When $E[X]=\mu$ and $\mathrm{Var}(X)=\sigma^2$ (hence normally distributed, given is that $E[Y \mid X] = a + bX$ then what is $E[XY]$?
From law of total expectation is know that: $$E[y] = E[E[Y \mid X]] = E[a+bX] = a + b\mu.$$ But now I don't know how to get $E[XY]$, the only thing I can think of is by using the formula for $\mathrm{Cov}[X,Y]$ but I don't have the covariance so i am stuck here.
Does somebody know what I am forgetting here? Thank You!
You can do something similar for $\Bbb E[XY]$: $$\Bbb E[XY] = \Bbb E[X\Bbb E[Y|X]] = \Bbb E[aX+bX^2] = a\mu + b(\sigma^2+\mu^2)$$ where we have used the fact that $\sigma^2 = \Bbb E[X^2] -\mu^2$.