I have this exercise:
An explosion was detected by two sensors. Each sensor is only able to output a noisy estimate of the location of the explosion due to measurement noise. Assuming the two sensor outputs are y1 and y2, and the likelihood of the exact location x given the sensor outputs is $$ p(y_1|x)p(y_2|x) = \mathcal{N}(y_1;x,σ_1^2)\mathcal{N}(y_2;x,σ_2^2) $$ where $σ_1^2$ and $σ_2^2$ are the measurement noise variances. Assuming a prior distribution over the location: $$ p(x) = \mathcal{N}(x; 0, σ_0^2), $$ where $σ_0^2$ is the prior variance.
1: Find the posterior distribution $p(x|y_1, y_2)$
When deriving the expression for the posterior, do you first find a new Gaussian Pdf for the likelihood $(p(y_1|x)p(y_2|x) = \mathcal{N}((y1,y1); μ3, σ3)$ and then multiply this with the prior to find yet another Gaussian? And if yes how does one go about doing that?
One important thing to note when working with posteriors and priors is that we assume they belong to the same class of probabilities. In other words, given a prior we already know the general form of the posterior.
Expanding on your problem here we have that (denoting the prior by $\pi(x)$) that; $$ \pi(x|y_1,y_2) \propto f(y|x)\pi(x)\propto \prod_{i=1}^{2}f(y_i|x)\pi(x) $$ which simplifies to $$\pi(x|y_1, y_2)\propto \frac{1}{\sqrt{2\pi}\sigma_1}e^{-\frac{(y_1 - x)^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi}\sigma_2}e^{-\frac{(y_2 - x)^2}{2\sigma_2^2}}\frac{1}{\sqrt{2\pi}\sigma_0}e^{-\frac{x^2}{2\sigma_0^2}}$$ This is possible because densities which are proportional have the same distribution.
You will find that this is again a normal distribution with $N(x;\mu',\sigma'^2)$ where $\mu', \sigma'^2$ are identified by examining how $\pi(x|y_1,y_2)$ depends on x up to some constant of proportionality not depending on x. (really recommend doing this step yourself as being comfortable with manipulating expressions in probability is soo useful)