基本概率分布Basic Concept of Probability Distributions 2: Poisson Distribution
PMF
A discrete random variable $X$ is said to have a Poisson distribution with parameter $\lambda > 0$, if the probability mass function of $X$ is given by $$f(x; \lambda) = \Pr(X=x) = e^{-\lambda}{\lambda^x\over x!}$$ for $x=0, 1, 2, \cdots$.
Proof:
$$ \begin{align*} \sum_{x=0}^{\infty}f(x; \lambda) &= \sum_{x=0}^{\infty} e^{-\lambda}{\lambda^x\over x!}\\ & = e^{-\lambda}\sum_{x=0}^{\infty}{\lambda^x\over x!}\\ &= e^{-\lambda}\left(1 + \lambda + {\lambda^2 \over 2!}+ {\lambda^3\over 3!}+ \cdots\right)\\ & = e^{-\lambda} \cdot e^{\lambda}\\ & = 1 \end{align*} $$
Mean
The expected value is $$\mu = E[X] = \lambda$$
Proof:
$$ \begin{align*} E[X] &= \sum_{x=0}^{\infty}xe^{-\lambda}{\lambda^x\over x!}\\ & = \sum_{x=1}^{\infty}e^{-\lambda}{\lambda^x\over (x-1)!}\\ & =\lambda e^{-\lambda}\sum_{x=1}^{\infty}{\lambda^{x-1}\over (x-1)!}\\ & = \lambda e^{-\lambda}\left(1+\lambda + {\lambda^2\over 2!} + {\lambda^3\over 3!}+\cdots\right)\\ & = \lambda e^{-\lambda} e^{\lambda}\\ & = \lambda \end{align*} $$
Variance
The variance is $$\sigma^2 = \mbox{Var}(X) = \lambda$$
Proof:
$$ \begin{align*} E\left[X^2\right] &= \sum_{x=0}^{\infty}x^2e^{-\lambda}{\lambda^x\over x!}\\ &= \sum_{x=1}^{\infty}xe^{-\lambda}{\lambda^x\over (x-1)!}\\ &= \lambda\sum_{x=1}^{\infty}xe^{-\lambda}{\lambda^{x-1}\over (x-1)!}\\ & = \lambda\sum_{x=1}^{\infty}(x-1+1)e^{-\lambda}{\lambda^{x-1}\over (x-1)!}\\ &= \lambda\left(\sum_{x=1}^{\infty}(x-1)e^{-\lambda}{\lambda^{x-1}\over (x-1)!} + \sum_{x=1}^{\infty} e^{-\lambda}{\lambda^{x-1}\over (x-1)!}\right)\\ &= \lambda\left(\lambda\sum_{x=2}^{\infty}e^{-\lambda}{\lambda^{x-2}\over (x-2)!} + \sum_{x=1}^{\infty} e^{-\lambda}{\lambda^{x-1}\over (x-1)!}\right)\\ & = \lambda(\lambda+1) \end{align*} $$ Hence the variance is $$ \begin{align*} \mbox{Var}(X)& = E\left[X^2\right] - E[X]^2\\ & = \lambda(\lambda + 1) - \lambda^2\\ & = \lambda \end{align*} $$
Examples
1. Let $X$ be Poisson distributed with intensity $\lambda=10$. Determine the expected value $\mu$, the standard deviation $\sigma$, and the probability $P\left(|X-\mu| \geq 2\sigma\right)$. Compare with Chebyshev's Inequality.
Solution:
The Poisson distribution mass function is $$f(x) = e^{-\lambda}{\lambda^x\over x!},\ x=0, 1, 2, \cdots$$ The expected value is $$\mu= \lambda=10$$ Then the standard deviation is $$\sigma = \sqrt{\lambda} = 3.162278$$ The probability that $X$ takes a value more than two standard deviations from $\mu$ is $$ \begin{align*} P\left(|X-\lambda| \geq 2\sqrt{\lambda}\right) &= P\left(X \leq \lambda-2\sqrt{\lambda}\right) + P\left(X \geq \lambda + 2\sqrt{\lambda}\right)\\ & = P(X \leq 3) + P(X \geq 17)\\ & = 0.03737766 \end{align*} $$ R code:
sum(dpois(c(0:3), 10)) + 1 - sum(dpois(c(0:16), 10)) # [1] 0.03737766
Chebyshev's Inequality gives the weaker estimation $$P\left(|X - \mu| \geq 2\sigma\right) \leq {1\over2^2} = 0.25$$
2. In a certain shop, an average of ten customers enter per hour. What is the probability $P$ that at most eight customers enter during a given hour.
