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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?
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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).

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you can find echos of your expressions for the mean and variance at so there probably is a probabilistic interpretation – Henry Feb 28 '12 at 15:34

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