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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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    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

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\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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    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