A very simple (looks like...) statistical problem, however I don't even know how to name it in a formal way...
Suppose in a Bayesian framework I have random variables $y, x_1,$ and $x_2$,
$$f(x) = p(x|y) = \frac{p(y|x)p(x)}{p(y)} = \frac{N(y|x_1+x_2,\sigma^2)N( \left( \begin{array}{c} x_1\\ x_2\end{array} \right) | \left( \begin{array}{c}m_1\\m_2\end{array} \right),C)}{p(y)}$$
Assume the likelihood $p(y|x)$ is Gaussian, as well as the prior $p(x)$.
$p(y|x)$ has it's mean to be a linear combination of $x$, which is $x_1+x_2$.
The x here is 2 dimension Gaussian and y is 1 dimension. It looks pretty complicated to reform it into a pdf of $x$ (will $f(x)$ be Gaussian?). Any idea to do this, or even in more complicated cases? Thanks!
Up to proportionality factors independent of $x$, $$ f(x)\propto\exp\left(-\frac1{2\sigma^2}(y-x^*u)^2-\frac12(x-m)^*C^{-1}(x-m)\right), $$ where $x=(x_1,x_2)^*$, $m=(m_1,m_2)^*$ and $u=(1,1)^*$. Thus, $$ f(x)\propto\exp\left(-\frac12(x-\mu)^*\Gamma^{-1}(x-\mu)\right), $$ for some vector $\mu$ and some matrix $\Gamma$ which depend a priori on $y$, that is, $f$ is the gaussian density $N(\mu,\Gamma)$. To identify $(\mu,\Gamma)$, one solves the identity $$ \sigma^{-2}(y-x^*u)^2+(x-m)^*C^{-1}(x-m)=(x-\mu)^*\Gamma^{-1}(x-\mu)+g(y), $$ where $g(y)$ can be any function independent on $x$. Both sides are quadratic forms with respect to $x$ hence the second degree terms must coincide. Using $(x^*\cdot u)^2=x^*uu^*x$, this yields $$ \sigma^{-2}x^*uu^*x+x^*C^{-1}x=x^*\Gamma^{-1}x. $$ Thus, $\Gamma$ does not depend on $y$ and $$ \Gamma^{-1}=C^{-1}+\sigma^{-2}uu^*. $$ Likewise, the first degree terms coincide, hence $$ 2y\sigma^{-2}x^*u+m^*C^{-1}x+x^*C^{-1}m=\mu^*\Gamma^{-1}x+x^*\Gamma^{-1}\mu, $$ that is, $$ y\sigma^{-2}u^*x+m^*C^{-1}x=\mu^*\Gamma^{-1}x. $$ One sees that $\mu$ depends on $y$ and $$ \mu=\Gamma (y\sigma^{-2}u+C^{-1}m)=m+\sigma^{-2}(y+u^*m)\Gamma u. $$