Sums of Independent Random Variables | 独立随机变量之和

📚 Sums of Independent Random Variables | 独立随机变量之和

In Edexcel A Level Mathematics, the topic of sums of independent random variables brings together expectation, variance and common distributions such as the normal, Poisson and binomial. It underpins many statistical applications, from combining measurement errors to modelling total demand.

在 Edexcel A Level 数学中，独立随机变量之和这一主题综合了期望、方差以及正态、泊松和二项等常见分布。它支撑着许多统计应用，从合并测量误差到模拟总需求。

1. Independent Random Variables: Quick Recap | 独立随机变量快速回顾

Two random variables X and Y are independent if knowing the value of X gives no information about the distribution of Y.

如果知道 X 的取值不能提供关于 Y 分布的任何信息，则称两个随机变量 X 和 Y 独立。

In probability notation, independence means P(X ∈ A and Y ∈ B) = P(X ∈ A) × P(Y ∈ B) for any sets A and B.

用概率符号表示，独立性意味着对于任意集合 A 和 B，都有 P(X ∈ A 且 Y ∈ B) = P(X ∈ A) × P(Y ∈ B)。

Many results in this topic only apply under independence, so always check this condition before using the variance and distribution rules.

本主题中的许多结论仅在独立条件下成立，因此在使用方差和分布规则之前，务必检查这一条件。

2. Expectation of a Sum | 和的期望

For any two random variables X and Y, whether independent or not, the expectation of their sum is the sum of their expectations.

对于任意两个随机变量 X 和 Y，无论它们是否独立，其和的期望等于期望的和。

E(X + Y) = E(X) + E(Y) always holds, and it extends to any finite number of variables.

E(X + Y) = E(X) + E(Y) 始终成立，并且可以推广到任意有限个变量。

This linearity property makes expectation very convenient when combining different sources of variation.

这种线性性质使得在合并不同变异来源时，期望的计算非常方便。

3. Variance of a Sum for Independent Variables | 独立变量和的方差

If X and Y are independent, then the variance of their sum is the sum of their variances.

如果 X 和 Y 独立，则它们和的方差等于方差的和。

Var(X + Y) = Var(X) + Var(Y) is valid only under independence.

Var(X + Y) = Var(X) + Var(Y) 仅在独立条件下成立。

If X and Y are not independent, the formula would require an extra covariance term, but Edexcel A Level focuses on the independent case.

如果 X 和 Y 不独立，该公式需要加入额外的协方差项，但 Edexcel A Level 主要考察独立情形。

Remember that variances add, not standard deviations, so you must square standard deviations first.

请记住，相加的是方差而不是标准差，因此必须先将标准差平方。

4. Linear Combinations aX + bY | 线性组合 aX + bY

For independent X and Y, the expectation of aX + bY is aE(X) + bE(Y).

对于独立的 X 和 Y，aX + bY 的期望是 aE(X) + bE(Y)。

Provided X and Y are independent, Var(aX + bY) = a²Var(X) + b²Var(Y).

只要 X 和 Y 独立，Var(aX + bY) = a²Var(X) + b²Var(Y)。

Constants do not enter the variance except through the multiplier squared, so for a single variable Var(aX + b) = a²Var(X).

常数项不会直接进入方差，只会通过乘数的平方产生影响，因此对于单个变量，Var(aX + b) = a²Var(X)。

This is one of the most useful results when scaling measurements or calculating totals from several components.

这是缩放测量值或计算多个组成部分总和时最有用的结论之一。

5. Sums of Independent Normal Variables | 独立正态变量之和

If X ~ N(μ_X, σ_X²) and Y ~ N(μ_Y, σ_Y²) are independent, then X + Y ~ N(μ_X + μ_Y, σ_X² + σ_Y²).

如果 X ~ N(μ_X, σ_X²) 和 Y ~ N(μ_Y, σ_Y²) 独立，则 X + Y ~ N(μ_X + μ_Y, σ_X² + σ_Y²)。

More generally, aX + bY ~ N(aμ_X + bμ_Y, a²σ_X² + b²σ_Y²).

更一般地，aX + bY ~ N(aμ_X + bμ_Y, a²σ_X² + b²σ_Y²)。

The sum of independent normal variables is exactly normal, not just approximately, which makes calculations very clean.

独立正态变量之和精确服从正态分布，而不仅仅是近似，这使得计算非常清晰。

You must add variances and then take the square root at the end when standardising.

在标准化时，必须先加方差，最后再开平方根。

6. Differences of Independent Normal Variables | 独立正态变量之差

For independent normal variables, X – Y is also normal, but the variances still add rather than subtract.

对于独立的正态变量，X – Y 也服从正态分布，但方差仍然相加，而不是相减。

Var(X – Y) = Var(X) + Var(Y), because subtracting Y is equivalent to adding -Y and Var(-Y) = Var(Y).

Var(X – Y) = Var(X) + Var(Y)，因为减去 Y 等价于加上 -Y，而 Var(-Y) = Var(Y)。

Therefore X – Y ~ N(μ_X – μ_Y, σ_X² + σ_Y²).

因此 X – Y ~ N(μ_X – μ_Y, σ_X² + σ_Y²)。

This is one of the most common error points in exams: never subtract variances for the difference of independent variables.

