Random Variables

Lessons

Senior 6 · Mathematics

Random Variables

5 lessons

Lesson 1 of 5

Discrete random variables and their distributions

35 min

Outcomes that are numbers

In S5 you found probabilities of events: "a head", "a total of 7", "a defective bulb". Often what matters is a number attached to the outcome: how many seeds in a packet germinate, how many customers reach a mobile-money agent in an hour, how long a phone battery lasts. A quantity like this, whose value depends on chance, is a random variable. Its distribution says which values it can take and how likely each is. Insurance premiums, quality control in factories and the planning of hospital beds all rest on distributions.

Definition

A random variable $X$ is a function that assigns a real number to each outcome of a random experiment. Capital letters ( $X, Y$ ) name random variables; small letters ( $x, y$ ) name their values.

$X$ is discrete if its values can be listed, $x_1, x_2, x_3, \dots$ (finitely or infinitely many), such as a count.
$X$ is continuous if it can take every value in an interval, such as a time, a mass or a length (lessons 4 and 5).

Example: toss three coins and let $X$ be the number of heads. The outcome HTH gives $X = 2$ ; the possible values are $0, 1, 2, 3$ .

Definition

The probability mass function (pmf) of a discrete random variable $X$ is $p (x) = P (X = x)$ . It satisfies

p(x_i) \ge 0 \text{ for every } i , \qquad \sum_i p(x_i) = 1 .

The cumulative distribution function (cdf) of any random variable $X$ is defined for every real $x$ by

F(x) = P(X \le x) .

For a discrete $X$ , $F(x) = \sum_{x_i \le x} p(x_i)$ : a staircase that jumps by $p (x_{i})$ at each value $x_{i}$ .

Worked example 1: three coins

Three fair coins are tossed and $X$ is the number of heads. Find the pmf and the cdf of $X$ , and $P(1 \le X \le 2)$ .

Solution. The 8 outcomes are equally likely. $X = 0$ only for TTT; $X = 1$ for HTT, THT, TTH; $X = 2$ for HHT, HTH, THH; $X = 3$ only for HHH.

$x$	0	1	2	3
$p (x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$
$F (x)$	$\frac{1}{8}$	$\frac{4}{8} = \frac{1}{2}$	$\frac{7}{8}$	1

In full, $F (x) = 0$ for $x \lt 0$ , $\frac{1}{8}$ for $0 \le x \lt 1$ , $\frac{1}{2}$ for $1 \le x \lt 2$ , $\frac{7}{8}$ for $2 \le x \lt 3$ , and 1 for $x \ge 3$ . Finally $P(1 \le X \le 2) = \frac{3}{8} + \frac{3}{8} = \frac{3}{4}$ .

The number of heads in three tosses of a fair coin: its probability mass function (left) and its cumulative distribution function (right). A filled dot marks the value of

F

at each jump.

Important

Properties of a cdf

$0 \le F(x) \le 1$ ; $F$ is non-decreasing; $F(x) \to 0$ as $x \to -\infty$ and $F(x) \to 1$ as $x \to +\infty$ .
$P(a \lt X \le b) = F(b) - F(a)$ for $a \lt b$ .
For a discrete $X$ with values $x_1 \lt x_2 \lt \cdots$ : $p(x_k) = F(x_k) - F(x_{k-1})$ , the height of the jump at $x_{k}$ .

Reason for the second: the events $X \le a$ and $a \lt X \le b$ are mutually exclusive and together make $X \le b$ , so $F(b) = F(a) + P(a \lt X \le b)$ .

Worked example 2: finding a constant

A random variable has $P (X = x) = k x$ for $x = 1, 2, 3, 4$ . Find $k$ , $P(X \ge 3)$ and $F (2.5)$ .

Solution. The probabilities must add up to 1: $k (1 + 2 + 3 + 4) = 10 k = 1$ , so $k = \frac{1}{10}$ . Then $P(X \ge 3) = \frac{3}{10} + \frac{4}{10} = \frac{7}{10}$ , and $F(2.5) = P(X \le 2.5) = P(X = 1) + P(X = 2) = \frac{3}{10}$ .

