Expectation of a binomial random variable

If the random variable \(X\) is such that \(X \sim B(n, p)\) then

\[\mathrm{E}(X) = np \quad \text{and} \quad \mathrm{Var}(X) = npq = np(1 - p)\]

where \(q = 1 - p\)

Recall that
\[ \mathrm{E}(X) = \sum_{\rm{all } x} x P(X =x), \]
and
\[ \mathrm{Var}(X) = \mathrm{E}(X^2) - [(\mathrm{E}(X)]^2. \]

With \(P(X = x) = {}^n C_{x} p^x (1-p)^{n-x }\), where \(x = 0, \ldots, n\), the derivations for the expected value and variance of a binomial random variable are non-trivial.

The definition of expectation is:
\[ \mathrm{E}(X) = \sum_{k=0}^{n} k \cdot P(X=k). \]

Substituting the Binomial probability distribution formula:
\[ \mathrm{E}(X) = \sum_{k=0}^{n} k \binom{n}{k} p^k q^{n-k}. \]

Since \( k=0 \) makes the term zero, we start from \( k=1 \):
\[ \mathrm{E}(X) = \sum_{k=1}^{n} k \frac{n!}{k!(n-k)!} p^k q^{n-k}. \]

Using the identity \( k \binom{n}{k} = n \binom{n-1}{k-1} \),
\[ \mathrm{E}(X) = \sum_{k=1}^{n} n \binom{n-1}{k-1} p^k q^{n-k}. \]

Factorinf out \( np \),
\[ \mathrm{E}(X) = np \sum_{k=1}^{n} \binom{n-1}{k-1} p^{k-1} q^{(n-1)-(k-1)}. \]

Let \( j = k-1 \) and \( m = n-1 \),
\[ \mathrm{E}(X) = np \sum_{j=0}^{m} \binom{m}{j} p^j q^{m-j}. \]

By the Binomial Theorem, the sum is \( (p+q)^m = 1^m = 1 \), and so
\[ \mathrm{E}(X) = np. \]

The derivation of \(\mathrm{Var(X)\) is also complex, and so we omit it.