# Darts and the Gaussian - Placing the bivariate in front of the univariate. Just an integration trick?

I have come across a proof of the Gaussian pdf that starts like a story - throwing darts on a Cartesian-ized target with $x$ and $y.$ The errors (it's all about the errors) will be independent in both directions and rotationally invariant.

From there, what follows is predictable - a change to polar coordinates, and integration over the circle, and in comes $\pi.$

Yes, this question spawns from this post on pi and the exponential that happened to be active today.

So in this dart set up we really start off from an uncorrelated bivariate normal to arrive at the pdf of the univariate normal.

Needless to say, the darts, the circle and pi provide with a wonderful intuition, but the question (understanding that there are more than one derivation of the Gaussian) is:

Why do we need to resort to the bivariate normal as a set up to arrive at the normal unvariate Gaussian pdf?

Is it purely practical? If so which step in the derivation necessitates this set up?

It is clear that the sooner we get to Pythagoras the smoother the path will be... Which reminds me of a recent tweet by @fermatslibrary that just starts point blank with the square of the integral:

• Your question is a little rambling. Are you asking (i) if it's possible to obtain the form of the Gaussian PDF without considering a bivariate Gaussian, (ii) if it's possible to obtain specifically the normalization constant $\sqrt\pi$, given the form $e^{-x^2}$, or (iii) something else?
– user856
Feb 27, 2018 at 4:16

And it doesn't stop in one dimension, either. Independent vectors from multivariate normal distributions also obey a generalized central limit theorem. This motivates not only spherically-symmetric Gaussian distributions like the derivation you reference, but also ones that are asymmetric and even have correlation between the variables. Cause the sum of $n$ independent and identically distributed vectors whose mean vector and covariance matrix exist tend in distribution to a multivariate normal.