Bayesian updating with normal signals and independence error Suppose I want to know a realization of $y\sim\ N(0,\sigma^2)$ and observe two signals,
$y+\varepsilon_1$ 
and $y+\varepsilon_2$. Where $\varepsilon_1 \sim\ N(0,\sigma^2_{\varepsilon_1})$ and $\varepsilon_2 \sim\ N(0,\sigma^2_{\varepsilon_2})$ and $y,\varepsilon_1,\varepsilon_2$ are independent.
Is there a simple characterization for the posterior distribution of $y$ conditional on observing both signals?
 A: I disagree with Tim in that the problem seems to me to fit within a Bayesian framework: Since $y∼ N(0,σ^2)$ is the mean of both observations, $x_1=y+ε_1$ and $x_2=y+ε_2$, where $ε_1∼ N(0,σ^2_{ε_1})$ and $ε_2∼ N(0,σ^2_{ε_2})$, the posterior distribution on the mean $y$ is with density
$$f(y|x_1,x_2) \propto \exp\{-y^2/2σ^2\}\times \exp\{-(y-x_1)^2/2σ^2_{ε_1}\}\times \exp\{-(y-x_2)^2/2σ^2_{ε_2}\}$$which is a normal pdf on $y$ since
\begin{align*}f(y|x_1,x_2) &\propto \exp\left\{-y^2/2[σ^{-2}+σ^{-2}_{ε_1}+σ^{-2}_{ε_2}] +y[σ^{-2}_{ε_1}x_1+σ^{-2}_{ε_2}x_2]\right\}\\
&\propto \exp\left\{-\frac{1}{2}\frac{(y-[σ^{-2}+σ^{-2}_{ε_1}+σ^{-2}_{ε_2}]^{-1}[σ^{-2}_{ε_1}x_1+σ^{-2}_{ε_2}x_2])^2}{σ^{-2}+\sigma^2_{ε_1}+σ^{-2}_{ε_2}}\right\}
\end{align*}
which means that the posterior on $y$ is a Gaussian distribution with mean
$$\mathbb{E}[y|x_1,x_2]=[σ^{-2}+σ^{-2}_{ε_1}+σ^{-2}_{ε_2}]^{-1}[σ^{-2}_{ε_1}x_1+σ^{-2}_{ε_2}x_2]$$and variance
$$\text{var}(y|x_1,x_2)=σ^{-2}+\sigma^2_{ε_1}+σ^{-2}_{ε_2}$$
A: There is nothing Bayesian in this problem since you did not define any model in terms of likelihood function and priors.
If you have independent $y_1, y_2 \sim \mathcal{N}(0, \sigma^2)$ and $\varepsilon_1 \sim \mathcal{N}(0, \sigma_{ \varepsilon_1 }^2)$, $\varepsilon_2 \sim \mathcal{N}(0, \sigma_{ \varepsilon_2 }^2)$ then
$$ y_i + \varepsilon_i \sim \mathcal{N}(0, \sigma^2 + \sigma_{ \varepsilon_i }^2) $$
given the property of variance that
$$\mathrm{Var}(X+Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$$
when $X$ and $Y$ are independent.
So you end up with two random variables that are not identically distributed.
A: Here's the way I see it. 
Let $x = (x_1, x_2)$.
The posterior distribution is proportional to the likelihood times the prior:
\begin{equation}
p(y|x) \propto p(x|y)\,p(y) .
\end{equation}
The prior is
\begin{equation}
p(y) = \textsf{N}(y|0,1) 
\end{equation}
and the likelihood is
\begin{equation}
p(x|y) = \prod_{i=1}^2 p(x_i|y) = \prod_{i=1}^2 = \textsf{N}(x_i|y,\sigma_i^2) .
\end{equation}
Therefore,
\begin{equation}
p(y|x) = \textsf{N}(y|m,s^2) , 
\end{equation}
where 
\begin{align*}
m & = s^2\left(\frac{x_1}{\sigma_1^2} + \frac{x_2}{\sigma_2^2}\right) \\
s^2 &= \left(1 + \frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}\right)^{-1} .
\end{align*}
The posterior mean is a weighted average from all three sources of information: the prior and the two signals. (Since the mean of the prior equals zero, this term does not appear explicitly in $m$.) The weights are determined by the precisions of the sources of information. (Precision is the inverse of variance.)
