) \(=\) \(\displaystyle \map \Pr {X = 0} e^0 + \map \Pr {X = 1} e^t\)
\(\displaystyle \) \(=\) \(\displaystyle q + p e^t\) Exponential of Zero

$\blacksquare$