Let X be a positive random variable such that for all $x,y>0$ we have that $$\mathbb{P}[X >x+y | X>x] = \mathbb{P}[X > y]$$
I need to show that X has exponential distribution, i.e, $\mathbb{P}[X>x]=e^{-\lambda x}$ for $x>0$. I was able to prove the converse statement, given that X has exponential distribution, I showed that the formula is true, which was pretty straightforward. However, I am not sure how to approach this problem. Any suggestions?
Denoting $r(x)=\mathbb P[X>x]$ we have:
$$r(x+y)=\mathbb{P}\left[X>x+y\right]=\mathbb{P}\left[X>x+y\mid X>x\right]\mathbb{P}\left[X>x\right]=\mathbb{P}\left[X>y\right]\mathbb{P}\left[X>x\right]=r\left(y\right)r\left(x\right)$$
Moreover $r$ is a non increasing functions taking values in $[0,1]$.
Now have a look at the answers on this question.
Look at characterization (5) in the following article:
http://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function
In your case, $f(x):= \mathbb{P}(X > x)$.
To get the formula in characterization (5) you need to integrate out the conditioning in $x$ in your formula.
Of course, you will need to make appropriate modifications of characterization (5) to account for the $-\lambda$!
