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?
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.