3
$\begingroup$

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 $6\cdot\displaystyle\frac{ n-5\choose m-3}{n\choose m}$?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber Mar 30 '14 at 17:53
  • $\begingroup$ 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? $\endgroup$ – Glen_b Mar 31 '14 at 3:40
2
$\begingroup$

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

$$\frac{\binom{4}{2}\binom{1}{1}\binom{n-5}{m-3}}{\binom{n}{m}} = 6\frac{\binom{n-5}{m-3}}{\binom{n}{m}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.