Mean & Variance of Combinations of Random Variables | 随机变量组合的均值与方差

📚 Mean & Variance of Combinations of Random Variables | 随机变量组合的均值与方差

In A-Level Further Mathematics, understanding how to find the mean and variance of combinations of random variables is essential. This topic extends your knowledge of single random variables to situations involving two or more variables, such as the sum or difference of independent measurements, or a linear combination of several variables.

在A-Level进阶数学中，理解如何求随机变量组合的均值与方差至关重要。本知识点将单一随机变量的知识扩展至涉及两个或多个变量的情形，例如独立测量值之和或差，以及多个变量的线性组合。

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

The expectation operator, E, is linear. This means that for any random variables X and Y, and any constants a, b and c, the following rule holds:

期望算子 E 具有线性性。这意味着对于任意随机变量 X 和 Y，以及任意常数 a、b 和 c，以下规则成立：

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

Importantly, this rule requires no assumption about independence between X and Y. The expectation of a sum is always the sum of the expectations, regardless of how the variables are related.

重要的是，该规则不需要对 X 与 Y 之间的独立性作任何假设。和的期望始终等于期望之和，无论变量之间存在怎样的关系。

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

For variance, the situation is more subtle. When X and Y are independent, the variance of their sum or difference is the sum of their variances:

对于方差，情况更为微妙。当 X 和 Y 独立时，它们之和或之差的方差等于各自方差之和：

Var(aX ± bY) = a²Var(X) + b²Var(Y)   (X, Y independent)

Note that the plus/minus sign inside the brackets does not affect the variance. Whether you add or subtract two independent random variables, the variances are always added. Also note the coefficients are squared: Var(aX) = a²Var(X). This is because variance involves squared deviations from the mean.

注意括号内的正负号不影响方差结果。无论是对两个独立随机变量做加法还是减法，方差始终相加。同时注意系数要平方：Var(aX) = a²Var(X)。这是因为方差涉及相对于均值的平方偏差。

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3. The Role of Independence | 独立性的作用

Independence is a crucial condition for the variance rule above. If X and Y are not independent, a covariance term must be included:

独立性是上述方差规则的关键条件。如果 X 和 Y 不独立，则必须加入协方差项：

Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y)

Var(X − Y) = Var(X) + Var(Y) − 2Cov(X, Y)

When X and Y are independent, Cov(X, Y) = 0, and both equations reduce to the simpler rule from Section 2. In Edexcel A-Level Further Mathematics, exam questions typically state that variables are independent, but you should always check.

当 X 和 Y 独立时，Cov(X, Y) = 0，上述两个等式均简化为第2节的简单规则。在Edexcel A-Level进阶数学考试中，题目通常会说明变量是独立的，但你应该时刻检查这一条件。

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4. Extension to n Random Variables | 推广至 n 个随机变量

The rules extend naturally to any finite number of independent random variables. If X₁, X₂, …, Xₙ are independent, and a₁, a₂, …, aₙ are constants, then:

这些规则可自然推广至任意有限个独立随机变量。若 X₁, X₂, …, Xₙ 相互独立，a₁, a₂, …, aₙ 为常数，则：

E(a₁X₁ + a₂X₂ + … + aₙXₙ) = a₁E(X₁) + a₂E(X₂) + … + aₙE(Xₙ)

Var(a₁X₁ + a₂X₂ + … + aₙXₙ) = a₁²Var(X₁) + a₂²Var(X₂) + … + aₙ²Var(Xₙ)

This general form appears frequently in exam questions involving the sum of several independent observations, such as measuring the total weight of multiple items or the total time of several independent processes.

这一通式在考试中频繁出现，常用于涉及多个独立观测值之和的问题，例如多个物品的总重量或数个独立过程的总时间。

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5. The Special Case: Mean of n Observations | 特殊情况：n 次观测的均值

A common exam scenario involves taking the average of n independent observations from a distribution. Let X̄ = (X₁ + X₂ + … + Xₙ)/n, where each Xᵢ has mean μ and variance σ². Then:

一个常见的考试场景涉及取自同一分布的 n 个独立观测值的平均值。设 X̄ = (X₁ + X₂ + … + Xₙ)/n，其中每个 Xᵢ 的均值为 μ、方差为 σ²，则：

E(X̄) = (1/n) × (nμ) = μ

Var(X̄) = (1/n)² × (nσ²) = σ²/n

The mean of the sample mean equals the population mean, but the variance is reduced by a factor of 1/n. This demonstrates the benefit of averaging: the more observations you take, the less variability there is in the sample mean.

