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Apr 2, 2013 at 21:27 answer added Hugh Thomas timeline score: 1
Mar 5, 2013 at 17:24 answer added user16215 timeline score: 1
Mar 5, 2013 at 17:16 history edited Fredrik CC BY-SA 3.0
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Mar 5, 2013 at 16:59 comment added Fredrik Yes, it's the Euclidean norm. I can't find where I missed a square (not saying I didn't :) ). Notice however that it's "chi distributed" not "chi-square distributed". $\|X_i-\mu_X\|^2$ is however chi-square.
Mar 5, 2013 at 15:54 comment added robin girard "will be chi distributed with d degrees of freedom" I think you forgot about a square somewhere ? are you talking about the euclidian norm ?
Mar 5, 2013 at 15:35 history edited Fredrik CC BY-SA 3.0
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Mar 5, 2013 at 15:25 history edited Fredrik CC BY-SA 3.0
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Mar 5, 2013 at 9:56 comment added Fredrik Ideally $Y$ and $X$ are independent, but other feasible dependence structures is ok. Thanks!
Mar 5, 2013 at 8:38 comment added SBF Do you assume a fixed probability space (=dependence structure between $X$ and $Y$), or you only know distributions?
Mar 4, 2013 at 18:14 history asked Fredrik CC BY-SA 3.0