Discrete random variables, expected value and variance (VCE Maths Methods Unit 2)

What this dot point is asking

VCAA wants you to define a discrete random variable, write down its probability distribution, and compute its expected value (mean) and variance.

Discrete random variable

A random variable $X$ is discrete if it takes a countable set of values (often integers).

The probability distribution of $X$ lists each possible value $x_i$ with its probability $p_i = P(X = x_i)$. Requirements:

• $0 \le p_i \le 1$ for all $i$.
• $\sum p_i = 1$.

Expected value (mean)

The expected value is the probability-weighted average. It is the long-run mean of many independent observations of $X$.

For a fair six-sided die: $E[X] = (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5$. Not one of the possible outcomes, but the long-run average.

Variance

Equivalent computational form:

where $E[X^2] = \sum x_i^2 p_i$.

Standard deviation: $\sigma = \sqrt{\text{Var}(X)}$.

Linearity of expected value

For constants $a, b$:

For two independent random variables $X, Y$: $E[X + Y] = E[X] + E[Y]$.

Variance is not linear in the same way: $\text{Var}(aX + b) = a^2 \text{Var}(X)$.

Finding unknown probabilities

A frequent exam task gives a distribution containing one or two unknown probabilities and a piece of information such as the value of $E[X]$. Two equations then pin down the unknowns: the probabilities must sum to $1$, and the expected value (or another given) supplies a second linear relation. Solving the pair simultaneously is routine algebra, but you must remember to apply the "sum to one" condition, which students often omit. Once the distribution is complete, variance and standard deviation follow from the usual formulas.

Functions of a random variable

For constants $a$ and $b$, the expected value is linear: $E[aX + b] = aE[X] + b$, because adding a constant shifts every value and scaling stretches them. Variance behaves differently: $\text{Var}(aX + b) = a^2\,\text{Var}(X)$, since adding a constant does not change spread (so $b$ disappears) while scaling by $a$ stretches deviations by $a$ and hence variance by $a^2$. The standard deviation scales by $|a|$. These transformation rules are examinable and let you read off the mean and spread of a rescaled variable without recomputing the whole distribution.

In one sentence

A discrete random variable has a probability distribution listing each value $x_i$ with probability $p_i$ (summing to $1$); the expected value is $E[X] = \sum x_i p_i$ (the probability-weighted mean) and the variance is $\text{Var}(X) = E[X^2] - (E[X])^2$ (with standard deviation $\sqrt{\text{Var}(X)}$).

Examples in context

Example 1. Expected payout on a carnival game. A stall charges $\$2$ to spin a wheel. The wheel pays $\$5$ with probability $0.2$, $\$1$ with probability $0.3$, and $\$0$ with probability $0.5$. Let $X$ be the payout. Then $E[X] = 5(0.2) + 1(0.3) + 0(0.5) = 1.0 + 0.3 = \$1.30$. Since the player pays $\$2$ but expects only $\$1.30$ back, the expected loss per play is $\$0.70$, the stall's average profit.

Example 2. Insurance expected cost. An insurer covers a $\$2000$ claim that occurs with probability $0.03$ in a year; otherwise the cost is $\$0$. Let $C$ be the claim cost. Then $E[C] = 2000(0.03) + 0(0.97) = \$60$. To break even the insurer must charge at least $\$60$ in premium (before expenses), the expected value of the random cost.

Try this

Q1. A variable $X$ has $P(X = 0) = 0.5$, $P(X = 1) = 0.3$, $P(X = 2) = 0.2$. Find $E[X]$. [2 marks]

• Cue. $0(0.5) + 1(0.3) + 2(0.2) = 0.7$.

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

• Cue. $E[X^2] = 0 + 1(0.3) + 4(0.2) = 1.1$; $\text{Var}(X) = 1.1 - 0.7^2 = 1.1 - 0.49 = 0.61$.

Q3. A game gives net winnings $X$: $+\$8$ with probability $0.25$, $-\$3$ with probability $0.75$. (a) Find $E[X]$. (b) Is the game favourable to the player? [2+1 marks]

• Cue. (a) $8(0.25) - 3(0.75) = 2 - 2.25 = -\$0.25$. (b) No; negative expected value.