Sweet spots of the bell curve

Continuing with the theme of "basic stuff I never learned," here is something interesting about the normal distribution. It turns out that the inflection points of the bell curve are the points where x = σ. I think that's pretty amazing. Let's see if we can prove it using a small bit o'calculus.

As we come over the top of the curve and head down, the slope is becoming increasingly negative. But at some point the slope reaches its maximum negative value and then starts to turn less negative (more positive). At one instance the slope of the slope or second derivative of the pdf is zero. So we need to differentiate the normal density function twice, set it equal to zero, and then solve.

I have to admit I got too confused in the middle of the calculation, so I needed help. I googled 'second derivative normal distribution' and found this.



The pdf for the normal distribution is equation (1). We're going to differentiate twice and set that result equal to zero, so the constant out front can be ignored. We rewrite the pdf as equation (2). To simplify the notation, we will define f(x) as in equation (3) and then we can rewrite equation (2) as equation (4).

We will use the result in equation (5) several times. This is just a generalization of what I mentioned the other day with respect to the exponential distribution and its cdf.

We consider the exponent part as f(x) in (3) and (4). We use the chain rule to find its derivative. Set x - μ / σ = g(x) and then do: df/dx = df/dg * dg/dx and we obtain equation (6).

We use the results from (5) and (6) in figuring out the derivative of φ(x) as shown in equation (7). As mentioned, the derivative of exp { f(x) } is just f '(x) exp { f(x) }. We calculated f'(x) in (6) and we put the results together in (7). Now we have something substantially more complicated but it is really just the product of two functions each of which we know how to differentiate.

Remember that the derivative of g(x) f(x) is g'(x) f(x) + g(x) f '(x). "This times the derivative of that plus that times the derivative of this."



In the second panel we restate (7). To do the second differentiation, we first pull the constant out front and do this (x-μ / σ) times the derivative of that (7), plus that (exp { f(x) }) times the derivative of this (1/sigma;). We pull out the common factors and obtain (9).

We set this equal to zero, and now it is easy to see that the solutions are as shown in (11). Pretty neat!

And if anybody can tell me how to make my beautiful images not look like crap in blogger I would appreciate it. I suppose I need to RTFM.