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:

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

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 $p + q = 1$

$\blacksquare$