Discrete random variables

CIE A-Level Mathematics· 9709 S1 Section 4: Discrete random variables· 15 min read

1. Definition of Discrete Random Variables★☆☆☆☆⏱ 4 min

A random variable is a quantity whose value depends on the outcome of a random, uncertain event. Discrete random variables take only distinct, separate values (most commonly whole numbers), unlike continuous random variables which can take any value in an interval.

📘 Definition

Discrete Random Variable

X

A random variable that can only take a countable number of distinct values, where each possible value has a non-zero probability of occurring.

Example:

Number of heads in 5 coin tosses, number of defective items in a batch, score on a die roll

📐 Worked Example

Identify which of the following are discrete random variables: (a) Number of customers arriving at a shop in 1 hour, (b) Weight of a randomly selected apple, (c) Number of correct answers on a 10-question multiple choice test.

1
Check if the variable takes countable distinct values:
2
(a) Number of customers can only be 0, 1, 2, ... (whole numbers), so it is countable. Conclusion: discrete.
3
(b) Weight can take any value in a range of possible weights, it is uncountable. Conclusion: not discrete.
4
(c) Number of correct answers is 0 to 10, distinct whole numbers, so countable. Conclusion: discrete.
5
Final answer: (a) and (c) are discrete random variables.

2. Probability Distributions★★☆☆☆⏱ 5 min

The probability distribution of a discrete random variable (X) specifies the probability (P(X=x)) for every possible value (x) that (X) can take. All valid probability distributions follow two core rules.

📐 Worked Example

A discrete random variable (X) has the probability distribution below. Find the value of the unknown constant (k):

x	1	2	3	4
P(X=x)	k	2k	0.3	0.1

1
Use the rule that the sum of all probabilities equals 1, to set up an equation:
2
$k + 2k + 0.3 + 0.1 = 1$
3
Simplify and solve for k:
4
$3k + 0.4 = 1 \rightarrow 3k = 0.6 \rightarrow k = 0.2$
5
Check all probabilities are between 0 and 1: 0.2, 0.4, 0.3, 0.1, all satisfy the condition. Final answer: (k=0.2)

3. Expectation and Variance★★★☆☆⏱ 6 min

The expected value (or population mean) of (X) is the long-run average value we would expect to get if we repeated the random experiment many times. Variance measures how spread out the values of (X) are around the expected value.

📘 Definition

Expected Value

E(X) or \(\mu\)

For a discrete random variable (X), the expected value is calculated as: (E(X) = \sum x P(X=x)), where the sum is over all possible values of (x).

📘 Definition

Variance

Var(X) or \(\sigma^2\)

Variance is calculated using the identity: (Var(X) = E(X^2) - [E(X)]^2), where (E(X^2) = \sum x^2 P(X=x)).

📐 Worked Example

Using the probability distribution from the previous example (x: 1,2,3,4; P(X=x): 0.2, 0.4, 0.3, 0.1), calculate (E(X)) and (Var(X)).

1
First calculate (E(X)) by multiplying each (x) by (P(X=x)) and summing:
2
$E(X) = (1 \times 0.2) + (2 \times 0.4) + (3 \times 0.3) + (4 \times 0.1) = 0.2 + 0.8 + 0.9 + 0.4 = 2.3$
3
Next calculate (E(X^2)) by multiplying each (x^2) by (P(X=x)) and summing:
4
$E(X^2) = (1^2 \times 0.2) + (2^2 \times 0.4) + (3^2 \times 0.3) + (4^2 \times 0.1) = 0.2 + 1.6 + 2.7 + 1.6 = 6.1$
5
Use the variance formula to get the final result:
6
$Var(X) = E(X^2) - [E(X)]^2 = 6.1 - (2.3)^2 = 6.1 - 5.29 = 0.81$

4. Linear Transformations of Discrete Random Variables★★★★☆⏱ 4 min

If we transform a discrete random variable (X) to create a new random variable (Y = aX + b) (where (a) and (b) are constants), we do not need to rebuild the full probability distribution to find (E(Y)) and (Var(Y)). We can use simple pre-derived rules.

Expectation rule: (E(aX + b) = aE(X) + b)
Variance rule: (Var(aX + b) = a^2 Var(X))

📐 Worked Example

Given (E(X) = 2.3) and (Var(X) = 0.81), find (E(3X - 2)) and (Var(3X - 2)).

1
Apply the expectation rule with (a=3) and (b=-2):
2
$E(3X - 2) = 3E(X) - 2 = 3(2.3) - 2 = 6.9 - 2 = 4.9$
3
Apply the variance rule (the constant -2 does not affect variance):
4
$Var(3X - 2) = 3^2 Var(X) = 9 \times 0.81 = 7.29$

5. Common Pitfalls

Wrong move:

Writing (Var(aX + b) = aVar(X)), forgetting to square the coefficient (a)

Why:

Variance is a measure of spread in squared units, so scaling (X) by (a) scales variance by (a^2)

Correct move:

Always square the coefficient of (X) when calculating variance of a linear transformation

Wrong move:

Writing variance as (Var(X) = [E(X)]^2 - E(X^2))

Why:

Simple sign error that almost always results in a negative variance, which is impossible

Correct move:

Remember variance is always non-negative, so use: (Var(X) = E(X^2) - [E(X)]^2)

Wrong move:

Calculating (E(X)) as (sum of x values) / (number of values), ignoring unequal probabilities

Why:

Confuses uniform distributions (where all probabilities are equal) with general discrete distributions

Correct move:

Always multiply each (x) by (P(X=x)) before summing, regardless of whether probabilities are equal or not

Wrong move:

Calling a rounded continuous variable (e.g. height to nearest cm) discrete

Why:

Discreteness depends on the underlying variable's possible values, not how it is measured

Correct move:

If the underlying variable can take any value in a range, it is continuous, even when rounded to whole units

Wrong move:

Forgetting to check that all probabilities are between 0 and 1 after solving for an unknown constant (k)

Why:

Sometimes multiple solutions for (k) exist, and one may give an invalid negative probability or probability greater than 1

Correct move:

Always verify that the value of (k) you found results in all probabilities between 0 and 1

6. Quick Reference Cheatsheet

Concept	Formula/Rule
Discrete Random Variable	Countable distinct values, each with non-zero probability
Probability Rule	$\sum P (X = x) = 1$ , $0 \leq P (X = x) \leq 1$
Expected Value	$E (X) = \sum x P (X = x)$
Expected Value of $X^{2}$	$E (X^{2}) = \sum x^{2} P (X = x)$
Variance	$V a r (X) = E (X^{2}) - [E (X)]^{2}$
Linear Transformation Expectation	$E (a X + b) = a E (X) + b$
Linear Transformation Variance	$V a r (a X + b) = a^{2} V a r (X)$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 5
Find expectation and variance
2022 · 5
Construct probability distribution
2021 · 5
Linear transformation of DRV

What's Next

Discrete random variables form the foundation for all specific discrete probability distributions heavily tested in CIE A-Level Statistics, including the binomial and geometric distributions you will study next. The general expectation and variance rules you learned here apply directly to these specific distributions, helping you derive and remember their standard formulas. This topic also builds the core concepts you will use for continuous random variables in later units, and underpins topics like sampling and hypothesis testing further into your A-Level studies. Mastery of the rules in this module will make all future statistics topics much easier to learn.

→Normal distribution
→Hypothesis Testing