3
$\begingroup$

I know that the product of two Gaussian functions is also Gaussian function. This is stated in Wikipedia, but I might need to cite a classical (text)book/paper stating this result. A book containing the proof is welcome. Could you name such a book?

  • 3
    This is really just a simple consequence of well-kown identities concerning [powers](http://en.wikipedia.org/wiki/Exponentiation).2010-08-28
  • 0
    @Rasmus: Thanks, I didn't think before asking :|. I maintain the question only because of @John's answer.2010-09-07

1 Answers 1

4

The product of two independent normal (Gaussian) random variables is not normal. That would be convenient, but it's not true. Here are a some related theorems that are true:

  • The sum of two independent normal random variables is normal.

  • The product of two independent log normal random variables is log normal.

  • The ratio of two independent normal random variables is Cauchy.

For more on distribution relationships, see this chart.

If you multiply two normal PDFs you get the joint PDF of a multivariate normal, i.e. if f(x) is a normal PDF then g(x, y) defined by f(x)g(y) is a multivariate normal PDF, but in that case you're multiplying densities and not random variables.

Update: I looked back at your question and I see you said Gaussian "function," rather than "random variable", so perhaps you had in mind the PDF result I mentioned above. If so, just write down the definition of multivariate normal and there it is.

Update 2: Here's an explicit formula for the the product of two normal pdfs being proportional to another normal pdf: blog post.

  • 1
    Unfortunately John's blog post doesn't contain the full derivation. Would anyone like to contribute a line-by-line proof? Or cite a textbook that has one?2014-09-12