Discrete random variables, their probability distributions, the expected value (mean) $E(X) = \sum x P(X = x)$, the variance $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$ and the standard deviation $\mathrm{sd}(X) = \sqrt{\mathrm{Var}(X)}$

What this dot point is asking

VCAA wants you to work with discrete random variables, their probability distributions, and the summary statistics (mean, variance, standard deviation) that describe them. These appear in Paper 1 calculations and underpin the binomial distribution.

Discrete random variables

A discrete random variable $X$ takes values from a countable set (typically a finite list of integers). Its probability distribution is a table or rule specifying $P(X = x)$ for each possible value $x$.

Two conditions every distribution must satisfy:

1. $P(X = x) \geq 0$ for all $x$.
2. $\sum_{x} P(X = x) = 1$ (the probabilities sum to one).

Solving for an unknown probability often comes down to using condition 2.

Example distribution

A spinner is divided into four unequal sectors numbered 1, 2, 3, 4 with respective probabilities $0.2, 0.3, 0.3, 0.2$.

The sum is 1. The variable $X$ representing the outcome is a discrete random variable.

Expected value (mean)

The expected value of $X$ is the long-run average outcome if the experiment is repeated many times:

It is a weighted average: each possible value is multiplied by its probability.

For the spinner above: $E(X) = 1(0.2) + 2(0.3) + 3(0.3) + 4(0.2) = 0.2 + 0.6 + 0.9 + 0.8 = 2.5$.

Expected value of a function of $X$

For any function $g$:

The most common application is $g(x) = x^2$, giving $E(X^2)$, which is needed for variance.

Variance and standard deviation

The variance measures the spread of $X$ around its mean. The computational formula (memorise this):

The equivalent definition is $\mathrm{Var}(X) = E\!\left((X - \mu)^2\right)$, but the computational formula is faster.

The standard deviation is the positive square root:

Standard deviation has the same units as $X$ itself, which makes it easier to interpret than variance.

For the spinner: $E(X^2) = 1(0.2) + 4(0.3) + 9(0.3) + 16(0.2) = 0.2 + 1.2 + 2.7 + 3.2 = 7.3$.

$\mathrm{Var}(X) = 7.3 - (2.5)^2 = 7.3 - 6.25 = 1.05$. So $\mathrm{sd}(X) = \sqrt{1.05} \approx 1.025$.

Linear transformations

For constants $a$ and $b$:

Key insight: an added constant shifts the mean but not the variance. A multiplicative constant scales the mean linearly and the variance by the square.

Worked example

A game pays \Y$where$Y = 10 X - 5$and$X$ is the spinner above.

E(Y) = 10 E(X) - 5 = 10(2.5) - 5 = \20$.

$\mathrm{Var}(Y) = 100 \mathrm{Var}(X) = 100(1.05) = 105$.

\mathrm{sd}(Y) = 10 \cdot 1.025 \approx \10.25$.

Cumulative distribution

The cumulative probability $P(X \leq x)$ is the sum of $P(X = k)$ for all $k \leq x$. Inequalities to watch:

• $P(X < x) = P(X \leq x - 1)$ for integer-valued $X$.
• $P(X \geq x) = 1 - P(X < x) = 1 - P(X \leq x - 1)$.
• $P(a \leq X \leq b) = P(X \leq b) - P(X \leq a - 1)$ for integer-valued $X$.

Read the inequality direction carefully. Strict versus non-strict matters when $X$ takes the boundary value with positive probability.

Examples in context

Example 1. Expected demand for a cafe. The number of cakes $X$ sold per hour has distribution $P(X=0)=0.1$, $P(X=1)=0.3$, $P(X=2)=0.4$, $P(X=3)=0.2$. The expected sales are $E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0.3 + 0.8 + 0.6 = 1.7$ cakes per hour. With $E(X^2) = 0 + 0.3 + 1.6 + 1.8 = 3.7$, the variance is $3.7 - 1.7^2 = 0.81$, so $\mathrm{sd}(X) = 0.9$.

Example 2. Scaling a payoff. A wholesaler's profit per cake is \4$minus a$\$1$ fixed handling cost per hour, so hourly profit is $Y = 4X - 1$ using the $X$ above. Then E(Y) = 4(1.7) - 1 = \5.80$and$\mathrm{Var}(Y) = 4^2 \times 0.81 = 12.96$, illustrating that the constant shifts the mean but not the variance.

Try this

Q1. A variable has $P(X=1)=0.2$, $P(X=2)=0.5$, $P(X=3)=0.3$. Find $E(X)$. [2 marks]

• Cue. $1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1$.

Q2. For the same distribution, find $\mathrm{Var}(X)$. [3 marks]

• Cue. $E(X^2) = 0.2 + 2.0 + 2.7 = 4.9$; $\mathrm{Var}(X) = 4.9 - 2.1^2 = 4.9 - 4.41 = 0.49$.

Q3. Given $E(X) = 6$ and $\mathrm{Var}(X) = 4$, find $E(3X + 2)$ and $\mathrm{Var}(3X + 2)$. [2+2 marks]

• Cue. $E(3X + 2) = 3(6) + 2 = 20$; $\mathrm{Var}(3X + 2) = 9 \times 4 = 36$.