OxfordAQA International AS and A-level Mathematics: Discrete random variables: expectation and variance (9660) – Revision Notes

These revision notes condense section S1.2 (Discrete random variables) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. This content is part of International AS, Unit PSM1 (section 3.2, S1). For full explanations and longer worked examples, read the study guide. Then use the practice questions, the course hub, the printable checklist and the free 10-minute diagnostics.

Key definitions

• A discrete random variable X takes a finite set of values x, each with probability P(X = x).
• A valid distribution has 0 ≤ P(X = x) ≤ 1 and Σ P(X = x) = 1.
• The mean E(X) = μ is the probability-weighted average of the values.
• The variance Var(X) = σ² measures spread about the mean. The standard deviation is σ = √Var(X).
• Independent variables: knowing the value of one tells you nothing about the other.

Formulae to learn

The specification states that knowledge of these is expected.

Result	Formula
Mean	E(X) = μ = Σ x p
Mean of a function	E(g(X)) = Σ g(x) p
Variance	Var(X) = E((X − μ)²) = E(X²) − μ²
Second moment	E(X²) = Σ x² p
Linear function	E(aX + b) = a E(X) + b
Linear function	Var(aX + b) = a² Var(X)
Two independent variables	E(aX ± bY) = a E(X) ± b E(Y)
Two independent variables	Var(aX ± bY) = a² Var(X) + b² Var(Y)
Sums	E(ΣXᵢ) = ΣE(Xᵢ) and Var(ΣXᵢ) = ΣVar(Xᵢ) for independent Xᵢ

Examples of functions named in the specification: E(2X + 3), E(5X²), E(10X⁻¹), E(100X⁻²), Var(3X), Var(4X − 5), Var(6X⁻¹).

Method boxes

Mean and variance from a table

1. Check or find the unknown so that Σ p = 1.
2. Add a row for x p and a row for x² p.
3. E(X) = Σ x p and E(X²) = Σ x² p.
4. Var(X) = E(X²) − (E(X))². Check it is not negative.
5. Square-root for the standard deviation.

E(g(X)) for a non-linear g

1. Apply g to each x in the table, keeping the same probabilities.
2. Multiply each g(x) by its probability and add.
3. Never put E(X) into g.

Var(g(X)) for a non-linear g

1. Find E(g(X)) as above.
2. Find E(g(X)²) as above.
3. Var(g(X)) = E(g(X)²) − (E(g(X)))².

Combining independent variables

1. Write down E and Var of each variable.
2. Means follow the signs and coefficients. Variances use the squared coefficients and are always added.
3. Take the square root last if a standard deviation is wanted.

Worked reminder

W takes the values 10, 20, 30 with probabilities 0.2, 0.5, 0.3.

E(W)   = 2 + 10 + 9 = 21
E(W²)  = 100(0.2) + 400(0.5) + 900(0.3) = 20 + 200 + 270 = 490
Var(W) = 490 − 21² = 490 − 441 = 49, so σ = 7
E(W/10) = 2.1, Var(W/10) = 49/100 = 0.49

Distributions given by a function

Write the table out first. For P(X = x) = kx for x = 1, 2, 3, the probabilities are k, 2k, 3k, so 6k = 1 and k = 1/6. Then E(X) = (1 + 4 + 9)/6 = 7/3 and E(X²) = (1 + 8 + 27)/6 = 6, so Var(X) = 6 − 49/9 = 5/9. The same steps work for any rule on a finite list of values.

Worked reminder: combining variables

A and B are independent. E(A) = 8, Var(A) = 2.5, E(B) = 5 and Var(B) = 1.5.

E(2A − B)    = 16 − 5 = 11
Var(2A − B)  = 4(2.5) + 1(1.5) = 11.5
E(A + B)     = 13
Var(A + B)   = 2.5 + 1.5 = 4, so sd = 2

Now let T be the total of 6 independent copies of A.

E(T)   = 6 × 8   = 48
Var(T) = 6 × 2.5 = 15

For comparison, 6A has mean 48 and variance 36 × 2.5 = 90.

