Projection theorem for conditional probability

$\def\cov{\mathop{\mathrm{cov}}}\def\var{\mathop{\mathrm{var}}}$My professor uses something that he calls the "projection theorem", to get rid of the condition in conditional probabilities (expectation and variance). I have not found anything about it on the internet, so I am wondering where it comes from, and if it is right.

Here is the so-called "projection theorem":

$$E[\tilde{x}\mid \tilde{y} = y] = E[\tilde{x}] + \frac{\cov(\tilde{x},\tilde{y})}{\var(\tilde{y})}\times(\tilde{y}-E(\tilde{y})),$$

and

$$\var[\tilde{x}\mid \tilde{y}] = \var(\tilde{x})-\frac{\cov^2(\tilde{x},\tilde{y})}{\var(\tilde{y})}.$$

Are these formulas correct?

• They are correct if $\tilde x$ and $\tilde y$ have a joint normal distribution. – John Dawkins Apr 3 '16 at 21:56
• @JohnDawkins thank you. Forgive my ignorance, but what does joint probability mean? And do you have any reference (like any online course/wikipedia) where I can find these? Thanks – MarinD Apr 4 '16 at 6:36
• You may want to look at en.wikipedia.org/wiki/Joint_probability_distribution and then google "bivariate nprmal distribution". – John Dawkins Apr 5 '16 at 16:43

I guess your professor used them in a finance class because the first time I heard (I tried to google) the term "projection theorem" was in my asset pricing class. As John said, they are correct if $x$ and $y$ are jointly normal.
$$(X,Y) \sim N \Big(\begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix} , \begin{bmatrix} \sigma_X^2 & \rho \sigma_X \sigma_Y \\ \rho \sigma_X \sigma_Y & \sigma_Y^2 \end{bmatrix} \Big)$$ , we have \begin{align} E(Y|X=x)= \mu_Y+\frac{\rho \sigma_Y}{\sigma_X}(x-\mu_X), \end{align} and \begin{align} Var(Y|X=x) &= E(Y^2|X=x)-[E(Y|X=x)]^2 =(1-\rho^2)\sigma_Y^2. \end{align} You can verify that equations above are consistent with the theorem.