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Let $X$ and $Y$ be two independent geometric random variables with common parameter $p$. Find $P(Y = y|X + Y = z)$ where $z ≥ 2$ and $y = 1, 2, · · · , z − 1$.

I am quite lost with this question. The correct solution is $1\over z-1$, however I have no idea how they got there. This is what I started with.

$$P(Y = y|X + Y = z)$$

$$ P(Y = y)\cap P(X + Y = z)\over P(X + Y = z)$$

$$ p(1-p)^{y-1}\cap\sum_{n=0}^zp(1-p)^{n-1}p(1-p)^{z-n-1}\over \sum_{n=0}^zp(1-p)^{n-1}p(1-p)^{z-n-1}$$

$$ p(1-p)^{y-1}\cap\sum_{n=0}^zp(1-p)^{n-1}p(1-p)^{z-n-1}\over \sum_{n=0}^zp(1-p)^{n-1}p(1-p)^{z-n-1}$$

$$ p(1-p)^{y-1}\cap\sum_{n=0}^zp^2(1-p)^{z-2}\over \sum_{n=0}^zp^2(1-p)^{z-2}$$

$$ p(1-p)^{y-1}\cap(z+1)p^2(1-p)^{z-2}\over (z+1)p^2(1-p)^{z-2}$$

As you can see, I have made quite a mess and am nowhere near the correct solution. Can somebody please help me out with this?

dlp
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1 Answers1

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You go wrong in the first step: writing $P(Y = y) \cap P(X + Y = z)$ is meaningless. The intersection of two numbers?

Instead, for the numerator you should write $$ P(Y = y \land X + Y = z) = P(Y = y \land X = z - y). $$ Now use the independence of $Y$ and $X$ and go from there.

Mees de Vries
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  • When I do that I get $$p(1-p)^{y-1}p(1-p)^{z-y-1}\over (z+1)p^2(1-p)^{z-2}$$

    $$p^2(1-p)^{z-2}\over (z+1)p^2(1-p)^{z-2}$$

    $$1\over z+1$$ What am I still doing wrong?

     – dlp Sep 04 '18 at 13:51
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    You are wrong in summing from $0$ to $z$, you should be summing from $1$ to $z-1$ and you get the correct answer. – Jan Sep 04 '18 at 13:59
  • @Jan Is that because $z\ge2$? – dlp Sep 04 '18 at 14:02
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    @agblt That is because $Y$ can only take values in ${1,2,\dots,z-1}$ under condition $X+Y=z$. Observe that $X,Y\geq1$. – drhab Sep 04 '18 at 14:06
  • @agblt Yes your final answer should be a discrete uniform distribution. Similar post: https://math.stackexchange.com/questions/1415259/conditional-probability-distribution-with-geometric-random-variables?noredirect=1&lq=1. – StubbornAtom Sep 04 '18 at 17:47