I'm trying to understand the origin of the formulas for the $\mu$ and the $\sigma^2$ from this side (stanford):


As I have understood actually the product of two Gaussian PDF's is not again a Gaussian PDF. But with the formula from the stanford website I again get a Gaussian PDF.

Where I need help is:

  1. Understand the origin of the two formulas $(\mu, \sigma^2)$
  2. Is the center of the product of two PDF's and the center of a PDF calculated via the two formulas in the same place?

1 Answer 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).

  • $\begingroup$ you can find echos of your expressions for the mean and variance at stats.stackexchange.com/questions/9071 so there probably is a probabilistic interpretation $\endgroup$
    – Henry
    Feb 28, 2012 at 15:34
  • $\begingroup$ @Henry No. Actually it seems difficult to understand what factual elements you base this strange conviction on. $\endgroup$
    – Did
    Nov 10, 2018 at 8:39
  • 9
    $\begingroup$ Well, for Bayesian statistics the product of two Gaussians being Gaussian and the formulas for the mean and variance of it are quite important. Essentially they give the solution for the posterior with Gaussian prior and Gaussian likelihood or for multiple observations with Gaussian likelihood. There also the normalization constant is often ignored anyway. I think this is quite clearly a probabilistic interpretation. $\endgroup$
    – Xenon
    Jun 18, 2019 at 21:57

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