Question

Identify the expression for calculating the mean of a binomial distribution.

Answer 1

A random variable $X$ following a binomial distribution with success probability $p$ across $n$ trials has PMF

$\mathbb P(X=x)=\begin{cases}\dbinom nxp^x(1-p)^{n-x}&\text{for }x\in\{0,1,\ldots,n\}\\\\0&\text{otherwise}\end{cases}$

where $\dbinom nx=\dfrac{n!}{x!(n-x)!}$.

The mean of the distribution is given by the expected value which is defined by

$\mathbb E(X):=\displaystyle\sum_xx\mathbb P(X=x)$

where the summation is carried out over the support of $X$. So the mean is

$\displaystyle\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$

Because this is a proper distribution, you have

$\displaystyle\sum_x\mathbb P(X=x)=1$

which is a fact that will be used to evaluate the sum above.

$\displaystyle\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\displaystyle\sum_{x=1}^nx\binom nxp^x(1-p)^{n-x}$
$\displaystyle\sum_{x=1}^n\frac{xn!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!(n-x)!}p^{x-1}(1-p)^{n-x}$
$\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$

Letting $y=x-1$, this becomes

$\displaystyle np\sum_{y=0}^{n-1}\frac{(n-1)!}{y!((n-1)-y)!}p^y(1-p)^{(n-1)-y}$

Observe that the remaining sum corresponds to the PMF of a new random variable $Y$ which also follows a binomial distribution with success probability $p$, but this time across $n-1$ trials. Therefore the sum evaluates to 1, and you're left with $np$ as the expression for the mean for $X$.