Distributions of Linear Combinations of Random Variables (9709 PS2 §6.2) — Mathematics

Introduction

When random variables are combined — scaled, shifted, or added together — a natural question arises: does the resulting variable follow a recognisable distribution? For the 9709 Paper 7 exam, two families are particularly important: the Normal distribution and the Poisson distribution. Both are closed under certain linear operations, meaning the combined variable stays within the same family. Knowing these closure results — and being able to apply them quickly and accurately — is essential for multi-part probability questions.

Proofs are not required. The syllabus asks only that you use these results confidently.

Core Concept

There are three results you must know and use.

Result 1 — Linear transformation of a Normal variable

If $X \sim \mathrm{N}(\mu, \sigma^2)$ and $a, b$ are constants ( $a \neq 0$ ), then:

aX + b \sim \mathrm{N}(a\mu + b,\ a^2\sigma^2)

The shape remains Normal; the mean and variance transform according to the rules you already know from linear combinations of expectations and variances.

Result 2 — Linear combination of two independent Normal variables

If $X \sim \mathrm{N}(\mu_X, \sigma_X^2)$ and $Y \sim \mathrm{N}(\mu_Y, \sigma_Y^2)$ are independent, and $a, b$ are constants, then:

aX + bY \sim \mathrm{N}(a\mu_X + b\mu_Y,\ a^2\sigma_X^2 + b^2\sigma_Y^2)

Independence is crucial: without it, the variance formula (which adds $a^2\sigma_X^2$ and $b^2\sigma_Y^2$ ) does not hold.

Result 3 — Sum of independent Poisson variables

If $X \sim \mathrm{Po}(\lambda)$ and $Y \sim \mathrm{Po}(\mu)$ are independent, then:

X + Y \sim \mathrm{Po}(\lambda + \mu)

This extends naturally: the sum of any finite number of independent Poisson variables is Poisson with rate equal to the sum of the individual rates.

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