TITLE: Why ignoring the function $f(x)=1/(1+x^2)$?
QUESTION [9 upvotes]: I heard that the function $$f(x)=e^{-x^2}$$
Is extremely important in probability and statistics, because it looks like the normal distribution (or something like that). But i noticed that the graph of this function is similar to the function $$g(x)=\frac{1}{1+x^2}$$
Furthermore, the area under this curve is pretty easy to calculate, since
$$\int_{0}^z \frac{\,dx}{1+x^2}=\arctan(z).$$
So why the first function is more important than the second one, although they look pretty similar?

REPLY [13 votes]: They are not very similar.
First let us note that $\displaystyle\int_{-\infty}^{+\infty} e^{-x^2/2} \, dx = \sqrt{2\pi}$ and $\displaystyle\int_{-\infty}^{+\infty} \frac{dx}{1+x^2} = \pi. $
But $\displaystyle\int_0^{+\infty} x^n \cdot e^{-x^2/2} \, dx<+\infty,$ no matter how big $n>0$ is, whereas $\displaystyle\int_0^{+\infty} x\cdot\frac{dx}{1+x^2} = +\infty,$ and a fortiori $\displaystyle \int_0^{+\infty} x^2 \cdot\frac{dx}{1+x^2} = \infty$
The reason for dividing by $2,$ so that $x^2/2$ appears in the exponent rather than just $x^2,$ is that that makes the variance, and hence the standard deviation, equal to $1.$
A large number of theorems characterize the standard normal distribution $\dfrac 1 {\sqrt{2\pi}} e^{-z^2/2}\,dz. $ Here are two of them:

Suppose $X_1,X_2,X_3,\ldots$ are independent and identically distributed random variables with expected value $\mu$ and variance $\sigma^2<+\infty.$ Then the distribution of $$ \frac{(X_1+\cdots+X_n)/n-\mu}{\sigma/\sqrt n} $$ approaches the standard normal distribution as $n\to\infty.$ In particular, the standard deviation of this sample mean is $\sigma/\sqrt n$ rather than $\sigma.$ Thus the dispersion of the sample mean goes to $0$ as the sample size grows.By contrast, if $X_1,X_2,X_3,\ldots$ are independent and are distributed as $\dfrac{dx}{\pi(1+x^2)},$ the Cauchy distribution then $(X_1+\cdots+X_n)/n$ also has that same Cauchy distribution. Its dispersion does not get smaller.

Suppose the vector $(X_1,\ldots,X_n)$ has a probability density that depends on the $n$ arguments only through the sum of their squares, i.e. the distribution is spherically symmetric and centered at the origin. Suppose further that these $n$ scalar components are mutually independent. Then each of these random variables separately is normally distributed.