# Expectation of Binomial Distribution

## 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 1

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 $p + q = 1$ |

$\blacksquare$

## Proof 2

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

- $\displaystyle X = \sum_{i \mathop = 1}^n Y_i$

Each of the Bernoulli trials is independent of each other, by definition of a Bernoulli process. It follows that:

\(\displaystyle E \left({X}\right)\) | \(=\) | \(\displaystyle E \left({ \sum_{i \mathop = 1}^n Y_i }\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^n E \left({Y_i}\right)\) | Sum of Expectations of Independent Trialsâ€Ž | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^n \ p\) | Expectation of Bernoulli Distribution | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n p\) | Sum of Identical Terms |

$\blacksquare$

## Proof 3

From the Probability Generating Function of Binomial Distribution, we have:

- $\displaystyle \Pi_X \left({s}\right) = \left({q + ps}\right)^n$

where $q = 1 - p$.

From Expectation of Discrete Random Variable from PGF, we have:

- $\displaystyle E \left({X}\right) = \Pi'_X \left({1}\right)$

We have:

\(\displaystyle \Pi'_X \left({s}\right)\) | \(=\) | \(\displaystyle \frac {\mathrm d} {\mathrm d s} \left({q + ps}\right)^n\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n p \left({q + ps}\right)^{n-1}\) | Derivatives of PGF of Binomial Distribution |

Plugging in $s = 1$:

- $\displaystyle\Pi'_X \left({1}\right) = n p \left({q + p}\right)$

Hence the result, as $q + p = 1$.

$\blacksquare$

## Proof 4

From Moment Generating Function of Binomial Distribution, the moment generating function of $X$, $M_X$, is given by:

- $\displaystyle \map {M_X} t = \paren {1 - p + pe^t}^n$

By Moment in terms of Moment Generating Function:

- $\displaystyle \expect X = \map {M_X'} 0$

We have:

\(\displaystyle \map {M_X'} t\) | \(=\) | \(\displaystyle \frac \d {\d t} \paren {1 - p + pe^t}^n\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map {\frac \d {\d t} } {1 - p + pe^t} \map {\frac \d {\map \d {1 - p + pe^t} } } {1 - p + pe^t}^n\) | Chain Rule for Derivatives | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n p e^t \paren {1 - p + pe^t}^{n - 1}\) | Derivative of Exponential Function, Derivative of Power |

Setting $t = 0$ gives:

\(\displaystyle \expect X\) | \(=\) | \(\displaystyle n p e^0 \paren {1 - p + p e^0}^{n - 1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n p \paren {1 - p + p}^{n - 1}\) | Exponential of Zero | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n p\) |

$\blacksquare$

## Sources

- 1986: Geoffrey Grimmett and Dominic Welsh:
*Probability: An Introduction*... (previous) ... (next): $\S 2.4$: Expectation: Exercise $9$