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Here's my attempt, and I'm not sure if it is even close to the right direction.

If $P(X>m+n|X>m)=P(X>n)$ then if F is the cumulative distribution function of X, we have $$\frac{P(X>m+n)}{P(X>m+n)+P(m<X<m+n)} = P(X>n)$$$$\frac{1-F(m+n)}{1-F(m+n)+[F(m+n)-F(m)]} = 1-F(n)$$$$ \frac{1-F(m+n)}{1-F(m)} = 1-F(n)$$$$ F(n) = 1-\frac{1-F(m+n)}{1-F(m)}$$$$ F(n) = \frac{F(m)+F(m+n)}{1-F(m)}$$

Is this even anything?

I was hoping to be able to show that this function must be something like $$F(n)=1-(1-p)^n$$ where $p$ is the probability of a success but I don't know how to do that. Is this a dead end? If so what would be a good way to attack this?

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up vote 2 down vote accepted

It is better to think in terms of $1-F(n) = \mathbb{P}(X>n)$. Then $$ \mathbb{P}(X>m+n |X>m) = \frac{\mathbb{P}(X>m+n,X>m)}{\mathbb{P}(X>m)} \stackrel{n\geqslant 0}= \frac{\mathbb{P}(X>m+n)}{\mathbb{P}(X>m)} = \frac{1-F(n+m)}{1-F(m)} $$ Thus the complementary cdf $1-F_X(n)$ satisfies a functional equation: $$ \left( 1-F(n+m)\right) = \left( 1-F(m)\right) \left( 1-F(n)\right) $$ Hence the solution is $ 1-F(n) = q^n$ for some $0 < q \leqslant 1$, i.e. $X$ is geometric with $p=1-q$.

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Ahh, perfect! I just didn't know what to look for to show that F(n) must be $q^x$ — I didn't think of taking advantage of the rules of adding exponents. One thing, how can you be sure that $0<q\le 1$? – crf Jul 31 '12 at 15:37
Being the probability $0 \leqslant 1-F_X(n) \leqslant 1$, giving $0 \leqslant q \leqslant 1$. But in order for the conditional probability to make sense you need to rule out $q=0$ case. – Sasha Jul 31 '12 at 15:44
Duh. Thank you so much! I really wasn't sure about the functional equation thing, I really just kind of pulled that idea out of nowhere. I'm glad it ends up working. – crf Jul 31 '12 at 15:52

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