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=(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(xn!)/(x!(n-x)!)p^x(1-p)^(n-x)`

`\displaystyle np\sum_(x=1)^n((n-1)!)/((x-1)!(n-x)!)p^(x-1)(1-p)^(n-x)`

`\displaystyle np\sum_(x=1)^n((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)((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`.