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Linear combinations of random variables

Edexcel IAL Mathematics · WST03 Statistics 3

S3 opens by combining random variables. You already know E(aX + b) and Var(aX + b) from S1; now two or more independent variables are added or subtracted, and the key result is that a linear combination of independent normal variables is itself normal.

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Expectation

For any random variables X and Y and constants a and b:

E(aX ± bY) = aE(X) ± bE(Y).

This holds whether or not X and Y are independent. For independent X and Y the formulae booklet also gives E(XY) = E(X)E(Y).

Variance: variances always add

If X and Y are independent:

Var(aX ± bY) = a² Var(X) + b² Var(Y).

Notice the plus sign even for a difference: subtracting a random quantity adds uncertainty, it never removes it. The constants are squared, so Var(3X) = 9Var(X). Standard deviations do not add; always combine variances and square-root at the end.

Combinations of normal variables

If X ~ N(μx, σx²) and Y ~ N(μy, σy²) are independent then

aX ± bY ~ N(aμx ± bμy, a²σx² + b²σy²).

The same applies to any number of independent normal variables. No proof is required: you use the result, then standardise with Z = (value - mean)/sd as in S1.

Combination	Mean	Variance
X + Y	μx + μy	σx² + σy²
X - Y	μx - μy	σx² + σy²
3X - 2Y	3μx - 2μy	9σx² + 4σy²

Worked example

X ~ N(30, 2²) and Y ~ N(20, 3²) are independent. Find P(X + Y > 55).

X + Y ~ N(30 + 20, 4 + 9) = N(50, 13).
z = (55 - 50)/sqrt(13) = 1.3868
P(Z > 1.39) = 1 - 0.9177

Answer: P(X + Y > 55) = 0.0823 (tables) or 0.0828 (calculator)

Key idea: Means combine with the same signs as the variables; variances of independent variables always add, with the constants squared. Normal in, normal out.

Check yourself

X ~ N(30, 2²) and Y ~ N(20, 3²) are independent. Find E(3X - 2Y).
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50. 3 × 30 - 2 × 20 = 50.
X ~ N(30, 2²) and Y ~ N(20, 3²) are independent. Find Var(3X - 2Y).
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72. 3² × 4 + 2² × 9 = 36 + 36 = 72.
X and Y are independent with Var(X) = 4 and Var(Y) = 9. What is Var(X - Y)?
- 13
- -5
- 5
- sqrt(13)
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13. Variances of independent variables add, even for a difference: 4 + 9 = 13.
X ~ N(30, 2²) and Y ~ N(20, 3²) are independent. Find P(X + Y > 55). Give your answer to 4 decimal places.
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0.0828. X + Y ~ N(50, 13); z = 5/sqrt(13) = 1.3868; P = 0.0823 from tables (0.0828 by calculator).
X ~ N(30, 2²) and Y ~ N(20, 3²) are independent. Find P(X - Y < 8). Give your answer to 4 decimal places.
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0.2895. X - Y ~ N(10, 13); z = -2/sqrt(13) = -0.5547; P = 0.2912 from tables.
X and Y are independent with standard deviations 4 and 3. Find the standard deviation of X + Y.
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5. Var = 16 + 9 = 25, so the standard deviation is 5. Standard deviations themselves do not add (4 + 3 = 7 is wrong).
X and Y are independent with E(X) = 30 and E(Y) = 20. Find E(XY).
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600. For independent variables E(XY) = E(X)E(Y) = 600.
Why must X and Y be independent to use Var(X + Y) = Var(X) + Var(Y)?
- If they are related, the way they vary together changes the variance of the sum
- Otherwise E(X + Y) would not equal E(X) + E(Y)
- Only independent variables can be normally distributed
- Independence makes the two variances equal
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If they are related, the way they vary together changes the variance of the sum. Dependent variables can move together (making the sum more variable) or against each other (less variable), so the simple rule fails.

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Last updated 8 November 2026. Spotted a mistake? Tell us.