📚 Investigation 4 - Poisson Mean and Variance | 探究4 - 泊松分布的均值与方差
irm that both the expectation and the variance are equal to the rate parameter λ. Understanding why E(X) = Var(X) = λ is a key insight that often appears in examinations and practical applications.
泊松分布是IB数学中最重要的离散概率模型之一。本探究将深入探讨泊松随机变量的两个基本性质：均值与方差。我们将通过解析推导、直观认识和实例分析，验证期望和方差都等于速率参数 λ。理解为什么 E(X)=Var(X)=λ 是考试和实际应用中常出现的关键洞察。
1. Introduction to the Poisson Distribution | 泊松分布简介
The Poisson distribution models the number of times a random event occurs in a fixed interval of time or space, provided the events happen independently and at a constant average rate λ. Typical examples include the number of phone calls received by a call centre per hour, the number of decay events from a radioactive source in a given period, or the number of typing errors on a page.
泊松分布用于描述在固定时间或空间区间内随机事件发生的次数，前提是这些事件独立发生且具有恒定的平均速率 λ。典型例子包括呼叫中心每小时接到的电话数、给定时间内放射源的衰变次数、或一页纸上的打字错误数。
2. Probability Mass Function | 概率质量函数
The random variable X follows a Poisson distribution with parameter λ > 0 if its probability mass function is given by:
如果随机变量 X 服从参数为 λ > 0 的泊松分布，其概率质量函数为：
P(X = x) = e-λ λx / x! , x = 0, 1, 2, …
Here e is the base of the natural logarithm, and x! denotes the factorial of x. Notice that the sum of all probabilities over x = 0, 1, 2, … equals 1, a consequence of the Maclaurin series for eλ.
其中 e 为自然对数的底数，x! 表示 x 的阶乘。注意所有概率对 x = 0, 1, 2, … 求和等于1，这是 eλ 的麦克劳林级数的结果。
3. Intuition Behind the Mean | 均值的直观理解
The parameter λ represents the average rate of occurrence. If events happen at a rate of λ per unit time, then over an interval of length 1 we would intuitively expect the average count to be λ. For example, if a shop serves customers at an average of 4 per hour, after many hours the mean number of customers per hour should approach 4.
参数 λ 代表事件发生的平均速率。如果事件以每单位时间 λ 的速率发生，那么在长度为1的时间区间内，我们直觉上会认为平均计数应为 λ。例如，如果一家商店平均每小时服务4名顾客，经过多个小时后每小时的顾客平均数应当趋近于4。
The Poisson distribution formalises this intuition. But we need to confirm mathematically that the expected value E(X) equals λ. The next section provides the formal derivation.
泊松分布将这种直觉严格形式化。但我们需要从数学上验证期望值 E(X) 是否等于 λ。下一节将给出正式推导。
4. Formal Calculation of the Mean | 均值的正式推导
By definition, the expected value of a discrete random variable is the sum of each possible value multiplied by its probability:
根据定义，离散随机变量的期望值是每一个可能取值乘以其概率的总和：
E(X) = ∑x=0∞ x · (e-λ λx / x!)
= 0, the term is zero, so we can start the sum at x = 1 and simplify the factorial: x / x! = 1 / (x-1)!. The constant e-λ does not depend on x and can be factored out:
当 x = 0 时该项为零，因此我们可以从 x = 1 开始求和，并化简阶乘：x / x! = 1 / (x-1)!。常数 e-λ 与 x 无关，可以提取出来：
E(X) = e-λ ∑x=1∞ λx / (x-1)!
Rewrite λx as λ · λx-1 to obtain:
将 λx 改写为 λ · λx-1，得到：
E(X) = λ e-λ ∑x=1∞ λx-1 / (x-1)!
Let k = x – 1. Then k runs from 0 to infinity, and the sum becomes Σk=0∞ λk/k!, which is the series expansion for eλ:
令 k = x – 1，k 从 0 变化到无穷大，求和式变为 Σk=0∞ λk/k!，这正是 eλ 的级数展开：
E(X) = λ e-λ · eλ = λ
Thus, we have proved analytically that the mean of a Poisson random variable equals its rate parameter λ.
