Additivity of the Poisson Distribution and Its Proof | 泊松分布的可加性及其证明

📚 Additivity of the Poisson Distribution and Its Proof | 泊松分布的可加性及其证明

The Poisson distribution is one of the most important discrete probability distributions in A-Level Further Mathematics, used to model the number of events occurring in a fixed interval of time or space. One of its most elegant and frequently examined properties is additivity: if two independent random variables follow Poisson distributions, then their sum also follows a Poisson distribution whose parameter is the sum of the original parameters.

泊松分布是 A-Level 进阶数学中最重要的离散型概率分布之一，常用于模拟在固定时间或空间区间内事件发生的次数。它最优雅且最常被考查的性质之一便是可加性：若两个独立的随机变量服从泊松分布，则它们的和也服从泊松分布，且其参数为原参数之和。

1. The Poisson Distribution: A Quick Recap | 泊松分布快速回顾

A random variable \(X\) is said to follow a Poisson distribution with parameter \(\lambda\) (where \(\lambda > 0\)) if its probability mass function is given by:

若随机变量 \(X\) 服从参数为 \(\lambda\)（其中 \(\lambda > 0\)）的泊松分布，则其概率质量函数为：

P(X = x) = e⁻ᵏ kˣ / x! , x = 0, 1, 2, …

Here, \(k\) is the average rate of occurrence, and both the mean and variance of the distribution are equal to \(k\). This is a defining feature: for a Poisson distribution, E(X) = Var(X) = k.

其中 \(k\) 为平均发生率，且该分布的均值与方差均等于 \(k\)。这是泊松分布的一个标志性特征：对于泊松分布，E(X) = Var(X) = k。

In Edexcel A-Level Further Mathematics, the Poisson distribution is denoted as X ~ Po(k). It is used to model rare events, such as the number of phone calls received by a call centre in one minute, or the number of defects in a metre of fabric.

在 Edexcel A-Level 进阶数学中，泊松分布记作 X ~ Po(k)。它用于模拟稀有事件，例如呼叫中心在一分钟内接到的电话数量，或一米布料上的瑕疵数量。

2. Statement of the Additivity Property | 可加性性质的陈述

Let X and Y be two independent random variables such that X ~ Po(k₁) and Y ~ Po(k₂). Then the sum T = X + Y also follows a Poisson distribution with parameter k₁ + k₂, i.e., T ~ Po(k₁ + k₂).

设 X 与 Y 为两个独立的随机变量，且 X ~ Po(k₁)，Y ~ Po(k₂)。则它们的和 T = X + Y 也服从泊松分布，参数为 k₁ + k₂，即 T ~ Po(k₁ + k₂)。

This property is known as the reproductive property or additivity (also called the convolution property) of the Poisson distribution. It is crucial to note that this holds only when X and Y are independent. If the variables are dependent, the result does not necessarily hold.

这一性质被称为泊松分布的可加性（也称再生性或卷积性）。需要特别强调的是，该性质仅在 X 与 Y 独立时成立。若变量之间存在相关性，则结论未必成立。

Intuitively, if one source of events produces on average k₁ events per unit time, and another independent source produces k₂ events per unit time, then the combined source produces on average k₁ + k₂ events per unit time.

直观上理解，如果一个事件源平均每单位时间产生 k₁ 个事件，另一个独立事件源平均每单位时间产生 k₂ 个事件，那么合并后的事件源平均每单位时间产生 k₁ + k₂ 个事件。

3. Proof: Method of Convolution | 证明：卷积法

The most direct proof of the additivity property uses the convolution formula for independent discrete random variables. For independent random variables X and Y, the probability that T = X + Y equals t is:

证明可加性最直接的方法是利用独立离散型随机变量的卷积公式。对于独立的随机变量 X 和 Y，T = X + Y = t 的概率为：

P(T = t) = Σ P(X = i) · P(Y = t − i)

where the summation runs over all possible values of i, namely i = 0, 1, 2, …, t. This formula arises from the Law of Total Probability, conditioning on all possible values of X.

其中求和遍历 i 的所有可能取值，即 i = 0, 1, 2, …, t。该公式由全概率公式导出，即对所有可能的 X 取值进行条件化。

Since X ~ Po(k₁) and Y ~ Po(k₂), we substitute the Poisson probability mass functions into the convolution formula:

由于 X ~ Po(k₁) 且 Y ~ Po(k₂)，我们将泊松概率质量函数代入卷积公式：

P(T = t) = Σᵢ₌₀ᵗ [ e⁻ᵏ¹ k₁ⁱ / i! ] · [ e⁻ᵏ² k₂ᵗ⁻ⁱ / (t − i)! ]

Here, the first term in each product corresponds to X, and the second to Y. Note that both exponential factors e⁻ᵏ¹ and e⁻ᵏ² do not depend on i, so they can be factored out of the summation.

