Discrete random variables and distributions

What you’ll learn

• Construct and check a discrete probability distribution.
• Calculate expectation, variance and cumulative probabilities.
• Use transformations and the discrete uniform model.

Understand the idea

A random variable assigns a numerical value to each outcome. A discrete distribution lists its possible values and their probabilities; every probability must be non-negative and the total must be one. The cumulative distribution function $F(x)=P(X\leq x)$ is a step function, including each mass at its jump.

Expectation is a probability-weighted average and need not be a possible observed value. Calculate variance from $E(X^2)-[E(X)]^2$, taking care to square the possible values inside the first expectation. For $aX+b$, the shift affects the mean but not the variance.

A discrete uniform model assigns equal probability to a finite list of outcomes. It is distinct from the continuous uniform distribution studied in S2: a particular discrete value can have positive probability, whereas a single value of a continuous variable has probability zero.

Key formulae

$$E(X)=\sum_x xp(x)$$

$$\operatorname{Var}(X)=\sum_x x^2p(x)-[E(X)]^2$$

$$E(aX+b)=aE(X)+b$$

$$\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)$$

Worked example

$X$ takes values 0, 1 and 2 with probabilities 0.2, 0.5 and 0.3. Find its mean and variance.

1. Check that $0.2+0.5+0.3=1$.
2. $E(X)=0(0.2)+1(0.5)+2(0.3)=1.1$.
3. $E(X^2)=0^2(0.2)+1^2(0.5)+2^2(0.3)=1.7$.
4. The variance is $1.7-1.1^2=0.49$.

Answer: $E(X)=1.1$ and $\operatorname{Var}(X)=0.49$.