Basic probability distribution basic Concept of probability distributions 2:poisson distribution

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A discrete random variable $X $ is said to has 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 $ is 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$$ the expected Valu E is $$\mu= \lambda=10$$ and the standard deviation are $$\sigma = \sqrt{\lambda} = 3.162278$$ the probability that $X $ ta Kes A value more than-deviations from $\mu$ is $$ \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) \ & = 0.03737766 \end{align*} $$ R Code:

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's the probability $P $ that for most eight customers enter during a given hour.

Solution:

Recall 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:

3. What's the probability $Q $ that for most customers enter the shop from the previous problem during a day's hours ?

Solution:

The number $Y $ of customers during an entire day was the sum of ten independent Poisson distribution with parameter $\LAMBD a=10$. $ $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) &= \sum_{y=0}^{80}e^{-100}{100^{y}\over y!} \ \ &= 0.02264918 \end{align*} $$ R Code:

Alternatively, we can use normal approximation (generally if $\LAMBDA > 9$) with $\mu = \lambda = 100$ and $\sigma = \sqrt{\lambda}=10$. $$ \begin{align*} P (Y \leq) &= \phi\left ({80.5-100\over)}\right \ \ &= \phi\left ({ -19.5\over10}\right) \ &am p;=0.02558806 \end{align*} $$ R Code:

4. At the 2006 FIFA World Championship, a total of all games were played. The number of goals per game is distributed as Follows:8 games with 0 goals, with 1 goal, with 2 goals 1 1 games with 3 goals games with 4 goals 2 games with 5 goals 2 games with 6 goals determine whether the number of goals Per game may is assumed to being Poisson distributed.

Solution:

We can use chi-squared test. The observations is in Table 1.

On the other hand, if this is a Poisson distribution then the parameter should be $$ \begin{align*} \lambda &= \mu\\ & Amp = {0\times8 + 1\times13 +\cdots + 6\times2 \over 8+13+\cdots+2}\\ & = {144\over 64}\\ &=2.25 \end{align*} $$ and T He Poisson point probabilities is listed in Table 2.

And hence the expected numbers is listed in Table 3.

Note that we had 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 be 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 * 64PROB; expect# [1] 0 .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  0.51 2O = C (8, 4) E = C (Expect[1:5], sum (expect[6:8)) O; e# [1]  8  4# [1]  6.720 15.168 17.088 12.800  7.232  

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's, it's reasonable to claim the number of goals per game is Poisson distributed.

Reference

1. Ross, S. (2010). A first Course in probability (8th Edition). Chapter 4. Pearson. Isbn:978-0-13-603313-4.
2. Brink, D. (2010). Essentials of Statistics:exercises. Chapter 5 & 9. isbn:978-87-7681-409-0.

Basic probability distribution basic Concept of probability distributions 2:poisson distribution