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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.

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  • $\begingroup$ If you can show the product of two Gaussian random variables is not Gaussian, that would be sufficient. $\endgroup$ – Zhanxiong Mar 20 at 19:14
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The goal of this passage in the textbook is to establish the following fact:

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.

The second point they make is the following:

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.

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