Does the property $\min_{c \in \mathbb{R}} E[ (X-c)^2] =E[ (X-E[X])^2]$ generalize to the covariance of $(X,Y)$ It is well known that 
\begin{align}
\min_{c \in \mathbb{R}} E[ (X-c)^2] =E[ (X-E[X])^2]= V(X)
\end{align}
My question does this property in some nontrivial way (I don't know which way) generalize to the covariance.  Where the covariance for a pair $(X,Y)$ is defined as
\begin{align}
Cov(X,Y)=E[(X-E[X]) (Y-E[Y])  ]. 
\end{align} 
 A: $$E[(X-a)^2] = \sigma_X^2 + (\mu_X-a)^2$$  where $\mu_X:=E[X]$ and $\sigma_X:=\sqrt{E[(X-\mu_X)^2]}$. So, $a=\mu_X$ is minimizes $f(a) = E[(X-a)^2].$
The "generalization" of this equation is 
$$E[(X-a)(Y-b)] = Cov(X,Y) + (\mu_X-a)(\mu_Y-b).$$
So, $(a,b)=(\mu_X, \mu_Y)$ is the saddle-point of the hyperbolic-paraboloid $z=g(a,b)=E[(X-a)(Y-b)]$.
A: It might be tempting to think that $$\text{cov}(X,Y) = \min_{a,b \in \mathbb{R}} \mathbb{E}((X-a)(Y-b)).$$ This is, however, not true. It's easy to see that "$\geq$" always holds (take $a=\mathbb{E}(X)$ and $b=\mathbb{E}(Y)$) but "$\leq$" is, in general, wrong.
Consider for instance two independent random variables $X$ and $Y$. Then
$$\text{cov}(X,Y)=  \mathbb{E}(X-\mathbb{E}(X)) \mathbb{E}(Y-\mathbb{E}Y)=0.$$
However,
$$\min_{a,b \in \mathbb{R}} \mathbb{E}((X-a)(Y-b)) = \min_{a,b} (\mathbb{E}(X)-a) (\mathbb{E}(Y)-b) =- \infty$$
(choose $a=\mathbb{E}(X)+n^2$ and $b=\mathbb{E}(Y)-1/n$).

This being said, I'm not sure whether there is some other way to generalize the variation principle to the covariance.
