What is the final result of $$I=\mathcal{N}_{x}(\mu_1,v_1)\,\mathcal{N}_{x}(\mu_2,v_2)\ldots\,\mathcal{N}_{x}(\mu_n,v_n)=\frac{1}{\sqrt{2\pi\,v_1} } e^{ -\frac{(x-\mu_1)^2}{2v_1} } \frac{1}{\sqrt{2\pi\,v_2} } e^{ -\frac{(x-\mu_2)^2}{2v_2} }\cdots \frac{1}{ \sqrt{2\pi\,v_n} } e^{ -\frac{(x-\mu_n)^2}{2v_n} }$$ in which $\mu_{1,\ldots,n}$ are mean and $v_{1,\ldots,n}$ are variance?
From the identity of two Gaussians multipliation, that is $$\mathcal{N}_{x}(\mu_1,v_1)\,\mathcal{N}_{x}(\mu_2,v_2)=\mathcal{N}_{\mu_1}(\mu_2,v_1+v_2)\,\mathcal{N}_{x}\left(\frac{\mu_1\,v_2+\mu_2\,v_1}{v_1+v_2},\frac{v_1\,v_2}{v_1+v_2}\right)$$ the $I$ can be computed one by one, but it is a daunting task. Is there a compact expression of the $I$?
Best Answer
The product of Gaussians $\prod_{i=1}^n \mathcal N(\mu_i,\sigma_i)$ simplifies to $S\, \mathcal N (\mu,\sigma)$, where $$ \sigma = \left(\sum_{i=1}^n \sigma_i^{-2} \right)^{-1/2} $$ $$ \mu = \sigma^2 \sum_{i=1}^n \sigma_i^{-2}\mu_i $$ and $$ S = (2\pi)^{(1-n)/2} \frac{\sigma}{\prod_{i=1}^n \sigma_i}\exp\left(\frac12 \sigma^{-2}\mu^2 - \frac12\sum_{i=1}^n \sigma_i^{-2}\mu_i^2 \right) $$ The derivation is given in the report Products and Convolutions of Gaussian Probability Density Functions by P.A. Bromiley, which was pointed out by jibounet.