How to show these random variables are independent?

[psie]

TL;DR Summary

I am studying order statistics in An Intermediate Course in Probability by Gut. First the author treats only continuous distributions. In a section on the joint distribution of the extreme order variables $X_{(n)}=\max\{X_1,\ldots,X_n\}$ and $X_{(1)}=\min\{X_1,\ldots,X_n\}$, the author derives the density of the range. that is $R_n=X_{(n)}-X_{(1)}$. Then there's an exercise which I simply do not understand why it's in that section.

The exercise that appears in the text is:

Exercise 2.5 The geometric distribution is a discrete analog of the exponential distribution in the sense of lack of memory. More precisely, show that if $X_1$ and $X_2$ are independent $\text{Ge}(p)$-distributed random variables, then $X_{(1)}$ and $X_{(2)}-X_{(1)}$ are independent.

What I find confusing about this exercise is that the author has, up until now, not derived any results for order statistics when it comes to discrete distributions. I know the formula for the density of $X_{(1)}$ and the range when the underlying distribution is continuous, but these do not apply for discrete distribution. I was thinking going back to an earlier chapter where the author derives distributions of transformations of random variables. I was thinking I could assume $X_2$ to be greater than $X_1$ and then compute the pmf of their difference, but this doesn't feel like a sensible assumption, since after all, $\max\{X_1,\ldots,X_n\}$ is understood pointwise.

How would you go about solving this exercise?

 

[psie]

This has been solved. It's a bit of work writing it all down, but basically you want to compute the probabilities $P\left(X_{(1)}=u\right)$, $P\left(X_{(1)}=u, X_{(2)}=u+d \right)$ and $P\left(X_{(2)}-X_{(1)}=d \right)$. The first two are fairly straightforward, splitting up the probability using indicator functions. The third probability is a bit more tricky and conditional expectation will come in handy (in addition to indicator functions). In particular, the identity $$P(A)=E[\mathbf1_A]=E[E[\mathbf1_A\mid X]]=E[P(A\mid X)].$$