Poisson Distribution Mean and Variance | 泊松分布的均值与方差

📚 Poisson Distribution Mean and Variance | 泊松分布的均值与方差

The Poisson distribution is one of the most important discrete probability distributions in A-Level Further Mathematics. Named after the French mathematician Siméon Denis Poisson, it models the number of events occurring in a fixed interval of time or space, given that these events happen with a known constant mean rate and independently of each other. In this article, we will derive the mean and variance of the Poisson distribution from first principles, explore their key properties, and work through exam-style examples to solidify your understanding.

泊松分布是进阶数学中最重要的离散概率分布之一。它得名于法国数学家西梅翁·德尼·泊松，用于描述在固定时间或空间区间内事件发生的次数，前提是事件以已知的恒定平均速率独立发生。在本文中，我们将从基本原理出发推导泊松分布的均值与方差，探讨它们的关键性质，并通过考试风格的例题来巩固你的理解。

1. The Poisson Distribution Formula | 泊松分布公式

Let X be a Poisson random variable with parameter λ, where λ (lambda) represents the average number of events in the interval. The probability mass function (PMF) is given by:

设 X 为参数为 λ 的泊松随机变量，其中 λ 表示区间内事件的平均发生次数。其概率质量函数为：

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

Here, e is Euler’s number (approximately 2.71828), λ > 0 is the rate parameter, and x! denotes the factorial of x. The distribution is denoted as X ~ Po(λ). Note that the parameter λ is both the mean and the variance of the distribution — a property we will prove rigorously in the following sections.

其中 e 是欧拉数（约等于 2.71828），λ > 0 是速率参数，x! 表示 x 的阶乘。该分布记为 X ~ Po(λ)。注意参数 λ 既是分布的均值也是方差——我们将在接下来的章节中严格证明这一性质。

2. Verifying the Total Probability | 验证总概率

Before deriving the mean, we must first confirm that the Poisson PMF is a valid probability distribution, meaning the sum of all probabilities equals 1. Using the Maclaurin series expansion of e^λ:

在推导均值之前，我们必须先确认泊松概率质量函数是一个有效的概率分布，即所有概率之和等于 1。利用 e^λ 的麦克劳林级数展开：

eᵏ = Σ (λˣ / x!) from x=0 to ∞

Therefore, summing the PMF over all x:

因此，对所有 x 求和泊松概率质量函数：

Σ P(X = x) = Σ e⁻ᵏ λˣ / x! = e⁻ᵏ · eᵏ = 1

This confirms that the probabilities sum to 1, validating the distribution. This summation technique — factoring out e⁻^λ and recognising the exponential series — is the cornerstone of all moment derivations that follow.

这证实了概率之和为 1，验证了分布的有效性。这种求和技巧——提出 e⁻ᵏ 并识别出指数级数——是后续所有矩推导的基石。

3. Defining the Mean (Expected Value) | 定义均值（期望值）

For a discrete random variable, the expected value (mean) is defined as:

对于离散随机变量，期望值（均值）定义为：

E(X) = Σ x · P(X = x) over all x

Applying this definition to the Poisson distribution, we write:

将此定义应用于泊松分布，我们写出：

E(X) = Σ (x · e⁻ᵏ λˣ / x!) from x=0 to ∞

The x = 0 term contributes zero to the sum, so we can start the summation from x = 1. Noting that x / x! = 1 / (x−1)! for x ≥ 1, we simplify:

x = 0 的项对总和的贡献为零，因此我们可以从 x = 1 开始求和。注意到当 x ≥ 1 时，x / x! = 1 / (x−1)!，我们化简得：

E(X) = e⁻ᵏ Σ (λˣ / (x−1)!) from x=1 to ∞

Now introduce a change of variable y = x − 1. As x goes from 1 to ∞, y goes from 0 to ∞, and λˣ = λʸ⁺¹ = λ · λʸ. Thus:

现在引入变量替换 y = x − 1。当 x 从 1 到 ∞ 时，y 从 0 到 ∞，且 λˣ = λʸ⁺¹ = λ · λʸ。因此：

E(X) = e⁻ᵏ · λ · Σ (λʸ / y!) from y=0 to ∞ = e⁻ᵏ · λ · eᵏ = λ

Hence, the mean of a Poisson distribution is exactly equal to its parameter λ. This elegant result aligns with the intuitive interpretation of λ as the average rate of occurrence.

