Minimum Mean Square Error Estimate Example We have data from 2D normal (gaussian) distribution.
$$\begin{bmatrix}y\\x\end{bmatrix}\,\text{~}\,\mathcal{N}\left(\begin{bmatrix}2\\4\end{bmatrix},\begin{bmatrix}10&2\\2&20\end{bmatrix}\right)$$
where $\mathcal{N}(\mu,P)$ and $\mu$ is mean and $P$ is covariance matrix in form
$$P=\begin{bmatrix}P_{yy}&P_{yx}\\P_{xy}&P_{xx}\end{bmatrix}$$
$x$ is the unknown and $y$ is the observation.
We have an observation $\textbf{y=1}$.
Compute mean square estimate $\hat{x}_{MS}$.
Compute covariance of the error $P_{\hat{x}_{MS}}$.

This is what my notes says. I can't find anywhere in the notes nor in the lectures any hint, how to compute this, so I'm not sure, if it makes sense. If it does, can you show me how to compute this?
 A: We have a prior $P(\vec{v}) \propto e^{-1/2(\vec{v} - \vec{\mu})^T P^{-1} (\vec{v} - \vec{\mu})}$.
We observe that $v_1 = 1$, and we are asked for (moments of) the posterior distribution of $v_2$. So plug $v_1 = 1$ into the above equation.  $P^{-1} = \begin{bmatrix} 20/14 & -2/14 \\ -2/14 & 10/14 \end{bmatrix}, \begin{bmatrix} 1 & x \end{bmatrix} P^{-1} \begin{bmatrix} 1 \\ x \end{bmatrix} = 20/14 - 4x/14 + 10x^2/14 = 10/14*(x-1/5)^2 + C$.
$P(x | y=1) \propto e^{-5/14 (x-1/5)^2}$
So $E[x] = 1/5, Var[x] = 14/10 = 7/5$.
A: This problem is described extensively in literature. One way to go would be by using the book "Pattern Recognition and Machine Learning" from Bishop, 2006. Equations 2.94 till 2.98 will do the trick for you.
So let the mean be noted by
$$
\begin{bmatrix}
\mu_y \\ \mu_x
\end{bmatrix}=
\begin{bmatrix}
2 \\ 4
\end{bmatrix}.
$$
The inverse of the covariance is:
$$
\Lambda=P^{-1}=\begin{bmatrix}
\Lambda_{yy} & \Lambda_{yx} \\
\Lambda_{xy} & \Lambda_{xx}
\end{bmatrix} = \frac{1}{196}\begin{bmatrix}
20 & -2 \\ -2 & 10
\end{bmatrix} 
$$
Then, according to Eq. (2.96) of the aforementioned book, we have
$$
p(x|y)=\mathcal{N}(x | \mu_{x|y}, \Lambda_{xx}^{-1})
$$
with
\begin{align}
\mu_{x|y} &= \mu_x - \Lambda_{xx}^{-1} \Lambda_{xy} (y - \mu_y) \\
&= 4 - \frac{196}{10}\cdot \frac{-2}{196}\cdot(1-2) \\
&=\frac{19}{5}
\end{align}
and
$$
\Lambda_{xx}^{-1}=\frac{196}{10}=\frac{98}{5}.
$$
So we have
\begin{align}
\hat{x}_{MS} &= \frac{19}{5}, \\
P_{\hat{x}_{MS}} &= \frac{98}{5}.
\end{align}
A: Another way - perhaps more intuitive - to look at this problem, is as follows. Suppose we can describe the stochastic variables $x$ and $y$ as follows:
\begin{equation}
\begin{bmatrix} y \\ x \end{bmatrix} = 
\begin{bmatrix} \mu_y \\ \mu_x \end{bmatrix} + 
\begin{bmatrix} p_{yy} & 0 \\ p_{xy} & p_{xx} \end{bmatrix} 
\begin{bmatrix} \epsilon_y \\ \epsilon_x \end{bmatrix},
\end{equation}
where $\epsilon_y$ and $\epsilon_x$ are stochastic variables that come from a normal distribution with mean $0$ and variance $1$, i.e. $\epsilon_y, \epsilon_x \sim \mathcal{N}(0,1)$. When computing the expectation of the above formula, it is easy to see that $\mu_y=2$ and $\mu_x=4$. Furthermore, when computing the covariance, we see that
\begin{equation}
\begin{bmatrix} p_{yy} & 0 \\ p_{xy} & p_{xx} \end{bmatrix}
\begin{bmatrix} p_{yy} & p_{xy} \\ 0 & p_{xx} \end{bmatrix} = 
\begin{bmatrix} p_{yy}^2 & p_{yy}p_{xy} \\ p_{yy}p_{xy} & p_{xy}^2 + p_{xx}^2 \end{bmatrix} = 
\begin{bmatrix} 10 & 2 \\ 2 & 20 \end{bmatrix} \tag{1}
\end{equation}
Now we can rewrite $x$, by using the fact that $\epsilon_y = \frac{y - \mu_y}{p_{yy}}$:
\begin{align}
x &= \mu_x + p_{xy}\epsilon_y + p_{xx}\epsilon_x \\
&=\mu_x + \frac{p_{xy}}{p_{yy}} (y - \mu_y) + p_{xx} \epsilon_2
\end{align}
Now, given that $y$ is known, it is easy to see that the expected value of $x$ (you call is $\hat{x}_{MS}$) is
\begin{equation}
E[x] = \mu_x + \frac{p_{xy}}{p_{yy}}(y-\mu_y) \tag{2}
\end{equation} and the covariance of the $x$ (you call is $P_{\hat{x}_{MS}}$) is simply $p_{xx}^2$. From Eq. $(1)$, it follows that
\begin{equation}
p_{yy}p_{xy}=2,
\end{equation}
so
\begin{equation}
\frac{p_{xy}}{p_{yy}}=\frac{2}{p_{yy}^2}=\frac{2}{10}=\frac{1}{5}.
\end{equation}
Filling this in Eq. $(2)$ gives 
$$
\hat{x}_{MS} = 4 + \frac{1}{5}(1-2)=4-\frac{1}{5}=\frac{19}{5}.
$$
Furthermore, we have
$$
P_{\hat{x}_{MS}} = p_{xx}^2 = 20 - p_{xy}^2 = 
20 - p_{yy}^2 \left(\frac{p_{xy}}{p_{yy}}\right)^2=20-\frac{10}{25}=\frac{98}{5},
$$
which corresponds to the other answer I gave :).
