Define a discrete random variable by its probability distribution, and calculate the expected value, variance and standard deviation

What this dot point is asking

NESA wants you to recognise a discrete random variable, check that its probability distribution is valid, compute the expected value and variance from the distribution, and apply the linear transformation rules to $a X + b$.

The answer

Discrete random variables and their distributions

A discrete random variable $X$ takes a countable list of values $x_1, x_2, \dots, x_n$ with probabilities $p_i = P(X = x_i)$. The list of values with their probabilities is the probability distribution of $X$. For it to be valid:

• IMATH_11 for every $i$,
• IMATH_13 .

The probability that $X$ falls in some set is the sum of $p_i$ for the values in that set. For example, $P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$ if $X$ takes integer values from $0$.

Expected value

The expected value (or mean) of $X$ is the long-run average value if we repeated the experiment many times. It is the weighted sum

The expected value need not be one of the values $X$ can actually take.

Expected value of a function of IMATH_21

For any function $g$,

The most common case is $g(x) = x^2$, which gives

Variance and standard deviation

The variance of $X$ measures spread around the mean. It is

which is algebraically equivalent (and usually easier to compute) as

The standard deviation is $\sigma = \sqrt{\text{Var}(X)}$, with the same units as $X$.

Linear transformations

If $Y = a X + b$ for constants $a$ and $b$,

Shifting $X$ by $b$ shifts the mean but not the spread. Scaling by $a$ multiplies the mean by $a$ and the standard deviation by $|a|$.

Worked examples

Checking a distribution and computing the mean

$X$ takes values $1, 2, 3, 4$ with $P(X = x) = c x$ for some constant $c$. Find $c$, then $E(X)$.

Sum of probabilities: $c(1 + 2 + 3 + 4) = 10 c = 1$, so $c = 0.1$.

$E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3$.

Variance via IMATH_44

With the same $X$, $E(X^2) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10$.

$\text{Var}(X) = 10 - 3^2 = 1$, so $\sigma = 1$.

A fair die

$X$ is the number rolled on a fair six-sided die. $E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$.

$E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6}$.

$\text{Var}(X) = \frac{91}{6} - 3.5^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12} \approx 2.92$.

Linear transformation

If $X$ has $\mu = 3.5$ and $\sigma^2 = \frac{35}{12}$, then $Y = 2 X + 1$ has $E(Y) = 2(3.5) + 1 = 8$ and $\text{Var}(Y) = 4 \cdot \frac{35}{12} = \frac{35}{3}$.

Common traps

Forgetting to check that probabilities sum to $1$. If a question gives a distribution in terms of a constant, solve $\sum p_i = 1$ first.

Using $\mu^2$ instead of $E(X^2)$. The formula is $\text{Var}(X) = E(X^2) - [E(X)]^2$, not $E(X^2) - E(X)$.

Squaring inside but not outside. $E(X)^2 = \mu^2$ is the square of a single number. $E(X^2) = \sum x^2 p$ is the weighted sum of squares. They are different.

Linear transformation on variance. $\text{Var}(a X + b) = a^2 \text{Var}(X)$, not $a \text{Var}(X)$, and the $+ b$ has no effect on variance.

Negative variance. If you get a negative variance, you have a calculation error. Variance is always non-negative.

In one sentence

A discrete random variable is summarised by its probability distribution; its mean is $E(X) = \sum x_i p_i$, its variance is $E(X^2) - \mu^2$, and linear transformations $a X + b$ scale the mean by $a$, shift it by $b$, and scale the variance by $a^2$.