Q.6.2 - A First Course in Probability

Question

Suppose that the number of events occurring in a given time period is a Poisson random variable with parameter $\lambda$. If each event is classified as a type $i$ event with probability$p_i,i=1,....,n,\sum p_i=1$, independently of other events, show that the numbers of type $i$ events that occur,$i=1,....,n,$ are independent Poisson random variables with respective parameters $\lambda p_i,i=1,.....,n.$

Step-by-Step Solution

Answer

Thus, the numbers of type events $i$ that occur $i=1,....,n,$ are independent Poisson random variables with corresponding parameters $\lambda p_i,i=1,....,n$, as seen on the right hand side of the formula that is given .

Step 1 : Poisson random variable :

A Poisson random variable is used to illustrate how many times an event will happen in a given amount of time.

Step 2 : Explanation :

Formula : 

For Poisson distribution of random variables $X_i,i=1,......,n,$ we have:

$P\left(\sum _{i=1}^n{X}_i=\sum _{i=1}^n{x}_i\right)=\frac{{}_{\left(\lambda \right)}\sum _{i=1}^n{x}_i}{\left(\sum _{i=1}^n{x}_i\right)!}{e}^{-\lambda }$

Proof :

Let $X_i,i=1,.....,n,$ be the Poisson random variables with respect to each event of type $i$.

Since, it is given that $\sum _{i=1}^n{X}_i~P\left(\lambda \right)$, then assume a set of non- negative integers $\left\{x_1,x_2,x_3,......,x_n\right\}$,

We have:

$$\begin{gathered} P(X_1=x_1, X_2=x_2,....X_n=x_n) \\ =P\left(X_1=x_1, X_2=x_2,....X_n=x_n\parallel \sum _{i=1}^n{X}_i=\sum _{i=1}^n{x}_i\right)·P\left(\sum _{i=1}^n{X}_i=\sum _{i=1}^n{x}_i\right) \end{gathered}$$

Assuming there were a total of $\sum _{i=1}^n{x}_i$ events, random variables $X_1,X_2,.....X_n$ will have a multinomial distribution with corresponding parameters $\sum _{i=1}^n{x}_i,p_1,p_2,...p_n$.

Thus we write:

$$\begin{gathered} P\left(X_1=x_1, X_2=x_2,....X_n=x_n\parallel \sum _{i=1}^n{X}_i=\sum _{i=1}^n{x}_i\right) \\ =\frac{\left(\sum _{i=1}^n{x}_i\right)!}{x_1!x_2!....x_n!}{p}_1^{x_1}{p}_2^{x_2}....p_n^{x_n}. \end{gathered}$$

And for Poisson distribution, we also know that:

$P\left(\sum _{i=1}^n{X}_i=\sum _{i=1}^n{x}_i\right)=\frac{{}_{\left(\lambda \right)}\sum _{i=1}^n{x}_i}{\left(\sum _{i=1}^n{x}_i\right)!}{e}^{-\lambda }$.

Step 3 : Explanation :

 Hence

$P(X_1=x_1,X_2=x_2,......,X_n=x_n)=\frac{{}_{\left(\lambda \right)}\sum _{i=1}^n{x}_i}{x_1!x_2!...x_n!}·(p_1^{x_1}{p}_2^{x_2}....p_n^{x_n})·e^{-\lambda }$

$$\begin{gathered} = \underset{i=1}{\overset{n}{\prod }}\frac{{\left(p_i\lambda \right)}^{x_i}}{x_i!}{e}^{-p_i\lambda } \\ =\underset{i=1}{\overset{n}{\prod }}P(X_i=x_i) \end{gathered}$$

$=P(X_1=x_1)·P(X_2=x_2)·····P(X_n=x_n)$