Product of two gaussian processes

Question

Given,

$\ {y}_{i} = N({\mu}_{i}, {\Sigma }_{i}) $

If we go by the link http://www.tina-vision.net/docs/memos/2003-003.pdf then we can understand that the product of many multivariate gaussians can be written as:

$ \prod {y}_{i} = {y}_{p} = N({\mu }_{p}, {\Sigma }_{p})$

Where,

$\Sigma_{p}^{-1} = \sum \Sigma_{i}^{-1}$

and $\Sigma_{p}^{-1}{\mu }_{p} = \sum \Sigma_{i}^{-1}{\mu }_{i}$

What can we say about the product $ \prod {Y}_{i}$ of gaussian processes given by:

$\ {Y}_{i} = GP({m}_{i}\left(x \right),{k}_{i}\left(x,x' \right))$

Answer 1

Let two independent normal (Gaussian) random variables,

$$X \sim N(\mu_x, \sigma^2_x),\;\;\; Y \sim N(\mu_y, \sigma^2_y)$$

with probability density functions $f_X(x)$ and $f_Y(y)$ respectively. Then the probability density function of the product of the two random variables, i.e. of the random variable $Z = XY$ is

$$ f_Z(z) = \int^{\infty}_{-\infty} f_X \left( x \right) f_Y \left( z/x \right) \frac{1}{|x|}\, dx$$

As you can see, the density of $Z$, $f_Z(z)$, is not the product of the densities. Informally, this is because the probaility density function does not determine what values the $Z$ variable takes, but how probability is allocated to the values that $Z$ takes (values that are determined by some other function, usually unspecified).

Answer 2

The product of Gaussian processes will not be a Gaussian process, unlike the sum of Gaussian processes. When you multiply to random variables, you don't simply multiply their PDFs, it's very easy to see why.