Discrete Random Variables

AS · S1 · 3 min

A discrete random variable takes separate values, each with a probability. Its probability distribution is a table of values and probabilities that add to $1$ . From the table you calculate the expectation (the long-run average) and the variance (the spread), which are the two numbers every later distribution is described by.

Probability distributions

Key result

For a discrete random variable $X$ with values $x_i$ and probabilities $p_i = P(X = x_i)$ :

\sum p_i = 1, \qquad E(X) = \mu = \sum x_i p_i, \qquad \operatorname{Var}(X) = \sigma^2 = \sum x_i^2 p_i - \mu^2.

$E(X)$ is also called the mean of $X$ . The standard deviation is $\sqrt{\operatorname{Var}(X)}$ .

Method

List every possible value of $X$ and find each probability from the situation (a sample space, a tree, or combinations).
Check the probabilities sum to $1$ .
Compute $\sum xp$ for the mean, then $\sum x^2 p$ and subtract $\mu^2$ for the variance.

Distribution from a game

A fair die is thrown. If it shows a $6$ , the player wins $5; if it shows a $4$ or $5$ , the player wins $1; otherwise the player wins nothing. Let $X$ be the winnings. Find the distribution, $E(X)$ and $\operatorname{Var}(X)$ .

Solution

$x$	0	1	5
$P(X = x)$	$\tfrac{1}{2}$	$\tfrac{1}{3}$	$\tfrac{1}{6}$

$E(X) = 0 + \tfrac{1}{3} + \tfrac{5}{6} = \tfrac{7}{6}$ . $\sum x^2 p = 0 + \tfrac{1}{3} + \tfrac{25}{6} = \tfrac{27}{6} = 4.5$ . $\operatorname{Var}(X) = 4.5 - \left(\tfrac{7}{6}\right)^2 = 4.5 - 1.3611 = 3.14$ .

Finding an unknown probability

$x$	1	2	3	4
$P(X = x)$	$0.2$	$k$	$0.3$	$2k$

Find $k$ , $E(X)$ and $\operatorname{Var}(X)$ .

Solution

$0.2 + k + 0.3 + 2k = 1 \Rightarrow k = \tfrac{1}{6}$ .

$E(X) = 0.2 + \tfrac{2}{6} + 0.9 + \tfrac{8}{6} = 0.2 + 0.9 + \tfrac{10}{6} = 2.767$ . $\sum x^2 p = 0.2 + \tfrac{4}{6} + 2.7 + \tfrac{32}{6} = 2.9 + 6 = 8.9$ . $\operatorname{Var}(X) = 8.9 - 2.767^2 = 1.25$ .

Distribution from selection

Two counters are taken without replacement from a bag with 3 red and 2 blue. Let $R$ be the number of red counters taken. Find the distribution of $R$ and $E(R)$ .

Solution

$P(R = 0) = \tfrac{2}{5} \cdot \tfrac{1}{4} = \tfrac{1}{10}$ ; $P(R = 2) = \tfrac{3}{5} \cdot \tfrac{2}{4} = \tfrac{3}{10}$ ; $P(R = 1) = 1 - \tfrac{4}{10} = \tfrac{6}{10}$ .

$E(R) = 0 + \tfrac{6}{10} + \tfrac{6}{10} = 1.2$ .

A probability given as a formula

$P(X = x) = \dfrac{x}{10}$ for $x = 1, 2, 3, 4$ . Show this is a valid distribution and find $E(X)$ and $\operatorname{Var}(X)$ .

Solution

$\tfrac{1 + 2 + 3 + 4}{10} = 1$ ✓.

$E(X) = \tfrac{1 + 4 + 9 + 16}{10} = 3$ . $\sum x^2 p = \tfrac{1 + 8 + 27 + 64}{10} = 10$ . $\operatorname{Var}(X) = 10 - 9 = 1$ .

Interpreting expectation

$E(X)$ is the average value of $X$ over many repetitions, not necessarily a value $X$ can take. A game is fair if the expected winnings equal the cost to play.

A fair game

In the die game above, how much should the player pay per go for the game to be fair?

Solution

$E(X) = \tfrac{7}{6} \approx 1.17$ dollars.

Watch out

Variance is $\sum x^2 p - \mu^2$ , not $\sum (xp)^2$ or $\left(\sum xp\right)^2$ . Square the values, weight by the probabilities, then subtract the square of the mean.

Exam tip

Present the distribution as a table with the probabilities as fractions or exact decimals. Keep $E(X)$ exact when it feeds into $\operatorname{Var}(X)$ ; rounding $\tfrac{7}{6}$ to $1.17$ before squaring changes the variance in the third significant figure.

Practice

Question

$X$ takes values $-1, 0, 2$ with probabilities $0.3, 0.5, 0.2$ . Find $E(X)$ and $\operatorname{Var}(X)$ .
A fair coin is tossed 3 times and $H$ is the number of heads. Tabulate the distribution and find $E(H)$ and $\operatorname{Var}(H)$ .
$P(Y = y) = ky$ for $y = 2, 3, 5$ . Find $k$ and $E(Y)$ .
Two fair dice are thrown and $M$ is the larger score (or the common score). Find $P(M = 4)$ and $E(M)$ .

Answers

$E(X) = 0.1$ ; $\operatorname{Var}(X) = 0.3 + 0.8 - 0.01 = 1.09$ .
Probabilities $\tfrac{1}{8}, \tfrac{3}{8}, \tfrac{3}{8}, \tfrac{1}{8}$ for $0, 1, 2, 3$ ; $E(H) = 1.5$ , $\operatorname{Var}(H) = 3 - 2.25 = 0.75$ .
$k = \tfrac{1}{10}$ ; $E(Y) = \tfrac{4 + 9 + 25}{10} = 3.8$ .
$P(M = m) = \dfrac{2m - 1}{36}$ : $P(M = 4) = \tfrac{7}{36}$ ; $E(M) = \dfrac{1 + 6 + 15 + 28 + 45 + 66}{36} = \tfrac{161}{36} \approx 4.47$ .

Probability distributions

Interpreting expectation

Practice

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