# Expectation of Binomial Distribution/Proof 1

## Theorem

Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$.

Then the expectation of $X$ is given by:

- $\expect X = n p$

## Proof

From the definition of expectation:

- $\displaystyle \expect X = \sum_{x \mathop \in \Omega_X} x \map \Pr {X = x}$

Thus:

\(\displaystyle \expect X\) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^n k \binom n k p^k q^{n - k}\) | Definition of Binomial Distribution, with $p + q = 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^n k \binom n k p^k q^{n - k}\) | since for $k = 0$, $k \dbinom n k p^k q^{n - k} = 0$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^n n \binom {n - 1} {k - 1} p^k q^{n - k}\) | Factors of Binomial Coefficient: $k \dbinom n k = n \dbinom {n - 1} {k - 1}$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n p \sum_{k \mathop = 1}^n \binom {n - 1} {k - 1} p^{k - 1} q^{\paren {n - 1} - \paren {k - 1} }\) | taking out $n p$ and using $\paren {n - 1} - \paren {k - 1} = n - k$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n p \sum_{j \mathop = 0}^m \binom m j p^j q^{m - j}\) | putting $m = n - 1, j = k - 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n p\) | Binomial Theorem and $p + q = 1$ |

$\blacksquare$