# A question in order statistics of continuous type distribution

Let $X_1,X_2,\dots$ be a sequence of random variables from a continuous type distribution and $m$ and $n$ be two integers such that $m<n$, and $2\le n-m$.

How can I show the probability that the third-order statistic of $X_1,\ldots,X_m$ is equal to the fifth-order statistic of $X_1,\dots,X_n$ is $6\cdot\displaystyle\frac{ n-5\choose m-3}{n\choose m}$?

• It would be more suggestive to write your formula as $\binom{4}{2}\binom{1}{1}\binom{n-5}{m-3}/\binom{n}{m}$. Note that you must assume the variables are exchangeable. – whuber Mar 30 '14 at 17:53
• if $2\le n-m$, then $m\le n-2$ (add $m-2$ to both sides). In which case the second inequality in your question implies the first, always, making the first one redundant. Did you mistype one of them? – Glen_b Mar 31 '14 at 3:40

When the $X_i$ are sorted in ascending order, the positions of $X_1, \ldots, X_m$ (which form a subset of the $n$ positions) are random and equidistributed (assuming, that is, either exchangeability of the $X_i$ or the stronger "iid" assumption of independence and equal distributions). This means that each possible subset of the $n$ positions has a constant chance of $1/\binom{n}{m}$ of occurring.
The event described in the question happens when two of these $m$ values are situated in the first four positions, one of these $m$ values is in the fifth position, and the remaining $m-3$ values are in the remaining $n-5$ positions.
There are $\binom{4}{2}=6$ ways to fill the first four positions, $\binom{1}{1}=1$ way to fill the fifth position, and $\binom{n-5}{m-3}$ ways to fill the last positions. The total number of subsets in the event is the product of these counts. Therefore the chance of the event is
$$\frac{\binom{4}{2}\binom{1}{1}\binom{n-5}{m-3}}{\binom{n}{m}} = 6\frac{\binom{n-5}{m-3}}{\binom{n}{m}}.$$