# Why is the product of two gaussian process $f1$ and $f2$ not a gaussian process?

In the book from Rasmussen/Williams on Gaussian Processes we have the following statement without proof (Page 95): "If f1 and f2 are Gaussian processes then the product f will not in general be a Gaussian process, but there exists a GP with this covariance function"

Unfortunalty I cannot find anything related to this statement in the book and I dont see a way of proving it.

• If you can show the product of two Gaussian random variables is not Gaussian, that would be sufficient. Mar 20, 2019 at 19:14

Fact 1: Suppose $$K_1$$ and $$K_2$$ are covariance functions (i.e., symmetric, positive definite, functions). Then $$K_1 K_2$$ is also a covariance function.
Their argument is this: suppose $$f$$ and $$g$$ are independent Gaussian processes with kernels $$K_1$$ and $$K_2$$ and mean function $$m(x) = 0$$. Then $$f(x) g(x)$$ is a stochastic process with covariance function $$K' = K_1 K_2$$. This is relatively easy to check just by direct calculation, and all we are really using is the fact that there exist mean-0 stochastic processes with the associated $$K_1$$ and $$K_2$$. The assumption that $$f(x)$$ and $$g(x)$$ are GPs is actually superfluous, it's just convenient to use them because you already know such an $$f$$ and $$g$$ exist.
Fact 2: There exists a Gaussian process with covariance function $$K_1 K_2$$.
Now, just because there exists a Gaussian process with this kernel, does not imply that $$f(x)g(x)$$ is a Gaussian process. We only considered this product to establish that $$K_1 K_2$$ is a valid covariance function. But, in general, if you take a product of Gaussian random variables, there is no reason to expect that the product will still remain Gaussian; for example, if $$X \sim N(0,1)$$ and $$Y \sim N(0,1)$$ then $$XY$$ is not Gaussian anymore. But, as long as $$K_1 K_2$$ is the covariance function for some stochastic process, it will immediately follow that there is a Gaussian process (different from the stochastic process we started with) which has this covariance.