Poisson Distribution Properties and Example Analysis | 泊松分布性质与例题分析

📚 Poisson Distribution Properties and Example Analysis | 泊松分布性质与例题分析

The Poisson distribution is one of the most important discrete probability distributions taught in IB Mathematics. It models the number of times an event occurs in a fixed interval of time or space when the events happen independently at a constant average rate. Understanding its properties and being able to apply them to examination problems is essential for high marks in IB.

泊松分布是 IB 数学中最重要的离散概率分布之一。它用于描述在固定的时间或空间间隔内，事件以恒定平均速率独立发生时该事件发生的次数。掌握其性质并能在考试题中灵活应用，是取得高分的关键。

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1. Definition and Probability Mass Function | 定义与概率质量函数

A random variable \(X\) is said to follow a Poisson distribution with parameter \(\lambda\), written as \(X \sim \text{Poisson}(\lambda)\), if its probability mass function is given by the following formula for \(x = 0, 1, 2, \dots\).

如果一个随机变量 \(X\) 服从参数为 \(\lambda\) 的泊松分布，记作 \(X \sim \text{Poisson}(\lambda)\)，那么当 \(x = 0, 1, 2, \dots\) 时，其概率质量函数由下面的公式给出。

P(X = x) = e^−λ λ^x / x!

Here, \(e\) is Euler’s number, approximately 2.71828, and \(\lambda\) is a positive real number representing the average number of events in the interval. The factorial \(x!\) is the product of all positive integers up to \(x\), with \(0! = 1\).

其中 \(e\) 是自然常数，约为 2.71828；\(\lambda\) 是正实数，表示该间隔内事件的平均发生次数。\(x!\) 是阶乘，表示从 1 到 \(x\) 的所有正整数的乘积，并规定 \(0! = 1\)。

Notice that the possible values of \(X\) are all non-negative integers. This makes the Poisson distribution suitable for counting phenomena, such as the number of cars passing a point in one hour or the number of typos on a page.

注意，\(X\) 的可能取值为所有非负整数。这使得泊松分布非常适合用于计数现象，例如一小时内经过某个地点的汽车数量，或者一页纸上的拼写错误数量。

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2. Mean and Variance | 期望与方差

One of the most distinctive properties of the Poisson distribution is that its mean and variance are both equal to the parameter \(\lambda\). In mathematical notation:

泊松分布最显著的性质之一是其期望与方差都等于参数 \(\lambda\)。用数学符号表示：

E(X) = λ   and   Var(X) = λ

This equality is a special feature that does not hold for most other distributions. For example, a binomial distribution has mean \(np\) and variance \(np(1-p)\), which are generally different.

这种相等是许多其他分布不具备的特殊性质。例如，二项分布的期望为 \(np\)，方差为 \(np(1-p)\)，两者通常并不相等。

These results can be derived using the moment generating function \(M(t) = e^{λ(e^t − 1)}\). Differentiating and evaluating at \(t = 0\) gives the mean, and the second derivative gives the variance. In IB problems, you may simply be expected to quote these formulas directly.

这些结果可以通过矩生成函数 \(M(t) = e^{λ(e^t − 1)}\) 推导得出。求导后在 \(t = 0\) 处取值即可得到期望，二阶导数可得到方差。在 IB 考试中，通常可以直接引用这些公式。

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3. Conditions and Applications | 条件与应用场景

Before applying the Poisson model, you must check that the underlying situation satisfies certain conditions. These conditions are often tested in IB examination questions.

在应用泊松模型之前，必须检查实际情况是否满足特定条件。这些条件在 IB 考试中经常被考查。

• Events occur independently of one another. The occurrence of one event does not affect the probability of another.

  事件相互独立。一个事件的发生不会影响另一个事件发生的概率。

• Events occur at a constant average rate. The expected number of events in any interval is proportional to the length of the interval.

  事件以恒定的平均速率发生。任何间隔内的事件期望数与间隔长度成正比。

• Two events cannot occur at exactly the same instant. In continuous time, this is equivalent to saying that the probability of more than one event in a very small interval is negligible.