因此，泊松分布的均值恰好等于其参数 λ。这一简洁的结果与 λ 作为平均发生速率的直观解释相吻合。

4. Deriving E(X²) | 推导 E(X²)

To find the variance, we first need E(X²). A standard trick in probability theory is to express x² as x(x−1) + x. This decomposition separates the factorial cancelation into two cleaner series:

为了求方差，我们首先需要 E(X²)。概率论中的一个标准技巧是将 x² 表示为 x(x−1) + x。这种分解将阶乘消去分成两个更清晰的级数：

E(X²) = E[X(X−1)] + E(X)

We already know E(X) = λ from Section 3. Now compute E[X(X−1)]:

我们从第 3 节已知 E(X) = λ。现在计算 E[X(X−1)]：

E[X(X−1)] = Σ [x(x−1) · e⁻ᵏ λˣ / x!] from x=0 to ∞

For x = 0 and x = 1, the term x(x−1) = 0, so the summation effectively starts at x = 2. Using the identity x(x−1) / x! = 1 / (x−2)! for x ≥ 2, we get:

当 x = 0 和 x = 1 时，项 x(x−1) = 0，因此求和实际上从 x = 2 开始。利用恒等式：当 x ≥ 2 时，x(x−1) / x! = 1 / (x−2)!，我们得到：

E[X(X−1)] = e⁻ᵏ Σ (λˣ / (x−2)!) from x=2 to ∞

Let z = x − 2. Then λˣ = λᶻ⁺² = λ² · λᶻ, and the series becomes:

令 z = x − 2。则 λˣ = λᶻ⁺² = λ² · λᶻ，级数变为：

E[X(X−1)] = e⁻ᵏ · λ² · Σ (λᶻ / z!) from z=0 to ∞ = e⁻ᵏ · λ² · eᵏ = λ²

Substituting back: E(X²) = λ² + λ. This intermediate result is essential for computing the variance.

代回：E(X²) = λ² + λ。这一中间结果对于计算方差至关重要。

5. Deriving the Variance | 推导方差

Using the fundamental variance formula:

使用方差的基本公式：

Var(X) = E(X²) − [E(X)]²

Substitute E(X²) = λ² + λ and E(X) = λ:

代入 E(X²) = λ² + λ 和 E(X) = λ：

Var(X) = (λ² + λ) − λ² = λ

Thus, the variance of a Poisson distribution is also exactly λ. This is a remarkable characteristic: for a Poisson distribution, the mean equals the variance, both being λ. This property is known as equidispersion and is a key diagnostic feature used to identify Poisson-distributed data in real-world statistical analysis.

因此，泊松分布的方差也恰好是 λ。这是一个非凡的特性：对于泊松分布，均值等于方差，都等于 λ。这一性质称为等离散性，是现实统计分析中识别泊松分布数据的关键诊断特征。

6. Moment Generating Function Approach (Optional) | 矩生成函数方法（选学）

For a deeper understanding, we can also derive both moments via the moment generating function (MGF), though this is not strictly required by the Edexcel A-Level syllabus. The MGF of X ~ Po(λ) is M(t) = e^(λ(eᵗ − 1)). Differentiating with respect to t and evaluating at t = 0:

为了更深入的理解，我们还可以通过矩生成函数来推导这两个矩，尽管这不是 Edexcel A-Level 考纲的严格要求。X ~ Po(λ) 的矩生成函数为 M(t) = e^(λ(eᵗ − 1))。对 t 求导并在 t = 0 处取值：

M'(t) = λeᵗ · e^(λ(eᵗ − 1)), M'(0) = λ

M”(t) = λeᵗ · e^(λ(eᵗ − 1)) + λ²e²ᵗ · e^(λ(eᵗ − 1)), M”(0) = λ + λ²

Since E(X) = M'(0) and E(X²) = M”(0), we immediately recover E(X) = λ and Var(X) = λ + λ² − λ² = λ. This method is elegant but relies on background knowledge beyond the standard syllabus; the direct summation method remains the expected exam technique.

