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Feb 19, 2014 at 17:50 vote accept Bjørn Kjos-Hanssen
Nov 9, 2010 at 22:03 comment added Michael Hardy So.... It seems that the answer given was correct, but maybe only shows why a different question should have been asked.
Nov 9, 2010 at 21:59 comment added Michael Hardy Correct. A STRICTLY bigger X entails a strictly bigger Y. I was hasty.
Nov 9, 2010 at 21:29 comment added Michael Hardy But $Y$ is not in general determined by $X$. Simplest case: $$ \Omega = \{ (0,0), (0,1), (1,0), (1,1)\}. $$ $$ X(i,j) = i $$ $$ Y(i,j) = 10i + j $$ If $X=0$ then $Y$ may be either 0 or 1; if $X=1$ then $Y$ may be either 10 or 11. A larger $X$ value entails a larger $Y$ value. But $X$ does not determine $Y$.
Nov 9, 2010 at 21:03 history answered Louigi Addario-Berry CC BY-SA 2.5