Inflection point of normal distribution

To be honest, I didn't know the fact that inflection point of normal distribution is one standard deviation above the mean and one standard deviation below the mean. Hence I will share with you what is in behind the scene of it here. From now on, when I write normal distribution by hand, I will pay attention so that it looks like what should be :)

0. What is inflection point in the first place ?¶

One of the feature pertains to curve we can consider is "concavity". Concavity is the direction of curvature. In a nutshell, curve is "concave down" if it's shaped like $\cap$. Curve is "concave upwards" if it's shaped like $\cup$. Then, "inflection point" is where curve change its concavity !

1. Second Derivatives¶

Next we can consider about "derivative". As you may know "derivative" can be used to determine slope of tangent line at given point. Then Second derivative capture the rate of change of slope. And what's worthy of special mention is if $y=f(x)$ has an inflection point at $x=a$, then the second derivative of f evaluated at $a$ is zero. Therefore we will utilize this property to derive inflection point of bell curve.

2. Inflectoin point of gaussian distribution¶

At first we're gonna calculate first derivative. "Chain rule" can be applied to get it.
$$\begin{eqnarray}\left\{\frac{1}{\sqrt2\pi\sigma^2}exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}\right\}^{'} &=& \frac{1}{\sqrt2\pi\sigma^2}exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}\left\{-\frac{(2x-2\mu)}{2\sigma^2}\right\}\end{eqnarray}$$
Then we can get second derivative, at this time, we can apply product rule .

$$\begin{eqnarray}\left\{\frac{1}{\sqrt2\pi\sigma^2}exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}\right\}^{''} &=& \frac{1}{\sqrt2\pi\sigma^2}exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}\left\{-\frac{(2x-2\mu)}{2\sigma^2}\right\}\left\{-\frac{(2x-2\mu)}{2\sigma^2}\right\}+\frac{1}{\sqrt2\pi\sigma^2}exp\left\{-\frac{(x-\mu)^2}{2\sigma^2}\right\}\left\{\frac{-2}{2\sigma^2}\right\}\end{eqnarray}$$

What we wanna find is the point where second derivative is zero.
$$\begin{eqnarray}\left\{\frac{(2x-2\mu)^2}{4\sigma^4}\right\} &=& \left\{\frac{2}{2\sigma^2}\right\}\\ (2x-2\mu)^2 &=& 4\sigma^2\\ 2x-2\mu &=& ^+_-2\sigma\\ \therefore \ \ x &=& \mu ^+_-2\sigma\end{eqnarray}$$
This consequence says inflection point resides one standard deviation above the mean and one standard deviation below the mean :)