If $X,Y,Z \sim $ Poisson($\lambda$) are independent, how do I calculate $P(X+Y=2,X+Z=3)$?

Question

I tried only with some algebric manipulation like $$P(X+Y=2,X+Z=3)=P(X=2-Y,Z=1+Y)=P(X=2-Y)P(Z=1+Y)$$ But it seems to me that this is not going anywhere. Maybe can be useful the fact that $X+Y,X+Z \sim $ Poisson($2\lambda$), but I don't know how to use it.

Answer 1

Note that $X+Y=2, X+Z=3$ requires that $X\leq 2.$ You can now use the following: $$P(X+Y =2, X+Z = 3) = \sum_{x=0}^2 P(X+Y = 2, X+Z=3|X=x)P(X=x) = \sum_{x=0}^2 P(Y=2-x)\cdot P(Z=3-x)\cdot P(X=x),$$ using independence of $X,Y,Z.$ This is ugly, but it is something that you can just calculate by hand, in principle.

Answer 2

Note that $X$, $Y$, and $Z$ can only take on non-negative integer values.

For $X + Y = 2$, the following are possible:

$$\begin{array}{c|c} X & Y \\ \hline 0 & 2 \\ 1 & 1 \\ 2 & 0 \end{array}$$

For $X + Z = 3$, the following are possible:

$$\begin{array}{c|c} X & Z \\ \hline 0 & 3 \\ 1 & 2 \\ 2 & 1 \\ 3 & 0 \end{array}$$

Thus, the entire list of outcomes you are interested in is:

$$\begin{array}{c|c|c} X & Y & Z \\ \hline 0 & 2 & 3 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \\ 3 & - & 0 \end{array}$$

and we drop the last outcome since no value of $Y$ is appropriate, so you are then left with

$$\begin{array}{c|c|c} X & Y & Z \\ \hline 0 & 2 & 3 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \end{array}$$

of which you can find the probabilities easily through independence.