由此，我们从解析上证明了泊松随机变量的均值等于其速率参数 λ。
5. Understanding Variance | 理解方差
gher rates λ naturally lead to greater variability. The formal definition is Var(X) = E[(X – μ)2], which is equivalent to E(X2) – [E(X)]2. Our task is to compute E(X2) and show that Var(X) also simplifies to λ.
方差衡量分布在其均值附近的离散程度。对于泊松分布，我们可能会猜想更高的速率 λ 自然地导致更大的变异性。方差的正式定义为 Var(X) = E[(X – μ)2]，它等价于 E(X2) – [E(X)]2。我们的任务是计算 E(X2) 并证明 Var(X) 同样简化为 λ。
6. Computing E(X2) | 计算 E(X²)
Start with the definition of the second moment:
从二阶矩的定义出发：
E(X2) = ∑x=0∞ x2 · (e-λ λx / x!)
The algebraic trick is to express x2 as x(x – 1) + x, which separates the sum into two parts that we can handle with known Poisson sums:
代数技巧是将 x2 写为 x(x – 1) + x，这样把求和分成两部分，从而可利用已知的泊松和式：
E(X2) = ∑ x(x – 1) P(X=x) + ∑ x P(X=x) = E[X(X–1)] + E(X)
We already know E(X) = λ. The term E[X(X–1)] requires only a slight extension of the earlier technique. For x = 0 or 1, x(x–1)=0, so we begin at x = 2:
我们已经知道 E(X) = λ。计算 E[X(X–1)] 只需略微扩展前面的技术。对 x = 0 或 1，x(x–1)=0，因此可以从 x = 2 开始：
E[X(X–1)] = e-λ ∑x=2∞ x(x–1) λx / x!
Simplify x(x–1)/x! = 1/(x–2)!, factor λ2, and shift the index with m = x – 2:
化简 x(x–1)/x! = 1/(x–2)!，提取 λ2，并用 m = x – 2 替换下标：
E[X(X–1)] = λ2 e-λ ∑m=0∞ λm / m! = λ2 e-λ · eλ = λ2
Therefore, the second moment is:
因此，二阶矩为：
E(X2) = λ2 + λ
7. Deriving the Variance | 推导方差
Substitute the results into the variance formula:
将结果代入方差公式：
Var(X) = E(X2) – [E(X)]2 = (λ2 + λ) – λ2 = λ
We have proved that a Poisson random variable has both mean and variance equal to λ. This property, known as equidispersion, is a hallmark of the Poisson distribution and is often used as a diagnostic check in data analysis.
我们证明了泊松随机变量的均值和方差都等于 λ。这一性质称为等离散，是泊松分布的标志，常被用作数据分析中的诊断检查。
8. Mean and Variance via the Probability Generating Function | 通过概率生成函数求均值与方差
An elegant alternative derivation uses the probability generating function (PGF). For a Poisson random variable, the PGF is:
一种更优雅的推导方法使用概率生成函数（PGF）。对于泊松随机变量，PGF为：
G(t) = E(tX) = eλ(t – 1)
Differentiating with respect to t gives:
对 t 求导得：
G′(t) = λ eλ(t – 1) , G″(t) = λ2 eλ(t – 1)
Evaluating at t = 1 yields the factorial moments: G′(1) = E(X) = λ, and G″(1) = E[X(X–1)] = λ2. The variance is then obtained from Var(X) = G″(1) + G′(1) – [G′(1)]2 = λ2 + λ – λ2 = λ. This method confirms the earlier result with remarkable compactness.
在 t = 1 处取值得到阶乘矩：G′(1) = E(X) = λ，G″(1) = E[X(X–1)] = λ2。然后由 Var(X) = G″(1) + G′(1) – [G′(1)]2 = λ2 + λ – λ2 = λ 求得方差。该方法极其简洁地验证了之前的结果。
9. Worked Examples | 实例分析
= 10. The table below summarises their theoretical mean, variance, and standard deviation.
考虑三个不同参数的泊松过程：λ = 2、λ = 5 和 λ = 10。下表总结了它们的理论均值、方差和标准差。
| λ | E(X) = λ | Var(X) = λ | Standard Deviation √λ |
|---|---|---|---|
| 2 | 2 | 2 | √2 ≈ 1.414 |
| 5 | 5 | 5 | √5 ≈ 2.236 |
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