Chapter 17.6, Problem 1RE

To determine

Prove that the random variable ${\displaystyle \sum _{i=1}^n{X}_i}$ is a Poisson random variable with mean ${\displaystyle \sum _{i=1}^n{\mu }_i}$ and variance ${\displaystyle \sum _{i=1}^n{\mu }_i}$.

Explanation of Solution

Calculation:

According to Theorem 7.10, it is known that for n random variables $X_1,X_2,\mathrm{....},X_n$, the moment generating function can be written as follows:

$M_{X_1+X_2+\mathrm{....}+X_n}=M_{X_1}\left(t\right)M_{X_2}\left(t\right)\mathrm{...}{M}_{X_n}\left(t\right)$, where $M_{X_i}\left(t\right)\text{'}\text{s}$ are moment generating function of random variable $X_i$ for all $i=1,2,\mathrm{...},n$.

The moment generating function of Poisson random variable X with mean and variance $\mu$ is known as $M_X\left(t\right)=e^{\mu \left(e^t-1\right)}$.

According to given question $X_1,X_2,\mathrm{...},X_n$ are independent Poison random variables with means ${\mu }_1,{\mu }_2,\mathrm{....},{\mu }_n$, respectively.

Now, using Theorem 7.10, the moment generating function of ${\displaystyle \sum _{i=1}^n{X}_i}$ is obtained as follows:

$\begin{array}{c}{M}_{{\displaystyle \sum _{i=1}^n{X}_i}}=M_{X_1+X_2+\mathrm{....}+X_n}\\ =M_{X_1}\left(t\right)M_{X_2}\left(t\right)\mathrm{...}{M}_{X_n}\left(t\right)\\ =e^{{\mu }_1\left(e^t-1\right)}{e}^{{\mu }_2\left(e^t-1\right)}\mathrm{....}{e}^{{\mu }_n\left(e^t-1\right)}\\ =e^{\left({\mu }_1+{\mu }_2+\mathrm{...}+{\mu }_2\right)\left(e^t-1\right)}\\ =e^{\left({\displaystyle \sum _{i=1}^n{\mu }_i}\right)\left(e^t-1\right)}\end{array}$

Hence, it is proved that the random variable ${\displaystyle \sum _{i=1}^n{X}_i}$ is a Poisson random variable with mean ${\displaystyle \sum _{i=1}^n{\mu }_i}$ and variance ${\displaystyle \sum _{i=1}^n{\mu }_i}$.