Calculate the product of two Gaussian PDF's

Solution 1:

As you noticed, the product of two gaussian PDFs is not a PDF. However, as any positive integrable function, it is proportional to another PDF, which happens to be itself gaussian. The rest is calculus.

Write $g_{\mu,\sigma^2}$ for the gaussian density with mean $\mu$ and variance $\sigma^2$, that is, $$ g_{\mu,\sigma^2}(x)=\frac1{\sqrt{2\pi\sigma^2}}\exp\left(-\frac1{2\sigma^2}(x-\mu)^2\right). $$ Then the function $g_{\mu_1,\sigma_1^2}\cdot g_{\mu_2,\sigma_2^2}$ is proportional to $g_{\mu,\sigma^2}$, where the parameters $\mu$ and $\sigma^2$ are uniquely determined by the two relations $$ (\sigma_1^2+\sigma_2^2)\mu=\mu_1\sigma_2^2+\mu_2\sigma_1^2,\qquad \frac1{\sigma^2}=\frac1{\sigma_1^2}+\frac1{\sigma_2^2}. $$ This stems from the fact that one can write $$ \frac1{\sigma_1^2}(x-\mu_1)^2+\frac1{\sigma_2^2}(x-\mu_2)^2=\frac1{\sigma^2}(x-\mu)^2+C, $$ for the values of $\mu$ and $\sigma^2$ given above, where $C$ is independent on $x$.

There is no probabilistic interpretation to this algebraic fact that I am aware of and, to tell you the truth, I wonder why this factoid was selected as noticeable on the website you link to (much more significant are the characteristic function of a gaussian PDF being a gaussian function and the convolution of gaussian PDFs being a gaussian PDF).