1. Proof Var[X | Y=2]

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?

2. Re: Proof Var[X | Y=2]

$P[X|Y=2]$ 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 $X$.

3. Re: Proof Var[X | Y=2]

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]

4. Re: Proof Var[X | Y=2] 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.