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

This study guide covers 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). It teaches the mean, variance and standard deviation of a discrete random variable, of a function of it, and of sums and differences of independent variables.

Related pages: the revision notes, the practice questions, the course hub, the printable checklist and the free 10-minute diagnostics. For the binomial distribution, see Bernoulli and binomial distributions.

What this unit covers

S1.2 content	What you must be able to do
Discrete random variables and distributions	Work with a finite distribution given as a table or a simple function
Mean, variance, standard deviation	Know and use E(X) = Σ x p, E(g(X)) = Σ g(x) p and Var(X) = E(X²) − μ²
Simple functions of X	Find E(2X + 3), E(5X²), E(10X⁻¹), E(100X⁻²), Var(3X), Var(4X − 5), Var(6X⁻¹)
Two independent variables	Know and use E(aX ± bY) and Var(aX ± bY)
Sums of independent variables	Know and use E(ΣXᵢ) = ΣE(Xᵢ) and Var(ΣXᵢ) = ΣVar(Xᵢ)

The specification says the number of outcomes is finite and that distributions are given or easily determined as a table or simple function. It states that knowledge of the formulae above is expected, so learn them.

1. Discrete random variables and their distributions

A discrete random variable X takes a finite list of values x with probabilities p = P(X = x). The distribution is a table:

x	x₁	x₂	…	xₙ
P(X = x)	p₁	p₂	…	pₙ

Two facts do most of the work:

• every probability is at least 0 and at most 1;
• the total of all the probabilities is 1, so Σ p = 1.

The second fact finds a missing probability or an unknown constant.

Worked example 1. The table gives the distribution of X. Find k.

x	1	2	3	4	5
P(X = x)	0.10	0.30	0.25	k	0.15

0.10 + 0.30 + 0.25 + k + 0.15 = 1
0.80 + k = 1
k = 0.20

We reuse this distribution below, so keep it in view.

2. Mean, variance and standard deviation

The mean (expected value) is the average of X over many repetitions:

E(X) = μ = Σ x p

Each value is weighted by its probability, so the mean need not be a possible value of X. The mean is the measure of central tendency used here; the mode is the most likely value, and the standard deviation and variance are the measures of spread. The variance measures spread about the mean:

Var(X) = σ² = E((X − μ)²) = E(X²) − μ²
E(X²) = Σ x² p

The standard deviation is σ = √Var(X). Variance is never negative, so a negative answer means an arithmetic slip. Use the second form, E(X²) − μ², for tables.

Worked example 2. Find E(X), Var(X) and the standard deviation for the distribution in example 1 (k = 0.20).

E(X)  = 1(0.10) + 2(0.30) + 3(0.25) + 4(0.20) + 5(0.15)
      = 0.10 + 0.60 + 0.75 + 0.80 + 0.75 = 3
E(X²) = 1(0.10) + 4(0.30) + 9(0.25) + 16(0.20) + 25(0.15)
      = 0.10 + 1.20 + 2.25 + 3.20 + 3.75 = 10.5
Var(X) = 10.5 − 3² = 1.5
σ = √1.5 = 1.22 (3 s.f.)

3. Linear functions: E(aX + b) and Var(aX + b)

If you change every value of X to aX + b, the specification expects you to know:

E(aX + b) = a E(X) + b
Var(aX + b) = a² Var(X)

The mean moves and scales like the values. The spread is only scaled, and by a² because variance is a squared quantity. Adding b shifts every value equally, so the spread does not change. A negative a gives a positive variance, since a² is positive. For example, Var(−3X) = 9 Var(X).

Worked example 3. Using example 2, find E(2X + 3), Var(2X + 3) and Var(4X − 5).

E(2X + 3)   = 2(3) + 3 = 9
Var(2X + 3) = 2² × 1.5 = 6
Var(4X − 5) = 4² × 1.5 = 24

4. E(g(X)) for any function of X

For a function that is not linear, go back to the table and use

E(g(X)) = Σ g(x) p

You apply g to each value of x and keep the probability unchanged. The mean of g(X) is not g applied to the mean. In particular E(10X⁻¹) is not 10/E(X), and E(5X²) is not 5(E(X))².

Worked example 4. The variable W has the following distribution. Find E(W), E(10/W), E(100/W²) and compare E(10/W) with 10/E(W).

w	2	4	5	10
P(W = w)	0.4	0.3	0.2	0.1

E(W)       = 0.8 + 1.2 + 1.0 + 1.0 = 4
E(10/W)    = 10/2(0.4) + 10/4(0.3) + 10/5(0.2) + 10/10(0.1)
           = 2 + 0.75 + 0.4 + 0.1 = 3.25
E(100/W²)  = 25(0.4) + 6.25(0.3) + 4(0.2) + 1(0.1)
           = 10 + 1.875 + 0.8 + 0.1 = 12.775
10/E(W)    = 10/4 = 2.5

So E(10/W) = 3.25, which is not equal to 2.5. For a reciprocal, always work from the table.

