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.

  • $\begingroup$ Are they independent? $\endgroup$ – APC89 Feb 11 at 15:40
  • $\begingroup$ Yes, sorry forgot to add it $\endgroup$ – ryszard eggink Feb 11 at 15:41
  • $\begingroup$ Then $X + Y \sim \text{Poisson}(\lambda + \nu)$. $\endgroup$ – APC89 Feb 11 at 15:42
  • $\begingroup$ I am aware, used it in my calculation $\endgroup$ – 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 $j=0$.

  • $\begingroup$ Ouch right thanks! $\endgroup$ – ryszard eggink Feb 11 at 17:28

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