Prussian Ascendancy: Poisson Distribution & the Prussian Army Horse-Kick Data | 普鲁士崛起：泊松分布与普鲁士军队马踢数据

📚 Prussian Ascendancy: Poisson Distribution & the Prussian Army Horse-Kick Data | 普鲁士崛起：泊松分布与普鲁士军队马踢数据

In Edexcel A-Level Mathematics, the Poisson distribution is a core topic in the statistics specification. It is used to model the number of times a rare event occurs in a fixed interval of time or space. One of the most famous historical illustrations comes from the Prussian army: the number of soldiers killed by horse kicks each year. This article uses that classic data set to build a rigorous understanding of the Poisson model, its assumptions, calculations, approximations, and hypothesis tests.

在爱德思 A-Level 数学中，泊松分布是统计学模块的核心主题。它用于建模稀有事件在固定时间或空间间隔内发生的次数。历史上最著名的实例之一来自普鲁士军队：每年被马踢死的士兵人数。本文利用这一经典数据集，系统构建对泊松模型、其假设、计算、近似和假设检验的严谨理解。

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1. Historical Context and the Prussian Data | 历史背景与普鲁士数据

In the late nineteenth century, the statistician Ladislaus Bortkiewicz studied the number of Prussian cavalry soldiers killed by horse kicks per army corps per year. His data covered 200 corps-years, with deaths ranging from 0 to 4. This data set became a landmark example for the Poisson distribution because the events were rare, independent, and occurred over a fixed time interval.

19 世纪末，统计学家拉迪斯劳斯·博特凯维奇研究了普鲁士骑兵军团每年被马踢死的士兵人数。他的数据涵盖 200 个军团年，死亡人数从 0 到 4 不等。该数据集成为泊松分布的里程碑式例子，因为这些事件稀有、独立且发生在固定时间间隔内。

The Prussian army kept detailed records of accidental deaths, which allowed Bortkiewicz to compare observed frequencies with theoretical probabilities. The close match between observation and theory convinced many scientists that the Poisson distribution was a practical tool, not just an abstract curiosity.

普鲁士军队保存了详细的意外死亡记录，这使博特凯维奇能够将观测频数与理论概率进行比较。观测与理论之间的高度吻合使许多科学家相信泊松分布是一种实用工具，而不仅仅是抽象的数学奇观。

In the Edexcel syllabus, this historical example often appears in exam questions as a context for calculating Poisson probabilities, estimating λ, or testing whether the Poisson model is appropriate.

在爱德思考试大纲中，这一历史实例经常作为考试题目的背景出现，用于计算泊松概率、估计 λ 或检验泊松模型是否合适。

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2. Defining the Poisson Distribution | 泊松分布的定义

The Poisson distribution models the number of times an event occurs in a fixed interval of time or space. A discrete random variable X is said to follow a Poisson distribution with parameter λ if its probability mass function is:

泊松分布用于建模事件在固定时间或空间间隔内发生的次数。若离散随机变量 X 服从参数为 λ 的泊松分布，其概率质量函数为：

P(X = k) = (λᵏ e⁻λ) ÷ k!, k = 0, 1, 2, …

The parameter λ represents the average number of occurrences in the given interval. It must be positive. In the Prussian horse-kick example, λ is the average number of deaths per corps-year.

参数 λ 表示给定区间内事件发生的平均次数。它必须为正。在普鲁士马踢例子中，λ 是每个军团年死亡人数的平均值。

The factorial k! in the denominator accounts for the number of ways k events can occur without regard to order. The term e⁻λ ensures that the probabilities sum to 1 over all possible values of k.

分母中的阶乘 k! 考虑了 k 个事件在不考虑顺序的情况下发生的方式数。项 e⁻λ 确保所有可能 k 值的概率之和为 1。

You will often see the notation X ~ Po(λ), which means ‘X follows a Poisson distribution with mean λ’.

