Some doubts about conditional expectation Suppose I have two random variable $X_t\sim NID(0,1)$ and $Y_t\sim NID(0,4)$ and $Cov(X_t,Y_t)=2$ .
Consider the random variable $Z_t = X_t + Y_t + Y_tX_t$. 
$E(Z_t) = E(X_t)+E(Y_t X_t)+E(Y_t) = 0 + 2 + 0  = 2$
because,
$E(X_t)=0$
$E(Y_t)=0$
$E(X_tY_t) = E(X_t)\times E(Y_t)+cov(X,Y) = 0 + 0 + 2$
But if I want to obtain $E(Z_t|X_t)$ what it is the reasoning behind? 
I am not able to figure out.
The same with variance:
$Var(Z_t) = Var(X_t)+Var(Y_t X_t)+Var(Y_t) = 1 + 2 + 4  = 7$
but what about $Var(Z_t|X_t)$ ?
 A: The basic approach is as follows.  Take the expression for $Z_t$, and find its expectation with respect to the "unfixed" variables $(Y_t)$ while holding the "fixed" variables $(X_t)$ constant:
$\mathbb{E}[Z_t|X_t] = \mathbb{E}[X_t + Y_t + Y_tX_t | X_t]$
As the expectation of a sum is the sum of the expectations, we get:
$\mathbb{E}[X_t + Y_t + Y_tX_t | X_t] = \mathbb{E}[X_t|X_t] + \mathbb{E}[Y_t|X_t] + \mathbb{E}[Y_tX_t|X_t]$
and, as I'm sure you know, $\mathbb{E}[X_t|X_t] = X_t$ and $\mathbb{E}[Y_tX_t|X_t] = X_t\mathbb{E}[Y_t|X_t]$.  
Given your covariance matrix, you should be able to write out $\mathbb{E}[Y_t|X_t]$ in terms of $X_t$, plug it in, and there you are.  
The conditional variance of $Z_t|X_t$ can be found in a similar manner by expanding the square and taking expectations of the result.
However, your professor provided you with something of a shortcut, hidden inside the covariance matrix of $(X_t,Y_t)$.  Unfortunately, I don't know enough about what your class has covered (except that it's late in the term) to know how to hint at it w/o giving it away.  Hence my comment about the determinant of the covariance matrix above.  It implies something about the relationship between $X_t$ and $Y_t$, which you can figure out in other ways as well.  No worries if you haven't covered determinants, the problem is straightforward enough the obvious way, once you get the hang of it.
