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I don't know if such question could be posed, anyway I'll try.

Let's consider the Gaussian distribution, and in particular the exponential factor (in its simplest form), $e^{-x^2}$. Its derivation, its importance via the Central limit theorem is perfectly sound to me, but conceptually I could not associate its specific form to its widespread importance: i.e. why the exponential decrease in probability density is quadratic, and not, say, $e^{-|x|}$? Why so much natural phenomena seem to obey to quadratic exponential decrease?

PS: If it could help, I've found this discussion on meta.math. I liked, in Terry Tao's answer, the link between the exponentiated version of a quadratic form and the Taylor expansion concept.

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  • $\begingroup$ One reason could be that $|x|$ is hard to deal with, and its not differentiable too at $x=0$ $\endgroup$ – Oswald Nov 21 '16 at 3:43
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    $\begingroup$ @TheGhostOfPerdition The normal distribution is something humans discovered, and its formula is something we deduced, not decided on out of whim. $\endgroup$ – arctic tern Nov 21 '16 at 3:45
  • $\begingroup$ I'm not sure I 100% agree arctic. While the normal distribution we have is good, we've decided it's form, and we've decided it's importance. Another culture might not have used "e", another culture might have constructed it with sinusoids, or presented it with a series solution. They could have produced it a the derivative of it's cdf. The way we've presented it shows off our willingness to like $e^{-x^2}$, and the answer to "why do we think about stats the way we do?" could be considered a soft question. $\endgroup$ – Kaynex Nov 21 '16 at 3:52
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    $\begingroup$ Well, the central limit theorem says that if you started out with sinusoids, you'd still end up with the normal distribution pretty quickly. So I'd say that "we decided its importance" (without the apostrophe) is probably wrong. We decided that probability and statistics were important. Our interest in the normal distribution was a consequence of that decision. $\endgroup$ – John Hughes Nov 21 '16 at 3:56
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I'm not sure if this actually helps, but maybe it would be easier to grasp intuitively going back to the simplest case, that of a fair coin? In particular, let's say we already know the peak probability that in $n$ flips we get exactly $\frac{n}{2}$ heads, we're only looking for its ratio to some $\frac{n}{2} \pm l$, which would be:

$$\frac{\frac{n}{2}}{\frac{n}{2}+1}\cdot\frac{\frac{n}{2}-1}{\frac{n}{2}+2}\cdot\cdots\cdot\frac{\frac{n}{2}-l+1}{\frac{n}{2}+l}$$

Now let's say that $l$ is much larger than one, and much less than $n$, in which case you can approximate the logarithm of this ratio as:

$$\int_0^l\left(\log\left(\frac{n}{2}-t\right) - \log\left(\frac{n}{2}+t\right)\right)dt = \int_0^l\left(\log\left(1-\frac{2t}{n}\right) - \log\left(1+\frac{2t}{n}\right)\right)dt$$ $$\approx-\int_0^l\frac{4t}{n}dt = -\frac{2l^2}{n}$$

All cases can ultimately be reduced to weighted iterations of this one, so it always ends up being $e^{-x^2}$.

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  • $\begingroup$ Well, your interesting answer didn't helped directly, but surely put me on the right path to find an answer my own! As soon as I'll elaborate it rigorously, I'll write it here. $\endgroup$ – Lo Scrondo Nov 22 '16 at 2:59

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