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I have a problem which asks me to find $\Bbb E[Y]$ and $Var(Y)$ given that $Y\text{~}Normal(x,1)$ conditional on $X=x$. $X$ is standard normal. So I have worked out that $\Bbb E[Y]=0$ using the law of iterated expectations; now I want to work out the variance. I figure I need to use the law of total expectation, which states that $$Var(Y)=\Bbb E[Var(Y|X)]+Var(\Bbb E[Y|X])$$

I have worked out that $\Bbb E[Var(Y|X)]=1$, so all I need is $Var(\Bbb E[Y|X])$, which is given by $$Var(\Bbb E[Y|X])=\Bbb E[\Bbb E[Y|X]^2]-(\Bbb E[\Bbb E[Y|X]])^2$$ I can further work out that $(\Bbb E[\Bbb E[Y|X]])^2=\Bbb E[Y]^2=0$; what I cannot figure out is $\Bbb E[\Bbb E[Y|X]^2]$.

I understand that $\Bbb E[Y|X=x]=x$, so I assume that means $\Bbb E[Y|X]=X$, so does that mean that $\Bbb E[Y|X]^2=X^2$? If so, I take it that $\Bbb E[\Bbb E[Y|X]^2]=0$, because $X$ is standard normal. Am I correct? If not, where is my error?

I am also wondering how to work out $\text{Covariance}(X,Y)$ once the variance is calculated. Namely, what will the distribution $XY$ be? Because $\Bbb E[XY]$ is needed. Is it bivariate normal? Any help is greatly appreciated.

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    $\begingroup$ How did you find $\mathbb EY=0$? We have $\mathbb E[Y\mid X]=X$ so $\mathbb EY=\mathbb E[\mathbb E[Y\mid X]]=\mathbb EX$. What exactly do you know about $X$? $\endgroup$ – drhab May 23 '19 at 13:29
  • $\begingroup$ Sorry, should have mentioned $X$ is standard normal. Will edit that in. $\endgroup$ – mathenthusiast May 23 '19 at 13:31
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Hint:

I think you can do it with:$$Y=X+U$$where $U$ has standard normal distribution and $X$ and $U$ are independent. In that case under condition $X=x$ random variable $Y$ has normal distribution with parameters $\mu=x$ and $\sigma^2=1$ (as is requested).

If moreover $X$ has standard normal distribution then it is immediate that $Y$ has normal distribution with parameters $\mu_Y=0$ and $\sigma_Y^2=2$.

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