  两个事件不能恰好同时发生。在连续时间中，这等价于在非常短的时间间隔内，多于一个事件发生的概率可以忽略不计。

Typical applications include the number of telephone calls received in a call centre per minute, the number of radioactive decays in a fixed period, the number of insurance claims per year, and the number of defects on a manufactured sheet of material.

典型的应用包括：呼叫中心每分钟接到的电话数量、固定时间内的放射性衰变次数、每年的保险理赔数量，以及一块制造材料上的缺陷数量。

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4. Poisson Process and the Parameter λ | 泊松过程与参数λ

The Poisson distribution is often introduced through the Poisson process. If events occur according to a Poisson process with rate \(\lambda\) per time unit, then the number of events \(N(t)\) occurring in \(t\) time units has a Poisson distribution with parameter \(\lambda t\).

泊松分布通常通过泊松过程引入。如果事件按照每单位时间 \(\lambda\) 的速率服从泊松过程发生，那么在 \(t\) 个时间单位内发生的事件数 \(N(t)\) 服从参数为 \(\lambda t\) 的泊松分布。

N(t) ~ Poisson(λt)

This means \(\lambda\) is the average number of events per single time unit, not necessarily the total interval length. If you are given the average for one hour and asked about a three-hour period, you must multiply by three.

这意味着 \(\lambda\) 是每个单位时间内的平均事件数，而不一定是总间隔长度。如果题目给出一个小时内的事件平均数，而问你三个小时的情况，你必须将平均数乘以三。

For example, if a shop averages 5 customers every hour, then in 20 minutes the expected number is \(\lambda = 5 \times \frac{1}{3} = \frac{5}{3}\). This adjustment is a common source of error in IB examinations.

例如，如果一家商店平均每小时有 5 名顾客，那么 20 分钟内的事件期望数为 \(\lambda = 5 \times \frac{1}{3} = \frac{5}{3}\)。这种调整是 IB 考试中常见的错误来源。

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5. Additive Property | 可加性

The Poisson distribution has a very useful additive property: if \(X\) and \(Y\) are independent Poisson random variables with parameters \(\lambda_1\) and \(\lambda_2\), then their sum \(X + Y\) also follows a Poisson distribution, with parameter \(\lambda_1 + \lambda_2\).

泊松分布具有非常有用的可加性：如果 \(X\) 和 \(Y\) 是相互独立的泊松随机变量，参数分别为 \(\lambda_1\) 和 \(\lambda_2\)，那么它们的和 \(X + Y\) 仍然服从泊松分布，参数为 \(\lambda_1 + \lambda_2\)。

If X ~ Poisson(λ₁), Y ~ Poisson(λ₂), X and Y independent, then X + Y ~ Poisson(λ₁ + λ₂)

This property allows you to combine information from two separate intervals or two independent sources. For instance, if Monday’s number of accidents follows Poisson(2) and Tuesday’s follows Poisson(3), then the total over both days follows Poisson(5).

这一性质允许你将两个独立间隔或两个独立来源的信息合并。例如，如果周一的事故数服从 Poisson(2)，周二的事故数服从 Poisson(3)，那么两天的事故总数服从 Poisson(5)。

In proofs, this property can be established using probability generating functions or moment generating functions. In exams, you are often expected to state and apply the result directly.

在证明中，这一性质可以通过概率生成函数或矩生成函数得到。在考试中，通常要求直接陈述并应用该结论。

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6. Approximation to the Binomial Distribution | 近似二项分布

When the number of trials \(n\) is large and the probability of success \(p\) is small, the binomial distribution \(\text{Bin}(n, p)\) can be approximated by a Poisson distribution with \(\lambda = np\). This is often called the law of rare events.

当试验次数 \(n\) 很大且成功概率 \(p\) 很小时，二项分布 \(\text{Bin}(n, p)\) 可以用参数为 \(\lambda = np\) 的泊松分布来近似。这通常被称为“稀有事件定律”。

Bin(n, p) ≈ Poisson(np) when n is large and p is small

A common rule of thumb is that the approximation is good when \(n \geq 50\) and \(p \leq 0.1\), or more conservatively when \(np \leq 5\). The approximation can also be justified by taking the limit of the binomial probability mass function as \(n \to \infty\) with \(np\) held constant.