样本均值的期望等于总体均值，但方差缩小了 1/n 倍。这体现了取平均的益处：观测次数越多，样本均值的变异性越小。

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6. Discrete Random Variables: Worked Example | 离散随机变量：例题讲解

Consider two independent discrete random variables X and Y. X has E(X) = 3 and Var(X) = 4. Y has E(Y) = 2 and Var(Y) = 1. Find E(2X − 3Y + 5) and Var(2X − 3Y + 5).

考虑两个独立离散随机变量 X 和 Y。X 满足 E(X) = 3，Var(X) = 4；Y 满足 E(Y) = 2，Var(Y) = 1。求 E(2X − 3Y + 5) 和 Var(2X − 3Y + 5)。

For the expectation:

对于期望：

E(2X − 3Y + 5) = 2E(X) − 3E(Y) + 5 = 2(3) − 3(2) + 5 = 6 − 6 + 5 = 5

For the variance, since X and Y are independent:

对于方差，由于 X 和 Y 独立：

Var(2X − 3Y + 5) = 2²Var(X) + (−3)²Var(Y) = 4(4) + 9(1) = 16 + 9 = 25

Notice that the constant +5 contributes nothing to the variance; only the coefficients of X and Y matter, and they are squared.

注意常数项 +5 对方差没有贡献；只有 X 和 Y 的系数起作用，且需取平方。

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7. Combining Two Independent Normal Distributions | 两个独立正态分布的组合

An important application is combining independent normal distributions. If X ~ N(μ₁, σ₁²) and Y ~ N(μ₂, σ₂²) are independent, then:

一个重要的应用是组合两个独立正态分布。若 X ~ N(μ₁, σ₁²) 和 Y ~ N(μ₂, σ₂²) 相互独立，则：

X + Y ~ N(μ₁ + μ₂, σ₁² + σ₂²)

X − Y ~ N(μ₁ − μ₂, σ₁² + σ₂²)

The distribution remains normal, the means combine additively (or subtractively), and the variances always add. This result is fundamental for many hypothesis-testing and confidence-interval problems.

分布仍为正态分布，均值按加（或减）法合并，而方差始终相加。该结果是许多假设检验和置信区间问题的基石。

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8. Worked Example: Word Problem | 例题讲解：应用题型

A machine fills bags of rice. The weight of rice in each bag, R, has mean 1.0 kg and standard deviation 0.05 kg. The weight of the bag itself, B, has mean 0.02 kg and standard deviation 0.003 kg. R and B are independent.

一台机器封装米袋。每袋米的重量 R 的均值为 1.0 kg，标准差为 0.05 kg。米袋自身的重量 B 的均值为 0.02 kg，标准差为 0.003 kg。R 和 B 相互独立。

(a) Find the mean and standard deviation of the total weight T = R + B.

（a）求总重量 T = R + B 的均值与标准差。

E(T) = E(R) + E(B) = 1.0 + 0.02 = 1.02 kg

Var(T) = Var(R) + Var(B) = 0.05² + 0.003² = 0.0025 + 0.000009 = 0.002509

SD(T) = √0.002509 ≈ 0.0501 kg

(b) Three bags are selected independently. Find the mean and variance of the total weight of the three filled bags.

（b）独立选取三袋米。求三袋总重量的均值与方差。

E(T₁ + T₂ + T₃) = 3 × 1.02 = 3.06 kg

Var(T₁ + T₂ + T₃) = 3 × 0.002509 = 0.007527

The mean scales by the factor 3, and the variance scales by 3² — wait, no. Here we are adding three independent but identically distributed variables, so the variance scales by 3, not by 3². The factor 3² would apply if we multiplied a single variable by 3.

