Is my conjecture correct that one standard deviation lies on the inflection points of the normal distribution curve (of the probability density function)? How can it be proved using the standard deviation formula?
Prove one standard deviation lies on inflection points
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0The question seems to be about the inflection points of the normal probability density function; the inflection point of the normal cumulative distribution function is at the mean. – 2012-09-02
1 Answers
We will find the inflection points of the density function of a general one-variable normal. This can be done in the usual calculus way, by examining the second derivative of the density function.
The density function $f(x)$ of the general one variable normal is given by $f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp((x-\mu)^2/2\sigma^2).$ Differentiate. We get a constant times $(x-\mu)\exp((x-\mu)^2/2\sigma^2).$ Differentiate the above expression. We get $\exp((x-\mu)^2/2\sigma^2)-\frac{(x-\mu)^2}{\sigma^2}\exp((x-\mu)^2/2\sigma^2).$ The second derivative is $0$ when $(x-\mu)^2=\sigma^2$, and changes sign there. That is the only place this happens, since $\exp(t)$ is never $0$. So the only inflection points are at $x=\mu\pm\sigma$, one "standard deviation unit" from the mean.
Remark: I do not know what you mean by the standard deviation formula. If it is $E(X-\mu)^2$, then the answer would be no, since there are plenty of density functions that do not even have an inflection point, and plenty that do but whose inflection point(s) are not $1$ standard deviation unit from the mean. So some special property of the normal needs to be used.
If you note that the density function of the general normal is just obtained by shifting and scaling the standard normal, it is enough to work with the simpler density function $\frac{1}{\sqrt{2\pi}}\exp(-t^2/2),$ and show that its inflection points are at $\pm 1$. The calculation is very similar to the one above, but looks simpler.
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0It is not a coincidence, it is a property of what you are taking the second derivative of, the Gaussian/normal probability density function. If you take the second derivative of a PDF of a different mathematical form (and equate the second derivative to zero as demonstrated by Andre Nicolas), you will nearly always find inflection points elsewhere (not one standard deviation away from the mean). The fact that we see this for the Gaussian function is one of many convenient properties of the normal distribution. – 2018-03-27