这里每个乘积中的第一项对应 X，第二项对应 Y。注意 e⁻ᵏ¹ 和 e⁻ᵏ² 这两个指数因子均不依赖于 i，因此可以提到求和号外面。

4. Simplifying the Expression | 化简表达式

Factoring out the exponential terms, we obtain:

将指数因子提出后，我们得到：

P(T = t) = e⁻⁽ᵏ¹⁺ᵏ²⁾ Σᵢ₌₀ᵗ [ k₁ⁱ k₂ᵗ⁻ⁱ ] / [ i! (t − i)! ]

The remaining summation needs to be simplified. The key is to recognise that the binomial theorem will allow us to combine the terms. Recall that the binomial expansion of (k₁ + k₂)ᵗ is:

剩下的求和需要进一步化简。关键在于利用二项式定理来合并这些项。回顾 (k₁ + k₂)ᵗ 的二项式展开：

(k₁ + k₂)ᵗ = Σᵢ₌₀ᵗ [ t! / (i! (t − i)!) ] k₁ⁱ k₂ᵗ⁻ⁱ

Therefore, the summation \(\Sigma_{i=0}^{t} \frac{k_1^{i} k_2^{t-i}}{i!(t-i)!}\) can be rewritten by multiplying and dividing by t!:

因此，求和 Σᵢ₌₀ᵗ k₁ⁱ k₂ᵗ⁻ⁱ / [i!(t−i)!] 可以通过乘以并除以 t! 来改写：

Σᵢ₌₀ᵗ k₁ⁱ k₂ᵗ⁻ⁱ / [i!(t−i)!] = (1/t!) Σᵢ₌₀ᵗ [t! / (i!(t−i)!)] k₁ⁱ k₂ᵗ⁻ⁱ = (k₁ + k₂)ᵗ / t!

This is the crux of the proof: the summation collapses neatly into a single binomial power divided by t!.

这是证明的核心所在：该求和巧妙地收缩为一个二项式幂除以 t! 的形式。

5. The Final Step: Assembling the Result | 最后一步：汇总结果

Substituting the simplified summation back into our expression for P(T = t), we obtain:

将化简后的求和代回 P(T = t) 的表达式，我们得到：

P(T = t) = e⁻⁽ᵏ¹⁺ᵏ²⁾ (k₁ + k₂)ᵗ / t!

This is precisely the probability mass function of a Poisson distribution with parameter k₁ + k₂. Hence T ~ Po(k₁ + k₂), and the additivity property is proved.

这正是参数为 k₁ + k₂ 的泊松分布的概率质量函数。因此 T ~ Po(k₁ + k₂)，可加性得证。

The proof is elegant because the discrete convolution of two Poisson probability mass functions reproduces the same functional form, with only the parameter changing from k₁ and k₂ to their sum.

这一证明十分优美，因为两个泊松概率质量函数的离散卷积重现了相同的函数形式，仅参数从 k₁ 和 k₂ 变为它们的和。

6. Generalisation to n Independent Poisson Variables | 推广至 n 个独立泊松变量

The additivity property extends naturally by induction to any finite number of independent Poisson variables. If X₁, X₂, …, Xₙ are independent and Xᵢ ~ Po(kᵢ) for each i, then:

可加性可以通过数学归纳法自然地推广到任意有限个独立的泊松变量。若 X₁, X₂, …, Xₙ 相互独立且对每个 i 有 Xᵢ ~ Po(kᵢ)，则：

X₁ + X₂ + … + Xₙ ~ Po(k₁ + k₂ + … + kₙ)

For example, if the number of cars passing a checkpoint in one hour follows Po(3), and the number of motorcycles independently follows Po(1), then the total number of vehicles in one hour follows Po(4).

例如，若一小时内经过检查站的汽车数量服从 Po(3)，而摩托车数量独立地服从 Po(1)，则一小时内车辆总数服从 Po(4)。

This generalisation is particularly useful in real-world modelling because many systems aggregate multiple independent Poisson processes. In queueing theory, for instance, the superposition of independent Poisson arrival streams is itself a Poisson process.

这一推广在实际建模中尤为有用，因为许多系统汇聚了多个独立的泊松过程。例如在排队论中，多个独立泊松到达流的叠加本身仍是一个泊松过程。

7. Worked Example 1: Direct Application | 例题 1：直接应用

Problem: The number of emails received by a server in one minute is modelled by X ~ Po(2.5), and the number of text messages received by the same server in one minute is independently modelled by Y ~ Po(1.5). Find P(X + Y ≥ 6).

