# How do I find this probability if $X$ and $Y$ are independent?

The problem says: Suppose that $X$ and $Y$ are independent Poisson random variables with parameters 1 and 2 respectively. Find: $$\Bbb P (X=1|\frac{X+Y}{2} = 2)$$ I have thought about this for a long time and I thought that since $X$ and $Y$ are independent we could make it say $$\Bbb P(X=1|Y=4-2)=\frac{\Bbb P(X=1) \Bbb P(Y=4-X)}{\Bbb P(Y=4-X)}=\Bbb P(X=1)$$ however, I realize that $X$ and $Y$ are not independent that way. What am I doing wrong? I know this cant be that trivial but something isn't making sense to me.

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$$\Bbb P \left(X=1|\frac{X+Y}{2} = 2\right) = \Bbb P (X=1|{X+Y}= 4)$$

$$= \frac{\Bbb P \left(X=1 \text{ and } X+Y=4\right)}{\Bbb P \left(X+Y=4\right)}= \frac{\Bbb P \left(X=1 \text{ and } Y=3\right)}{\Bbb P \left(X+Y=4\right)} = \frac{\Bbb P (X=1) \Bbb P (Y=3) }{\Bbb P \left(X+Y=4\right)}$$

$$= \tfrac{\Bbb P (X=1) \Bbb P (Y=3) }{\Bbb P (X=0) \Bbb P (Y=4) + \Bbb P (X=1) \Bbb P (Y=3) +\Bbb P (X=2) \Bbb P (Y=2)+\Bbb P (X=3) \Bbb P (Y=1)+\Bbb P (X=4) \Bbb P (Y=0)}$$

which you can calculate.

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Let me mention that the denominator is P(Z=4) for Z Poisson with parameter 1+2=3, hence it equals e^(-3)3^4/4! (which might be simpler than the summation of five products). The exponentials cancel out hence one is left with (1^1/1!)(2^3/3!)(4!/3^4)=32/81. –  Did Oct 30 '12 at 23:06

Recall that $$P(A \vert B) = P(A \cap B)/P(B) = P(B/A) \times P(A)/P(B)$$ Hence, $$P(X = 1 \vert X+Y = 4) = P(X+Y=4 \,\,\& \,\, X=1)/P(X+Y = 4)$$ So we need to evaluate is $P(X+Y=4 \,\,\& \,\, X=1)$ and $P(X+Y = 4)$ $$P(X+Y=4 \,\,\& \,\, X=1) = P(X=1 \,\,\& \,\, Y=3) = P(X=1)P(Y=3)$$ $$P(X+Y = 4) = \sum_{k=0}^4P(X=k)P(Y=4-k)$$

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The sum at the end is not useful, see my comment on the other answer. –  Did Oct 30 '12 at 23:12
@did True. But the way I would have gone about simplifying is to multiply and divide by $4!$, which would have given $e^-(1+2) (1+2)^4/4!$. (essentially proving that sum of two poisson is again possion) –  user17762 Oct 30 '12 at 23:15