Minimum of expected value of deviation

Question

It is widely known that for any random variables $X, Y$, the minimum of $f(g) = \mathbb{E}(Y - g(X))^2$ over functions $g: R\to R$ is reached at $g(X) = \mathbb{E}(Y|X)$.

What can we say in the $n$-dimensional case?

Let $X, Y,$ be the $n$-dimensional random vectors and, for instance, $$f(g) = \mathbb{E}||Y-g(X)||^2.$$ Is it true that for the Euclidean norm the minimum of $f(g)$ over functions $g: R^n\to R^n$ is reached at $a=\mathbb{E}(Y|X)$? Or maybe we can say something about other norms?

I represent $f(g)$ as $$f(g)=\mathbb{E}[(Y-g(X))^T(Y-g(X))]\\ = \mathbb{E}(Y^TY)-g(X)^T\mathbb{E}Y - (\mathbb{E}Y)^T g(X) + \mathbb{E}(g(X)^T g(X)) \\= \mathbb{E}(||Y||^2)-2g(X)^T \mathbb{E} Y + ||g(X)||^2,$$ but the next steps are not obvious for me.

Also, we may represent $f(g)$ as $$f(g)=\mathbb{E}[(Y - \mathbb{E}(Y|X) + \mathbb{E}(Y|X) -g(X))^T(Y - \mathbb{E}(Y|X) + \mathbb{E}(Y|X) -g(X))]$$ and then expand this?

In this case I should prove that the middle summand is equal to 0, but I have no idea how to deal with this.

Any help would be appreciated. Thanks in advance.

Answer 1

The middle summand is \begin{align}E[(Y-E[Y \mid X])(E[Y \mid X] - g(X))] &= E[E[(Y-E[Y \mid X])(E[Y \mid X] - g(X)) \mid X]]\\ &= E[(E[Y \mid X] - g(X)) \cdot \underbrace{E[Y-E[Y \mid X] \mid X]}_{=0}]\\ &= 0. \end{align}