TITLE: What is $\operatorname{Cov}(\widehat{Y},Y)$?
QUESTION [6 upvotes]: If $\hat{Y}$ is the OLS linear regression model for $Y$, what can I say about $\operatorname{Cov}(\hat{Y},Y)$? Is this value $0$?

REPLY [8 votes]: $\newcommand{\var}{\operatorname{var}}$
$\newcommand{\cov}{\operatorname{cov}}$
If you know matrix algebra, one often writes
$$
\begin{array}{ccccccccccc}
Y & = & X & \beta & + & \varepsilon \\  \\
(n\times1) & & (n\times p) & (p\times1) & & (n\times1)
\end{array}
$$
where $X$ is observable and "fixed" (i.e. not random), $\beta$ is unobservable and fixed, $\varepsilon$ is unobservable and random, and $Y$ is observable and random.  The $n\times n$ matrix $H = X(X^T X)^{-1}X^T$ projects orthogonally onto the column space of $X$, and
$$
\hat Y = HY.
$$
Recall that if $Y$ is an $n\times 1$ random column vector, then
$$
V=\var(Y) = \mathbb E\Big( (Y-\mathbb E Y)(Y - \mathbb E Y)^T  \Big)
$$
is an $n\times n$ matrix.  And
$$
\begin{array}{cccccccccccccccc}
\cov\Big( & A & Y & , & B & Y & \Big) & = & A & \var(Y) & B^T   \\  \\
& (j\times n) & (n\times1) & & (k\times n) & (n\times1) & & & (j\times n) & (n\times n) & (n\times k)
\end{array}
$$
is a $j\times k$ matrix.
So
$$
\cov(\hat Y, Y) = \cov(HY, Y) = H \cov(Y,Y) = H\sigma^2 I_{n\times n} = \sigma^2 H.
$$
You could also write
$$
\cov(\hat Y, Y) = \cov(\hat Y, \hat Y) + \cov(\hat Y, \hat\varepsilon)
$$
$$
= \cov(HY, HY) + 0 = H\cov(Y,Y) H^T = H(\sigma^2 I_{n\times n})H^T = \sigma^2 HH^T.
$$
But, being the matrix of an orthogonal projection, $H$ is both its own transpose and its own square, so this reduces to the same thing we got by the other method.