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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?

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    $\begingroup$ There’s a vast chasm separating “looks like” and “is equal to” or “is similar enough to have the same properties as.” Your question is a little like asking why $e$ is so important if we could just round it to $3$. The reason that $e^{-x^2}$ is important isn’t that it “looks like” a bell curve. It’s that it has exactly the right properties with respect to certain mathematical structures (for example, the Fourier transform). $\endgroup$ May 12, 2021 at 22:01
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    $\begingroup$ The central limit theorem is the reason. Normal distribution appears everywhere. Your $g(x)$ is also useful though, it's basically the Cauchy distribution. $\endgroup$ May 12, 2021 at 22:02
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    $\begingroup$ Also those distributions are rather different - the Cauchy distribution has much thicker tails. $\endgroup$ May 12, 2021 at 22:34
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    $\begingroup$ @Jair Taylor Even worse, this distribution has no second order moment ($E(X^2)$ doesn't exist) $\endgroup$
    – Jean Marie
    May 14, 2021 at 22:38

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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.

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  • $\begingroup$ i wonder...is saying that they are similar similar to saying hyperbolas and parabolas are similar? $\endgroup$
    – BCLC
    May 13, 2021 at 3:45
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I don't think this function is being ignored. It was studied by an Italian mathematician Agnesi, and before that Fermat, Newton and Grandi. It has been called the "Witch of Agnesi". With no apparent connection to the circle, the area under it is $\pi$. Try searching. Apparently it is also useful in probability and statistics.

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That function is also important in statistics. It is (a scaled version) of the $t$-distribution with one degree of freedom. See this Wikipedia article.

As the number of degrees of freedom approaches $\infty$, the $t$-distribution approaches the Normal distribution, which is a scaled version of $e^{-x^2}$. So the two functions are at different ends of a certain spectrum of functions.

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    $\begingroup$ Actually, the Cauchy distribution $\dfrac{dx}{\pi(1+x^2)}$ is the t-distribution with one degree of freedom. That parameter—the degrees of freedom—is not a scale parameter. You cannot get from just one degree of freedom to more than that by rescaling. $\endgroup$ May 13, 2021 at 2:03
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    $\begingroup$ @MichaelHardy I wrote "a scaled version", meaning $\frac{1}{1+x^2}$ (OP's function) gets scaled by $\frac{1}{\pi}$. And then with $e^{-x^2}$, I am referring to a simultaneous vertical scaling and horizontal compression to get the Gaussian bell curve. $\endgroup$
    – 2'5 9'2
    May 13, 2021 at 2:24

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