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Nov 27, 2023 at 14:56 history edited whuber
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Nov 27, 2023 at 14:55 answer added whuber timeline score: 3
Nov 26, 2023 at 17:55 history edited kjetil b halvorsen CC BY-SA 4.0
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Mar 31, 2014 at 15:36 comment added Revin Ah, so $$f_{X_1,X_{(n)}}(u,v)=\sum_i^n \frac{1}{n}f_{X_{(i)},X_{(n)}}(u,v)$$?
Mar 31, 2014 at 15:34 history edited Revin CC BY-SA 3.0
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Mar 31, 2014 at 14:39 comment added whuber Assuming the $X_i$ are exchangeable, for each $j$ there is a $1/n$ chance that $X_1$ is $X_{[j]}$. Thus the joint distribution of $(X_1,X_{[n]})$ is a mixture of the joint distributions of the $(X_{[j]}, X_{[n]})$. The rest is straightforward... .
Mar 31, 2014 at 4:32 comment added Revin Would it be true that $f_{X_1,X(n)}=f_{X_1,X(n)|X_1=X(n)}+f_{X_1,X(n)|X_1\neq X(n)}$?
Mar 31, 2014 at 4:29 review First posts
Mar 31, 2014 at 4:30
Mar 31, 2014 at 4:24 comment added Glen_b One possibility is to consider two cases; first where $(n)=1$ (i.e. where $X_1$ is the biggest value) and then where $(n)\neq 1$.
Mar 31, 2014 at 4:23 comment added Dilip Sarwate What if $X_1$ is the largest sample value?
Mar 31, 2014 at 4:12 history asked Revin CC BY-SA 3.0