Variance of Binomial Distribution

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:
 * $\operatorname{var} \left({X}\right) = n p \left({1-p}\right)$

Proof 1
From the definition of Variance as Expectation of Square minus Square of Expectation:
 * $\operatorname{var} \left({X}\right) = E \left({X^2}\right) - \left({E \left({X}\right)}\right)^2$

From Expectation of Function of Discrete Random Variable:
 * $\displaystyle E \left({X^2}\right) = \sum_{x \in \operatorname{Im} \left({X}\right)} x^2 \Pr \left({X = x}\right)$

To simplify the algebra a bit, let $q = 1 - p$, so $p+q = 1$.

So:

Then:

as required.

Proof 2
From Variance of Discrete Random Variable from P.G.F., we have:
 * $\operatorname{var} \left({X}\right) = \Pi''_X \left({1}\right) + \mu - \mu^2$

where $\mu = E \left({x}\right)$ is the expectation of $X$.

From the Probability Generating Function of Binomial Distribution, we have:
 * $\Pi_X \left({s}\right) = \left({q + ps}\right)^n$

where $q = 1 - p$.

From Expectation of Binomial Distribution, we have:
 * $\mu = n p$

From Derivatives of PGF of Binomial Distribution, we have:
 * $\Pi''_X \left({s}\right) = n \left({n-1}\right) p^2 \left({q + ps}\right)^{n-2}$

Putting $s = 1$ using the formula $\Pi''_X \left({1}\right) + \mu - \mu^2$:
 * $\operatorname{var} \left({X}\right) = n \left({n-1}\right) p^2 + np - n^2p^2$

and hence the result.

Proof 3
Alternatively, we can derive this directly from the Variance of Bernoulli Distribution.

From Bernoulli Process as Binomial Distribution, we see that $X$ as defined here is the sum of the discrete random variables that model the Bernoulli distribution.

Each of the Bernoulli trials is independent of each other.

Hence we can use Sums of Variances of Independent Trials.

The Variance of Bernoulli Distribution is $p \left({1-p}\right)$ so the variance of $B \left({n, p}\right)$ is $n p\left({1-p}\right)$.