Variance of Bernoulli Distribution/Proof 5
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Theorem
Let $X$ be a discrete random variable with the Bernoulli distribution with parameter $p$:
- $X \sim \Bernoulli p$
Then the variance of $X$ is given by:
- $\var X = p \paren {1 - p}$
Proof
From Moment Generating Function of Bernoulli Distribution, the moment generating function $M_X$ of $X$ is given by:
- $\map {M_X} t = q + p e^t$
From Variance as Expectation of Square minus Square of Expectation, we have:
- $\var X = \expect {X^2} - \paren {\expect X}^2$
From Moment in terms of Moment Generating Function:
- $\expect {X^2} = \map {M_X''} 0$
We have:
\(\ds \map {M_X''} t\) | \(=\) | \(\ds \frac {\d^2} {\d t^2} \paren {q + p e^t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p \frac \d {\d t} \paren {e^t}\) | Derivative of Constant, Derivative of Exponential Function | |||||||||||
\(\ds \) | \(=\) | \(\ds p e^t\) | Derivative of Exponential Function |
Setting $t = 0$ gives:
\(\ds \expect {X^2}\) | \(=\) | \(\ds p e^0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p\) | Exponential of Zero |
In Expectation of Bernoulli Distribution, it is shown that:
- $\expect X = p$
So:
\(\ds \var X\) | \(=\) | \(\ds p - p^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds p \paren {1 - p}\) |
$\blacksquare$