How to show the range is independent of the smallest order variable?

Question

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 the following exercise:

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 approach solving this exercise?

Answer 1

Assuming Geometric$(p)$ is supported on $\mathcal{N}=\lbrace 0,1,2,\dots \rbrace$, and denoting $\mathcal{Z}^+=\lbrace 1,2,\dots\rbrace$,

$$\begin{align} P\left[X_{(1)}=u\right] &=P(X_1=X_2=u)+P(X_1=u,X_2>u)+P(X_2=u,X_1>u)\\ &=P(X_1=u)^2 +2 P(X_1=u)P(X_2>u) \\ &=\left(p^2(1-p)^{2u} + 2 p(1-p)^u\ (1-p)^{u+1} \right)I_{\mathcal{N}}(u) \\ &=(1-p)^{2u}p(2-p)\ I_{\mathcal{N}}(u) . \tag{1}\label{1} \end{align}$$

$$\begin{align} &\phantom{=}P\left[X_{(1)}=u, X_{(2)}=u+d \right] \\ &=\left[ P(X_1=u, X_2=u+d) + P(X_1=u+d, X_2=u)\right]I(d>0) \\ &\quad\quad + P(X_1=X_2=u)I(d=0) \\ &= 2P(X_1=u+d, X_2=u)I(d>0)+P(X_1=X_2=u)I(d=0) \\ &= p^2(1-p)^{2u}\left[ 2(1-p)^d I_{\mathcal{Z}^+}(d) + I(d=0)\right] \ I_{\mathcal{N}}(u) \tag{2}\label{2} \end{align}$$

$$\begin{align} P\left[X_{(2)}-X_{(1)}=d \right] &=\left[ P(X_1-X_2=d) + P(X_2-X_1=d)\right]I_{\mathcal{Z}^+}(d) \\ &\qquad + P(X_1=X_2)I(d=0) \\ &= 2\mathbb{E}\left[ P(X_1=X_2+d\mid X_2)\right] I_{\mathcal{Z}^+}(d) \\ &\qquad +\mathbb{E}\left[ P(X_1=X_2\mid X_2) \right]I(d=0) \\ &= 2\mathbb{E}\left[ p(1-p)^{X_2+d}\right] I_{\mathcal{Z}^+}(d) + \mathbb{E}\left[p(1-p)^{X_2}\right]I(d=0)\\ &=p\left[2(1-p)^d I_{\mathcal{Z}^+}(d)+I(d=0)\right]\mathbb{E}\left[ (1-p)^{X_2}\right]\\ &= p\left[2(1-p)^d I_{\mathcal{Z}^+}(d)+I(d=0)\right] (2-p)^{-1} \tag{3}\label{3} \end{align}$$

Now equation \eqref{2} is a product of \eqref{1} and \eqref{3} for all $u$ and $d$, implying independence of $X_{(1)}$ and $X_{(2)}-X_{(1)}$.

Answer 2

For $d\geq0$, the memoryless property is usually stated as $$P(X>u+d\mid X\geq u) = \frac{P(X>u+d)}{P(X\geq u)}=\frac{(1-p)^{u+d+1}}{(1-p)^{u}}=(1-p)^{d+1}=P(X>d),$$ which is free of $u$. Thus, $P(X_2>X_1+d,\ X_2\geq X_1)=P(X_2>d)P(X_2\geq X_1)$.

Therefore, $$\begin{align} P\left[X_{(2)}-X_{(1)}>d\right]&=2P(X_2-X_1>d) \\ &=2\left[P(X_2>X_1+d,\ X_2\geq X_1)+ P(X_2>X_1+d,\ X_2 < X_1) \right]\\ &=2\left[ P(X_2>d)P(X_2\geq X_1) + 0 \right]\\ &=2 (1-p)^{d+1} \frac{1+P(X_1=X_2)}2\\ &=\frac{2(1-p)^{d+1}}{2-p}. \end{align}$$ But $$P\left[X_{(1)}>u\right]=P(X_1>u)P(X_2>u)=(1-p)^{2(u+1)}. $$

Their joint probility is $$\begin{align} P\left[X_{(1)}>u,\ X_{(2)}-X_{(1)}>d \right]&= 2P\left[X_1>u,\ X_2-X_1>d \right] \\ &=2\sum_{k=u+1}^\infty P(X_1=k, X_2>k+d)\\ &=2p(1-p)^{d+1} \sum_{k=u+1}^\infty(1-p)^{2k} \\ &=2p(1-p)^{d+1} \frac{(1-p)^{2(u+1)}}{2p-p^2}\\ &=\frac{2(1-p)^{2u+d+3}}{2-p}. \end{align} $$

Thus, $$P\left[X_{(1)}>u,\ X_{(2)}-X_{(1)}>d \right]=P\left[X_{(1)}>u\right]\ P\left[X_{(2)}-X_{(1)}>d\right].$$