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Why is the standard deviation $\displaystyle\sigma$ defined in such a way that in the exponent of the normal distribution,

$\displaystyle f{{\left({x}\right)}}=\frac{1}{{\sigma\sqrt{{{2}\pi}}}}{e}^{{-{\left(\frac{{{x}-\mu}}{{\sigma\sqrt{{{2}}}}}\right)}^{2}}}$

$\displaystyle \sigma$ needs to be scaled up by an additional factor of $\displaystyle\sqrt{{{2}}}$?

Because intuitively, I would define the normal distribution like this, namely simply as the normalized Gaussian integral:

$\displaystyle {\int_{{-\infty}}^{{+\infty}}}{e}^{{-{x}^{2}}}{\left.{d}{x}\right.}=\sqrt{{\pi}}\quad\Rightarrow\quad\displaystyle\ f{{\left({x}\right)}}:\:=\frac{1}{\sqrt{{\pi}}}{e}^{{-{x}^{2}}}$

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    $\begingroup$ Standard deviation makes sense for other distributions besides normal ones. Even finite data sets have a standard deviation. Variance is just the average squared distance to the mean...then standard deviation just takes the square root to get the units right. $\endgroup$
    – lulu
    Aug 29, 2020 at 17:27
  • $\begingroup$ @lulu: Yes, I've learned that the standard deviation is basically the RMS (or rather, the square root of the variance) and - for physicists - the square root takes care of the units. But why the factor of $\displaystyle\sqrt{{{2}}}$? $\endgroup$
    – david
    Aug 29, 2020 at 17:44
  • $\begingroup$ The square root of $2$ is an artifact of the form of the normal. You could certainly write the normal without it, but then you would need to supply the factor of $\sqrt 2$ when you tried to read off the standard deviation. Think how much worse it would be if, instead of the usual variance, statistics had built up around the average absolute difference! In many ways, that's the more natural measure (why sum the squares and take the square root?)...but, analytically, it is a lot less friendly. $\endgroup$
    – lulu
    Aug 29, 2020 at 17:51
  • $\begingroup$ To stress: the Central Limit Theorem makes the normal distribution "natural" or even unavoidable, but a priori I don't know why one should have expected that this "limiting" distribution would be conveniently expressed in terms of the standard statistical measures. A few stray constants here and there is a small price. $\endgroup$
    – lulu
    Aug 29, 2020 at 17:53
  • $\begingroup$ The thing is that $\displaystyle \sigma$ is a general expression, so there is no specific scaling inherent to it and this factor of $\displaystyle\sqrt{{{2}}}$ must somehow originate from the normal distribution itself. There is also this "68-95-99.7 rule" and I always thought that these numbers are rather arbitrary - i.e. one could have also just scaled the $\displaystyle{1}\sigma$, $\displaystyle{2}\sigma$, $\displaystyle{3}\sigma$ bands differently (e.g. by said factor of $\displaystyle\sqrt{{{2}}}$) $\endgroup$
    – david
    Aug 29, 2020 at 18:26

2 Answers 2

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The form $\frac{x^2}{2}$ is rather common in mathematics, often arising from the fact that it's the integral of $x$. For example the formula for kinetic energy is $m\frac{v^2}{2}$, the distance fallen in time $t$ is $g\frac{t^2}{2}$, and the Taylor expansion of $\exp(x)$ is $1+x+\frac{x^2}{2}\dots$. So we shouldn't be afraid when we see

$$\frac{1}{\sqrt{2\pi}\sigma}\exp\left(\frac{\left(\frac{x-\mu}{\sigma}\right)^2}{2}\right).$$

The expression $\frac{x-\mu}{\sigma}$ is x after being 'normalised' by subtracting off the mean and scaling by the standard deviation, and then squaring and dividing by two is a very standard thing to do.

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  • $\begingroup$ Exactly. The pattern $\frac12 z^2$ is very common in math. If you integrate $z$ you get $\frac12 z^2$. $\endgroup$
    – littleO
    Aug 29, 2020 at 18:57
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Because the normal distribution is a probability density function. Therefore the integral from $-\infty$ to $+\infty$ has to equal one.

$$\displaystyle {\int_{{-\infty}}^{{+\infty}}}\frac{1}{{\sigma\sqrt{{{2}\pi}}}}{e}^{{-{\left(\frac{{{x}-\mu}}{{\sigma\sqrt{{{2}}}}}\right)}^{2}}} = 1 $$

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  • $\begingroup$ Yes, but normalization to 1 is already guaranteed by the general form $\displaystyle\frac{1}{{\alpha\sqrt{{\pi}}}}{e}^{{-{\left(\frac{x}{\alpha}\right)}^{2}}}$ :) The question here is, why the $\displaystyle\sqrt{{{2}}}$...maybe Daniel Fischer's comment gives an explanation for this (gotta check this) $\endgroup$
    – david
    Aug 29, 2020 at 19:33
  • $\begingroup$ @david If you want the standard deviation to be $\sigma$, then you are forced to choose $\alpha = \sigma \sqrt{2}$. $\endgroup$
    – littleO
    Aug 29, 2020 at 20:40
  • $\begingroup$ Yes, but why? I'm not looking for the definition of the normal distribution (which I posted in the question), but for an explanation why there is this factor of $\displaystyle\sqrt{{{2}}}$ (and not e.g. $\displaystyle\sqrt{{{3}}}$ or $\displaystyle\sqrt{{\pi}}$ or some other constant) $\endgroup$
    – david
    Aug 30, 2020 at 7:00

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