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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?$$\int_{-\infty}^{+\infty} \left(f(x)\cdot\prod_i g_i(x)\right) \, \mathrm{d} x$$

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Are you aware that the product of Gaussians is again a Gaussian? –  joriki Nov 28 '12 at 13:40
    
Sorry, I'm learning how this works. I'm aware of it. –  Medical physicist Nov 28 '12 at 13:50
    
Then it would help if you explained where you see an obstacle to combining the Gaussians into a single Gaussian and integrating it. –  joriki Nov 28 '12 at 13:51
    
When $i=1$, I obtain $$ \frac{1}{\sqrt{2\pi((b\sigma_x)^2+\sigma_i^2)}}e^{-\frac{1}{2}(\frac{a^2}{(b\sig‌​ma_x)^2+\sigma_i^2})}$$ I would expect something similar for any $i$. –  Medical physicist Nov 28 '12 at 13:58
    
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: google.com/… –  Gwyn W Nov 28 '12 at 14:00
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up vote 3 down vote accepted

\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)

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Nice! I suppose that the next step is to try to convolve it to obtain a normal distribution a la $i=1$. –  Medical physicist Nov 28 '12 at 14:46
    
@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 $1/A$. –  Michael Hardy Nov 28 '12 at 17:34
    
Ok. I don't find any nicer form. I accept your answer, but you have a mistake in the term B. Thanks. –  Medical physicist Nov 29 '12 at 11:17
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