4
$\begingroup$

Considering: $$f(x) = \frac{1}{\sigma_x\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x}{\sigma_x})^2}$$

$$g_i(x) = \frac{1}{\sigma_i\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{a_i+b_ix}{\sigma_i})^2}$$

Is there a closed-form expression for this integral?$$\int_{-\infty}^{+\infty} \left(f(x)\cdot\prod_i g_i(x)\right) \, \mathrm{d} x$$

  • 1
    Are you aware that the product of Gaussians is again a Gaussian?2012-11-28
  • 0
    Sorry, I'm learning how this works. I'm aware of it.2012-11-28
  • 0
    Then it would help if you explained where you see an obstacle to combining the Gaussians into a single Gaussian and integrating it.2012-11-28
  • 0
    When $i=1$, I obtain $$ \frac{1}{\sqrt{2\pi((b\sigma_x)^2+\sigma_i^2)}}e^{-\frac{1}{2}(\frac{a^2}{(b\sigma_x)^2+\sigma_i^2})}$$ I would expect something similar for any $i$.2012-11-28
  • 0
    Isn't the question general enough leaving out $f(x)$ and throwing that as a special case of the $g_i(x)$, with $a_i=0$, $b_i=1$, and $\sigma_i=\sigma_x$? So you're really just asking whether there's a closed form for the product of Gaussians with different parameters. See this article for the case where you've only got two: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&ved=0CDEQFjAA&url=http%3A%2F%2Fwww.tina-vision.net%2Fdocs%2Fmemos%2F2003-003.pdf&ei=tRi2ULzYHbSP0QHo_oGICQ&usg=AFQjCNH2kIfaP98gv7phLFQJRRUwUWr_aQ2012-11-28
  • 0
    Thank you. I already knew it. Actually, I obtained the result when $i = 1$ using the convolution of two gaussian functions. But, is there any easy way also for $i>i$?2012-11-28
  • 0
    That gives the closed form for the new parameters $\sigma$, $a$, and $b$ that you'd get by taking the product of two Gaussians. I think from examining that formula that you might be out of luck writing a "nice" closed form solution for the parameters of your new distribution (the product of arbitrarily many Gaussian distributions), but maybe I'm just being pessimistic.2012-11-28

1 Answers 1

4

\begin{align} \sum_{i=1}^n \left(\frac{a_i+b_ix}{\sigma_i}\right)^2 & = \left(\sum_{i=1}^n \left( \frac{b_i}{\sigma_i} \right)^2 \right) x^2 + 2\left(\sum_{i=1}^n \frac{a_i b_i}{\sigma_i^2}\right) x + \left( \sum_{i=1}^n \left(\frac{a_i}{\sigma_i}\right)^2 \right) \\[12pt] & = Ax^2 + Bx + C \\[12pt] & = A\left(x+\frac{B}{2A}\right)^2 + C - \frac{B^2}{4A}. \end{align} So you get another Gaussian function.

(Update: corrected B term)

  • 0
    Nice! I suppose that the next step is to try to convolve it to obtain a normal distribution a la $i=1$.2012-11-28
  • 0
    @Medicalphysicist : The exponential of $-1/2$ times the function above is a scalar multiple of a normal density function centered at $-B/(2A)$, with variance $1/A$.2012-11-28
  • 0
    Ok. I don't find any nicer form. I accept your answer, but you have a mistake in the term B. Thanks.2012-11-29