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!

  • $\begingroup$ From $E[X]=\mu$ and $\mathrm{Var}(X)=\sigma^2$ you most certainly cannot conclude "hence normally distributed"--all you have been given is that the expectation and variance exist, and many other distributions have expectation and variance. Of course you can have it be an additional given fact that $X$ is normally distributed, though I think it is unnecessary for the solution of this problem. $\endgroup$ – David K Feb 13 '18 at 11:47
  • $\begingroup$ If you have any questions or need additional clarification regarding my answer, please let me know! Also, don't forget to choose a best answer to your questions (now and in the future) once they have been answered to your satisfaction. $\endgroup$ – bames Feb 14 '18 at 9:28

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 $\sigma^2 = \Bbb E[X^2] -\mu^2$.

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