# Let $\mathbf{R}^{*}$ be rank vector of a random sample and $V=R_{1}^{*}-R_{N}^{*}$. Then $\mathsf{P}(V=k)= \frac{N-|k|}{N(N-1)}$

Let $Z_{1},\ldots, Z_{N}$ be i.i.d. continuous random variables with rank vector $\mathbf{R}^{*}=(R^{*}_{1},\ldots,R_{N}^{*})$, where $R_{i}^{*}$ denotes the rank of $Z_{i}$ among $Z_{1},\ldots, Z_{N}$, that is, if $Z_{(1)}\leq \cdots \leq Z_{N}$ are the orders statistics of $Z_{1},\ldots, Z_{N}$, then $Z_{(R_{i}^{*})}=Z_{i}$.

Assume $N\geq 2$. Consider $V=R_{1}^{*}-R_{N}^{*}$. Show that $$\mathsf{P}(V=k)=\left\{\begin{array}{ll} \frac{N-|k|}{N(N-1)} & \mbox{if }|k|=1,\ldots,N-1 \\ 0 & \mbox{elsewhere} \end{array}\right.$$

Well if all the $(Z_{i})_{1 \leq i \leq N}$ are i.i.d. then the $(R_{i}^{*})_{1 \leq i \leq N}$ are identitcally uniformly distributed over $\{1,.., N\}$. Now since, $P(R_{N}^{*} = i | R_{1}^{*} = j) = \frac{1}{(N-1)}$ if $i \neq j$ and $0$ otherwise, what you are looking for is the following sum if $N > k > 0$:

$P(V = k) = \sum_{i=1}^{N-K} P(R_{1}^{*} = K + i)*P(R_{N}^{*} = i | R_{1}^{*} = K + i) = \frac{N - k}{N(N-1)}$

Obviously $P(V = k) = P(V = -k)$