Consider the distribution of [$X\vert X>Y$][1] or the [memorylessness][2]. Then see if it is easy to change it to $X-Y\vert X>Y$

> But here Y is also a random variable. Does it matter?


If you know $\mathbb{P}(X \vert Y)$ and $Y$ is itself a random variable, then you can find the probability of $\mathbb{P}(X)$ as a [compound distribution][3] or by using the [law of total probability][4] 


$$\mathbb{P}(X) = \sum_{\forall Y}  \mathbb{P}(X \vert Y)\mathbb{P}(Y) \underbrace{ = \mathbb{P}(X \vert Y) \sum_{\forall Y}  \mathbb{P}(Y) = \mathbb{P}(X \vert Y)}_{\text{if $ \mathbb{P}(X \vert Y)$ is independent of $Y$}} $$


  [1]: https://stats.stackexchange.com/questions/48496/conditional-expectation-of-exponential-random-variable
  [2]: https://en.wikipedia.org/wiki/Exponential_distribution#Memorylessness
  [3]: https://en.wikipedia.org/wiki/Compound_probability_distribution
  [4]: https://en.wikipedia.org/wiki/Law_of_total_probability