Question

Problem 6. [Poisson is Pronounced 'Pwah-ssohn] (a) Suppose that X is a random variable following the Poisson distribution with rate parameter A. Show that E[x]-A Hint: You may find the following fact useful: at k! (b) Suppose that we obtained the following count data: Count Frequency 24 30 17 19 Fit a Poisson distribution to the data using the Method of Moments (c) Suppose that X is a random variable that follows the Poisson distribution that you fit in part (b). Compute P(X0). P(X1), and P(x 2). Are these similar to the sample probabilities from the data in part (b)? Based on these probabilities alone, comment on whether the Poisson distribution is a good fit to the above data

Answer 1

a) Let X is a random variable following the Poisson distribution with rate parameter $\lambda$.

The probability mass function (pmf) of X is

$\begin{align*} P(X=x)&=\frac{\lambda^xe^{-\lambda}}{x!},\quad x=0,1,... \end{align*}$

the expected value of X is

$\begin{align*} E(X)&=\sum_{x=0}^{\infty}xP(X=x)\\ &=\sum_{x=0}^{\infty}x\frac{\lambda^xe^{-\lambda}}{x!}\quad \text{we can start from x=1 as the term is zero for x=0}\\ &=\sum_{x=1}^{\infty}x\frac{\lambda^xe^{-\lambda}}{x!}\\ &=e^{-\lambda}\sum_{x=1}^{\infty}\frac{\lambda^x}{(x-1)!}\\ &=e^{-\lambda}\sum_{x=1}^{\infty}\frac{\lambda\lambda^{x-1}}{(x-1)!}\\ &=\lambda e^{-\lambda}\sum_{x=1}^{\infty}\frac{\lambda^{x-1}}{(x-1)!}\\ \end{align*}$

Now we substitute k=x-1 and get

$\begin{align*} E(X)&=\lambda e^{-\lambda}\sum_{x=1}^{\infty}\frac{\lambda^{x-1}}{(x-1)!}\quad\text{ by replacing }k=x-1\\ &=\lambda e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k}}{k!}\\ &\text{but we know that }\sum_{k=0}^{\infty}\frac{\lambda^{k}}{k!}=e^{\lambda}\\ &=\lambda e^{-\lambda}e^{\lambda}\\ &=\lambda \end{align*}$
b) We will equate the first moment to the sample mean and get

$\begin{align*} &E(X)=\bar{x}\\ \implies&\lambda=\bar{x} \end{align*}$

The method of moment estimator of rate parameter $\lambda$ is

$\begin{align*} \hat{\lambda}=\bar{x} \end{align*}$

The sample mean is

$\begin{align*} \bar{x}&=\frac{\sum f \times x}{\sum f} = \frac{0\times24+ 1\times30+ 2\times17+ 3\times19+ 4\times7+ 5\times0+ 6\times3}{24+ 30+ 17+ 19+ 7+ 0+ 3}=1.67 \end{align*}$

The method of moment estimate of rate parameter $\lambda$ is

$\begin{align*} \hat{\lambda}=\bar{x}=1.67 \end{align*}$

The Poisson distribution of count data (where X is the count) is

$\begin{align*} P(X=x)&=\frac{1.67^xe^{-1.67}}{x!},\quad x=0,1,... \end{align*}$

c) We calculate the required probabilities using

$\begin{align*} P(X=0)&=\frac{1.67^0e^{-1.67}}{0!}=0.19\\ P(X=1)&=\frac{1.67^1e^{-1.67}}{1!}=0.31\\ P(X=2)&=\frac{1.67^2e^{-1.67}}{2!}=0.26\\ \end{align*}$

The relative frequency for the data is

count	Frequency	Relative frequency
0	24	0.24
1	30	0.3
2	17	0.17
3	19	0.19
4	7	0.07
5	0	0
6	3	0.03
sum	100

That is the sample probability of

$\begin{align*} P(\text{count}=0) &= \frac{\text{frequency}}{\text{total number of observations}}=24/100=0.24\\ P(\text{count}=1) &= \frac{\text{frequency}}{\text{total number of observations}}=30/100=0.30\\ P(\text{count}=2) &= \frac{\text{frequency}}{\text{total number of observations}}=17/100=0.17\\ \end{align*}$

The following is the comparison

count	Frequency	Poisson Probability P(x)	Relative frequency
0	24	0.19	0.24
1	30	0.31	0.3
2	17	0.26	0.17

The difference in probability for count=0 is 0.05 and count=2 is 0.09. But even with these differences, the probabilities are similar.

Hence we can say that the Poisson distribution is a good fit to the data.

Of course we need to goodness of fit analysis to ascertain if the fit is good.