由于 E(X) = M'(0) 且 E(X²) = M”(0)，我们立即得到 E(X) = λ 和 Var(X) = λ + λ² − λ² = λ。这种方法很优雅，但依赖于超出标准考纲的背景知识；直接求和法仍然是考试中预期的方法。

7. Key Properties and Their Meanings | 关键性质及其含义

The equality of mean and variance has several important implications in exams and applications:

均值与方差的相等在考试和应用中有几个重要含义：

Single-parameter distribution: Knowledge of the mean automatically determines the variance, and vice versa. The entire distribution is characterized by one number λ.
Diagnostic tool: When analysing data, if the sample mean and sample variance are approximately equal, a Poisson model is plausible. If the variance is much larger than the mean (overdispersion), alternative models such as the negative binomial should be considered.
Calculating probabilities: For exam problems, once you estimate λ from data (usually the sample mean), you can compute any probability using the PMF.

单参数分布：知道均值就自动确定方差，反之亦然。整个分布由单个数字 λ 刻画。

诊断工具：在分析数据时，如果样本均值和样本方差大致相等，则泊松模型是合理的。如果方差远大于均值（过离散），则应考虑负二项分布等替代模型。

计算概率：对于考试题目，一旦你从数据中估计出 λ（通常是样本均值），你就可以使用概率质量函数计算任何概率。

8. Worked Example 1: Direct Calculation | 例题 1：直接计算

Suppose X ~ Po(5). Find E(X), Var(X), and P(X = 4).

设 X ~ Po(5)。求 E(X)、Var(X) 和 P(X = 4)。

Solution: Since the mean and variance of a Poisson distribution both equal λ, we have E(X) = 5 and Var(X) = 5 immediately. For the probability:

解答：由于泊松分布的均值和方差都等于 λ，我们立即有 E(X) = 5 和 Var(X) = 5。对于概率：

P(X = 4) = e⁻⁵ · 5⁴ / 4! = e⁻⁵ · 625 / 24 ≈ 0.1755

Notice that the standard deviation is √5 ≈ 2.236, illustrating that the spread of the distribution grows with the mean — larger λ leads to greater variability.

注意标准差是 √5 ≈ 2.236，这说明分布的范围随均值增长——更大的 λ 导致更大的变异性。

9. Worked Example 2: Estimating λ from Data | 例题 2：从数据估计 λ

A call centre receives an average of 2.4 calls per minute. Let Y be the number of calls received in a randomly selected minute. Assuming a Poisson model, find Var(Y) and P(Y ≥ 3).

某呼叫中心平均每分钟接到 2.4 个电话。设 Y 为随机选取的一分钟内接到的电话数。假设泊松模型，求 Var(Y) 和 P(Y ≥ 3)。

Solution: Here λ = 2.4, so Var(Y) = 2.4. For P(Y ≥ 3), it is easier to use the complement rule:

解答：这里 λ = 2.4，所以 Var(Y) = 2.4。对于 P(Y ≥ 3)，使用补事件法则更方便：

P(Y ≥ 3) = 1 − P(Y = 0) − P(Y = 1) − P(Y = 2)

= 1 − e⁻²·⁴ − 2.4e⁻²·⁴ − (2.4² / 2)e⁻²·⁴ = 1 − e⁻²·⁴ (1 + 2.4 + 2.88)

= 1 − 6.28e⁻²·⁴ ≈ 1 − 6.28 × 0.0907 ≈ 0.4305

This example shows how the variance directly informs the spread of probability across possible outcomes.

这个例子展示了方差如何直接反映概率在各种可能结果之间的分布范围。

10. Common Exam Pitfalls | 常见考试陷阱

Students often make the following mistakes when working with the mean and variance of the Poisson distribution. Avoid them to secure full marks:

学生在处理泊松分布的均值和方差时经常犯以下错误。避免它们以确保满分：

Confusing λ with the standard deviation: The standard deviation is √λ, not λ. For example, if λ = 9, the standard deviation is 3.
Forgetting the condition for Poisson approximation: When using the Poisson distribution to approximate a binomial distribution B(n, p), we require n large (typically n ≥ 50) and p small (typically p ≤ 0.1), with λ = np.
Using the wrong units: If the average rate is given per hour but the question asks for a 30-minute interval, you must rescale λ. For example, 10 events per hour implies λ = 5 for 30 minutes. The variance rescales in exactly the same way.
Assuming independence without checking: The Poisson model assumes events occur independently. If events cluster, the Poisson distribution is invalid, and the variance would not equal the mean.

