Construct probability distributions for discrete random variables and compute the expectation and variance, including for functions of the variable

What this dot point is asking

SEAB wants you to construct a probability distribution for a discrete random variable, verify it, and compute the expectation (mean) and variance, including for linear functions of the variable.

The answer

A discrete probability distribution

A discrete random variable $X$ takes a countable set of values, each with a probability $\mathrm{P}(X = x)$. A valid distribution satisfies:

The total-probability condition is often used to find an unknown constant.

Expectation

The expectation (mean) is the long-run average value:

For a function, $\mathrm{E}(\mathrm{g}(X)) = \sum_x \mathrm{g}(x)\,\mathrm{P}(X = x)$; in particular $\mathrm{E}(X^2) = \sum_x x^2 \mathrm{P}(X = x)$.

Variance

The variance measures spread:

This computational form is almost always easier than the definition $\mathrm{E}\big((X - \mu)^2\big)$. The standard deviation is $\sqrt{\operatorname{Var}(X)}$.

Linear transformations

For constants $a$ and $b$:

Adding a constant shifts the mean but leaves the spread unchanged; scaling by $a$ multiplies the variance by $a^2$.

Why the computational variance formula works

The form $\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$ is not a separate definition but an algebraic rearrangement of $\operatorname{Var}(X) = \mathrm{E}\big((X - \mu)^2\big)$. Expanding the square gives $\mathrm{E}(X^2 - 2\mu X + \mu^2) = \mathrm{E}(X^2) - 2\mu\mathrm{E}(X) + \mu^2$, and since $\mathrm{E}(X) = \mu$, the last two terms combine to $-\mu^2$, leaving $\mathrm{E}(X^2) - \mu^2$. Knowing this derivation explains why you must compute $\mathrm{E}(X^2)$ separately and why it is almost always less work than summing $(x - \mu)^2$ term by term, especially when $\mu$ is not a whole number.

Setting up a distribution from a scenario

Many H2 questions describe a situation and ask you to build the distribution table before computing anything. The routine is: list every value the variable can take, find the probability of each from the scenario (using counting or basic probability), tabulate them, and check the probabilities sum to $1$. For the number of heads in two coin tosses, the values are $0, 1, 2$ with probabilities $\tfrac{1}{4}, \tfrac{1}{2}, \tfrac{1}{4}$. Constructing the table correctly is the foundation everything else rests on, because a single wrong probability throws off both the expectation and the variance that follow.

Examples in context

Example 1. Expected winnings. A game paying out according to a die roll has an expected value computed as $\sum x\,\mathrm{P}(X = x)$; comparing it to the stake tells a player whether the game is favourable in the long run.

Example 2. Insurance pricing. An insurer sets premiums above the expected claim $\mathrm{E}(X)$ and uses the variance to gauge risk, the everyday actuarial use of expectation and variance.

Try this

Q1. A variable has $\mathrm{P}(X = 0) = 0.5$, $\mathrm{P}(X = 1) = 0.5$. Find $\mathrm{E}(X)$. [1 mark]

• Cue. $0(0.5) + 1(0.5) = 0.5$.

Q2. Given $\mathrm{E}(X) = 4$, find $\mathrm{E}(2X + 3)$. [2 marks]

• Cue. $2(4) + 3 = 11$.

Q3. Given $\operatorname{Var}(X) = 5$, find $\operatorname{Var}(2X - 1)$. [2 marks]

• Cue. $2^2 \times 5 = 20$.