Obvious demonstration of the posterior density of normal laws in bayesian inference

Question

Given the observations' density $f(X|\theta) = \mathcal{N}(\theta,1)$ and the prior density $f(\theta) = \mathcal{N}(0,\sigma^2)$, it it is said that it is obvious that the posterior density is equal to : $\mathcal{N}(\frac{X}{1+\sigma^{-2}},\frac{1}{1+\sigma^{-2}})$.

I use following relation to compute the posterior density : $f(\theta|X) = \frac{f(X|\theta)f(\theta)}{\int f(X|\theta)f(\theta)d\theta}$, but I dont find it obvious at all...

Any help appreciated

Answer 1

You are on the right track. Substitute expressions for the densities of normals into the Bayes law (and don't keep track of constants !!)

\begin{align} f(\theta|X) &\propto \exp(-\frac{1}{2}(X-\theta)^2)\cdot \exp(-\frac{1}{2}\sigma^{-2}\theta^2)\\ & \propto \exp(-\frac{1}{2} ((1+\sigma^{-2})\theta^2 - 2 \theta X + X^2)) \\ & \propto \exp(-\frac{1}{2} (1+\sigma^{-2}) (\theta^2 - 2 \theta \frac{X}{1+\sigma^{-2}})) \\ & \propto \exp(-\frac{1}{2} (1+\sigma^{-2}) (\theta - \frac{X}{1+\sigma^{-2}})^2) \end{align}

The latter is the kernel of a normal $\mathcal N (\frac{X}{1+\sigma^{-2}}, \frac{1}{1+\sigma^{-2}})$.