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Oct 11, 2019 at 8:41 review Suggested edits
Oct 11, 2019 at 11:29
Mar 27, 2019 at 16:15 vote accept Anon
Mar 27, 2019 at 15:55 answer added dlnB timeline score: 6
Mar 27, 2019 at 15:41 comment added Dilip Sarwate If you write $Z$ for $XY$ (note $Z$ has nothing to do with normal random variables), then would you agree that $E[Z\mid X=x] = E[xY\mid X=x] = xE[Y\mid X=x]$? If so, then note that when $X$ equals $x$, the random variable $E[Z\mid X] = E[XY\mid X]$ takes on value $xE[Y\mid X=x]$ and so the random variable $E[XY \mid X]$ is $XE[Y\mid X]$ (and is a function of $X$ as it shoud be),.
Mar 27, 2019 at 15:29 history asked Anon CC BY-SA 4.0