0
$\begingroup$

Let's assume we have 2 gaussian functions Allowing some notation abuse they would look like:

$$G_i(x) = Ce^{\frac{-1}{2}[(x-\mu_i) / \sigma_i]^2}$$

$$G_j(x) = Ce^{\frac{-1}{2}[(x-\mu_j) / \sigma_j]^2}$$

Where $C$ is the constant factor of the Gaussians which I encapsulate here for simplicity.

I claim that $G_i \cdot G_j$ Is another Gaussian with parameters $\mu_{ij}, \sigma_{ij}$

But I am struggling at expressing the product as a new Gaussian to find the new parameters.

So far I have this:

$$G_i \cdot G_j(x) = Ce^{(\frac{-1}{2}[(x-\mu_j) / \sigma_j]^2) + (\frac{-1}{2}[(x-\mu_j) / \sigma_j]^2)}$$

If we focus on just the exponent and remove the constant factor we get: $$[(x-\mu_j) / \sigma_j]^2 + [(x-\mu_j) / \sigma_j]^2$$

My goal is to reorder the terms in that expression to get something of the form $[(x-\mu_{ij})/\sigma_{ij}]^2$

However I got stuck after getting the following:

$$\frac{(\sigma_j^2 + \sigma_i^2)x^2 + (\mu_i+\mu_j)(-2x) + (\mu_i^2 + \mu_j^2)}{\sigma_i^2\sigma_j^2}$$

I could keep going by dividing both elements of the fraction by the coefficient in front of the $x^2$ term but after that I am not finding an easy way to factor everything back into a singular square term on the numerator (i.e i don't know what do with $\mu_i^2 + \mu_j^2$).

Can this be done? Am I wrong about my hypothesis? Is there a theorem I can quote to avoid doing the entire expansion myself?

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ Complete the square; you'll have a constant term left over, but since that's in an exponent you can take that out of the exponent and multiply the constant factor $C$ by something to compensate $\endgroup$ – alphacapture Apr 4 at 1:46
  • $\begingroup$ Oh right! I forgot that I could kill that term that way. I'll try it tomorrow and see where that leads me. Thank you for the tip. $\endgroup$ – Makogan Apr 4 at 1:48
1
$\begingroup$

If, as @alphacapture commented, you write $$\frac{(x-\mu_1)^2}{\sigma_1^2}+\frac{(x-\mu_2)^2}{\sigma_2^2}=\frac{(x-\mu)^2}{\sigma^2}+\tau$$ completing the square or identifying the coefficients, you should arrive to $$\mu=\frac{\mu_2 \sigma_1^2+\mu_1 \sigma_2^2}{\sigma_1^2+\sigma_2^2}\qquad \sigma=\frac{\sigma_1 \sigma_2}{\sqrt{\sigma_1^2+\sigma_2^2}}\qquad \tau=-\frac{(\mu_1-\mu_2)^2}{\sigma_1^2+\sigma_2^2}$$ and then $$e^{-\frac{(x-\mu_1)^2}{2 \sigma_1^2}}+e^{-\frac{(x-\mu_2)^2}{2 \sigma_2^2}}=e^{-\frac \tau 2}\,e^{\frac{(x-\mu)^2}{2\sigma^2} }$$

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.