将 λ 与标准差混淆：标准差是 √λ，而不是 λ。例如，若 λ = 9，标准差是 3。

忘记泊松近似的条件：当使用泊松分布近似二项分布 B(n, p) 时，需要 n 大（通常 n ≥ 50）且 p 小（通常 p ≤ 0.1），其中 λ = np。

使用错误的单位：如果平均速率以每小时给出，但题目问的是 30 分钟区间，你必须重新调整 λ。例如，每小时 10 个事件意味着 30 分钟内 λ = 5。方差也以完全相同的方式重新调整。

不检查独立性就假设成立：泊松模型假设事件独立发生。如果事件聚集出现，泊松分布无效，方差不会等于均值。

11. Sum of Independent Poisson Variables | 独立泊松变量之和

An important extension: if X ~ Po(λ₁) and Y ~ Po(λ₂) are independent, then X + Y ~ Po(λ₁ + λ₂). This can be proved using the mean and variance properties:

一个重要扩展：如果 X ~ Po(λ₁) 和 Y ~ Po(λ₂) 独立，则 X + Y ~ Po(λ₁ + λ₂)。这可以使用均值和方差性质来证明：

E(X + Y) = E(X) + E(Y) = λ₁ + λ₂

Var(X + Y) = Var(X) + Var(Y) = λ₁ + λ₂

Since the mean equals the variance for X + Y, it must be Poisson with parameter λ₁ + λ₂. This result is extremely useful for exam questions that combine multiple independent Poisson processes, such as counting the total number of cars passing two different junctions or total arrivals from two independent sources. Note that this property relies on independence; if the variables are correlated, the variance of the sum includes a covariance term.

由于 X + Y 的均值等于方差，它必然是参数为 λ₁ + λ₂ 的泊松分布。这个结果在结合多个独立泊松过程的考试题中极为有用，例如计算通过两个不同路口的汽车总数或来自两个独立来源的总到达数。注意该性质依赖独立性；如果变量相关，和的方差将包含协方差项。

12. Practice Questions and Revision Tips | 练习题与复习建议

To master this topic, attempt the following questions and then check the key steps below:

为了掌握这个主题，尝试以下问题，然后核对下面的关键步骤：

Question 1: If X ~ Po(λ) and Var(X) = 16, find P(X > 12).
Question 2: The number of errors on a page follows a Poisson distribution with mean 0.5. Find the probability that a page has at least one error, and state the variance.
Question 3: Two independent Poisson variables U ~ Po(3) and V ~ Po(7). Find the mean and variance of U + V, and compute P(U + V = 5).

问题 1：如果 X ~ Po(λ) 且 Var(X) = 16，求 P(X > 12)。

问题 2：一页上的错误数服从均值为 0.5 的泊松分布。求一页至少有一个错误的概率，并说明方差。

问题 3：两个独立泊松变量 U ~ Po(3) 和 V ~ Po(7)。求 U + V 的均值和方差，并计算 P(U + V = 5)。

Hints and answers:

提示与答案：

For Question 1, since Var(X) = λ = 16, we have λ = 16 and P(X > 12) = 1 − P(X ≤ 12). Using a calculator or tables, this is approximately 0.868. For Question 2, P(at least one error) = 1 − e⁻⁰·⁵ ≈ 0.3935, and the variance is 0.5. For Question 3, U + V ~ Po(10), so mean = 10 and variance = 10; P(U + V = 5) = e⁻¹⁰ · 10⁵ / 5! ≈ 0.0378.

对于问题 1，由于 Var(X) = λ = 16，我们有 λ = 16 且 P(X > 12) = 1 − P(X ≤ 12)。使用计算器或查表，约为 0.868。对于问题 2，P(至少一个错误) = 1 − e⁻⁰·⁵ ≈ 0.3935，方差为 0.5。对于问题 3，U + V ~ Po(10)，所以均值 = 10 且方差 = 10；P(U + V = 5) = e⁻¹⁰ · 10⁵ / 5! ≈ 0.0378。

When revising, focus on the derivation steps rather than memorising the final results. The ability to derive E(X) = λ and Var(X) = λ using the factorial cancellation and exponential series technique is a frequent examination requirement that tests your algebraic fluency. Practise writing the derivation on blank paper until it becomes automatic.

复习时，专注于推导步骤而不是记住最终结果。使用阶乘消去和指数级数技巧推导 E(X) = λ 和 Var(X) = λ 的能力是考试中频繁考查的，它测试你的代数熟练度。练习在白纸上写出推导过程，直到成为本能反应。

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Poisson Distribution Mean and Variance | 泊松分布的均值与方差