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Let X and Y are independent random variables that follow geometric distribution.

Then distribution of X given that X+Y=k is:

a) Binomial b) Poisson c) Uniform d) None of these

My attempt: I took both X and Y to have the same distribution, ie f(x) = f(y) = q^(x-1).p then tried to proceed but nothing really came out of it...

M31
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  • Well? any ideas? Add your tries to the question. – drhab Dec 01 '15 at 17:37
  • $X$ and $Y$ are also identically distributed or only independent r.v? – V. Vancak Dec 01 '15 at 17:46
  • Only independent... But I took them to be identically distributed trying to make it simpler.. but I just got a weird expression... – M31 Dec 01 '15 at 17:50
  • Related: http://math.stackexchange.com/questions/548525/how-to-compute-the-sum-of-random-variables-of-geometric-distribution – Brian Tung Dec 01 '15 at 17:57
  • @M31 If they are identically distributed, you can use the relation $X+Y \sim NB(2, p).$ Now, try to expend $P(X=x|X+Y=k)$ using the definition $P(A| B) = P(A\cap B)/P(B), P(B)>0$. – V. Vancak Dec 01 '15 at 18:36

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