Solution:
Recall that the Poisson distribution mass function is $$P(X=x) = e^{-\lambda}{\lambda^x\over x!}$$ and $\lambda=10$. So we have $$ \begin{align*} P(X \leq 8) &= \sum_{x=0}^{8}e^{-10}{10^{x}\over x!}\\ &= 0.3328197 \end{align*} $$ R code:
sum(dpois(c(0:8), 10)) # [1] 0.3328197 ppois(8, 10) # [1] 0.3328197
3. What is the probability $Q$ that at most 80 customers enter the shop from the previous problem during a day of 10 hours?
Solution:
The number $Y$ of customers during an entire day is the sum of ten independent Poisson distribution with parameter $\lambda=10$. $$Y = X_1 + \cdots + X_{10}$$ Thus $Y$ is also a Poisson distribution with parameter $\lambda = 100$. Thus we have $$ \begin{align*} P(Y \leq 80) &= \sum_{y=0}^{80}e^{-100}{100^{y}\over y!}\\ &= 0.02264918 \end{align*} $$ R code:
sum(dpois(c(0:80), 100)) # [1] 0.02264918 ppois(80, 100) # [1] 0.02264918
Alternatively, we can use normal approximation (generally when $\lambda > 9$) with $\mu = \lambda = 100$ and $\sigma = \sqrt{\lambda}=10$. $$ \begin{align*} P(Y \leq 80) &= \Phi\left({80.5-100\over 10 }\right)\\ &= \Phi\left({-19.5\over10}\right)\\ &=0.02558806 \end{align*} $$ R code:
pnorm(-19.5/10) # [1] 0.02558806
4. At the 2006 FIFA World Championship, a total of 64 games were played. The number of goals per game was distributed as follows: 8 games with 0 goals 13 games with 1 goal 18 games with 2 goals 11 games with 3 goals 10 games with 4 goals 2 games with 5 goals 2 games with 6 goals Determine whether the number of goals per game may be assumed to be Poisson distributed.
Solution:
We can use Chi-squared test. The observations are in Table 1.
![](http://images0.cnblogs.com/blog/696400/201501/030011227006810.png)
On the other hand, if this is a Poisson distribution then the parameter should be $$ \begin{align*} \lambda &= \mu\\ & = {0\times8 + 1\times13 +\cdots + 6\times2 \over 8+13+\cdots+2}\\ & = {144\over 64}\\ &=2.25 \end{align*} $$ And the Poisson point probabilities are listed in Table 2.
![](http://images0.cnblogs.com/blog/696400/201501/030011436228452.png)
And hence the expected numbers are listed in Table 3.
![](http://images0.cnblogs.com/blog/696400/201501/030011565594999.png)
Note that we have merged some categories in order to get $E_i \geq 3$. The statistic is $$ \begin{align*} \chi^2 &= \sum{(O-E)^2\over E}\\ &= {(8-6.720)^2 \over 6.720} + \cdots + {(4-4.992)^2 \over 4.992}\\ &= 2.112048 \end{align*} $$ There are six categories and thus the degree of freedom is $6-1 = 5$. The significance probability is 0.8334339. R code:
prob = c(round(dpois(c(0:6), 2.25), 3), + 1 - round(sum(dpois(c(0:6), 2.25)), 3)) expect = prob * 64 prob; expect # [1] 0.105 0.237 0.267 0.200 0.113 0.051 0.019 0.008 # [1] 6.720 15.168 17.088 12.800 7.232 3.264 1.216 0.512 O = c(8, 13, 18, 11, 10, 4) E = c(expect[1:5], sum(expect[6:8])) O; E # [1] 8 13 18 11 10 4 # [1] 6.720 15.168 17.088 12.800 7.232 4.992 chisq = sum((O - E) ^ 2 / E) 1 - pchisq(chisq, 5) # [1] 0.8334339
The hypothesis is $$H_0: \mbox{Poisson distribution},\ H_1: \mbox{Not Poisson distribution}$$ Since $p = 0.8334339 > 0.05$, so we accept $H_0$. That is, it is reasonable to claim that the number of goals per game is Poisson distributed.
Reference
- Ross, S. (2010). A First Course in Probability (8th Edition). Chapter 4. Pearson. ISBN: 978-0-13-603313-4.
- Brink, D. (2010). Essentials of Statistics: Exercises. Chapter 5 & 9. ISBN: 978-87-7681-409-0.
作者:赵胤
出处:http://www.cnblogs.com/zhaoyin/
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