Probability distributions of discrete random variables, the expectation and variance, the effect of linear coding, and expectation and variance of functions of a random variable.

What this dot point is asking

AQA wants you to work with the probability distribution of a discrete random variable, compute its expectation (mean) and variance, apply linear coding to find the mean and variance of $aX + b$, and find the expectation and variance of a function of the variable such as $X^2$.

Probability distributions

A discrete random variable takes a finite or countable set of values, each with an assigned probability. Two conditions define a valid distribution: every probability is non-negative, and the probabilities sum to exactly $1$. The summing condition is the lever for almost every opening part of an exam question, because it lets you solve for an unknown constant in a probability formula or fill in a missing entry in a table. Once the distribution is fully known, every other quantity (expectation, variance, the probability of an event) follows by summation over the values.

Expectation and variance

The expectation is the long-run average value, a weighted mean of the outcomes with the probabilities as weights. The variance measures how spread out the values are about that mean. The formula $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$ is almost always preferred over the defining form $E[(X - \mu)^2]$ because it needs only the two sums $E(X)$ and $E(X^2)$, which are quick to tabulate. A common rearrangement, $E(X^2) = \operatorname{Var}(X) + [E(X)]^2$, lets you recover $E(X^2)$ when the mean and variance are given.

Linear coding and functions of a variable

Linear coding turns an awkward set of values into convenient ones, then transforms the answers back. The key insight in the variance rule is that adding a constant slides the whole distribution along without changing how spread out it is, so $b$ drops out, while multiplying by $a$ stretches the spread by a factor of $a$, which squares to $a^2$ in the variance. For a more general function $g(X)$ the expectation is found directly as $E(g(X)) = \sum g(x) P(X = x)$, summing the transformed values against the original probabilities. A frequent special case is $g(X) = X^2$, whose expectation is exactly the $E(X^2)$ used in the variance formula.

A subtle point that examiners probe is that the expectation of a non-linear function is not the function of the expectation. In general $E(g(X)) \neq g(E(X))$, and the variance formula is the clearest example: $E(X^2)$ is almost never equal to $[E(X)]^2$, and their difference is precisely the variance, which is positive for any genuinely random variable. The linear rules are special because a straight line is the one shape of function for which working through the expectation gives the same answer either way, which is why $E(aX + b) = aE(X) + b$ holds exactly. For any curved function you must average the transformed values, not transform the average.

Putting the steps together

A typical exam question chains these ideas in a fixed order, and recognising the chain saves time. First use the total-probability condition to find any unknown constant or missing probability. Next tabulate the products $xP(X = x)$ and sum them for $E(X)$. Then tabulate $x^2 P(X = x)$ and sum them for $E(X^2)$, from which the variance follows as $E(X^2) - [E(X)]^2$. Finally apply any coding or function in the last part, using the rules $E(aX + b) = aE(X) + b$ and $\operatorname{Var}(aX + b) = a^2\operatorname{Var}(X)$, or the direct sum for a non-linear function. Laying the work out as a table with a row for each value and columns for $x$, $P(X = x)$, $xP(X = x)$ and $x^2 P(X = x)$ keeps the arithmetic organised and makes slips easy to spot.