We assume that $X_1\mid Y=y\sim \operatorname{Bin}(y,p_1)$ and $X_2\mid Z=z\sim \operatorname{Bin}(z,p_2)$, $Z$ and $Y$ are independent, and also $Y\sim \operatorname{Pois}(\lambda_1)$ , $Z\sim \operatorname{Pois}(\lambda_2)$. What is the conditional probability distribution of $$ P(X_1+X_2=x\mid Z,Y)=\text{?}$$ I actually wrote as the following but at last I could not find any well behaved distribution. (The values of $x_1+x_2$ vary from $0$ to $x=z+y$):
\begin{align} & P(X_1+X_2=x\mid Z,Y) = \sum_{i=0}^x P(X_1=i, X_2=x-i\mid Z,Y) \\[10pt] = {} & \sum_{i=0}^x P(X_1=i, X_2=x-i\mid Z,Y) \\[10pt] = {} & \sum_{i=0}^x P(X_1=i\mid Z,Y) P(X_2=x-i\mid Z,Y) \\[10pt] = {} & \sum_{i=0}^x P(X_1=i\mid Y=y) P(X_2=x-i\mid Z=z) \\[10pt] = {} & \sum_{i=0}^x \binom{y}{i} p_1^i (1-p_1)^{(y-i)} \binom{z}{x-i} p_2^{x-i} (1-p_2)^{z-(x-i)} \\[10pt] = {} & \sum_{i=0}^x \binom{y}{i} \left( \frac{p_1}{1-p_1}\right)^i (1-p_1)^y \binom{z}{x-i} \left(\frac{p_2}{1-p_2}\right)^{(x-i)} (1-p_2)^z \\[10pt] = {} & \frac{z!y!}{x!}(1-p_1)^y (1-p_2)^z \sum_{i=0}^x \frac{x!}{i! (x-i)!} \left( \frac{p_1}{1-p_1}\right)^i \left(\frac{p_2}{1-p_2}\right)^{x-i} \frac{1}{(y-i)!(z-(x-i))!}\ \end{align}
as based on the binomial rule we would have: $$ \sum_{i=0}^x \frac{x!}{i! (x-i)!} \left(\frac{p_1}{1-p_1}\right)^i \left(\frac{p_2}{1-p_2}\right)^{x-i} = \left(\frac{p_1}{1-p_1}+\frac{p_2}{1-p_2}\right)^x$$
but the remainders are $\frac{1}{(y-i)!(z-(x-i))!}$ and also $\frac{z!y!}{x!}(1-p_1)^y (1-p_2)^z $! and in this way I could not find any well defined distribution.
Can anybody help me with this problem please? Many thanks
$\require{cancel}$ \begin{align} \text{wrong: } & & \xcancel {\sum_{i=0}^x } & \binom y i p_1^i (1-p_1)^{y-i} \binom z {x-i} p_2^{x-i} (1-p_2)^{z-(x-i)} \\[10pt] \text{right: } & & \sum_{i\,=\,\max\{\,0,\,x-z\,\}}^{\min\{\,x,\,y\,\}} & \binom y i p_1^i (1-p_1)^{y-i} \binom z {x-i} p_2^{x-i} (1-p_2)^{z-(x-i)} \end{align} I.e. we cannot allow $x-i$ or $y-i$ or $x-(x-i)$ to be negative.
For example, if $x=8, y=6, z=5,$ then we have $\displaystyle \sum_{i\,=\,3}^6,$ and not $\displaystyle \sum_{i\,=\,0}^8.$
I doubt there is a simple closed form, but the probabilities can all be computed.