marginal distribution poisson distributed variables Suppose  $Y_1,Y_2$  both has poisson distribution with mean values $e^{\lambda}$ and $e^{\lambda+ \psi}$ respectively.both are independent
I want to show that the  marginal distribution
$f(Y_1+Y_2 = u) = \sum_v f(Y_1+ Y_2 = u, Y_2 = v ) $ has a $Po(\mu_u)$ with $ \mu_u= E[Y_1 + Y_2] =  e^{\lambda}(1+e^\psi)$
here  is  where I am:
\begin{align}
& \sum_v f(Y_1+ Y_2 = u, Y_2 = v ) = \sum_{v=0}^u f(Y_1 = u-v, Y_2 = v ) \\[10pt]
= {} & \sum_{v=0}^u  f(Y_1 = u-v)f(Y_2=v) = \sum_{v=0}^u \dfrac{(e^{\lambda})^{(u-v)} }{(u-v)!}e^{-e^\lambda} \dfrac{(e^{\lambda + \psi})^v }{v!} e^{-e^{(\lambda + \psi)}}
\end{align}
I don't see how this can be a poisson , the  $\sum_{v=0}^{v=u}$ is actually confusing me. Can I get some help how to proceed?
 A: $$\sum_{v=0}^{u} \dfrac{(e^{\lambda})^{(u-v)} }{(u-v)!}e^{-e^{\lambda}} \dfrac{(e^{\lambda + \psi})^{v} }{v!}e^{-e^{(\lambda + \psi)}} = \dfrac{e^{u\lambda}e^{-e^{\lambda}}e^{-e^{(\lambda + \psi)}}}{u!}\sum_{v=0}^{u} \binom{u}{v} (e^{\psi})^{v} \\ = \dfrac{e^{u\lambda}e^{-e^{\lambda}}e^{-e^{(\lambda + \psi)}}(1+e^\psi)^u}{u!} = \dfrac{e^{-(e^{\lambda}(1+e^{\psi}))}(e^{\lambda}(1+e^\psi))^u }{u!}$$
Possibly easier method would be to use moment generating functions.
A: \begin{align}
\huge &  \sum_{v=0}^u \frac 1 {(u-v)! v!} \Big(\cdots\cdots\Big) = \frac 1 {u!} \sum_{v=0}^u \frac{u!}{(u-v)!v!} \Big(\cdots\cdots\Big) = \frac 1 {u!} \sum_{v=0}^u \binom u v \Big(\cdots\cdots\Big)
\end{align}
The reason the $\dfrac 1 {u!}$ can be outside the summation is that it does not change as $v$ goes from $0$ to $u$.
The next thing to look at is whether the binomial theorem can be applied to the sum that remains.
A: Your question reduces to showing that the sum of 2 independent Poisson random variables is a Poisson random variable. This can be done as a one-liner using pgf's: see, for instance, the wiki page:  
https://proofwiki.org/wiki/Sum_of_Independent_Poisson_Random_Variables_is_Poisson
