How to prove relationship between inverse covariance matrix and linear regression coefficients? Edited:
I would like to work out the above relationship, more precisely:

Let $(Y_{1}, ..., Y_{m})$ be a zero-mean vector with covariance matrix
  $\Sigma$, and let $S \subset \{1, ..., m\}.$ The best linear predictor
  of $Y_{S}$ upon $Y_{\setminus{S}}$, defined as $\tilde{\beta}_{S} =
 \Sigma^{-1}_{\setminus{S}, \setminus{S}} \Sigma_{\setminus{S}, S}$, is
  a matrix multiple of $\Sigma^{-1}_{\setminus{S}, S}$.

It is assumed that$(Y_{1}, ..., Y_{m})$ is jointly normally distributed.
The above statement seems to imply that there is a number of steps from
$\tilde{\beta} = [X^{T}X]_{\setminus{S}, \setminus{S}}^{-1} [X^{T}Y]_{\setminus{S}, S}$
to 
$[X^{T}X]_{\setminus{S}, S}^{-1}  c$, where c is some scalar.
I would be interested in those steps.
Thanks!
 A: Here is the solution:
Starting with regression equation:
$y = Xb$
$X^{T} y = X^{T} X b$
$[X^{T}X]^{-1} X^{T} y = [X^{T}X]^{-1} [X^{T} X] b $
$[X^{T}X]^{-1} X^{T} y = Ib$
$[X^{T}X]^{-1} X^{T} y = b$
Partitioning the covariance matrix: 
Let $\{X_{1}, ... ,X_{p}\}$ be zero-mean random vectors. 
Divide the set of random vectors in two subsets $X_{s}, X_{\setminus{s}}$ with $X_{s} \in \mathbb{R} $ and $X_{\setminus{s}} \in \mathbb{R}^{p-1}$
$\Sigma_{11}$ is a $1\times1$ matrix and the variance of $X_{s}$, $\Sigma_{22}$ is a $(p-1)\times (p-1)$ matrix and the the covariance matrix of  $X_{\setminus{s}}$ and $\Sigma_{21} = \Sigma_{12}^{T}$ is a $(p-1)\times1$ matrix and the covariance matrix of $X_{s}$ and $X_{\setminus{s}}$.
$\Sigma =
\left[\begin{array}{cc}
\Sigma_{11}&\Sigma_{12}\\
\Sigma_{21}&\Sigma_{22}
\end{array}\right]
$
Getting inverse covariance by matrix inversion:
using: http://en.wikipedia.org/wiki/Schur_complement
Inverse covariance matrix:
$\Sigma^{-1} =
\left[\begin{array}{cc}
I&0\\
-\Sigma_{21}\Sigma^{-1}_{11}&I
\end{array}\right]
\left[\begin{array}{cc}
\Sigma_{11}&\Sigma_{12}\\
\Sigma_{21}&\Sigma_{22}
\end{array}\right]
\left[\begin{array}{cc}
I&-\Sigma^{-1}_{11}\Sigma_{12}\\
0&I\\
\end{array}\right]
$
$=
\left[\begin{array}{cc}
\Sigma^{-1}_{11.2}&-\Sigma^{-1}_{11.2} \Sigma_{12}\Sigma^{-1}_{22}\\
-\Sigma^{-1}_{22} \Sigma_{21}\Sigma^{-1}_{11.2}&\Sigma^{-1}_{22}+\Sigma^{-1}_{22}\Sigma_{21}\Sigma^{-1}_{11.2}\Sigma_{12}\Sigma^{-1}_{22}\\
\end{array}\right]
$
where $\Sigma_{11.2} = \Sigma_{11} - \Sigma_{12} \Sigma^{-1}_{22} \Sigma_{21} $
Show equivalence up to scalar multiple: inverse cov &  regression coefs
As all vectors have mean equal to zero:
$[X^{T}X]^{-1} = \Sigma^{-1}_{22}$ and $X^{T} y = \Sigma_{21}$ (Sorry for defining X in two different ways in the regression setup and the definition of the random vectors, but I think it's still clear)
... and we can write: $b = \Sigma^{-1}_{22} \Sigma_{21}$.
Let $K = \Sigma^{-1}$.
Then $K_{21} = -\Sigma^{-1}_{22} \Sigma_{21} \Sigma^{-1}_{11.2}$
As $\Sigma_{11}$ is a $1\times 1$ matrix $\Sigma_{11}$ is a scalar and $\Sigma^{-1}_{11.2}$ is also a scalar.
Therefore we can rearrange $K_{21} = [- \Sigma^{-1}_{11.2}] \Sigma^{-1}_{22} \Sigma_{21}$
We can now write $K_{21} = [- \Sigma^{-1}_{11.2}] b$.
As $[- \Sigma^{-1}_{11.2}]$ is a scalar we showed what was claimed in the question.
