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Why does the variance in the Normal density have a $2$ in it? I can make sense of the rest of the function, but I do not understand what the $2$ adds to the equation or why is it there. Wouldn't the version on the left give a proper density, with about the same properties as the normal density on the right?

$$ \underbrace{\int_{-\infty}^{\infty} \frac{1}{\sqrt{\pi\sigma^2}}\exp{\frac{-(x-\mu)^2}{\sigma^2}}}_{\text{(without 2)}}\qquad\qquad\qquad \underbrace{\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi\sigma^2}}\exp{\frac{-(x-\mu)^2}{2\sigma^2}}}_{\text{(standard version)}}$$

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  • $\begingroup$ I believe he's referring to the the 2's present in the normal p.d.f., i.e. under the sqrt and in the denominator of the exponentiated term. $\endgroup$
    – klumbard
    Oct 12, 2017 at 19:24
  • $\begingroup$ Without the 2 there, $\sigma^2$ would be "twice-the-variance", which is an awkward name, isn't it? $\endgroup$
    – Xi'an
    Oct 12, 2017 at 19:57
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    $\begingroup$ @Xi'an, well that is supposing you want to conserve the normal pdf by adjusting sigma. However if you suppose we do not do that adjustment and use the pdf without the multiplication by 2, we get another proper density, which although not as concentrated on the expectation, still seems like a pretty good pdf to me. I do know of the efficacity of the the normal distribution, and thus wonder of the meaning behind the 2, which leads to the normal distribution instead of the distribution I specified (without the 2 coefficients). $\endgroup$
    – Gian Soldo
    Oct 12, 2017 at 21:10
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    $\begingroup$ This would just be changing the parameterisation and why do that? I think this is @Xian's main point. Note that as Dynamic Stardust also spells out $\sigma$ has a very simple and natural interpretation for the normal as the distance between the mean and each inflection of the density function. Parameterisations are just conventions but it's still true that some conventions are better than others and that an established convention should not be fixed if it's not broken. $\endgroup$
    – Nick Cox
    Oct 13, 2017 at 11:18
  • $\begingroup$ @NIckCox I completely agree with you on the pointlessness of performing such change, I was trying to understand the origins or meaning of the standard normal parameterization and not propose a change. Luckily Stardust's answer explains it pretty clearly. $\endgroup$
    – Gian Soldo
    Oct 15, 2017 at 0:06

1 Answer 1

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You can find a good derivation of the univariate normal function using calculus in this document $[1]$. (I did try to reproduce the essential excerpt of the derivation here, but its still too lengthy, so you're better off having a look at the linked pdf file itself.)

Keep a lookout for the part, in the linked document, where there is an integral of x, which we know is: $\int x dx = \frac{x^{2}}{\mathbf{2}} + c$. That is where the 2 'enters the picture' in the final form of the univariate normal distribution.

One fairly good reason not to substitute for the standard deviation with $\sigma\prime = \sqrt{2}\sigma$, is that the points of inflection of normal distribution are exactly 1 standard deviation ($\sigma$) from mean ($\mu$) on both sides. (Point of inflection is where the second order derivative is zero). I'd rather have the inflection point at $\sigma$ and keep the 2 as 'variance coefficient'.

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The mention of 'coefficient of parameter' got me confused and curious. Have a look at coefficient and parameter.

inflection of gaussian

$[1]$ www.planetmathematics.com

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    $\begingroup$ It's OK not to quote the full document, if it's too much; you have provided a good summary of what the idea is. But I wonder if you could provide at least a full citation in case the link goes dead in the future. $\endgroup$
    – gung - Reinstate Monica
    Oct 13, 2017 at 0:54

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