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 of $X$, $M_X$, is given by:

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

From Variance as Expectation of Square minus Square of Expectation, we have:

$\displaystyle \var X = \expect {X^2} - \paren {\expect X}^2$

From Moment in terms of Moment Generating Function:

$\displaystyle \expect {X^2} = \map {M_X''} 0$

We have:

\(\displaystyle \map {M_X''} t\) \(=\) \(\displaystyle \frac {\d^2} {\d t^2} \paren {q + p e^t}\)
\(\displaystyle \) \(=\) \(\displaystyle p \frac \d {\d t} \paren {e^t}\) Derivative of Constant, Derivative of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle p e^t\) Derivative of Exponential Function

Setting $t = 0$ gives:

\(\displaystyle \expect {X^2}\) \(=\) \(\displaystyle p e^0\)
\(\displaystyle \) \(=\) \(\displaystyle p\) Exponential of Zero

In Expectation of Bernoulli Distribution, it is shown that:

$\displaystyle \expect X = p$

So:

\(\displaystyle \var X\) \(=\) \(\displaystyle p - p^2\)
\(\displaystyle \) \(=\) \(\displaystyle p \paren {1 - p}\)

$\blacksquare$