Expected value and Variance of a random variable

Beyond the full probability distribution, it is often enough to summarise a random variable by a single representative number computed from that distribution. The most important such summary is the mean (or expected value): if $X$ takes values $x_1, x_2, \ldots, x_n$ with probabilities $p_1, p_2, \ldots, p_n$, the expected value of $X$, written $E(X)$ or $\mu$, is $E(X) = \mu = \sum_{i=1}^n x_i p_i = x_1p_1 + x_2p_2 + \cdots + x_np_n$ — the sum of every possible value weighted by its probability. It roughly locates the 'middle' or long-run average value of the random variable.

The natural companion measure is spread: the variance of $X$, written $Var(X)$ or $\sigma_x^2$, is $Var(X) = \sum_{i=1}^n (x_i - \mu)^2 p_i$, the probability-weighted average squared distance from the mean; its non-negative square root $\sigma_x = \sqrt{Var(X)}$ is the standard deviation. In practice the defining formula is rarely the easiest to compute with, so a simplified (shortcut) form is used instead: $Var(X) = \sum_{i=1}^n x_i^2 p_i - \left(\sum_{i=1}^n x_i p_i\right)^2 = E(X^2) - [E(X)]^2$, where $E(X^2) = \sum_{i=1}^n x_i^2 p_i$.

Worked examples.

Example 1. Three coins are tossed simultaneously; $X$ is the number of heads. Find $E(X)$ and $Var(X)$.

With $S$ the usual 8 equally likely outcomes, $X \in \{0,1,2,3\}$ with $p_0=1/8, p_1=3/8, p_2=3/8, p_3=1/8$. Then $\sum x_ip_i = 0(1/8)+1(3/8)+2(3/8)+3(1/8) = 12/8$, so $E(X) = 12/8 = 1.5$. Also $\sum x_i^2 p_i = 0+3/8+12/8+9/8 = 24/8 = 3$, so $Var(X) = 3 - (1.5)^2 = 3 - 2.25 = 0.75$.

Example 2. A pair of dice is thrown; $X$ is the sum shown. Find $E(X)$ and $Var(X)$.

Using the same distribution as Table 7.1 (36 equally likely pairs, $X \in \{2,\ldots,12\}$), summing $x_ip_i$ over all 11 values gives $\sum x_ip_i = 252/36 = 7$, so $E(X) = 7$. Summing $x_i^2p_i$ gives $\sum x_i^2p_i = 1974/36 \approx 54.83$, so $Var(X) = 54.83 - 7^2 = 54.83 - 49 = 5.83$ (exactly $210/36 = 35/6$).

Example 3. Find the mean and variance of a number selected at random from 1 to 15.

Here $S=\{1,2,\ldots,15\}$, and each value is equally likely, $P(k)=1/15$. $\mu = E(X) = (1+2+\cdots+15)/15 = \dfrac{15\times16/2}{15} = 8$. Also $E(X^2) = (1^2+2^2+\cdots+15^2)/15 = \dfrac{15\times16\times31/6}{15} = 82.67$, so $Var(X) = 82.67 - 8^2 = 82.67-64 = 18.67$. …