# $P(X>0|X+Y)$ where X,Y have poisson distribution

X,Y both have poisson distributtion with parameters $$\lambda$$ $$\nu$$ accordingly: I have to calculate $$P(X>0|X+Y)$$, my first question is, does this mean I have to calculate it for every $$j$$ where $$X+Y=j$$ , because I am not sure if I understand this correctly.

If I am, $$\frac{P(X+Y=j\space\cap X>0)}{P(X+Y=j)}=1-\frac{P(X+Y=j\space\cap X=0)}{P(X+Y=j)}=1-\frac{\nu^{j}}{(\nu+\lambda)^{j}}e^{\lambda}$$ , but something must be wrong with this calculations, since when j=0, the probability should equal to zero, but it is not.

• Are they independent? – APC89 Feb 11 at 15:40
• Yes, sorry forgot to add it – ryszard eggink Feb 11 at 15:41
• Then $X + Y \sim \text{Poisson}(\lambda + \nu)$. – APC89 Feb 11 at 15:42
• I am aware, used it in my calculation – ryszard eggink Feb 11 at 15:45

You lost $$\mathbb P(X=0)$$ in last equality: $$1-\frac{\mathbb P(X+Y=j, X=0)}{\mathbb P(X+Y=j)}=1-\frac{\mathbb P(Y=j)\cdot\mathbb P(X=0)}{\mathbb P(X+Y=j)} = 1-\frac{\nu^{j}}{(\nu+\lambda)^{j}}.$$ This probability is equal to zero when $$j=0$$.