A-Level Further Mathematics: Distribution of the Sum of Independent Random Variables

📚 A-Level Further Mathematics: Distribution of the Sum of Independent Random Variables | 独立随机变量之和的分布

When two (or more) independent random variables are added together, the probability distribution of their sum is not simply the sum of the original distributions. Instead, we must use a technique called convolution. This article explains the key results and methods for the Edexcel A-Level Further Mathematics syllabus.

当两个（或多个）独立随机变量相加时，它们之和的概率分布并不是简单地把原来的分布相加。相反，我们需要使用一种称为“卷积”的技巧。本文为 Edexcel A-Level 进阶数学考纲讲解核心结论与求解方法。

1. Discrete Case: The Convolution Formula | 离散情形：卷积公式

Let X and Y be independent discrete random variables. Their sum is S = X + Y. The probability that S equals a particular value s is found by summing over all possible pairs (x, y) that satisfy x + y = s.

设 X 与 Y 为独立离散随机变量，其和记为 S = X + Y。S 取某个特定值 s 的概率，需要对所有满足 x + y = s 的配对 (x, y) 求和。

P(S = s) = Σₓ P(X = x) × P(Y = s − x)

Here the sum is taken over all values of x for which the probabilities are nonzero. This is the discrete convolution of the two probability mass functions. Independence is essential: the joint probability factorises as the product of the two marginals.

这里对 x 的所有非零概率取值求和。这就是两个概率质量函数的离散卷积。独立性至关重要：联合概率可以分解为两个边缘概率的乘积。

2. Sum of Independent Binomial Variables | 独立二项分布变量之和

Suppose X ~ B(n₁, p) and Y ~ B(n₂, p), with X and Y independent. Because both variables count successes in independent trials with the same success probability p, adding them counts the total number of successes in n₁ + n₂ trials. Therefore:

设 X ~ B(n₁, p)、Y ~ B(n₂, p)，且 X 与 Y 独立。由于两个变量都统计成功概率相同 p 的独立试验中的成功次数，把它们相加就是在 n₁ + n₂ 次试验中统计总成功次数。因此：

X + Y ~ B(n₁ + n₂, p)

This result holds only when the probability p is the same for both binomial variables. If the probabilities differ, the sum is not binomial.

只有当两个二项分布的成功概率 p 相同才能这样合并。如果概率不同，则和不再服从二项分布。

3. Sum of Independent Poisson Variables | 独立泊松分布变量之和

Let X ~ Po(λ₁) and Y ~ Po(λ₂) be independent. The convolution formula gives

设 X ~ Po(λ₁)、Y ~ Po(λ₂) 且相互独立。由卷积公式可得

P(X + Y = s) = e⁻⁽λ₁⁺λ₂⁾ × (λ₁ + λ₂)ˢ / s!

and hence X + Y ~ Po(λ₁ + λ₂). This is derived using the binomial expansion of (λ₁ + λ₂)ˢ. The parameter simply adds. This result is used frequently in modelling counts over combined time intervals.

因此 X + Y ~ Po(λ₁ + λ₂)。推导过程中用到了 (λ₁ + λ₂)ˢ 的二项展开。参数直接相加。这一结论常用于合并时间区间内计数的建模。

Original distributions	Distribution of the sum
X ~ Po(λ₁), Y ~ Po(λ₂)	X + Y ~ Po(λ₁ + λ₂)
X ~ B(n₁, p), Y ~ B(n₂, p)	X + Y ~ B(n₁ + n₂, p)
X ~ N(μ₁, σ₁²), Y ~ N(μ₂, σ₂²)	X + Y ~ N(μ₁ + μ₂, σ₁² + σ₂²)
X ~ U(0,1), Y ~ U(0,1)	Triangular distribution on (0,2)

4. Continuous Case: Convolution Integral | 连续情形：卷积积分

For independent continuous random variables X and Y with probability density functions fₓ(x) and fᵧ(y), the density of S = X + Y is

对于具有概率密度函数 fₓ(x) 和 fᵧ(y) 的独立连续随机变量 X 与 Y，其和 S = X + Y 的密度为

fₛ(s) = ∫ fₓ(x) fᵧ(s − x) dx

with the integral taken over all real x. Intuitively, we fix one variable at x and require the other to equal s − x. The convolution measures the total “overlap” of the two densities as they slide past each other.

积分对全体实数 x 进行。直观上，固定一个变量为 x，要求另一个变量等于 s − x。卷积衡量两个密度函数彼此滑动时的总“重叠量”。

5. Example: Sum of Two Uniform Variables | 示例：两个均匀分布变量之和

Let X ~ U(0,1) and Y ~ U(0,1) be independent. Applying the convolution integral gives a piecewise triangular density:

设 X ~ U(0,1)、Y ~ U(0,1) 且相互独立。应用卷积积分得到分段三角形的密度函数：

fₛ(s) = s for 0 ≤ s ≤ 1, and fₛ(s) = 2 − s for 1 ≤ s ≤ 2

Outside the interval [0,2], the density is zero. The density peaks at s = 1, where fₛ(1) = 1. This triangular shape is a classic illustration of the convolution of two uniform rectangular densities.

