Whille working in Bayesian statistics I am taught that the identity: $$\int N(z|\mu,\sigma_0^2)N(\mu|\mu_0,\sigma_1^2) d\mu = N(z|\mu_0,\sigma_0^2+\sigma_1^2)$$ holds where $N(x|m,v)$ is a normal distribution in the variable $x$, with mean $m$ and variance $v$.
I see there are some related questions on SE but they require a lot of algebraic work. Is there any other way I can prove this fact? Has anybody encountered this before?
First notice that the left hand side is equal to the convolution of the functions $N(z|0,\sigma_0^2)$ and $N(z|\mu_0,\sigma_1^2)$. Thus the left hand side is the density of the random variable X+Y where X and Y are independent and are distributed according to these two respective densities. So, CF(X+Y)=CF(X)*CF(Y) where CF denotes the characteristic function. Now use the explicit form of the characteristic function of a normal distribution and the uniqueness result for characteristic functions.
