This question comes out of the following proof.
Two independent random variables, $X_1, X_2$. $X_1$ occurs with probability $p$ and $X_2$ with $(1-p)$.
The sum, $N = X_2+X_1$ has $P(N=i+j) \sim Poisson(\lambda)*$.
*(see below re. dodgy notation)
If $N=i+j$ then,
$$ P(X_1=i, X_2=j) = P(X_1=i, X_2=j, N = j+i) = P(X_1=i, X_2=j\ |\ N = j+i)P(N=i+j) $$
This is where I have issues. I'm not sure why I can't do the following:
$$ \tag{1} P(X_1=i, X_2=j\ |\ N = j+i) \sim Binomial(N,p) $$
and by independence,
$$ \tag{2} P(X_1=i, X_2=j\ |\ N = j+i) = P(X_1=i\ |\ N = j+i)P(X_2=j|\ N = j+i)) $$
this gives me the wrong answer though as I get $Binomial(N,p)^2$ as a result.
- A side question: What is the proper notation for saying that $P(N=i+j)$ follows a Poisson distribution.
If $N=X_1+X_2$ for independent Bernoulli random variables $X_1, X_2$ then $N$ is not Poisson distributed, yet nor is it Binomial.
It is a Poisson Binomial Distribution; being the sum of two non-identical independent Bernoulli trials.
$$\begin{align} \mathsf P(N=n)\; & =\; \mathsf P(X_1=0)\mathsf P(X_2=n)+\mathsf P(X_1=1)\mathsf P(X_2=n-1) & \big[n\in\{0,1,2\}\big] \\[1ex] & = \; \begin{cases} (1-p)p & : n=0 \\ (1-p)^2+p^2 & : n=1 \\ (1-p)p & : n=2\end{cases} \end{align} $$