The Poisson distribution · 泊松分布
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Poisson distribution/ˈpɔɪsn ˌdɪstrɪˈbjuːʃn/ | 泊松分布 | pō sōng fēn bù |
| interval/ˈɪntəvl/ | 区间 | qū jiān |
| mean/miːn/ | 平均数 | píng jūn shù |
| variance/ˈveərɪəns/ | 方差 | fāng chà |
| overdispersion/ˌəʊvədɪˈspɜːʃn/ | 过度离散 | guò dù lí sàn |
| parameter/pəˈræmɪtə/ | 参数 | cān shù |
The queue that never ends
- Customers arrive at a shop at an average rate of 3 per minute. What's the probability that exactly 5 arrive in the next minute?
- The Poisson distribution 泊松分布 models rare events happening at a steady rate — from phone calls to radioactive decay.
永不结束的队列
- 顾客以平均每分钟 3 个的速率到达一家商店。下一分钟恰好有 5 个到达的概率是多少?
- 泊松分布(Poisson distribution)为以稳定速率发生的罕见事件建模——从电话到放射性衰变。
The Poisson distribution
- $X \sim \text{Po}(\lambda)$ models the number of random events in a fixed interval 区间 at a steady rate $\lambda$:
Worked example. Calls arrive at a rate of $\lambda = 4$ per hour. P(exactly 2 calls) $= e^{-4} \dfrac{4^2}{2!} = 0.0183 \times \dfrac{16}{2} = 0.1465$.
People arriving at random in a queue follow a Poisson distribution
泊松分布
- $X \sim \text{Po}(\lambda)$ 为一个固定区间内以稳定速率 $\lambda$ 发生的随机事件的数量建模:
算例。 电话以每小时 $\lambda = 4$ 的速率到达。P(恰好 2 个电话)$= e^{-4} \dfrac{4^2}{2!} = 0.0183 \times \dfrac{16}{2} = 0.1465$。
随机到达一个队列的人遵循一个泊松分布
The Poisson distribution · 泊松分布
X ~ Po(λ)
The Poisson counts random events in an interval. Its mean · 平均值 and variance are both λ. · 泊松计数一个区间内的随机事件。它的平均数和方差都是 λ。
For X ~ Po(3), find P(X = 2) = e⁻³ × 3²/2!. Give your answer to 3 decimal places. · 对 X ~ Po(3),求 P(X = 2) = e⁻³ × 3²/2!。答案给到 3 位小数。
P(X=2) = e⁻³ × 9/2 = 0.0498 × 4.5 ≈ 0.224. · P(X=2) = e⁻³ × 9/2 = 0.0498 × 4.5 ≈ 0.224。
For X ~ Po(2), P(X = 0) = e⁻². Find it (3 dp). · 对 X ~ Po(2),P(X = 0) = e⁻²。求它(3 位小数)。
P(X=0) = e⁻² × 2⁰/0! = e⁻² = 0.1353 ≈ 0.135. · P(X=0) = e⁻² × 2⁰/0! = e⁻² = 0.1353 ≈ 0.135。
Match each Poisson idea to its result. · 把每个泊松分布的概念与它的结果配对。
For a Poisson distribution the mean and the variance are both lambda. · 对于泊松分布,均值和方差都是 λ。
Properties of the Poisson
- For a Poisson variable, the mean 平均数 and variance 方差 both equal $\lambda$: $E(X) = \text{Var}(X) = \lambda$.
- This is unique — no other common distribution has this property.
For Po(3) the mean and variance are both 3; the bars give P(X = k).
Mean = variance is the test. If your data has mean ≈ variance, the Poisson model may be appropriate. If variance is much larger than the mean, the events are not independent (overdispersion 过度离散) and Poisson is wrong.
泊松分布的性质
- 对一个泊松变量,平均数和方差都等于 $\lambda$:$E(X) = \text{Var}(X) = \lambda$。
- 这是独特的——没有别的常见分布有这个性质。
对 Po(3),平均数和方差都是 3;条形给出 P(X = k)。
平均数 = 方差是检验。 如果你的数据平均数 ≈ 方差,泊松模型可能合适。如果方差远大于平均数,事件不独立(过度离散),泊松就是错的。
For X ~ Po(3), what is the mean of X? · 对 X ~ Po(3),X 的平均数是多少?
For a Poisson variable, the mean equals λ = 3 (and so does the variance). · 对一个泊松变量,平均数等于 λ = 3(方差也是)。
For X ~ Po(3), what is the variance of X? · 对 X ~ Po(3),X 的方差是多少?
For a Poisson variable, the variance also equals λ = 3. · 对一个泊松变量,方差也等于 λ = 3。
For a Poisson distribution, the mean and variance are always equal. · 对一个泊松分布,平均数和方差总是相等。
This is a defining property of the Poisson distribution: E(X) = Var(X) = λ. · 这是泊松分布的一个定义性性质:E(X) = Var(X) = λ。
Poisson approximation to the binomial
- The Poisson approximates the binomial when $n$ is large and $p$ is small.
- Use $\lambda = np$ as the Poisson parameter 参数.
- Example: $B(200, 0.01) \approx \text{Po}(2)$.
泊松对二项的近似
- 当 $n$ 大且 $p$ 小时,泊松近似二项。
- 用 $\lambda = np$ 作为泊松参数。
- 例子:$B(200, 0.01) \approx \text{Po}(2)$。
X ~ B(200, 0.01). Approximate with Poisson: λ = np. Find λ. · X ~ B(200, 0.01)。用泊松近似:λ = np。求 λ。
λ = np = 200 × 0.01 = 2. · λ = np = 200 × 0.01 = 2。
Worked example — approximation
- $X \sim B(100, 0.02)$. Approximate with Poisson: $\lambda = np = 100 \times 0.02 = 2$.
- $P(X = 0) \approx e^{-2} \dfrac{2^0}{0!} = e^{-2} = 0.1353$.
- The normal distribution approximates the binomial distribution (with a continuity correction) when appropriate.
算例——近似
- $X \sim B(100, 0.02)$。用泊松近似:$\lambda = np = 100 \times 0.02 = 2$。
- $P(X = 0) \approx e^{-2} \dfrac{2^0}{0!} = e^{-2} = 0.1353$。
- 在合适时正态分布(normal distribution)近似二项分布(binomial distribution,需连续性校正 continuity correction)。
You've got it
- Poisson: $P(X = r) = e^{-\lambda}\dfrac{\lambda^r}{r!}$
- mean and variance both equal $\lambda$
- approximates the binomial for large $n$, small $p$ (use $\lambda = np$)
你掌握了
- 泊松:$P(X = r) = e^{-\lambda}\dfrac{\lambda^r}{r!}$
- 平均数和方差都等于 $\lambda$
- 对大的 $n$、小的 $p$ 近似二项(用 $\lambda = np$)