Choosing the right result

The question says	Use
“each value is increased by 4” or “converted to new units”	E(aX + b), Var(aX + b)
“the average of the reciprocal”, “the mean of the square”	E(g(X)) = Σ g(x) p
“the variance of the reciprocal”	E(g²) − (E(g))²
“X and Y are independent” and a combination is asked	E(aX ± bY), Var(aX ± bY)
“n independent plays/readings, total”	nE(X) and n Var(X)
“the same reading multiplied by n”	nE(X) and n² Var(X)

Must-know distinctions

• E(X²) is Σ x² p. It is not (E(X))². The gap between them is the variance.
• Var(aX + b) ignores b and squares a. Var(−X) = Var(X).
• Var(X − Y) = Var(X) + Var(Y) for independent X and Y. The variance never subtracts.
• nX is one observation scaled: Var(nX) = n² Var(X). X₁ + … + Xₙ is n independent observations added: variance n Var(X). The means are both nE(X).
• E(10/X) is not 10/E(X). Work from the table.
• The mean formula for aX ± bY holds always. The variance formula needs independence.
• Variance is in squared units. Standard deviation is in the original units.

Quick self-test

1. X takes the values 0, 1, 2 with probabilities 0.5, k, 0.2. Find k.
2. Find E(X) for the variable in question 1.
3. Find Var(X) for the variable in question 1.
4. E(X) = 5 and Var(X) = 4. Find E(3X − 1) and Var(3X − 1).
5. Var(X) = 4. Find Var(−2X).
6. Y takes the values 1, 2, 4 with probabilities 0.5, 0.25, 0.25. Find E(8/Y) and compare it with 8/E(Y).
7. For Y in question 6, find E(3Y²).
8. X and Y are independent with E(X) = 6, E(Y) = 2, Var(X) = 3 and Var(Y) = 5. Find E(X − Y) and Var(X − Y).
9. For X and Y in question 8, find Var(2X + 3Y).
10. Each of 10 independent readings has mean 2.5 and variance 0.5. Find the mean and variance of the total of the 10 readings, and the variance of 10 times a single reading.
11. P(X = x) = k(x + 1) for x = 0, 1, 2. Find k and E(X).
12. For a random variable X, E(X²) = 13 and E(X) = 3. Find Var(X), the standard deviation, and Var(5 − X).

Answers

1. 0.5 + k + 0.2 = 1, so k = 0.3.
2. 0(0.5) + 1(0.3) + 2(0.2) = 0.7.
3. E(X²) = 0.3 + 4(0.2) = 1.1, so Var(X) = 1.1 − 0.7² = 0.61.
4. E = 3(5) − 1 = 14; Var = 9 × 4 = 36.
5. (−2)² × 4 = 16.
6. E(Y) = 0.5 + 0.5 + 1 = 2, so 8/E(Y) = 4. The values of 8/Y are 8, 4 and 2, so E(8/Y) = 8(0.5) + 4(0.25) + 2(0.25) = 4 + 1 + 0.5. E(8/Y) = 5.5, which differs from 8/E(Y) = 4.
7. E(Y²) = 0.5 + 1 + 4 = 5.5, so E(3Y²) = 16.5.
8. E(X − Y) = 6 − 2 = 4; Var(X − Y) = 3 + 5 = 8.
9. 4(3) + 9(5) = 12 + 45 = 57.
10. Mean 10 × 2.5 = 25; variance 10 × 0.5 = 5; variance of 10 times one reading = 100 × 0.5 = 50.
11. The probabilities are k, 2k, 3k, so 6k = 1 and k = 1/6. E(X) = (0 + 2 + 6)/6 = 4/3.
12. Var(X) = 13 − 9 = 4; σ = 2; Var(5 − X) = (−1)² × 4 = 4.

How to present working

Examiners can only reward what they can see, so set the working out in the order of the method boxes.

• Draw the table with extra rows for x p and x² p. Errors are easy to spot in a table and hard to spot in a long line of arithmetic.
• Write the formula you are using before substituting. A correct formula with an arithmetic slip is easier to follow than a bare number.
• Show E(X²) separately before subtracting μ². State the answer as a fraction or exact decimal where you can.
• For combinations, list E and Var of each variable first, then substitute into the formula with the coefficients visible.
• Finish with a sentence naming what you found, such as “the variance of the total is 15”.

Where marks are usually lost

• Squaring E(X) instead of summing x² p, giving a wrong or negative variance.
• Writing Var(aX + b) = a Var(X) + b, or forgetting to square a.
• Subtracting variances for a difference of independent variables.
• Replacing E(10/X) by 10/E(X), or E(5X²) by 5(E(X))².
• Confusing 2X with X₁ + X₂ when the question describes separate observations.
• Rounding E(X) early and carrying an inexact value into the variance. Keep fractions or full decimals until the end.
• Giving the variance when the standard deviation is asked for, or the reverse.
• Not stating independence when you add variances; say it in your working.
• Not checking that probabilities add to 1 before finding E(X).

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section S1.2 Discrete random variables. Published by OxfordAQA; International AS exams from May/June 2018 and International A-level exams from May/June 2019.