Linear Combination of Random Variables in A LEVEL Maths: The Complete Rules (CIE 9709)

Linear Combinations of Random Variables in A Level Maths: The Complete Rules (CIE 9709).

What Is a Linear Combination of Random Variables?

A linear combination of random variables is an expression like aX ± bY, where a and b are fixed numbers (constants) and X and Y are random variables, for example 3X − 2Y. In A Level Maths, "finding the linear combination" means working out the mean (expectation) and variance of that new combined quantity, not just adding X and Y's values directly. Cambridge International (CIE) 9709 tests this explicitly under Probability & Statistics, usually alongside the Normal distribution.

What Are the Formulas for E(aX ± bY) and Var(aX ± bY)?

For any constants a, b and random variables X, Y: E(aX ± bY) = aE(X) ± bE(Y). For variance, provided X and Y are independent: Var(aX ± bY) = a²Var(X) + b²Var(Y). Note the plus sign even when the expression itself has a minus. Variance can never be negative, so you never subtract variances, regardless of whether the original combination adds or subtracts the variables. Forgetting to square the constants a and b is the single most common mark loss here.

What's the Difference Between "Sum of n Observations" and "n Times X"?

This is the classic exam trap. If X₁, X₂, …, Xₙ are n independent observations of the same variable X, then E(X₁+X₂+…+Xₙ) = nE(X) and Var(X₁+X₂+…+Xₙ) = nVar(X). But if you instead take one single observation and multiply it by n, E(nX) = nE(X) is the same, yet Var(nX) = n²Var(X), not nVar(X). Five separately weighed apples summed is not the same random variable as one apple's weight multiplied by five, and examiners write questions specifically to test whether you can tell the two apart.

How Do You Combine Normal Random Variables?

If X~N(μ₁,σ₁²) and Y~N(μ₂,σ₂²) are independent, then X+Y and X−Y are both still Normally distributed: X+Y ~ N(μ₁+μ₂, σ₁²+σ₂²) and X−Y ~ N(μ₁−μ₂, σ₁²+σ₂²). Variance adds in both cases. More generally, aX+bY ~ N(aμ₁+bμ₂, a²σ₁²+b²σ₂²). This is what lets you turn a two-part, two-variable problem (e.g. "time for part 1" plus "time for part 2") into a single Normal probability calculation using the usual Z = (x−μ)/σ standardisation.

Combination

Mean

Variance

Still Normal if X, Y Normal?

aX ± bY

aE(X) ± bE(Y)

a²Var(X) + b²Var(Y)

Yes

X₁+X₂+…+Xₙ (n independent observations of X)

nE(X)

nVar(X)

Yes

nX (one observation scaled by n)

nE(X)

n²Var(X)

Yes

Worked Example: Combining Two Independent Normal Variables (CIE 9709/2018/M/J/P73)

The times, in minutes, taken to complete the two parts of a task are independently Normally distributed: Part 1 has mean 4.5 and standard deviation 1.1; Part 2 has mean 2.3 and standard deviation 0.7.

(i) Find P(total time < 8.5 minutes). Let P₁~N(4.5, 1.1²) and P₂~N(2.3, 0.7²). E(P₁+P₂) = 4.5+2.3 = 6.8. Var(P₁+P₂) = 1.1²+0.7² = 1.21+0.49 = 1.7. Standardise: Z = (8.5−6.8)/√1.7 = 1.7/1.3038 = 1.304. From tables, Φ(1.304) ≈ 0.904, so P(total < 8.5) ≈ 0.904.

(ii) Find P(Part 1 takes more than twice as long as Part 2). This means P(P₁ > 2P₂), rewritten as P(P₁ − 2P₂ > 0). Let W = P₁ − 2P₂: E(W) = 4.5 − 2(2.3) = −0.1. Var(W) = 1.1² + 2²(0.7²) = 1.21 + 4(0.49) = 1.21+1.96 = 3.17. Notice the 2² here: the constant gets squared, but the sign in front of P₂ doesn't carry through to the variance at all. Standardise: Z = (0−(−0.1))/√3.17 = 0.1/1.780 = 0.0562. P(W > 0) = 1 − Φ(0.0562) ≈ 1 − 0.522 = 0.478.

Both figures match the published mark scheme for this question. That's proof that once you have the right mean and variance, the rest is routine Normal-distribution work you've already practised for other CIE 9709 topics, including the Normal approximation used in our guide to the Poisson distribution.

FAQ

What is a linear combination of random variables in A Level Maths? It's an expression of the form aX ± bY, where a and b are constants and X, Y are random variables. You're asked to find its mean, E(aX±bY) = aE(X)±bE(Y), and, if X, Y are independent, its variance, Var(aX±bY) = a²Var(X)+b²Var(Y).

Why does variance always add, even for Var(aX − bY)? Because variance measures spread, and spread cannot cancel out. Subtracting two independent random quantities makes the combined outcome more uncertain, not less. So Var(aX−bY) = a²Var(X)+b²Var(Y), with a plus sign, never a minus.

What's the difference between Var(X₁+X₂+X₃) and Var(3X)? X₁+X₂+X₃ is three separate, independent observations of X summed, so Var(X₁+X₂+X₃) = 3Var(X). 3X is one observation scaled by 3, so Var(3X) = 3²Var(X) = 9Var(X). Mixing these up is the most common error CIE examiners report on this topic.

Can you add two independent Normal random variables together? Yes. If X~N(μ₁,σ₁²) and Y~N(μ₂,σ₂²) are independent, X+Y is also Normally distributed, with mean μ₁+μ₂ and variance σ₁²+σ₂² (variance still adds, even though the means are being summed).

Is "linear combinations of random variables" tested on AQA A Level Maths too, not just CIE? The explicit aX±bY notation and terminology used here is most associated with CIE 9709's Probability & Statistics component. AQA A-level Maths tests the same underlying principle, that variances add when independent Normal variables are combined, inside its Normal distribution content, usually without a separate named "linear combinations" unit, so check your own specification's formula booklet for the exact notation expected.