Problem
Let $X \sim \text{Binomial}(n,p)$. What is $E(X)$ and $V(X)$?
Solution
Expectation
We could attempt to compute the $E(X)$ directly using the formula for expectation and the pmf for Binomial:
$E(X) = \int x dF_x(x) = \sum_x x f_X(x) = \sum_{x=0}^{n} x \binom{n}{x} p^x (1-p)^{n-x}$
This isn't an easy sum to compute, however. Instead we'll rely on the theorem:
$$E(\sum X_i) = \sum E(X_i)$$
In the case of a Binomial, each $X_i$ is a Bernoulli coin flip. The expectation is simply:
$E(X_i) = 1 \times p + 0 \times (1-p) = p$.
thus making $E(X) = \sum_{x=i}^{n} E(X_i) = \sum_{x=i}^{n} p = np$.
Variance
Using the theorems:
$$V(X) = E(X^2) - E(X)^2$$
and
$$V(\sum X_i) = \sum V(X_i) \text{(for independent } X_i\text{)}$$
we start by computing $V(X_i)$:
$V(X_i) = E(X_i^2) - E(X_i)^2$
$E(X_i^2) = 1^2 \times p + 0^2 \times (1-p) = p$
plugging this back in $V(X_i) = E(X_i^2) - E(X_i)^2 = p - np = p(1-p)$
finally, $V(X) = \sum_{i=1}^n V(X_i) = np(1-p)$.
Note that $V(X) = 0$ for $p=1$ and $p=0$, which makes intuitive sense considering flipping all heads or all tails does not have any variation in outcomes.
Sources
- Examples 3.12 and 3.16