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.

$\blacksquare$