Expectation of Binomial Distribution

Theorem

Let $X$ be a discrete random variable with the binomial distribution with parameters $n$ and $p$.

Then the expectation of $X$ is given by:

$E \left({X}\right) = n p$

Proof 1

From the definition of expectation:

$\displaystyle E \left({X}\right) = \sum_{x \in \Omega_X} x \Pr \left({X = x}\right)$

Thus:

	\(\displaystyle \)	\(\displaystyle E \left({X}\right)\)	\(=\)	\(\displaystyle \sum_{k = 0}^n k \binom n k p^k q^{n - k}\)	\(\displaystyle \)	Definition of binomial distribution, with $p + q = 1$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k = 0}^n n \binom {n - 1} {k - 1} p^k q^{n - k}\)	\(\displaystyle \)	Factors of Binomial Coefficients: $\displaystyle k \binom n k = n \binom {n - 1} {k - 1}$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n p \sum_{k = 1}^n \binom {n - 1} {k - 1} p^{k - 1} q^{\left({n - 1}\right) - \left({k - 1}\right)}\)	\(\displaystyle \)	When $k = 0$ we have that $\displaystyle \binom {n - 1} {k - 1} = 0$.
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n p \sum_{j = 0}^m \binom m j p^j q^{m - j}\)	\(\displaystyle \)	Putting $m = n - 1, j = k - 1$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n p\)	\(\displaystyle \)	Binomial Theorem and $p + q = 1$

$\blacksquare$

Proof 2

Alternatively, we can derive this directly from the Expectation of Bernoulli Distribution.

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:

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

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

	\(\displaystyle \)	\(\displaystyle E \left({X}\right)\)	\(=\)	\(\displaystyle E \left({ \sum_{i = 1}^n Y_i }\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i = 1}^n E \left({Y_i}\right)\)	\(\displaystyle \)	Sum of Expectations of Independent Trials‎
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i = 1}^n \ p\)	\(\displaystyle \)	Expectation of Bernoulli Distribution
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n p\)	\(\displaystyle \)

$\blacksquare$

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 P.G.F., we have:

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

We have:

	\(\displaystyle \)	\(\displaystyle \Pi'_X \left({s}\right)\)	\(=\)	\(\displaystyle \frac d {ds} \left({q + ps}\right)^n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n p \left({q + ps}\right)^{n-1}\)	\(\displaystyle \)	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 $q + p = 1$.

$\blacksquare$

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 2.4$: Exercise $9$