derive the formula to a multivariate normal form I have this formula
$e^{-\frac{1}{2}\sum_{i=1}^N(z_i-x_i^T\beta)^2}\times e^{-\frac{1}{2}\beta^T\Sigma^{-1}\beta}$
$z_i$s are some known constants, $x_i$ and $\beta$ are $1\times n$ vectors ($x_i$ known)
$\Sigma$ is a known $n \times n$ matrix 
I believe this can be reorganized to be a multivariate normal distribution about $\beta$ by adding some constant.. yet I don't know how....
Can somebody help me with this?
Thanks!
Appendix: Given $z_i\in\mathbb R,\, i=1,\ldots,N$ and $x_i\in\mathbb R^{n\times 1}$ for $i=1,\ldots,N,$ and positive-definite symmetric $\Sigma\in\mathbb R^{n\times n}$ how does one find nonsingular $V\in\mathbb R^{n\times n}$ and $\mu\in\mathbb R^{n\times 1}$ such that for all $\beta\in\mathbb R^{n\times 1}$ we have $$\begin{align} & \sum_{i=1}^N (x_i^T\beta - z_i)^T(x_i^T\beta - z_i) + \beta^T\Sigma^{-1}\beta \\  \\ = {} & (\beta-\mu)^T V^{-1} (\beta-\mu) + \big(\text{constant not depending on $\beta$}\big). \end{align} $$ That is the question. $\qquad$
 A: Let $z=(z_1,\dots,z_n)^T$.
$$
\begin{equation}
\|z - X\beta\|^2 + \beta^T \Sigma^{-1} \beta = \|z\|^2 - 2z^TX\beta + \beta^T(X^TX+\Sigma^{-1})\beta \tag{1}
\label{e:1}
\end{equation}
$$
Let $L = (X^TX+\Sigma^{-1})^{\frac{1}{2}}$ be a symmetric square root, and $\beta_{*} = L\beta$. 
Then $\eqref{e:1}$ can be written as 
$$
\begin{align}
\|z\|^2 -2z^TXL^{-1}\beta_{*} +\|\beta_{*}\|^2 &= \|\beta_* - L^{-1}X^Tz\|^2+\|z\|^2-z^TXL^{-2}X^Tz\\
\ &= \|L\beta - L^{-1}X^Tz\|^2 + \|z\|^2-z^TXL^{-2}X^Tz\\
\ &= (\beta - L^{-2}X^Tz)^T L^TL (\beta - L^{-2}X^Tz)+ \|z\|^2-z^TXL^{-2}X^Tz
\\
\ &= (\beta - \mu)^TV^{-1}(\beta - \mu)^T + c
\end{align}
$$
Where $V^{-1} = L^TL = X^TX +\Sigma^{-1}\,,\, \mu = L^{-2}X^Tz = (X^TX+\Sigma^{-1})^{-1}X^Tz$ and $c =  \|z\|^2-z^TXL^{-2}X^Tz = \|z\|^2-z^TX (X^TX+\Sigma^{-1})^{-1}X^Tz$.
A: okay now I've got
$$
\beta^T \left( \sum_{i=1}^N x_ix_i^T+\Sigma^{-1} \right) \beta-\sum_{i=1}^N z_i x_i^T \beta-\beta^T\sum_{i=1}^Nz_ix_i^T=\beta^TV^{-1}\beta-\mu^TV^{-1}\beta-\beta^TV^{-1}\mu\\
$$
this means:
$V=\left(\sum_{i=1}^Nx_ix_i^T+\Sigma^{-1}\right)^{-1}$
$$\mu=\sum_{i=1}^Nz_ix_i^TV$$
is that right?
