Expectation of Bernoulli Distribution

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $X$ be a discrete random variable with a Bernoulli distribution with parameter $p$.


Then the expectation of $X$ is given by:

$\expect X = p$


Proof 1

From the definition of expectation:

$\displaystyle \expect X = \sum_{x \mathop \in \Img X} x \, \map \Pr {X = x}$

By definition of Bernoulli distribution:

$\expect X = 1 \times p + 0 \times \paren {1 - p}$

Hence the result.

$\blacksquare$


Proof 2

Follows directly from Expectation of Binomial Distribution, putting $n = 1$.

$\blacksquare$


Proof 3

From the Probability Generating Function of Bernoulli Distribution, we have:

$\map {\Pi_X} s = q + p s$

where $q = 1 - p$.


From Expectation of Discrete Random Variable from PGF, we have:

$\expect X = \map { {\Pi_X}'} 1$


From Derivatives of PGF of Bernoulli Distribution:

$\map { {\Pi_X}'} s = p$

Hence the result.

$\blacksquare$


Proof 4

From Moment Generating Function of Bernoulli Distribution, the moment generating function of $X$, $M_X$, is given by:

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

where $q = 1 - p$.

By Moment in terms of Moment Generating Function:

$\expect X = \map {M_X'} 0$

We have:

\(\displaystyle \map {M_X'} t\) \(=\) \(\displaystyle \frac \d {\d t} \paren {q + p e^t}\)
\(\displaystyle \) \(=\) \(\displaystyle p e^t\) Derivative of Constant, Derivative of Exponential Function

Setting $t = 0$ gives:

\(\displaystyle \expect X\) \(=\) \(\displaystyle p e^0\)
\(\displaystyle \) \(=\) \(\displaystyle p\) Exponential of Zero

$\blacksquare$