Order statistics, sample median

anja

New Member
#1
Hi! Can you help me with this:
we have X=(X1, X2, ... Xn) - iid, Xi ~ N(m,1), I need to prove that distribution of med(X)-X(j) is the same as distribution of X(n-j+1)-med(X), where med(X) is a sample median and X(j) is j-th order statistic.
 
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Dason

Ambassador to the humans
#4
Are you saying you figured out the answer or just that you know how to get the joint distribution of any two order statistics? If it's the second part do you have any idea what to do next?
 

anja

New Member
#5
2nd part :)
well I tried something like this:
n=2k-1, j<k
med(X)=X(k), then get joint distribution of (X(j), X(k)) and through linear transformation get distribution of (X(k)-X(j), X(j)) and then use a fact that X(k)-X(j) and X(j) are independent to get the distribution of X(k)-X(j) (but I dont know how to prove it that they are independent ) and then do this same with X(k) and X(n-j+1).
And it didn't work! (maybe I did something wrong or this is a bad idea)...
but I think it works only for n-odd...