Define a discrete random variable and its probability distribution, and compute expected value (mean) and variance for given distributions

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:

• IMATH_8 for all $i$.
• IMATH_10 .

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: $\text{Var}(aX + b) = a^2 \text{Var}(X)$.

Worked example (lottery)

A lottery ticket costs \10$. Probability of winning the prize of$\$1000$ is $0.005$. Net profit $X$: gain \990$with probability$0.005$, lose$\$10$ with probability $0.995$.

$E[X] = 0.005 \cdot 990 + 0.995 \cdot (-10) = 4.95 - 9.95 = -5.00$.

Expected loss of \5$ per ticket. Hence the term "negative-expected-value game".

Common traps

Confusing $E[X^2]$ with $(E[X])^2$. They differ by the variance.

Forgetting probabilities must sum to $1$. Check before computing.

Treating $E[X]$ as a typical outcome. It is the long-run average, not necessarily one of the observed values.

Negative variance. Variance is always non-negative; a negative result indicates a calculation error.

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 $\sqrt{\text{Var}(X)}$).