What is the reason behind the seemingly arbitrary factor of $\displaystyle\frac{1}{{2}}$ in the exponent of the Gaussian function?

Sometimes, as in this Wikipedia article on the Gaussian integral, the factor $\displaystyle\frac{1}{{2}}$ is omitted and it's not obvious why one should include it at all. It's just another factor one has to take care of (i.e. a possible source of error when doing calculations by hand).

As far as I can tell, it just scales the width parameter (e.g. the standard deviation $\displaystyle\sigma$ or the full width at half-maximum) by a "strange" factor of $\displaystyle\sqrt{{{2}}}$ and seems to serve no real purpose.

  • 2
    $\begingroup$ It appears naturally if you want to write the pdf of the normal distribution in terms of the standard deviation $\sigma$: en.wikipedia.org/wiki/Normal_distribution . Perhaps it would help psychologically to think of $- \frac{1}{2} x^2$ as the integral of $-x$. Similar factors of $\frac{1}{2}$ show in e.g. the definition of kinetic energy in physics. $\endgroup$ – Qiaochu Yuan Aug 28 at 6:36
  • $\begingroup$ Ultimately this is a convention though. You can see the difference between the two conventions in this discussion of the probabilists' vs. the physicists' Hermite polynomials: en.wikipedia.org/wiki/Hermite_polynomials#Definition $\endgroup$ – Qiaochu Yuan Aug 28 at 6:54
  • $\begingroup$ Hi, could you briefly elaborate on the idea of explaining the factor with the integration of $\displaystyle-{x}$? Also, how is it natural to appear when using the standard deviation and how does it connect to the integral of $\displaystyle-{x}$? I understand that the general Gaussian integral is finite and normalized to 1 in the case of the normal distribution, which explains the normalization constants $\endgroup$ – david Aug 28 at 11:15
  • $\begingroup$ ...differently put: Why don't we just absorb $\displaystyle\frac{1}{{2}}$ into $\displaystyle\sigma$? $\endgroup$ – david Aug 28 at 11:22
  • $\begingroup$ You mean into the standard deviation? I mean, you could, but then you'd introduce a factor of $2$ into the statement of theorems like Chebyshev's inequality: en.wikipedia.org/wiki/Chebyshev%27s_inequality $\endgroup$ – Qiaochu Yuan Aug 28 at 11:48

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