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


$$\mathbb{P}(A) = \sum_{\forall B}  \mathbb{P}(A \vert B)\mathbb{P}(B) $$

if $ \mathbb{P}(A \vert B) = f(A)$ is a function independent of $B$ then it can be taken out of the sum and you get 

$$\mathbb{P}(A) = \sum_{\forall B}  \mathbb{P}(A \vert B)\mathbb{P}(B)  = f(A)\sum_{\forall B}  \mathbb{P}(B) = f(A)  $$

Similarly when $\mathbb{P}(X-Y\vert X>Y, Y)$ is independent from $Y$ then you know  $\mathbb{P}(X-Y\vert X>Y)$


  [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