Covariance of random variable and absolute of the same random variable

305 Views Asked by At

Suppose $Y \sim N(0,1)$, and $X=|Y|$, what would the covariance of of $X,Y$ i.e $Cov(X,Y)$ be ? I seem to be getting zero as an answer and it doesn't quite sit right with me. This is my reasoning :

$ Cov(X,Y) = E[(X-\mu_X)(Y - \mu_Y)] = E[XY] - \mu_XE[Y] - \mu_YE[X] -\mu_X\mu_Y = E[XY] $

$ E[XY] = E[|Y|Y] $

$ = E[-Y^2|Y<0]P(Y<0) + E[Y^2|Y\ge 0]P(Y\ge 0) $

$ = E[-(-Y)^2|Y>0](1/2) + E[Y^2|Y>0](1/2) $

$ = -E[Y^2|Y>0](1/2) + E[Y^2|Y>0](1/2) = 0$

1

There are 1 best solutions below

0
On

Your calculations are correct, and in fact the same result will hold for any random variable $Y$ with a symmetric distribution (assuming that $Y$ has a finite second moment so that the covariance is defined).