Let $X$ be a random poisson variable with the parameter $5$, let $Y$ be a random poisson variable with the parameter $10$. $X, Y$ are independent of each other.
What is the probability that $2(X+Y) = 28?$
My Approach Let $Z=2(X+Y)$
Since $X$ and $Y$ are indepenent variables and and indepedent on each other, $Z$ is a poisson random variable with the parameter $\lambda_Z = \lambda_X + \lambda_Y = 15$.
Therefore we can use the poisson probability function -
$$P(Z=28)=P(2X+2Y=28) = e^{-15} \cdot \frac{15^{28}}{28!} = 0.00085$$
However I'm afraid I'm wrong here. What do I do with the $2$ here? If I had only $X+Y$ I would be pretty sure I'm right, but I am not sure.
Should I multiplie the result on $2$?
Thank you!
If $X$ and $Y$ have Poisson-distribution with rates $\lambda_X$, $\lambda_Y$ and are independent then $X+Y$ has Poisson-distribution with rate $\lambda_X+\lambda_Y$.
This can be exploited to find: $$P(2(X+Y)=28)=P(X+Y=14)$$
Observe for $Z:=2(X+Y)$ we have $P(Z=n)=0$ if $n$ is odd, showing that it has not Poisson-distribution.