Variance of Binomial Distribution/Proof 1
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Theorem
Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.
Then the variance of $X$ is given by:
- $\var X = n p \paren {1 - p}$
Proof
From the definition of Variance as Expectation of Square minus Square of Expectation:
- $\var X = \expect {X^2} - \paren {\expect X}^2$
From Expectation of Function of Discrete Random Variable:
- $\ds \expect {X^2} = \sum_{x \mathop \in \Img X} x^2 \Pr \paren {X = x}$
To simplify the algebra a bit, let $q = 1 - p$, so $p + q = 1$.
So:
\(\ds \expect {X^2}\) | \(=\) | \(\ds \sum_{k \mathop \ge 0}^n k^2 \binom n k p^k q^{n - k}\) | Definition of Binomial Distribution: $p + q = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^n k n \binom {n - 1} {k - 1} p^k q^{n - k}\) | Factors of Binomial Coefficient: $k \dbinom n k = n \dbinom {n - 1} {k - 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n p \sum_{k \mathop = 1}^n k \binom {n - 1} {k - 1} p^{k - 1} q^{\paren {n - 1} - \paren {k - 1} }\) | Change of limit: term is zero when $k = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n p \sum_{j \mathop = 0}^m \paren {j + 1} \binom m j p^j q^{m - j}\) | putting $j = k - 1, m = n - 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {\sum_{j \mathop = 0}^m j \binom m j p^j q^{m - j} + \sum_{j \mathop = 0}^m \binom m j p^j q^{m - j} }\) | splitting sum up into two | |||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {\sum_{j \mathop = 0}^m m \binom {m - 1} {j - 1} p^j q^{m - j} + \sum_{j \mathop = 0}^m \binom m j p^j q^{m - j} }\) | Factors of Binomial Coefficient: $j \dbinom m j = m \dbinom {m - 1} {j - 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {\paren {n - 1} p \sum_{j \mathop = 1}^m \binom {m - 1} {j - 1} p^{j - 1} q^{\paren {m - 1} - \paren {j - 1} } + \sum_{j \mathop = 0}^m \binom m j p^j q^{m - j} }\) | Change of limit: term is zero when $j = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {\paren {n - 1} p \paren {p + q}^{m - 1} + \paren {p + q}^m}\) | Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {\paren {n - 1} p + 1}\) | as $p + q = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n^2 p^2 + n p \paren {1 - p}\) | by algebra |
Then:
\(\ds \var X\) | \(=\) | \(\ds \expect {X^2} - \paren {\expect X}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {1 - p} + n^2 p^2 - \paren {n p}^2\) | Expectation of Binomial Distribution: $\expect X = n p$ | |||||||||||
\(\ds \) | \(=\) | \(\ds n p \paren {1 - p}\) |
as required.
$\blacksquare$