Why can the :$\mathbf W^*_{MMSE} $ be written as $E[\mathbf s \mathbf y^*]E[\mathbf y \mathbf y^*]^{-1}$

Question

I read this paper: https://arxiv.org/pdf/1305.2460.pdf

And there is a formula about MMSE:$\mathbf W^*_{MMSE}= E[\mathbf s \mathbf y^*]E[\mathbf y \mathbf y^*]^{-1}$

$\mathbf W _{BB}$:Baseband combiner

$\mathbf W _{RF}$:RF combiner

$\mathbf W $:The combination of $\mathbf W_{RF}$ and $\mathbf W _{BB}$

$\mathbf W^*$ : the hermitian of $\mathbf W$

$\mathbf s$:the signal on the left

$\mathbf y$:the signal when $\mathbf s$ goes through the channel

Why can the :$\mathbf W^*_{MMSE} $ be written as $E[\mathbf s \mathbf y^*]E[\mathbf y \mathbf y^*]^{-1}$.why should we multiply the these two signal ,$\mathbf s$ and $ \mathbf y$,and average them ?Does anyone have information about this formula ?

Answer 1

The mean squared error equalizer is one that minimizes the mean squred error

$$J(\mathbf{W})=\mathbb{E}\left[\|\mathbf{s}-\mathbf{W}\mathbf{y}\|^2\right]$$

which is the equalizer that minimizes the Euclidean distance between the transmitted signal $\mathbf{s}$ and the equalized received signal $\mathbf{Wy}$. To find the value of $\mathbf{W}$ the minimizes $J(\mathbf{W})$, take the derivative with respect to $\mathbf{W}$ and put it equal to $\mathbf{0}$, which will result in

$$-2\mathbb{E}\left[\left(\mathbf{s}-\mathbf{Wy}\right)\mathbf{y}^H\right]=\mathbf{0}$$ Solve the above equation with respect to $\mathbf{W}$ you get the MMSE equalizer.

Note: $(.)^H$ is the conjugate transpose, which is equivalent to $(.)^*$ in the paper.