TITLE: Where Gaussian Random Distribution came from?
QUESTION [7 upvotes]: I'm studying an introduction to probability, and before the explanation of the Central Limit Theorem the author presents the Gaussian Normal Distribution.
It has a complex and a non-intuitive formula, even if it's very important in the Probability field.
Was the gaussian distribution found before the deduction of the CLT ? How was this distribution discovered ? And what are the properties that makes it so special ?

REPLY [4 votes]: I want to comment on "It has a complex and a non-intuitive formula".
Let us first look at the reduced equation,
$$p(x)\propto e^{-x^2/2}.$$
This is simply the exponential of a square, nothing really difficult. The coefficient $2$ serves as a scaling factor to ensure that the standard deviation is $1$. (I won't give the technical details here as this involves improper integrals.)
Now we need to normalize the function so that the area under the curve is $1$ and the final formula is 
$$p(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}.$$
If we consider general mean $\mu$ and standard deviation $\sigma$, the formula is simply adapted by applying the linear transform $x\to(x-\mu)/\sigma$ that shifts $\mu$ to $0$ and rescales $\sigma$ to $1$. To ensure that the area remains $1$, the normalization factor needs to be multiplied by $\sigma$ to compensate. We have
$$p(x)=\frac1{\sqrt{2\pi}\sigma}e^{-(x-\mu)/2\sigma^2}.$$
If you look at the shape of the function, is very smooth, it is symmetrical, it has a single maximum and quickly decays on both sides. This behavior describes a random variable that remains "concentrated" around the mean.


To better understand the shape, you can look at the distributions of the sums of $n$ uniform random variables in the range $0,1$, for increasing $n$:

You are smoothing and smoothing the curve, which quickly converges to a Gaussian.
Now observe that with $x+y=z$, where $x,y$ follow a reduced Gaussian
$$\int_{-\infty}^\infty e^{-x^2/2}e^{-y^2/2}dx=\int_{-\infty}^\infty e^{-x^2/2}e^{-(z-x)^2/2}dx=\int_{-\infty}^\infty e^{-(x-z/2)^2/2-z^2/4}dx
\\=e^{-z^2/4}\int_{-\infty}^\infty e^{-x^2/2}dx\propto e^{-z^2/4}.$$
This proves that the distribution of the sum of two Gaussians is another Gaussian, with variance $2$. By a more complicated argument (based on the Fourier transform of convolutions), one can show the reciprocal: an additive distribution must be Gaussian.