Linear Combinations of Random Variables | 随机变量的线性组合

📚 Linear Combinations of Random Variables | 随机变量的线性组合

In Edexcel A Level Mathematics, combinations of random variables show how the expectation and variance of a new variable such as X + Y, X – Y or aX + bY depend on the original random variables.

在 Edexcel A Level 数学中，随机变量的组合研究的是新变量（如 X + Y、X – Y 或 aX + bY）的期望和方差如何依赖于原随机变量。

1. What Is a Linear Combination? | 什么是线性组合？

A linear combination of random variables has the form aX + bY + c, where X and Y are random variables and a, b, c are constants.

随机变量的线性组合形式为 aX + bY + c，其中 X 和 Y 是随机变量，a、b、c 是常数。

For example, if X is the mass of an apple and Y is the mass of a pear, then T = 2X + 3Y – 5 models a bag containing two apples and three pears with a 5 g packaging adjustment.

例如，若 X 表示一个苹果的质量，Y 表示一个梨的质量，则 T = 2X + 3Y – 5 可以表示一个装有两个苹果和三个梨、再扣除 5 克包装后的质量。

2. Expectation of a Linear Combination | 线性组合的期望

For any random variables X and Y, the expected value operator is linear, so the mean of a linear combination is found by applying the same coefficients to the individual means.

对任意随机变量 X 和 Y，期望运算具有线性性，因此线性组合的均值可以通过对各自的均值使用相同系数来求得。

E(aX + bY + c) = aE(X) + bE(Y) + c

This rule does not require X and Y to be independent, so it works for dependent variables as well.

这条规则不要求 X 和 Y 独立，因此对相关变量同样成立。

If X has mean 10 and Y has mean 6, then E(3X – 2Y + 4) = 3(10) – 2(6) + 4 = 22.

若 X 的均值为 10，Y 的均值为 6，则 E(3X – 2Y + 4) = 3(10) – 2(6) + 4 = 22。

3. Variance of a Linear Combination | 线性组合的方差

When X and Y are independent, the variance of a linear combination is found by squaring the coefficients and adding the variance contributions.

当 X 与 Y 独立时，线性组合的方差由系数的平方乘以各自方差后相加得到。

Var(aX + bY + c) = a²Var(X) + b²Var(Y)

The constant c disappears because adding a constant shifts all values but does not change spread. Coefficients a and b must be squared.

常数 c 消失，因为加一个常数只会平移所有取值而不改变离散程度；系数 a 和 b 必须平方。

If X and Y are not independent, an extra covariance term is needed: Var(aX + bY) = a²Var(X) + b²Var(Y) + 2abCov(X,Y). Edexcel questions usually state independence unless they expect this formula.

若 X 与 Y 不独立，则需要额外加上协方差项：Var(aX + bY) = a²Var(X) + b²Var(Y) + 2abCov(X,Y)。Edexcel 题目通常会说明变量独立，除非要求使用此公式。

4. The General Two-Variable Formulas | 两个随机变量的一般公式

For independent X and Y, the key results are E(X + Y) = E(X) + E(Y) and Var(X + Y) = Var(X) + Var(Y).

对于独立的 X 和 Y，关键结果是 E(X + Y) = E(X) + E(Y) 以及 Var(X + Y) = Var(X) + Var(Y)。

For the difference, E(X – Y) = E(X) – E(Y), but Var(X – Y) = Var(X) + Var(Y), not Var(X) – Var(Y).

对于差，有 E(X – Y) = E(X) – E(Y)，但 Var(X – Y) = Var(X) + Var(Y)，而不是 Var(X) – Var(Y)。

This is one of the most frequently tested ideas: subtracting independent variables still adds variances.

这是最常考的思想之一：独立变量相减时，方差仍然相加。

5. Linear Transformation aX + b | 线性变换 aX + b

For a single random variable X, the expectation and variance transform as follows.

对于单个随机变量 X，期望和方差的变化如下。

E(aX + b) = aE(X) + b

Var(aX + b) = a²Var(X)

Multiplying by a negative constant does not make variance negative; the square ensures the result is non-negative.

乘以负数不会使方差为负；平方保证了结果非负。

Example: If X has mean 8 and variance 4, then Y = 5 – 2X has mean E(Y) = 5 – 2(8) = -11 and variance Var(Y) = (-2)² × 4 = 16.

示例：若 X 的均值为 8，方差为 4，则 Y = 5 – 2X 的均值为 E(Y) = 5 – 2(8) = -11，方差 Var(Y) = (-2)² × 4 = 16。

6. Combining Independent Normal Variables | 合并独立正态变量

If X ~ N(μ₁, σ₁²) and Y ~ N(μ₂, σ₂²) are independent, then any linear combination aX + bY + c also follows a normal distribution.

若 X ~ N(μ₁, σ₁²) 与 Y ~ N(μ₂, σ₂²) 独立，则任意线性组合 aX + bY + c 也服从正态分布。

aX + bY + c ~ N(aμ₁ + bμ₂ + c, a²σ₁² + b²σ₂²)

Example: If X ~ N(10, 4) and Y ~ N(20, 9) are independent, then X + Y ~ N(30, 13) and X – Y ~ N(-10, 13).

示例：若 X ~ N(10, 4) 与 Y ~ N(20, 9) 独立，则 X + Y ~ N(30, 13)，X – Y ~ N(-10, 13)。

You can then use this normal distribution to calculate probabilities such as P(X + Y > 32) by standardising.

随后可以通过标准化使用该正态分布计算概率，例如 P(X + Y > 32)。

7. Sums of Independent Identically Distributed Variables | 独立同分布变量的和

If X₁, X₂, …, Xₙ are independent observations from the same distribution with mean μ and variance σ², then their sum Sₙ = X₁ + X₂ + … + Xₙ has mean nμ and variance nσ².

若 X₁、X₂、…、Xₙ 是来自同一分布且独立的观测值，其均值为 μ，方差为 σ²，则它们的和 Sₙ = X₁ + X₂ + … + Xₙ 的均值为 nμ，方差为 nσ²。

The sample mean X̄ = Sₙ / n therefore has mean μ and variance σ² / n.

因此样本均值 X̄ = Sₙ / n 的均值为 μ，方差为 σ² / n。

This distinction between the sum and the mean is a common exam trap, especially in sampling questions.

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Linear Combinations of Random Variables | 随机变量的线性组合