How to calculate probability distribution of a function of two independent Poisson random variables? I can't figure out how to determine the probability distribution function of
$$aX + bY,$$ 
where $X$ and $Y$ are independent Poisson random variable. Basically, I want to check whether $aX+ bY$ follows a Poisson distribution. Here, $a$ and $b$ are constants coefficients. 
I tried this way but can't work it out...

\begin{align*}
\mathbb{P}(aX + bY= N ) &= \mathbb{P}(aX =k)\mathbb{P}(bY = N - k)\\
&=  \mathbb{P}\left(X= \frac ka\right)\mathbb{P}\left(Y = \frac{N-k}{b}\right)
\end{align*}

But in this way , I got to determine $(k/a)!$ 
And,  I can't calculate ahead.
Is there a general rule for tackling problems like this? 
What if I wanted to determine the probability distribution of $aX^2 + bY^2$ ?
Thanks in advance. 
 A: We show that if $X$ and $Y$ are independent Poisson, and $aX+bY$ is Poisson, then $a=1,b=0$ or $a=0,b=1$, or $a=b=1$.  We give an elementary argument. It is nicer, but less elementary, to use moment generating functions. 
Let $W=aX+bY$. With positive probability we have $X=1$ and $Y=0$. So with positive probability we have that $W=a$. It follows that $a$ is a non-negative integer. Similarly, $b$ is a non-negative integer.
We have $W=1$ with positive probability. Since $a$ and $b$ are non-negative integers, it follows that one of $a$ or $b$ is $1$. Without loss of generality we may assume $a=1$. We show that we cannot have $b\ge 2$.
Let $X$ and $Y$ have parameters $\lambda$ and $\mu$ respectively. Then since $\Pr(W=0)=e^{-(\lambda+\mu)}$, it follows that $W$ has parameter $\lambda+\mu$, so 
$$\Pr(W=1)=(\lambda+\mu)e^{-(\lambda+\mu)}.$$
 But 
$$\Pr(W=1)=\Pr(X=1)\Pr(Y=0)=\lambda e^{-(\lambda+\mu)}.$$
Cancelling, we find that $\lambda+\mu=\lambda$, so $\mu=0$, which is impossible. 
Added, a better way: Suppose that $aX+bY$ is Poisson, where $X,Y$ respectively have parameters $\lambda$ and $\mu$. Then $aX+bY$ is Poisson with mean, and therefore variance, $a\lambda+b\mu$. But $aX+bY$ has variance $a^2\lambda+b^2\mu$. It follows that $(a^2-a)\lambda+(b^2-b)\mu=0$. Now we use a little of the previous proof, the part where we showed $a$ and $b$ are non-negative integers. Since $a^2\ge a$ and $b^2\ge b$, we conclude that $a^2=a$ and $b^2=b$, so $a$ and $b$ have to be one of $0$ or $1$, and we are finished.
This method generalizes nicely. We can obtain the following result. Let $X_1,X_2,\dots, X_n$ be independent Poisson. If $a_1X_1+a_2X_2+\cdots +a_nX_n$ is Poisson, then for any $i$ we have $a_i=0$ or $a_i=1$.
A: The calculation would be a special case of the more general discrete convolution formula.

Assuming $X$ and $Y$ are as above, we have that the probability mass function of $aX+bY$ is 
$$\mathbb{P}(aX+bY=z)= \sum_{i=0}^z \mathbb{P}(aX=i)\mathbb{P}(bY=z-i)$$ 

This becomes complicated in general because (for $Z \sim Poisson(\lambda)$, $m$ a non-negative integer, and $c$ some arbitrary real constant)

$$\mathbb{P}(cZ=m)=\mathbb{P}(Z=\frac{m}{c}) = \begin{cases} 0 \quad \text{if }\frac{m}{c}\text{ is not a non-negative integer} \\ e^{-\lambda} \displaystyle\frac{\lambda^{\frac{m}{c}}}{\left(\frac{m}{c}\right)! } \quad \text{if }\frac{m}{c} \text{is a non-negative integer}\end{cases}$$

As one can see in the answer for this question, the probability mass function of such random variables is therefore difficult to give in closed form in general. I will try to do so below:

$$\mathbb{P}(aX+bY=z)= \begin{cases} 0 \text{ if }z \not=am\text{ and } z \not=bn \text{ for some positive integer $m$ or $n$}  \\ \\ \mathbb{P}(X=m)\mathbb{P}(Y=0) \quad \text{if } z=am \text{ for some positive integer } m \\ \\ \mathbb{P}(X=0)\mathbb{P}(Y=n) \quad \text{if z = $bn$ for some positive integer $n$} \\ \\ \mathbb{P}(X= m)\cdot \mathbb{P}(Y=n) \quad \text{if $z=am=bn$ for some positive integers $m,n$.} \end{cases}$$

Without more specific conditions on $a$ and $b$ it is difficult to simplify this in general.
Also: a quick check on this answer -- if it does not sum to $1$ over all possible values of $aX+bY$, then it is wrong.