在区间 [0,2] 之外，密度为零。密度在 s = 1 处达到峰值，此时 fₛ(1) = 1。这个三角形形状是说明两个均匀矩形密度卷积的经典示例。

6. Sum of Independent Normal Variables | 独立正态分布变量之和

If X ~ N(μ₁, σ₁²) and Y ~ N(μ₂, σ₂²) are independent, their sum is again normal:

若 X ~ N(μ₁, σ₁²) 与 Y ~ N(μ₂, σ₂²) 独立，则它们的和仍为正态分布：

X + Y ~ N(μ₁ + μ₂, σ₁² + σ₂²)

Notice that the means add and the variances add, but standard deviations do not add. The standard deviation of the sum is √(σ₁² + σ₂²). This result is central to many exam problems involving the sum or difference of independent measurements.

注意：均值相加、方差相加，但标准差并不直接相加。和的标准差是 √(σ₁² + σ₂²)。这一结论是考试中处理独立测量值之和或差的核心依据。

7. Linear Combinations of Independent Normal Variables | 独立正态变量的线性组合

More generally, for constants a and b, if X and Y are independent normal variables then

更一般地，对常数 a 与 b，若 X、Y 为独立正态变量，则

E(aX + bY) = aμ₁ + bμ₂

Var(aX + bY) = a²σ₁² + b²σ₂²

The distribution of aX + bY is normal. This generalisation appears in risk analysis, portfolio theory and engineering tolerances. When subtracting two independent normal variables, the variance still adds because the coefficient b = −1 gives b² = 1.

aX + bY 服从正态分布。该推广出现在风险分析、投资组合理论和工程公差中。当两个独立正态变量相减时，方差仍然相加，因为系数 b = −1 时 b² = 1。

8. Moment Generating Functions | 矩生成函数

The moment generating function (MGF) of a random variable X is defined as Mₓ(t) = E(eᵗˣ), provided the expectation exists. A key property is that the MGF of the sum of independent variables is the product of their MGFs:

随机变量 X 的矩生成函数定义为 Mₓ(t) = E(eᵗˣ)，前提是该期望存在。一个关键性质是：独立变量之和的矩生成函数等于它们各自矩生成函数的乘积：

Mₓ₊ᵧ(t) = Mₓ(t) Mᵧ(t)

Because the MGF uniquely determines the distribution, this provides a powerful proof technique. For instance, the MGF of a normal distribution N(μ, σ²) is exp(μt + ½σ²t²); multiplying two such MGFs gives another normal MGF with mean μ₁ + μ₂ and variance σ₁² + σ₂². The same method proves the Poisson addition rule.

由于矩生成函数唯一决定分布，这成为强大的证明工具。例如，正态分布 N(μ, σ²) 的矩生成函数为 exp(μt + ½σ²t²)；将两个这样的矩生成函数相乘，就得到另一个正态矩生成函数，其均值为 μ₁ + μ₂，方差为 σ₁² + σ₂²。同样的方法也可证明泊松分布的可加性。

9. Central Limit Theorem | 中心极限定理

For a random sample X₁, X₂, …, Xₙ of independent observations each with mean μ and variance σ², the sum Sₙ = X₁ + X₂ + … + Xₙ has mean nμ and variance nσ². The Central Limit Theorem states that for large n, Sₙ is approximately normal. Standardising gives

对于一组独立观测值 X₁, X₂, …, Xₙ，每个观测的均值为 μ、方差为 σ²，其和 Sₙ = X₁ + X₂ + … + Xₙ 的均值为 nμ、方差为 nσ²。中心极限定理表明：当 n 足够大时，Sₙ 近似服从正态分布。标准化后得到

Z = (Sₙ − nμ) / (√n σ) 近似服从 N(0,1)

This theorem justifies using the normal approximation for sums of many small independent effects, even when the individual variables are not normally distributed. In exams, check that n is large enough before applying this approximation.

该定理为用正态近似处理许多微小独立效应之和提供了依据，即使单个变量本身并非正态分布。考试中应用此近似前，务必检验 n 是否足够大。

10. Exam Tips and Common Mistakes | 考点分析与常见错误

A common error is to add standard deviations instead of variances when combining normal variables. Always square the standard deviations first, then add, then take the square root. Another frequent mistake is forgetting to check independence before using the product rule for MGFs or the convolution formula.

常见错误之一：合并正态变量时把标准差直接相加，而不是先求方差再相加。正确做法是先将标准差平方、相加、再开方。另一个常见错误：使用矩生成函数乘积法则或卷积公式前，忘记确认独立性。

Check if the sum of two binomial variables requires the same p.

核对两个二项分布变量相加时是否要求相同 p。
For Poisson sums, simply add the parameters λ.

泊松分布之和，直接相加参数 λ。
When using the normal approximation to the binomial, verify np > 5 and n(1 − p) > 5.

用正态近似二项分布时，验证 np > 5 且 n(1 − p) > 5。
Remember that Var(aX + bY) uses squared coefficients.

记住 Var(aX + bY) 中的系数要平方。

11. Summary | 总结

The distribution of the sum of independent random variables is a core topic in A-Level Further Mathematics. Use convolution for general discrete or continuous variables, remember the addition rules for binomial, Poisson and normal distributions, and apply the MGF or Central Limit Theorem where appropriate.

独立随机变量之和的分布是 A-Level 进阶数学的核心内容。对一般离散或连续变量使用卷积；牢记二项、泊松和正态分布的可加规则；在合适场合使用矩生成函数或中心极限定理。

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A-Level Further Mathematics: Distribution of the Sum of Independent Random Variables | 独立随机变量之和的分布