Consider $n$ r.vs ${X_1, X_2,...,X_n}$. Each is i.i.d drawn from some distribution $f(.)$. What is the probability that $X_i$ is the $k^{th}$ order statistic in any two consecutive trials?
3
-
$\begingroup$ What do you mean with "in two consecutive trials"? When you say "$X_i$ is the kth order statistic", you mean "there are exactly $k-1$ elements of $\{X_1,\dots,X_n\}$ that are smaller than $X_i$"? Can you assume that the distribution is continuous, so that ties happen with probability 0? $\endgroup$– Federico PoloniCommented Jun 16, 2019 at 16:22
-
$\begingroup$ Assume $X_{i,j}$ are continuous random variables, so no ties, and are i.i.d. across i=1 ..n and j=1,2.. Let $X_{i,1}$ be the random variables in the 1st trial, and $X_{i,2}$ be the random variables in the 2nd trial,. Assume exactly 2 trials are conducted, each consisting of n i.i.d. variables being drawn..Specify a value of $i$. Let k be the value such that $X_{i,1}$ in the 1st trial is the kth order statistic of the first trial. Then the probability that $X_{i,2}$ is the kth order statistic of the 2nd trial is $1/n$. $\endgroup$– Mark L. StoneCommented Jun 17, 2019 at 23:20
-
$\begingroup$ Dear Federico, Let us say there are $n$ r.v.s each of which can take some values between range [a,b]. Yes, $k^{th}$ statistics mean there are exactly $k-1$ smaller elements than $X_i$. So what is the probability that in two consecutive trials $X_i$ will have the same position in the order/arrangement? $\endgroup$– Suvadip BatabyalCommented Jul 22, 2019 at 5:29
Add a comment
|