# Whether the statement is true or false.

Whether the statement is true or false.
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Corben Pittman

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.