Can you prove to me that Var[X | Y=2] = E[X^2 | Y=2] - (E[X | Y=2])^2 I see the relation of just say, var[x], but why is Y=2 not squared in the first part?
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$P[X|Y=2]$ is just a distribution like any other. so the formulas for expectation and variance are applied as usual. The $Y=2$ is just a conditioning expression. It's not part of the random object which is $X$.
I see thanks. Also, why when say random variables X and Y are independent, the covariance between X and Y is 0? Cov[X,Y]= E[XY] -E[X]E[Y] so it is just saying if they are independent there is no E[XY]
Originally Posted by math951 I see thanks. Also, why when say random variables X and Y are independent, the covariance between X and Y is 0? Cov[X,Y]= E[XY] -E[X]E[Y] so it is just saying if they are independent there is no E[XY] no. It says that $E[XY]=E[X]E[Y]$ Note the reverse is not always true. Zero covariance does not imply independence.