# 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:

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

Thus:

 $\ds \expect X$ $=$ $\ds \sum_{k \mathop = 0}^n k \binom n k p^k q^{n - k}$ Definition of Binomial Distribution, with $p + q = 1$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n k \binom n k p^k q^{n - k}$ since for $k = 0$, $k \dbinom n k p^k q^{n - k} = 0$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^n n \binom {n - 1} {k - 1} p^k q^{n - k}$ Factors of Binomial Coefficient: $k \dbinom n k = n \dbinom {n - 1} {k - 1}$ $\ds$ $=$ $\ds n p \sum_{k \mathop = 1}^n \binom {n - 1} {k - 1} p^{k - 1} q^{\paren {n - 1} - \paren {k - 1} }$ taking out $n p$ and using $\paren {n - 1} - \paren {k - 1} = n - k$ $\ds$ $=$ $\ds n p \sum_{j \mathop = 0}^m \binom m j p^j q^{m - j}$ putting $m = n - 1, j = k - 1$ $\ds$ $=$ $\ds n p$ 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:

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

Each of the Bernoulli trials is independent of each other, by definition of a Bernoulli process.

It follows that:

 $\ds \expect X$ $=$ $\ds \expect {\sum_{i \mathop = 1}^n Y_i }$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \expect {Y_i}$ Sum of Expectations of Independent Trials‎ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n p$ Expectation of Bernoulli Distribution $\ds$ $=$ $\ds n p$ Sum of Identical Terms

$\blacksquare$

## Proof 3

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

$\map {\Pi_X} s = \paren {q + p s}^n$

where $q = 1 - p$.

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

$\expect X = \map {\Pi'_X} 1$

We have:

 $\ds \map {\Pi'_X} s$ $=$ $\ds \map {\frac \d {\d s} } {q + p s}^n$ $\ds$ $=$ $\ds n p \paren {q + p s}^{n - 1}$ Derivatives of PGF of Binomial Distribution

Plugging in $s = 1$:

$\map {\Pi'_X} 1 = n p \paren {q + p}$

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:

$\ds \map {M_X} t = \paren {1 - p + p e^t}^n$
$\ds \expect X = \map {M_X'} 0$

We have:

 $\ds \map {M_X'} t$ $=$ $\ds \frac \d {\d t} \paren {1 - p + p e^t}^n$ $\ds$ $=$ $\ds \map {\frac \d {\d t} } {1 - p + p e^t} \map {\frac \d {\map \d {1 - p + p e^t} } } {1 - p + p e^t}^n$ Chain Rule for Derivatives $\ds$ $=$ $\ds n p e^t \paren {1 - p + p e^t}^{n - 1}$ Derivative of Exponential Function, Derivative of Power

Setting $t = 0$ gives:

 $\ds \expect X$ $=$ $\ds n p e^0 \paren {1 - p + p e^0}^{n - 1}$ $\ds$ $=$ $\ds n p \paren {1 - p + p}^{n - 1}$ Exponential of Zero $\ds$ $=$ $\ds n p$

$\blacksquare$