Let $X,Z$ and $W$ be independent random variables with $ X \sim Bin(1,p) $ and $Z,W \sim Pois(\lambda)$.
We define $$ Y:= XZ + W$$ I would like to determine the densities of $(X,Y)$ and $Y$.
I've shown that : $$ p_{X,Y}(0,n) = P(X=0,Y=n) = $$ $$ = P(X=0,W=n) = p_X(0)*p_W(n)= $$ $$=(1-p)\frac{e^{-\lambda}\lambda^n}{n!} $$ by taking into account the independence assumption.
Then I've shown that :
$$ p_{X,Y}(1,n)=P(X=1,Y=n)=P(X=1,Z+W=n)=p_X(1)*p_{Z+W}(n)=$$
$$=p\frac{e^{-2\lambda}(2\lambda)^n}{n!}$$
Now It remains to compute the density for $ Y=XZ+W$. So $$ p_Y(n)=P(XZ+W=n) $$ If $X=0$ we have that $W=n$ and when $X=1$ we get $Z+W=n$.
Can the density $p_Y(n)$ be written directly in terms of the densities previously defined? I'm not sure how to procede from here but I believe that $ p_Y(n) $ can be written as the sum of $p_{X,Y}(1,n)$ and $p_{X,Y}(0,n)$, but I'm not so sure.
Any support for this question would be appreciated.