常用的经验法则是，当 \(n \geq 50\) 且 \(p \leq 0.1\) 时近似效果良好；更保守的条件是 \(np \leq 5\)。该近似可以通过在 \(n \to \infty\) 且保持 \(np\) 不变时取二项分布概率质量函数的极限来证明。

In practice, this approximation simplifies calculations dramatically. Instead of evaluating a binomial coefficient with a large \(n\), you only need the simple Poisson formula.

在实践中，这种近似可以大大简化计算。无需计算包含很大 \(n\) 的二项系数，只需使用简单的泊松公式即可。

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7. Example 1: Basic Probability | 例题1：基础概率计算

Question: In a certain bookshop, the number of copies of a textbook sold per hour follows a Poisson distribution with mean 3. Find the probability that at least 2 copies are sold in a randomly selected hour.

题目：某书店每小时售出某教材的数量服从均值为 3 的泊松分布。求在随机选定的一个小时内至少售出 2 本的概率。

Solution: Let \(X\) be the number sold in one hour. Then \(X \sim \text{Poisson}(3)\). We need \(P(X \geq 2)\). Using the complement rule,

解答：设 \(X\) 为一小时内售出的数量，则 \(X \sim \text{Poisson}(3)\)。要求 \(P(X \geq 2)\)。利用补事件公式，

P(X ≥ 2) = 1 − P(X = 0) − P(X = 1)

Using the probability mass function,

代入概率质量函数，

P(X = 0) = e⁻³,   P(X = 1) = 3e⁻³

Therefore,

因此，

P(X ≥ 2) = 1 − e⁻³ − 3e⁻³ = 1 − 4e⁻³ ≈ 0.8009

So the probability that at least two copies are sold in the hour is about 0.801, or 80.1%.

因此，一小时内至少售出两本的概率约为 0.801，即 80.1%。

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8. Example 2: Using the Additive Property | 例题2：利用可加性

Question: At a customer service desk, calls arrive at an average rate of 4 per hour. Assuming a Poisson process, find the probability that exactly 10 calls arrive in a 2-hour period.

题目：在客户服务台，电话平均每小时到达 4 次。假设这是一个泊松过程，求在 2 小时内恰好有 10 通电话到达的概率。

Solution: The rate per hour is 4, so in 2 hours the expected number is \(\lambda = 4 \times 2 = 8\). Let \(Y\) be the number of calls in 2 hours. Then \(Y \sim \text{Poisson}(8)\).

解答：每小时的速率为 4，因此 2 小时内期望数为 \(\lambda = 4 \times 2 = 8\)。设 \(Y\) 为 2 小时内的电话数量，则 \(Y \sim \text{Poisson}(8)\)。

We require \(P(Y = 10)\):

我们需要计算 \(P(Y = 10)\)：

P(Y = 10) = e⁻⁸ × 8¹⁰ / 10!

Evaluating this gives approximately 0.0993. Thus, there is about a 9.93% chance of receiving exactly 10 calls in two hours.

计算得到约为 0.0993。因此，在两小时内恰好接到 10 通电话的概率约为 9.93%。

Notice that we did not need to add separate Poisson probabilities for each hour. The additive property allows us to treat the whole two-hour interval as a single Poisson event with parameter 8.

注意，我们不需要分别计算两个小时的泊松概率再相加。可加性允许我们将整个两小时间隔视为参数为 8 的单个泊松事件。

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9. Example 3: Poisson Approximation to Binomial | 例题3：泊松近似二项分布

Question: A rare disease affects 0.1% of the population. In a random sample of 1000 people, find the probability that at least 3 people have the disease.

题目：一种罕见疾病影响人口的 0.1%。在 1000 人的随机样本中，求至少有 3 人患病的概率。

Solution: The exact model is \(X \sim \text{Bin}(1000, 0.001)\). Since \(n\) is large and \(p\) is small, we can approximate \(X\) by a Poisson distribution with \(\lambda = np = 1\). Thus \(X \approx \text{Poisson}(1)\).