这是考试中最常见的错误点之一：独立变量之差的方差绝不能相减。

7. Sums of Independent Poisson Variables | 独立泊松变量之和

If X ~ Po(λ) and Y ~ Po(μ) are independent, then X + Y ~ Po(λ + μ).

如果 X ~ Po(λ) 和 Y ~ Po(μ) 独立，则 X + Y ~ Po(λ + μ)。

The condition of independence is essential; if events are not independent, the sum is not necessarily Poisson with parameter λ + μ.

独立性条件至关重要；如果事件不独立，其和不一定服从参数为 λ + μ 的泊松分布。

This result extends to more than two independent Poisson variables, so the total count remains Poisson.

这一结论可推广到两个以上独立的泊松变量，因此总计数仍然是泊松分布。

This is especially useful for modelling total arrivals from independent sources.

这对模拟来自独立来源的总到达次数特别有用。

8. Sums of Independent Binomial Variables | 独立二项变量之和

If X ~ B(n, p) and Y ~ B(m, p) are independent with the same success probability p, then X + Y ~ B(n + m, p).

如果 X ~ B(n, p) 和 Y ~ B(m, p) 独立，且成功概率 p 相同，则 X + Y ~ B(n + m, p)。

If the probabilities p are different, the sum is not binomial; you would need to use the probability generating function or treat the variables separately.

如果概率 p 不同，其和不是二项分布；你需要使用概率生成函数或分别处理各个变量。

This result makes sense because the total count is the number of successes in n + m independent trials with probability p.

这一结果是合理的，因为总次数是在 n + m 次独立试验中成功次数，每次成功概率为 p。

Always check that both binomial variables have exactly the same p before adding them.

在相加之前，务必检查两个二项变量是否具有完全相同的 p。

9. Sample Mean and Distribution of X̄ | 样本均值与 X̄ 的分布

Suppose X₁, X₂, …, Xₙ are independent and identically distributed as N(μ, σ²).

假设 X₁, X₂, …, Xₙ 独立同分布于 N(μ, σ²)。

The sample mean X̄ = (X₁ + X₂ + … + Xₙ)/n has expectation μ and variance σ²/n.

样本均值 X̄ = (X₁ + X₂ + … + Xₙ)/n 的期望为 μ，方差为 σ²/n。

Therefore X̄ ~ N(μ, σ²/n), and the standard deviation σ/√n is called the standard error.

因此 X̄ ~ N(μ, σ²/n)，标准差 σ/√n 称为标准误。

This is a direct consequence of the linear combination rules and is central to confidence intervals and hypothesis tests.

这是线性组合规则的直接结果，也是置信区间和假设检验的核心。

10. Central Limit Theorem and Large Samples | 中心极限定理与大样本

When the population is not normal, the central limit theorem states that for sufficiently large n, the sample mean X̄ is approximately normally distributed with mean μ and variance σ²/n.

当总体不服从正态分布时，中心极限定理指出，对于足够大的 n，样本均值 X̄ 近似服从均值为 μ、方差为 σ²/n 的正态分布。

Similarly, the sum Sₙ = X₁ + … + Xₙ is approximately N(nμ, nσ²).

类似地，和 Sₙ = X₁ + … + Xₙ 近似服从 N(nμ, nσ²)。

A common rule of thumb is n ≥ 30, but in practice the shape of the population matters.

常用的经验法则是 n ≥ 30，但实际上总体的形态也很重要。

This theorem justifies using normal approximations for sums and means even when the original distribution is unknown or skewed.

该定理证明了即使原始分布未知或偏斜，也可以对总和及均值使用正态近似。

11. Worked Example: Packaging Weights | 例题：包装重量

Example: The weights of two components A and B are independent normal variables. A ~ N(120, 5²) and B ~ N(80, 4²). Find the probability that the total weight exceeds 215 g.

例题：两个部件 A 和 B 的重量是独立的正态变量。A ~ N(120, 5²)，B ~ N(80, 4²)。求总重量超过 215 g 的概率。

Let T = A + B. Then T ~ N(120 + 80, 5² + 4²) = N(200, 41).

设 T = A + B。则 T ~ N(120 + 80, 5² + 4²) = N(200, 41)。

Standardising: P(T > 215) = P(Z > (215 – 200)/√41) = P(Z > 2.34) ≈ 0.0096.

标准化：P(T > 215) = P(Z > (215 – 200)/√41) = P(Z > 2.34) ≈ 0.0096。

This shows how combining variances, not standard deviations, is essential.

这说明合并方差而不是标准差是至关重要的。

12. Exam Tips and Common Mistakes | 考试提示与常见错误

Do not add standard deviations; add variances for independent variables.

不要将标准差相加；对于独立变量，应将方差相加。

Remember that Var(X – Y) = Var(X) + Var(Y), never Var(X) – Var(Y).

记住 Var(X – Y) = Var(X) + Var(Y)，绝不是 Var(X) – Var(Y)。

Check the conditions for Poisson and binomial sums: independence and, for binomial, equal p.

检查泊松和二项之和的条件：独立性，以及二项分布中 p 相等。

For linear combinations, square any coefficients in the variance formula.

对于线性组合，在方差公式中要将系数平方。

When standardising a sum or mean, use the combined variance, then take the square root once at the end.

在对总和或均值进行标准化时，先使用合并后的方差，最后再取一次平方根。

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Sums of Independent Random Variables | 独立随机变量之和