均值按因子 3 放大，而方差按因子 3 放大——注意，此处是三个独立同分布变量相加，方差按因子 3 放大，而非 3²。若将单个变量乘以 3，方差才按 3² 放大。

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9. The Connection: Var(aX) vs Var(X₁ + X₂) | 联系：Var(aX) 与 Var(X₁ + X₂) 之区别

A frequent source of confusion is the difference between multiplying a single random variable by a constant and adding several independent copies. If X₁ and X₂ are independent and identically distributed with variance σ²:

一个常见的混淆来源是：将单个随机变量乘以常数与将若干个独立副本相加之间的区别。若 X₁ 和 X₂ 是独立同分布的随机变量，方差均为 σ²：

Var(2X₁) = 4σ²   but   Var(X₁ + X₂) = 2σ²

In the first case, the same value is doubled; all deviations from the mean are also doubled, hence variance quadruples. In the second case, two independent values are added; deviations partially cancel, so variance merely doubles.

在第一种情况下，同一个值被翻倍；所有偏离均值的偏差也随之翻倍，因此方差变为原来的四倍。在第二种情况下，两个独立值相加；偏差部分相互抵消，所以方差仅为原来的两倍。

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10. Summary of Key Formulas | 关键公式汇总

The following table consolidates the essential formulas for this topic:

下表汇总了本专题的关键公式：

Operation | 运算	Mean | 均值	Variance | 方差
aX + b	aE(X) + b	a²Var(X)
X ± Y (independent)	E(X) ± E(Y)	Var(X) + Var(Y)
X + Y (dependent)	E(X) + E(Y)	Var(X) + Var(Y) + 2Cov(X, Y)
X − Y (dependent)	E(X) − E(Y)	Var(X) + Var(Y) − 2Cov(X, Y)
X̄ = (X₁+…+Xₙ)/n (i.i.d.)	μ	σ²/n

Remember: the expectation operator is always linear, but the variance operator is only additive under independence (or when the covariance is zero).

请记住：期望算子始终是线性的，但方差算子只有在独立（或协方差为零）时才具有可加性。

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11. Common Exam Mistakes | 常见考试失误

Students frequently make the following errors in exams. Avoid them:

学生在考试中经常犯以下错误，请务必避免：

• Forgetting to square coefficients inside variance: writing Var(3X) = 3Var(X) instead of 9Var(X).

  忘记对方差中的系数取平方：将 Var(3X) 写为 3Var(X) 而非 9Var(X)。

• Adding variances for X − Y without checking independence — if variables are dependent, the covariance term must be included.

  计算 X − Y 的方差时直接相加而不检查独立性——若变量相关，必须包含协方差项。

• Applying the variance rule when variables are not independent and no covariance is given.

  在变量不独立且未给出协方差时，仍套用方差相加规则。

• Confusing Var(X₁ + X₂) with Var(2X₁) as shown in Section 9.

  如第9节所示，混淆 Var(X₁ + X₂) 与 Var(2X₁)。

• Forgetting that the variance of a constant is zero: Var(c) = 0.

  忘记常数的方差为零：Var(c) = 0。

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12. Exam Strategy and Final Takeaways | 考试策略与最终要点

When tackling exam questions on this topic, follow this structured approach:

解答本专题的考题时，请遵循以下结构化步骤：

• Step 1 — Identify all random variables and their given means and variances.

  第一步——确定所有随机变量及其给定的均值与方差。

• Step 2 — Determine whether independence holds; if not stated, consider whether it is a reasonable assumption from the context.

  第二步——判断独立性是否成立；若未说明，根据题目情境判断该假设是否合理。

• Step 3 — Write down the linear combination in the form aX + bY + c, identifying a, b, c explicitly.

  第三步——将线性组合写成 aX + bY + c 的形式，明确写出 a、b、c。

• Step 4 — Apply the mean formula first, then the variance formula, checking that each coefficient is squared in the variance calculation.

  第四步——先应用均值公式，再应用方差公式，并确认方差计算中每个系数都已平方。

Mastering these rules will allow you to solve a wide range of problems involving sums, differences, and linear combinations of random variables with confidence.

熟练掌握这些规则，你将能够自信地解决涉及随机变量之和、差以及线性组合的各类问题。

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