I have three random variables: $x$, $y$, $z$ in $\mathbf{R}$. I know the following about their distributions: $x \sim \text{unif}[-\infty, \infty]$, $y \sim \mathcal{N}(x, \sigma)$, $z \sim \mathcal{N}(x, \sigma)$, $y-x$ and $z-x$ are independent.
From this, using Bayes' theorem I infer that
$x|y \sim \mathcal{N}(y, \sigma)$ and $x|z \sim \mathcal{N}(z, \sigma)$.
What I need to find is conditional density of $y$ given $z$, $f(y|z)$.
Many thanks!
On
Assuming that $x$ has density $g$, the density $h$ you look for is $$ h(y,z) dy= \frac{P(Y\in dy, Z\in dz)}{P(Z\in dz)} = \frac 1{\sqrt{2\pi}\sigma}\frac{\displaystyle\int_x g(x) e^{-(y-x)/2\sigma}\times e^{-(z-x)/2\sigma} dx\times dydz}{\displaystyle\int_x g(x) e^{-(z-x)/2\sigma} dx\times dz} $$ leading to the expression (with $g=1$) $$ h(y,z) = \frac 1{\sqrt{2\pi}\sigma} \frac{\displaystyle\int_x e^{-(y-x)/2\sigma}\times e^{-(z-x)/2\sigma} dx}{\displaystyle\int_x e^{-(z-x)/2\sigma} dx} $$ now it remains to compute the integrals.
If $x$ has density $g$ and $y=x+u$, $z=x+v$ with $(x,u,v)$ independent, $x$ with density $g$ and $u$ and $v$ with density $h$ then $(y,z)$ has density $$f(s,t)=\int g(r)h(s-r)h(t-r)\mathrm dr,$$ hence the conditional density of $y$ conditionally on $z$ is $$f_{y\mid z}(s\mid t)=\left(\int g(r)h(s-r)h(t-r)\mathrm dr\right)\cdot\left(\int g(r)h(t-r)\mathrm dr\right)^{-1}.$$ We are told that $x$ is uniform on the real line, which is (impossible and is a conventional way of) saying that $g=1$, hence $$f_{y\mid z}(s\mid t)=\int h(s-r)h(t-r)\mathrm dr.$$ It remains to plug in the value of $h$, namely, $$h(r)=\frac{\mathrm e^{-r^2/2\sigma^2}}{\sqrt{2\pi\sigma^2}},$$ to deduce that $$f_{y\mid z}(s\mid t)=\frac{\mathrm e^{-(s-t)^2/4\sigma^2}}{\sqrt{4\pi\sigma^2}},$$ that is, conditionally on $z$, $y$ is normal with mean $z$ and variance $2\sigma^2$.
The method above is general. In the case of gaussian random variables, shortcuts are possible.