Why π is in the normal distribution (beyond integral tricks)
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it when someone asks for the intuition behind the Gaussian density's form, its 1/2 factor and normalizing constant, and the rotational symmetry of the 2D Gaussian.
Recommended by 4 people across 4 different months, 2023 to 2025.
What it is
A 3Blue1Brown explainer on why the constant pi appears in the normal distribution. It walks through the classic polar-coordinates proof of the Gaussian integral, then gives the Herschel-Maxwell derivation based on rotational symmetry and independence, with a bonus problem at the end.
Cited in 1 Wikipedia article, including Maxwell's theorem.
What people said
Recommends tyogi · 2023-04 · Mathematics Stack Exchange · on “Intuition on why the density function of a normal law is$\frac{1}{\sqrt{2\pi \sigma^2 }}e^{-\frac{(x-\mu)^2}{2\sigma^2 }}$”
Although this post has been around for a while, let's have another go. This answer is organized as follows: Motivation as to why I focus on the standard Normal; Illustration of the problem using dart coordinates on a dartboard; Key assumptions which are intuitively desirable properties for the density; Solution to the functional equation which must be satisfied; Non-formal justification for the normalizing constants; Conclusion (and TLDR). Let me start by disclosing that all of this can be understood thanks to a wonderful video here [LINK] , from which I learned much of what I am about to write. (The video itself follows the work of John Herschel in 1850.) That said, the video is a bit long,
Recommends Carlos Pinzón · 2023-07 · Mathematics Stack Exchange · on “Pdf of the radius of a 2D Gaussian for a fixed angle”
The 2D Gaussian distribution is known for its rotational symmetry (as shown in this 3blue1brown video [this video] ), and it's also known that the radius $R = \sqrt{X^2 + Y^2}$ in polar coordinates follows a Rayleigh distribution [LINK] . Let me try to put these valid facts against each other towards a contradiction: Consider $p(R\mid\theta)$ for a given $\theta\in(-\pi,\pi)$ . From the rotational symmetry, we know that if we rotate the bell, we still get the normal distribution, so the radius, which is constrained to be positive, must follow a half-normal distribution [LINK] . For instance, if $\theta=0$ , we would be considering the distribution of the radius $R$ constrained to the positive x a
Recommends Mathemagician314 · 2024-11 · Mathematics Stack Exchange · on “Why, intrinsically, do the factorials underlie the perfect powers?”
It is a known fact that if you repeatedly take the forward finite difference [LINK] of a series of $n$ th powers $n$ times, you obtain a sequence whose only elements are $n!$ ( $n$ factorial). For example, the series of cubes: \begin{align*} &0, 1, 8, 27, 64, 125, 216, 343...\\ &1, 7, 19, 37, 61, 91, 127...\\ &6, 12, 18, 24, 30, 36...\\ &6, 6, 6, 6, 6... \end{align*} I recently rediscovered this for myself, and my searches to see what was known about it led me to this stack exchange question [LINK] (where there are many more examples and demonstrations of this fact). That question has two answers, both of which give entirely satisfying rigorous proofs of why this pattern exists. However, I s
Recommends φπατηαρ · 2025-02 · Mathematics Stack Exchange · on “Why does the Gaussian function has an additional factor of 1/2 in the exponent?”
If the Standard Normal PDF of a random variable Z is defined as $f(z)=\frac{1}{\sqrt{\pi}}e^{-z^2}$ then $$Var(Z)= E[Z^2],\quad since\,E[Z]=0$$ $$=\int_{-\infty}^{\infty}z ^2 \frac{1}{\sqrt{\pi}}e^{-z^2} dz $$ $$=\frac{1}{2}.$$ So, to have $\sigma^2=1$ , we have to define the PDF as $f(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}$ . For the intuition behind the PDF, I highly recommend this video [this video] .
Cites John · 2023-04 · Mathematics Stack Exchange · on “The proof of a function being factorable into the product of functions of two of it variables given that they are independent of each other.”
In this [this video] video at 14:32, Grant says that if a there are two variables in a function and these functions are independent then the function can be factored into the product of functions of the individual variables. What's the proof of this? For eg: $$ f(x, y) = \sqrt{x^2 + y^2} $$ is a function where x and y are independent of each other. But is this factorable into the product $g(x)h(y)$ ? What's the general proof of this?