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 $Z_t = X_t + Y_t + Y_tX_t$.

$E(Z_t) = E(X_t)+E(Y_t X_t)+E(Y_t) = 0 + 2 + 0 = 2$




$E(X_tY_t) = E(X_t)\times E(Y_t)+cov(X,Y) = 0 + 0 + 2$

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 $Var(Z_t|X_t)$ ?

  • $\begingroup$ There are some errors here. $E(X_tY_t) = E(X_t)+E(Y_t)+cov(X_t,Y_t)$ isn't true unless $E(X_t)+E(Y_t)$ happens to equal $E(X_t)E(Y_t)$. Also, $Var(Z_t) = Var(X_t)+Var(Y_t X_t)+Var(Y_t)$ is not true, you're missing some covariance terms, since $Y_t, X_t$ are not uncorrelated with $X_tY_t$. $\endgroup$ – Macro May 11 '12 at 16:33
  • $\begingroup$ You're right my mistake. I fix it immediately. But I don't get the point on the variance. $\endgroup$ – Marco May 11 '12 at 17:04
  • 1
    $\begingroup$ Look very carefully at the covariance matrix. Very, very carefully. Look at the numerical values... what do they tell you about the relationship between $X$ and $Y$? If you know about determinants, that will help. $\endgroup$ – jbowman May 11 '12 at 18:17
  • $\begingroup$ $Var(Z_t)=Var(X_t)+Var(Y_tX_t)+Var(Y_t)$ where $Var(Y_tX_t)=Var(Y_t)*Var(X_t)+COV(X,Y)$ therefore $Var(Z_t)=1 + 1 \times 4 + 2 + 4 = 11$ ? $\endgroup$ – Marco May 11 '12 at 19:57
  • $\begingroup$ det(Cov Matrix) = 1*4-2*2 = 0. I realize the importance of this may never have come up in your classes, though... $\endgroup$ – jbowman May 12 '12 at 0:06

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.

  • $\begingroup$ Thank you very much. So I know that a covariance matrix has to be positive definite. Can you provide me some reference on conditional expectation and variance for this kind of exercises? $\endgroup$ – Marco May 12 '12 at 21:14
  • $\begingroup$ Actually, covariance matrices don't have to be positive definite, just positive semidefinite. If the matrix isn't of full rank, that implies the variables span a subspace of dimension less than the number of variables. In other words, you can write the variables as linear functions of a smaller number of variables... I hope this helped w/o giving it away totally. $\endgroup$ – jbowman May 12 '12 at 21:44

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