Trouble with Adding poisson RV Is this how one could find distribution for Poisson random variables,
say $X$ is Poisson with $\lambda=9$ and $Y$ is with $\lambda=16$
Say I want to find distribution of $Z=X+2Y$
I want to use moment generating function method,
$$Mx=e^{-9(1-e^{t})}$$ and $$My=e^{-16(1-e^{t})}$$
then isn't $$M2y=e^{-16(1-e^{2t})}$$
Which would tell me $$Mz=e^{-25(1-e^{t})(1-e^{3t})}$$
Which isn't in the form I wanted.
So how do I fix this? 
 A: The problem is that $2Y$ is not Poisson, since $\Pr[2Y = 1] = 0$, for example.  The most straightforward method of calculation is to use convolution:  $$\Pr[X + 2Y = n] = \sum_{k=0}^{\lfloor n/2 \rfloor} \Pr[Y = k]\Pr[X = n-2k] = \sum_{k=0}^{\lfloor n/2 \rfloor} e^{-(\lambda_1 + \lambda_2)} \frac{\lambda_1^{n-2k}}{(n-2k)!} \frac{\lambda_2^k}{k!}.$$  There may be a way to write this more cleverly, but at this time I don't have any further manipulation of this expression.  What I can suggest is to consider the odd and even cases for $n$ separately; i.e., consider $n = 2m$ separately from $n = 2m+1$.
A: From a simple simulation in R one can get an idea of the shape
of this distribution. The histogram of a million simulated
values of $S = X + 2Y$ shows that the distribution is too
strongly skewed for a normal approximation to be accurate.
Also, comparing $E(S)$ and $Var(S),$ we see that $S$ cannot be Poisson.
 lam.x = 9;  lam.y = 16;  m = 10^6
 x = rpois(m, lam.x);  y = rpois(m, lam.y)
 s = x + 2*y
 > mean(s);  sd(s);  sqrt(9 + 4*16)
 [1] 41.0018  # approx. 9 + 32 = 41
 [1] 8.539994
 [1] 8.544004 # exact SD(S)
 > mean(s > 20);  mean(s > 60)
 [1] 0.995679
 [1] 0.01545