解答：精确模型为 \(X \sim \text{Bin}(1000, 0.001)\)。由于 \(n\) 很大且 \(p\) 很小，我们可以用参数 \(\lambda = np = 1\) 的泊松分布来近似 \(X\)。因此 \(X \approx \text{Poisson}(1)\)。

We want \(P(X \geq 3)\). Using the complement rule:

我们要求 \(P(X \geq 3)\)。利用补事件公式：

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

With \(\lambda = 1\), we have

在 \(\lambda = 1\) 时，有

P(X = 0) = e⁻¹,   P(X = 1) = e⁻¹,   P(X = 2) = e⁻¹ / 2

Therefore,

因此，

P(X ≥ 3) = 1 − e⁻¹(1 + 1 + 0.5) = 1 − 2.5e⁻¹ ≈ 0.0803

The probability that at least 3 people in the sample have the disease is approximately 0.0803, or 8.03%.

样本中至少有 3 人患病的概率约为 0.0803，即 8.03%。

Using the exact binomial distribution would require summing \({1000 \choose 0}\), \({1000 \choose 1}\), and \({1000 \choose 2}\), which is far more tedious. The Poisson approximation is both acceptable and efficient in IB contexts when the conditions are satisfied.

使用精确二项分布需要计算 \({1000 \choose 0}\)、\({1000 \choose 1}\) 和 \({1000 \choose 2}\) 的和，这要冗长得多。在 IB 中，当条件满足时，泊松近似既被接受又高效。

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10. Exam Tips and Common Pitfalls | 考点与常见易错点

When solving Poisson distribution problems in IB exams, always begin by identifying the interval and the corresponding mean. Write the distribution clearly using the notation \(X \sim \text{Poisson}(\lambda)\).

在 IB 考试中解答泊松分布问题时，始终先确定时间或空间间隔及对应的均值。用记号 \(X \sim \text{Poisson}(\lambda)\) 清晰地写出分布。

• Check whether the interval matches the given rate. If not, scale the mean proportionally. For example, “3 per hour” and a 20-minute interval means \(\lambda = 1\).

  检查间隔是否与给定速率匹配。如果不匹配，要按比例调整均值。例如，“每小时 3 次”而问题是 20 分钟内，则 \(\lambda = 1\)。

• Use the complement rule for “at least” or “more than” problems. For \(P(X \geq k)\), often \(1 – P(X \leq k-1)\) is easier.

  对“至少”或“多于”类型的问题使用补事件公式。例如求 \(P(X \geq k)\) 时，通常 \(1 – P(X \leq k-1)\) 更容易计算。

• Remember that \(0! = 1\). A surprising number of students mistakenly set \(0! = 0\) and lose simple marks.

  记住 \(0! = 1\)。很多学生误认为 \(0! = 0\)，从而丢失简单分。

• Do not confuse the Poisson distribution with the binomial distribution. Check whether there is a fixed number of trials or a fixed interval. Poisson is for counts in continuous or large intervals, while binomial is for counts in a fixed number of independent trials.

  不要将泊松分布与二项分布混淆。检查是否有固定的试验次数或固定的间隔。泊松适用于连续或较大间隔内的计数，而二项分布适用于固定次数的独立试验中的成功次数。

• When using a normal approximation to the Poisson (valid when \(\lambda\) is large, e.g., \(\lambda > 10\)), apply a continuity correction. For example, \(P(X = 5)\) becomes \(P(4.5 < Y < 5.5)\) in the normal model.

  当使用正态近似泊松分布（在 \(\lambda\) 较大时有效，例如 \(\lambda > 10\)）时，需要进行连续性修正。例如，正态模型中 \(P(X = 5)\) 应变为 \(P(4.5 < Y < 5.5)\)。

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11. Summary | 总结

The Poisson distribution is a versatile model for counting rare or random events in a fixed interval. Its key properties are the probability mass function, the equality of mean and variance, the additive property, and its role as an approximation to the binomial distribution.

泊松分布是描述固定间隔内稀有或随机事件计数的多功能模型。其关键性质包括概率质量函数、期望与方差相等、可加性，以及作为二项分布近似的作用。

In IB examinations, mastery of these properties allows you to solve a wide range of problems, from basic probability calculations to more complex applications involving interval scaling and distribution approximations. Always remember to verify the conditions, choose the correct parameter \(\lambda\), and express your answer with appropriate rounding.

在 IB 考试中，掌握这些性质可以帮助你解决各种问题，从基础概率计算到涉及间隔缩放和分布近似的复杂应用。始终记得验证条件、选择

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