题目：服务器在一分钟内收到的电子邮件数量用 X ~ Po(2.5) 建模，同一服务器在一分钟内收到的短信数量独立地用 Y ~ Po(1.5) 建模。求 P(X + Y ≥ 6)。

Solution: By the additivity property, T = X + Y ~ Po(2.5 + 1.5) = Po(4). Therefore:

解答：根据可加性，T = X + Y ~ Po(2.5 + 1.5) = Po(4)。因此：

P(T ≥ 6) = 1 − P(T ≤ 5)

Using the Poisson cumulative distribution function with parameter 4:

使用参数为 4 的泊松累积分布函数：

P(T ≤ 5) = e⁻⁴ Σᵗ₌₀⁵ 4ᵗ / t!

This gives P(T ≤ 5) ≈ 0.7851. Hence P(T ≥ 6) ≈ 1 − 0.7851 = 0.2149.

计算得 P(T ≤ 5) ≈ 0.7851。因此 P(T ≥ 6) ≈ 1 − 0.7851 = 0.2149。

This example demonstrates the power of the additivity property: instead of summing over many combinations of X and Y values, we simply transform the problem into a single Poisson calculation.

此例展示了可加性的威力：无需对 X 和 Y 的多种取值组合求和，只需将问题转化为一次单独的泊松计算即可。

8. Worked Example 2: Reversing the Property | 例题 2：逆向应用

Problem: A pooled sample of rainfall measurements shows that the total number of rainy days in two consecutive months is T ~ Po(9). The first month’s rainy days are known to follow Po(5), independent of the second month. Find the distribution of rainy days in the second month.

题目：某地连续两个月的降雨天数总和为 T ~ Po(9)。已知第一个月的降雨天数服从 Po(5)，且与第二个月独立。求第二个月降雨天数的分布。

Solution: Let X be the first month’s rainy days and Y the second month’s. Since T = X + Y ~ Po(5 + k₂) = Po(9), we have 5 + k₂ = 9, so k₂ = 4. Thus Y ~ Po(4).

解答：设 X 为第一个月降雨天数，Y 为第二个月降雨天数。由于 T = X + Y ~ Po(5 + k₂) = Po(9)，可得 5 + k₂ = 9，故 k₂ = 4。因此 Y ~ Po(4)。

This reverse application is often tested in Edexcel exams. It relies on the fact that the Poisson distribution is uniquely determined by its parameter: if T is Poisson and one component is Poisson, the other component must also be Poisson if independence holds.

这种逆向应用在 Edexcel 考试中经常出现。它依赖于泊松分布由参数唯一确定这一事实：若总和服从泊松分布且其中一个分量服从泊松分布，那么在独立性成立的前提下，另一个分量也必然服从泊松分布。

9. Common Pitfalls and Exam Tips | 常见误区与考试提示

The additivity property requires independence. If X and Y are not independent, X + Y may not be Poisson. Always check the independence condition in exam questions.
可加性要求独立。若 X 与 Y 不独立，则 X + Y 未必服从泊松分布。做题时务必检查题中是否给出独立性条件。
Do not confuse additivity of Poisson variables with the scaling property. If X ~ Po(k), then 2X is not Po(2k). The property only applies to the sum of independent variables, not to multiplying a single variable by a constant.
不要将泊松变量的可加性与缩放性质混淆。若 X ~ Po(k)，则 2X 并不服从 Po(2k)。该性质只适用于独立变量的和，而非将单个变量乘以常数。
When calculating probabilities for sums of Poisson variables, always combine the parameters first and then use the cumulative tables or the formula. Applying the convolution directly is unnecessarily time-consuming and error-prone.
计算泊松变量和的概率时，务必先合并参数，再使用累积分布表或公式。直接使用卷积计算既费时又容易出错。

In Edexcel A-Level Further Mathematics Paper 2 (or Paper 3 depending on route), Poisson additivity is frequently tested alongside the normal approximation to the Poisson distribution. A common question type asks students to find the probability that the sum of two or more independent Poisson variables exceeds a threshold, which is solved by first combining parameters and then applying the normal approximation with continuity correction.

在 Edexcel A-Level 进阶数学 Paper 2（或根据路线不同为 Paper 3）中，泊松可加性常与泊松分布的正态近似结合考查。常见题型是要求学生求多个独立泊松变量之和超过某一阈值的概率，解法为先合并参数，再应用带连续性修正的正态近似。

10. Connection to the Normal Distribution | 与正态分布的联系

Since T ~ Po(k₁ + k₂), when k₁ + k₂ is large (typically > 10), a normal approximation can be used: T ≈ N(k₁ + k₂, k₁ + k₂). This is because for a Poisson distribution, both mean and variance equal the parameter.

