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Multiplying two univariate Gaussian PDFs

$$ X \sim \mathcal{N}(\mu_X,\sigma_X) \\ Y \sim \mathcal{N}(\mu_Y,\sigma_Y) \\ Z = XY $$

results in closed form equations for $\mu_Z$ and $\sigma_Z^2$:

$$ \mu_Z = \frac{\sigma^2_X*\mu_Y+\sigma^2_Y*\mu_X}{\sigma^2_X+\sigma^2_Y} $$ and $$ \sigma_Z^2 = \frac{\sigma_X^2*\sigma_Y^2}{\sigma_X^2+\sigma_Y^2} $$

some derivations of this here

This is a direct consequence of the exponential functions and the addition of exponents.

But, it is said by a lot on here that it is "obvious" that the multiplication of two Gaussian random variables is not a Gaussian random variable.

I just don't get why it's obvious. There seems to be a move to jump to the integral to combine two random variables. Why? Why is multiplying two PDFs not the same as multiplying two random variables?

My motivation: there is a literature within psychophysics to use the combinations of gaussian functions to approximate sensory cue combination in humans. An example. I want to extrapolate this to a similar case, but I'm running into this theoretical distinction people are making that I don't understand.

Links to other posts about similar things that don't quite answer this question:

Is the product of two Gaussian random variables also a Gaussian?

Texbook with proof that product of two Gaussian functions is also Gaussian

What's the densitiy of the product of two independent Gaussian random variables?

Product of two Gaussian PDFs is a Gaussain PDF, but Produt of two Gaussan Variables is not Gaussian

Did
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bcmcmahan
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  • This is a very confused question. In particular, the paper you refer to derives the pdf of the product of 2 pdf's, say $f(x)$ * $f(y)$. By contrast, you refer to the product of two random variables $Z = X*Y$ ... which is entirely different. To be clear, if $X$ and $Y$ are independent, then $E[X Y] = E[X]E[Y]$. – wolfies Mar 09 '15 at 08:15
  • The formulas given for $\mu_Z$ and $\sigma^2_Z$ are pure fantasy. – Did Mar 09 '15 at 08:20
  • Thank you for the comments. To Wolfies: I am very very confused. I think the question I want to know can be summarized as: What is the difference? Why is multiplying two random variables not just the multiplication of the PDFs? Is it because of the normalization constant in the resultant? (edited to separate responses) – bcmcmahan Mar 09 '15 at 17:35
  • To Did: Can you explain why? I also found this blog post – bcmcmahan Mar 09 '15 at 17:35
  • @Did Actually, they seem to be very standard: http://www.bzarg.com/p/how-a-kalman-filter-works-in-pictures/ – Felix Goldberg Aug 15 '17 at 08:09
  • @FelixGoldberg Of course, multiplying PDFs $f$ and $g$ can have its uses, as the authors show, but these authors should be chastised for failing to be more explicit about the fact that they consider the renormalized product of $f$ and $g$, that is, the PDF $h$ defined by $$h(x)=\frac{f(x)g(x)}{c}\qquad c=\int_\mathbb R f(z)g(z)dz$$ Once again, it is "very standard", as was already mentioned on this page, that, in general, $$\int_\mathbb R f(z)g(z)dz\ne1$$ – Did Aug 15 '17 at 08:17
  • @bcmcmahan While I am at it... "Explain why" what? Surely you noted that the blog post you mention in your comment is making exactly the same point as I made here. Hint: Read the title of the blog post. – Did Aug 15 '17 at 08:19

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A random variable and a density function are two very different things.

Suppose $X$ and $Y$ are random variables and $f_X$ and $f_Y$ are their density functions.

The density function of $X+Y$ is not $f_X + f_Y$. The latter is not the density of anything because it integrates to 2!

Similarly, the density of $X*Y$ is not $f_X*f_Y$.

The density of $X+Y$ is the convolution of $f_X$ and $f_Y$. There is no way to relate the density of $X*Y$ to the densities of $X$ and $Y$ that I know of.

John D. Cook
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    "There is no way to relate the density of $X*Y$ to the densities of $X$ and $Y$ that I know of. " Hmmm... $$f_{XY}(z)=\int f_X(x)f_Y\left(\frac{z}x\right)\frac{\mathrm dx}{|x|}$$ – Did Mar 11 '15 at 17:02