你经常会看到记号 X ~ Po(λ)，表示“X 服从均值为 λ 的泊松分布”。

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3. Conditions for Using a Poisson Model | 使用泊松模型的条件

Before applying the Poisson distribution, you must check that the following conditions are satisfied:

在应用泊松分布之前，你必须检查是否满足以下条件：

• Events occur randomly and independently in a fixed interval. 事件在固定区间内随机且独立地发生。
• The average rate λ is constant throughout the interval. 平均发生率 λ 在整个区间内保持不变。
• Two events cannot occur at exactly the same instant. 两个事件不能在同一瞬间发生。
• The probability of an event occurring in a very short subinterval is proportional to the length of that subinterval. 在极短子区间内事件发生的概率与该子区间的长度成正比。

In the Prussian army data, horse-kick deaths were rare, independent, and the average rate of 0.61 per corps-year was stable across different years and corps. This makes the Poisson model highly appropriate.

在普鲁士军队数据中，马踢死亡事件稀有、独立，且每个军团年 0.61 的平均发生率在不同年份和军团之间保持稳定。这使得泊松模型高度适用。

In exam questions, you should always identify at least two conditions explicitly before using the Poisson distribution. For example, ‘deaths occur independently’ and ‘the rate is constant’ are enough to justify the model.

在考试题目中，你应该在使用泊松分布之前至少明确说明两个条件。例如，“死亡事件独立发生”和“发生率恒定”就足以证明使用该模型是合理的。

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4. Mean and Variance of the Poisson Distribution | 泊松分布的均值与方差

If X ~ Po(λ), then the expected value and variance are both equal to λ:

若 X ~ Po(λ)，则期望和方差都等于 λ：

E(X) = λ and Var(X) = λ

This equality is a key identifying feature of the Poisson distribution. It is often used in exam questions to justify a Poisson model when the sample mean and sample variance are approximately equal.

这一相等关系是泊松分布的关键识别特征。它经常在考试题目中用于在样本均值和样本方差近似相等时为泊松模型提供依据。

For the Prussian data, the sample mean was 0.61 and the sample variance can be calculated as follows. The sum of squares is 0²×109 + 1²×65 + 2²×22 + 3²×3 + 4²×1 = 196. The variance estimate is (196/200) − 0.61² ≈ 0.6079, which is very close to the mean. This supports the use of a Poisson distribution.

对于普鲁士数据，样本均值为 0.61，样本方差可以计算如下。平方和为 0²×109 + 1²×65 + 2²×22 + 3²×3 + 4²×1 = 196。方差估计为 (196/200) − 0.61² ≈ 0.6079，非常接近均值。这支持了泊松分布的使用。

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5. Worked Example: Prussian Horse-Kick Data | 例题：普鲁士马踢数据

The table below shows the observed frequencies of horse-kick deaths across 200 corps-years.

下表展示了 200 个军团年中马踢死亡人数的观测频数。

Deaths per corps-year (k)	Observed frequency
0	109
1	65
2	22
3	3
4	1
Total	200

The total number of deaths is 0×109 + 1×65 + 2×22 + 3×3 + 4×1 = 122. The sample mean λ̂ = 122/200 = 0.61 deaths per corps-year.

总死亡人数为 0×109 + 1×65 + 2×22 + 3×3 + 4×1 = 122。样本均值 λ̂ = 122/200 = 0.61 人/军团年。

Using λ = 0.61, the expected probabilities are calculated from the Poisson formula. For example, P(X = 2) = (0.61² × e⁻⁰·⁶¹) ÷ 2 ≈ 0.101. Multiplying by 200 gives an expected frequency of about 20.2 corps-years with exactly two deaths.

利用 λ = 0.61，根据泊松公式计算期望概率。例如，P(X = 2) = (0.61² × e⁻⁰·⁶¹) ÷ 2 ≈ 0.101。乘以 200 得到恰好发生两起死亡事件的期望频数约为 20.2 个军团年。

The full set of expected frequencies is: for k = 0, about 108.6; for k = 1, about 66.2; for k = 2, about 20.2; for k = 3, about 4.1; and for k = 4, about 0.6. These are remarkably close to the observed counts, which strongly supports the Poisson model.

完整的期望频数为：k = 0 时约为 108.6；k = 1 时约为 66.2；k = 2 时约为 20.2；k =

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