My question arises from the proof of the following result.
Let $X$ and $Y$ be two independent random variables, where $X$ has a geometric distribution and $Y$ takes only integer values. Then \begin{equation}\label{memoryless} \boldsymbol{\operatorname{P}}(X-Y>n|X>Y) = \boldsymbol{\operatorname{P}}(X>n), \end{equation} for all $n \in \mathbb{N}$.
At one step of the proof, the author concludes that it follows from the memoryles property \begin{equation*} \boldsymbol{\operatorname{P}}(X-n=k|X>n) = \boldsymbol{\operatorname{P}}(X=k), \end{equation*} for all $k,n \in \mathbb{N}$ that \begin{equation*} \boldsymbol{\operatorname{P}}(X-Y=k|X>Y) = \boldsymbol{\operatorname{P}}(X=k). \end{equation*}
Here the integer $n$ is replaces by the random variable $Y$. Why is it the case that this can be done? It seems obvious but how can you argue that this implication is true? When considering how the memoryless property is shown, that is by the calculation
\begin{align} \boldsymbol{\operatorname{P}}(X-n>k|X>n) &= \frac{\boldsymbol{\operatorname{P}}(X>k+n,X>k)}{\boldsymbol{\operatorname{P}}(X>n)} \\ &=\frac{\boldsymbol{\operatorname{P}}(X>k+n)}{\boldsymbol{\operatorname{P}}(X>k)} \\ &= \frac{(1-p)^{k+n}}{(1-p)^k} \\ &= (1-p)^k \\ &=\boldsymbol{\operatorname{P}}(X>k), \end{align}
it is not obvious to me how the third equality can hold, when $n$ is replaced by $Y$.
I would appreciate if you could help me clarifying this.
Because $Y$ only take values in integers, it holds $$\begin{align*} P(X−Y>k|X>Y) &= \sum_{n\in\Bbb N} P(X−Y>k, Y=n|X>Y) \\ &= \sum_{n\in\Bbb N} P(X - n > k, Y=n | X > n) \\ &=\sum_{n\in\Bbb N}P(X-n>k|X>n)P(Y=n) \\ &= \sum_{n\in\Bbb N}P(X=k)P(Y=n) \\ &= \sum_{n\in\Bbb N}P(X=k,Y=n) \\ &= P(X=k)\end{align*}$$
Where the third and fifth equality holds by independency assumption of $X$ and $Y$.