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Mar 2, 2012 at 0:59 history edited John Jiang CC BY-SA 3.0
Corrected the claimed formula
Mar 2, 2012 at 0:25 comment added John Jiang @Didier: you are absolutely right. I didn't see your comment from Jan 28. Please see my modified question above. Thanks!
Mar 1, 2012 at 8:16 comment added Did Not interested in a reformulation, or at least some explanations?
Jan 28, 2012 at 15:23 comment added Did Since the support of $X$, $Y$ and $Z$ is $[-1,1]$ and these are independent, the support of $XZ(Y+1)$ is $[-2,2]$ hence it cannot be distributed as $Y$.
Nov 10, 2011 at 18:55 history asked John Jiang CC BY-SA 3.0