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$\lambda p_i,i=1,.....,n.$

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

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:

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

Hence

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

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

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