I cam across an interesting claim that:
$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$
In trying to understand it I consulted Bromiley:
http://www.tina-vision.net/docs/memos/2003-003.pdf
Bromiley concludes that:
if
$f(x) = \frac{1}{\sqrt{2\pi\sigma_f^2}} e^{-\frac{(x-\mu_f)^2}{2 \sigma_f^2}}$
and
$g(x) = \frac{1}{\sqrt{2\pi\sigma_g^2}} e^{-\frac{(x-\mu_g)^2}{2 \sigma_g^2}}$
then:
$f(x)g(x) = D_{fg} \frac{1}{\sqrt{2\pi\sigma_{fg}^2}}
e^{-
\frac
{ (x – \mu_{fg})^2 }
{2 \sigma_{fg}^2 }
}$
where:
$\mu_{fg} = \frac { \sigma_g^2\mu_f + \sigma_f^2 \mu_g } {\sigma_f^2 + \sigma_g^2}$
and
$\sigma_{fg}^2 = \frac {\sigma_f^2 \sigma_g^2} {\sigma_f^2 + \sigma_g^2}$
$S_{fg} = \frac {1} {\sqrt{2\pi(\sigma_f^2+\sigma_g^2)}}
e^{
-\frac{(\mu_f-\mu_g)^2}{2(\sigma_f^2+\sigma_g^2)}
}$
Note that if $\mu_f$, $\mu_g$ , $\sigma_f$ and $\sigma_f$ are known constants then the $S_{fg}$ is a known constant too.
To wit, if I cast Bromiley's result in the format of the claim I'm exploring:
$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = S_{fg} \; \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$
In short there is a constant scaling factor $S_{fg}$. In fact Bromiley describes the product as a scaled Gaussian.
Given $f(x)$ and $g(x)$ are both functions of $x$ the original claim, which reads (as a reminder):
$\mathcal N (\mu_f, \sigma_f^2) \; \mathcal N (\mu_g, \sigma_g^2) = \mathcal N \left(\frac{\mu_f\sigma_g^2+\mu_g\sigma_f^2}{\sigma_f^2+\sigma_g^2}, \frac {\sigma_f^2\sigma_g^2}{\sigma_f^2+\sigma_g^2}\right)$
implies that:
$\int_{-\infty}^{\infty} f(x) g(x) \;dx = 1$
But Bromiley's result suggests this implication is false. I presume it inetgrates to S_{fg}, or:
$\int_{-\infty}^{\infty} f(x) g(x) \;dx = S_{fg}$
My tentative conclusion is that the claim I am exploring is false, and my questions would be:
- Is my tentative conclusion true? (is the explored claim false?)
- Am I right in concluding the integral would be $S_{fg}$?
Those are the areas I'm a little shakey on at present and seek some review on I guess.
Best Answer
My (subjective) opinion:
Do write
$$ \mathcal{N}(x \mid \mu_1,\sigma_1^2)\cdot\mathcal{N}(x \mid \mu_2,\sigma_2^2) $$
Don't write
$$ \mathcal{N}(\mu_1,\sigma_1^2)\mathcal{N}(\mu_2, \sigma^2) $$
The notation $\mathcal{N}(x|\mu_1,\sigma_1^2)$ makes it clear that this is a function, this particular function can have a particular significance in a statistical/probability setting question but that does not need to be the case, it is simply a function and the product of functions is a well defined and familiar operation with little ambiguity.
On the other hand the notation $\mathcal{N}(\mu, \sigma^2)$ is to be used as short hand to make $X \sim \mathcal{N}(\mu, \sigma^2)$ equivalent to "the random variable $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$".
In this setting $\mathcal{N}(\mu, \sigma^2)$ is not a function, it isn't even a distribution function it is a symbolic representation of a particular statement, a sign post that allows us to retrieve certain information should we wish, but it is not a mathematical object for which we have a well defined product.