Expectation of Bernoulli Distribution
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:
- $\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$
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:
| \(\ds \map {M_X'} t\) | \(=\) | \(\ds \frac \d {\d t} \paren {q + p e^t}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds p e^t\) | Derivative of Constant, Derivative of Exponential Function |
Setting $t = 0$ gives:
| \(\ds \expect X\) | \(=\) | \(\ds p e^0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds p\) | Exponential of Zero |
$\blacksquare$