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Let $X$ and $Y$ be iid random variables distributed geometrically with probability of success $p$ and support $\mathbb{N}=\{0,1,2,\cdots\}.$ So in particular, letting $q=1-p$, we have that $$\mathbf{P}(X=k)=\mathbf{P}(Y=k)=pq^k\quad\forall k \in \mathbb{N}.$$

I'm trying to find $$\mathbf{P}(X=x\mid X+Y=k)$$ for $x,k \in \mathbb{N}.$ As an intermediate goal, I'm trying to find $$\mathbf{P}(X=x,\, X+Y=k)$$

But the result I'm getting is independent of $x$ for some reason. Here's the logic:

$\mathbf{P}(X=x,\, X+Y=k)=$

$\mathbf{P}(X=x,\, Y=k-x)=$

$\mathbf{P}(X=x)\mathbf{P}(Y=k-x)=$

$pq^x pq^{k-x}=$

$p^2 q^k$

Surely the expression of interest isn't independent of $x$, so why is it vanishing?

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That the expression should indeed be independent of x might be seen by answering the following: in a heads-or-tail game, which event amongst THHHT, HTHHT, HHTHT and HHHTT is the most probable? –  Did Jun 8 '13 at 12:58

1 Answer 1

up vote 1 down vote accepted

The result is perhaps at first a little surprising. But your calculation is perfectly correct. Continue on the path you are on, dividing by the easily computed $\Pr(X+Y=k)$ to find the conditional probability. You will end up with a familiar distribution.

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