Question about inverse-variance weighting Suppose we want to make inference on an unobserved realization $x$ of a random variable $\tilde x$, which is normally distributed with mean $\mu_x$ and variance $\sigma^2_x$. Suppose there is another random variable $\tilde y$ (whose unobserved realization we'll similarly call $y$) that is normally distributed with mean $\mu_y$ and variance $\sigma^2_y$. Let $\sigma_{xy}$ be the covariance of $\tilde x$ and $\tilde y$. 
Now suppose we observe a signal on $x$,
\begin{align}a=x+\tilde u,\end{align}
where $\tilde u\sim\mathcal{N}(0,\phi_x^2)$, and a signal on $y$,
\begin{align}b=y+\tilde v,\end{align}
where $\tilde v\sim\mathcal{N}(0,\phi_y^2)$. Assume that $\tilde u$ and $\tilde v$ are independent.
What is the distribution of $x$ conditional on $a$ and $b$?
What I know so far:
Using inverse-variance weighting,
\begin{align}\mathbb{E}(x\,|\,a)=\frac{\frac{1}{\sigma_x^2}\mu_x+\frac{1}{\phi_x^2}a}{\frac{1}{\sigma_x^2}+\frac{1}{\phi_x^2}},\end{align}
and
\begin{align}
\mathbb{V}\text{ar}(x\,|\,a)=\frac{1}{\frac{1}{\sigma_x^2}+\frac{1}{\phi_x^2}}.
\end{align}
Since $x$ and $y$ are jointly drawn, $b$ should carry some information about $x$. Other than realizing this, I'm stuck. Any help is appreciated!
 A: I'm not sure whether the inverse-variance weighting formulas apply here. However I think you might compute the conditional distribution of $x$ given $a$ and $b$ by assuming that $x$, $y$, $a$ and $b$ follow a joint multivariate normal distribution.
Specifically, if you assume (compatibly with what specified in the question) that
\begin{equation}
\left[\begin{matrix}x \\ y \\ u \\ v \end{matrix}\right] \sim N\left( 
\left[\begin{matrix}\mu_x \\ \mu_y \\ 0 \\ 0 \end{matrix}\right],
\left[\begin{matrix} 
\sigma^2_x & \sigma_{xy} & 0 & 0 \\
\sigma_{xy} & \sigma^2_y & 0 & 0 \\
0 & 0 & \phi^2_x & 0 \\
0 & 0 & 0 & \phi^2_y 
 \end{matrix}\right]
\right)
\end{equation}
then, letting $a=x+u$ and $b=y+v$, you can find that 
\begin{equation}
\left[\begin{matrix}x \\ a \\ b \end{matrix}\right] \sim N\left( 
\left[\begin{matrix}\mu_x \\ \mu_x \\ \mu_y \end{matrix}\right],
\left[\begin{matrix} 
\sigma^2_x & \sigma^2_x & \sigma_{xy} \\
 \sigma^2_x & \sigma^2_x + \phi^2_x & \sigma_{xy} \\
\sigma_{xy} & \sigma_{xy} & \sigma^2_y + \phi^2_y  
 \end{matrix}\right]
\right).
\end{equation}
(Note that in the above it is implicitly assumed that $u$ and $v$ are independent between each other and also with $x$ and $y$.)
From this you could find the conditional distribution of $x$ given $a$ and $b$ using standard properties of the multivariate normal distribution (see here for example: http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions). 
