kuCAu

## Answered question

2021-02-06

Whether the statement is true or false.

### Answer & Explanation

Corben Pittman

Skilled2021-02-07Added 83 answers

We have to determine the sum of two independent Poisson random variables has a Poisson distribution.
Let X and Y are two Poisson random variables with parameters ${\lambda }_{1}$ and ${\lambda }_{2}$.
Hence, the moment-generating functions are:
${M}_{X}\left(t\right)={e}^{-{\lambda }_{1}+{\lambda }_{1}{e}^{t}}$
${M}_{Y}\left(t\right)={e}^{-{\lambda }_{2}+{\lambda }_{2}{e}^{t}}$
By theorem,
${M}_{X}+Y\left(t\right)=Mx\left(t\right)My\left(t\right)$
${M}_{X}+Y\left(t\right)={e}^{-{\lambda }_{1}+{\lambda }_{1}{e}^{t}}{e}^{-{\lambda }_{2}+{\lambda }_{2}{e}^{t}}$
${M}_{X}+Y\left(t\right)={e}^{\left(-{\lambda }_{1}+{\lambda }_{2}\right)}{e}^{\left(-{\lambda }_{2}+{\lambda }_{2}\right){e}^{t}\right)}$
Hence, the moment-generating function of X+Y is a moment:
generating function of Poisson distribution with parameter $\lambda 1+\lambda 2$.
That means X+Yis alsofollows a Poisson distribution with parameter $\lambda 1+\lambda 2$.
Therefore, the statement sum of two independent Poisson random variables has a Poisson distribution is true.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?