Worked example 3: from the cdf to the pmf

Two fair dice are rolled and $X$ is the larger of the two scores (the common score if they are equal). Find $F (x)$ for $x = 1, \dots, 6$ , and hence the pmf.

Solution. $X \le x$ means both dice show at most $x$ . The dice are independent, so $F(x) = \frac{x}{6} \times \frac{x}{6} = \frac{x^2}{36}$ . Then

P(X = x) = F(x) - F(x - 1) = \frac{x^2 - (x - 1)^2}{36} = \frac{2x - 1}{36} , \quad x = 1, \dots, 6 .

Check: $\frac{1 + 3 + 5 + 7 + 9 + 11}{36} = \frac{36}{36} = 1$ . The largest score is most likely to be 6, with probability $\frac{11}{36}$ .

Warning

$F$ is defined between the values too. In worked example 1, $F(1.7) = F(1) = \frac{1}{2}$ , not "undefined" and not 0.

Watch $< \lt$ and $\leq \le$ for discrete variables. $P(X \lt 2) = F(1)$ but $P(X \le 2) = F(2)$ ; they differ by $P (X = 2)$ . In worked example 1 they are $\frac{1}{2}$ and $\frac{7}{8}$ .

Tip

Before any other work, check that the table of a pmf adds up to exactly 1. Most errors in this topic, a missing outcome or a counting slip, show up at once in that sum.

Khan Academy: Random variables (about 10 min). Watch how a random variable maps each outcome to a number, and why a count is discrete while a measurement such as a mass is continuous.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks you to explain.

Classify each random variable as discrete or continuous: (a) the number of pupils absent today; (b) the time taken to run 100 m; (c) the number of tosses needed to get the first head.
A fair die is rolled, and $X = 1$ if the score is a multiple of 3 and $X = 0$ otherwise. Give the pmf of $X$ .
$P(X = x) = \frac{x^2}{c}$ for $x = 1, 2, 3$ . Find $c$ and $P(X \gt 1)$ .
Two fair dice are rolled and $S$ is the sum of the scores. Find $P (S = 7)$ , $P(S \le 4)$ and $F (3.5)$ .
A discrete $X$ has cdf $F (0) = 0.2$ , $F (1) = 0.5$ , $F (2) = 0.9$ , $F (3) = 1$ , and takes only the values 0, 1, 2, 3. Find its pmf and $P(0 \lt X \le 2)$ .
A bag holds 3 red and 2 blue balls. Two are drawn without replacement and $R$ is the number of red balls drawn. Find the pmf of $R$ .
A table claims that a random variable has $F (2) = 0.6$ and $F (3) = 0.5$ . Explain why this cannot be a cumulative distribution function.

Show answersHide answers

(a) discrete (b) continuous (c) discrete (values $1, 2, 3, \dots$ )
$P(X = 1) = \frac{2}{6} = \frac{1}{3}$ , $P(X = 0) = \frac{2}{3}$ .
$\frac{1 + 4 + 9}{c} = 1$ , $c = 14$ ; $P(X \gt 1) = \frac{4 + 9}{14} = \frac{13}{14}$ .
$P(S = 7) = \frac{6}{36} = \frac{1}{6}$ ; $P(S \le 4) = \frac{1 + 2 + 3}{36} = \frac{1}{6}$ ; $F(3.5) = P(S \le 3) = \frac{3}{36} = \frac{1}{12}$ .
$p (0) = 0.2$ , $p (1) = 0.3$ , $p (2) = 0.4$ , $p (3) = 0.1$ ; $P(0 \lt X \le 2) = F(2) - F(0) = 0.7$ .
$P(R = 0) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$ , $P(R = 1) = 2 \times \frac{3}{5} \times \frac{2}{4} = \frac{3}{5}$ , $P(R = 2) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ .
A cdf never decreases: $F(3) - F(2) = P(2 \lt X \le 3)$ , which would be $- 0.1$ , and no probability is negative.

Sources and credits

Written by Ganzaa: the explanations, the diagram (drawn from the computed probabilities), the worked examples, the practice items and the quiz questions. Every distribution was checked to sum to 1.
Video: Random variables, Khan Academy, embedded from YouTube.

1Outcomes that are numbers

Outcomes that are numbers