Expected value and variance of a random variable

The expected value is the long-run average of a random quantity and its variance the long-run average squared distance from that mean.

\E(X) = \mu = \sum_x x\,P(X = x), \qquad \Var(X) = \sum_x (x - \mu)^2 P(X = x) = \E(X^2) - \mu^2

Discrete random variable.

$x$: Each possible value of $X$ .
$P(X = x)$: Its probability (they sum to 1).

\E(aX + b) = a\,\E(X) + b, \qquad \Var(aX + b) = a^2 \Var(X)

Linear transformation.

$a,\ b$: Constants.

\E(X + Y) = \E(X) + \E(Y), \qquad \Var(X \pm Y) = \Var(X) + \Var(Y) \ \ (X, Y \text{ independent})

Sums and differences.

$X,\ Y$: Random variables; the variance rule needs independence (otherwise add $\pm 2\Cov(X, Y)$ ).

When to use

Fair games, insurance premiums, expected profit of a decision, expected cost of errors.
Summarising any probability distribution (continuous versions replace sums with integrals).

How to interpret

$\E(X)$ is what the average outcome approaches over many independent repetitions (law of large numbers) — it need not be a possible outcome. $\SD(X) = \sqrt{\Var(X)}$ measures risk: two bets with the same expected value can have very different spreads. A game is fair when the expected net gain is 0.

Common mistakes

Adding standard deviations instead of variances.
Subtracting variances for a difference: $\Var(X - Y)$ is $\Var(X) + \Var(Y)$ for independent variables.
Forgetting to square $a$ in $\Var(aX)$ .
Judging a one-off decision only by its expected value, ignoring risk.

Example

A promotion loses 2 with probability 0.5, breaks even with probability 0.3 and gains 10 with probability 0.2. $\E(X) = -1 + 0 + 2 = 1$ ; $\E(X^2) = 2 + 0 + 20 = 22$ ; $\Var(X) = 22 - 1 = 21$ , $\SD(X) = 4.58$ . Positive on average, but highly variable.