由于 T ~ Po(k₁ + k₂)，当 k₁ + k₂ 较大时（通常 > 10），可以使用正态近似：T ≈ N(k₁ + k₂, k₁ + k₂)。这是因为泊松分布的均值与方差均等于其参数。

For example, if X ~ Po(20) and Y ~ Po(30) are independent, then X + Y ~ Po(50), which can be approximated by N(50, 50). A continuity correction is applied when computing probabilities such as P(X + Y ≥ 60) ≈ P(W ≥ 59.5) where W ~ N(50, 50).

例如，若 X ~ Po(20) 与 Y ~ Po(30) 独立，则 X + Y ~ Po(50)，可用 N(50, 50) 近似。计算 P(X + Y ≥ 60) 等概率时需应用连续性修正，即 P(W ≥ 59.5)，其中 W ~ N(50, 50)。

This connection is essential in hypothesis testing and confidence interval estimation in A-Level further statistics papers. Candidates should be comfortable switching between exact Poisson calculations and normal approximations.

这种联系在 A-Level 进阶统计试卷中的假设检验和置信区间估计中至关重要。考生应熟练掌握在精确泊松计算与正态近似之间灵活转换。

11. Mean and Variance Verification | 均值与方差的验证

We can also verify the additivity property through the moment-generating function or directly through mean and variance. Since E(X) = k₁ and E(Y) = k₂, by linearity of expectation:

我们还可以通过矩母函数或直接通过均值与方差来验证可加性。由于 E(X) = k₁ 且 E(Y) = k₂，根据期望的线性性质：

E(T) = E(X) + E(Y) = k₁ + k₂

Similarly, because X and Y are independent, their covariance is zero, so:

类似地，由于 X 与 Y 独立，其协方差为零，因此：

Var(T) = Var(X) + Var(Y) = k₁ + k₂

The fact that E(T) = Var(T) = k₁ + k₂ is consistent with T being Poisson with parameter k₁ + k₂. However, equality of mean and variance alone is necessary but not sufficient to prove Poissonity; the full distributional proof via convolution is required.

E(T) = Var(T) = k₁ + k₂ 这一事实与 T 为参数 k₁ + k₂ 的泊松分布相吻合。然而，均值与方差相等仅是必要而非充分条件；要证明泊松性还需完整的卷积证明。

For completeness, using the moment-generating function, the MGF of a Poisson variable is M(t) = exp(k(eᵗ − 1)). For independent variables, the MGF of the sum is the product of the individual MGFs:

为完整起见，泊松分布的矩母函数为 M(t) = exp(k(eᵗ − 1))。对于独立变量，和的矩母函数为各矩母函数之积：

M_T(t) = M_X(t) · M_Y(t) = exp(k₁(eᵗ − 1)) · exp(k₂(eᵗ − 1)) = exp((k₁ + k₂)(eᵗ − 1))

This is exactly the MGF of a Po(k₁ + k₂) distribution, providing an alternative proof.

这恰好是 Po(k₁ + k₂) 分布的矩母函数，提供了另一种证明方法。

12. Summary and Final Advice | 总结与最终建议

The additivity of the Poisson distribution is a fundamental result in A-Level Further Mathematics. It states that the sum of independent Poisson variables is Poisson, with parameters added together. The proof via convolution is straightforward and relies on the binomial theorem to simplify the summation.

泊松分布的可加性是 A-Level 进阶数学中的基本结论。它指出独立泊松变量之和仍服从泊松分布，参数相加。通过卷积法进行的证明步骤清晰，依赖于二项式定理来化简求和。

To succeed in exams, students should: (1) memorise the statement of the additivity property, (2) be able to apply it to find probabilities efficiently, (3) recognise when independence is required, and (4) understand how the property links to the normal approximation for large parameters.

为在考试中取得成功，学生应：(1) 牢记可加性性质的表述；(2) 能高效运用它求解概率；(3) 识别何时需要独立性条件；(4) 理解该性质如何与参数较大时的正态近似相联系。

The proof presented here also demonstrates a broader mathematical skill: using convolution, the binomial theorem, and algebraic manipulation to prove distributional results. These techniques recur throughout Further Mathematics and are worth mastering thoroughly.

这里展示的证明过程也体现了一种更广泛的数学技能：运用卷积、二项式定理和代数运算来证明分布性质的结论。这些技巧在进阶数学中反复出现，值得深入掌握。

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Additivity of the Poisson Distribution and Its Proof | 泊松分布的可加性及其证明