5. Variance of a non-linear function

For Var(6X⁻¹) and similar, the linear shortcut does not apply. Use the definition with g(X) in place of X:

Var(g(X)) = E(g(X)²) − (E(g(X)))²

Worked example 5. For W in example 4, find Var(6/W).

E(6/W)    = 3(0.4) + 1.5(0.3) + 1.2(0.2) + 0.6(0.1)
          = 1.2 + 0.45 + 0.24 + 0.06 = 1.95
E(36/W²)  = 9(0.4) + 2.25(0.3) + 1.44(0.2) + 0.36(0.1)
          = 3.6 + 0.675 + 0.288 + 0.036 = 4.599
Var(6/W)  = 4.599 − 1.95² = 4.599 − 3.8025 = 0.7965

6. Distributions given by a simple function

Sometimes P(X = x) is a formula. Write out the table first, then proceed as before.

Worked example 6. P(V = v) = k(6 − v) for v = 1, 2, 3, 4, 5. Find k, E(V) and Var(V).

Probabilities: 5k, 4k, 3k, 2k, k, so 15k = 1 and k = 1/15
E(V)  = (1×5 + 2×4 + 3×3 + 4×2 + 5×1)/15 = 35/15 = 7/3
E(V²) = (1×5 + 4×4 + 9×3 + 16×2 + 25×1)/15 = 105/15 = 7
Var(V) = 7 − (7/3)² = 7 − 49/9 = 14/9

7. Two independent variables: E(aX ± bY) and Var(aX ± bY)

Independent means the value of one variable gives no information about the other. For independent X and Y the specification expects:

E(aX ± bY)   = a E(X) ± b E(Y)
Var(aX ± bY) = a² Var(X) + b² Var(Y)

The mean follows the sign. The variance always adds, even for a difference, because uncertainty from each variable accumulates. The mean formula holds whether or not the variables are independent. The variance formula needs independence.

Worked example 7. X is the variable from example 2 (mean 3, variance 1.5). Y = W from example 4 (mean 4, variance 5.4). They are independent. Find E(3X − 2Y), Var(3X − 2Y) and the standard deviation of X + Y.

First Var(W). E(W²) = 4(0.4) + 16(0.3) + 25(0.2) + 100(0.1) = 1.6 + 4.8 + 5 + 10 = 21.4, so Var(W) = 21.4 − 16 = 5.4.

E(3X − 2Y)   = 3(3) − 2(4) = 1
Var(3X − 2Y) = 9(1.5) + 4(5.4) = 13.5 + 21.6 = 35.1
Var(X + Y)   = 1.5 + 5.4 = 6.9, so sd = √6.9 = 2.63 (3 s.f.)

8. Sums of independent variables

For independent X₁, X₂, …, Xₙ the specification expects:

E(ΣXᵢ) = ΣE(Xᵢ)
Var(ΣXᵢ) = ΣVar(Xᵢ)

If the Xᵢ are independent copies of X, then E(X₁ + … + Xₙ) = nE(X) and Var(X₁ + … + Xₙ) = n Var(X). Compare this with nX, where one value is multiplied: E(nX) = nE(X) but Var(nX) = n² Var(X).

Worked example 8. X has mean 3 and variance 1.5 (example 2). Compare 2X with X₁ + X₂, where X₁ and X₂ are independent copies of X.

2X:       E = 6,  Var = 4 × 1.5 = 6
X₁ + X₂:  E = 6,  Var = 1.5 + 1.5 = 3

The means agree but the spreads do not. Doubling one reading doubles the error, whereas adding two independent readings lets some errors cancel. Choose the formula from the wording: “the same value doubled” means aX, “two separate observations added” means X₁ + X₂.

Common errors

• Forgetting E(X²) = Σ x² p and instead squaring E(X). This turns the variance into zero or a negative number.
• Writing Var(aX + b) = a Var(X) + b, or leaving a unsquared.
• Subtracting variances for X − Y. The variance of a difference is Var(X) + Var(Y).
• Taking E(10/X) as 10/E(X), or E(5X²) as 5(E(X))².
• Using Var(X + Y) = Var(X) + Var(Y) without saying X and Y are independent.
• Mixing up 2X with X₁ + X₂ (see example 8).
• Reporting a standard deviation when asked for variance, or the reverse.
• Not checking that the probabilities add to 1 before starting.

Next steps

Condense this into the revision notes, then try the practice questions. Check progress against the checklist.

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.