Can Uniform Points on a Circle be Derived from the Standard Normal Distribution?

[rabbed]

Is it possible to derive the standard normal distribution from using polar form p = r*e^(i*v) to distribute uniform points along the contour of a circle?

I've read that it is possible to randomize points like that using X and Y values with normal distribution by normalizing each point, but I haven't found a good reference which explains this in simple terms.

 

[Svein]

Yes. You just set Θ=Random([0,2Π>). That way you get a random distribution of angles = points on the circle.

 

[rabbed]

Yes, or in my case - v.
But I would like to start with the polar exponential form and end up with the standard normal distribution for the x coordinate for example.
It should be possible, right?

 

[Svein]

But I would like to start with the polar exponential form and end up with the standard normal distribution for the x coordinate for example.
It should be possible, right?

I am not quite sure what you mean. But if you start with a rectangular distribution of points on a circle, the projections on the x-axis is not going to have a rectangular distribution.

 

[rabbed]

For example, the top answer on this page explains a method to generate points on the surface of a sphere in any dimension, which in 2D should be the contour of a circle.
http://stats.stackexchange.com/ques...ed-points-on-the-surface-of-the-3-d-unit-sphe
In the comments there are also hints at a proof using matrices, but if the method is just used to distribute uniform points on a circle, it should be possible to show this using polar exponential form also? I figured the polar exponential form shouldn't be too far away from the standard normal distribution since they are already pretty similar.

 

[rabbed]

Hm, maybe the surface of a sphere in 2D is the area of the circle. In that case I would like to start with p = r*sqrt(u)*e^(i*2*pi*v)
U_PDF(u) = 1 (0 < u < 1)
V_PDF(v) = 1 (0 < v < 1)
The end result should be
X = some function
Y = some other function
where
X_PDF(x) = 1/sqrt(2*pi) * e^(-0.5*x^2)
Y_PDF(y) = 1/sqrt(2*pi) * e^(-0.5*y^2)

But if it's easier to go the other way around, starting from the standard normal distributed X and Y and show how this relates to polar exponential form, that's OK too.

 

[mathman]

https://en.wikipedia.org/wiki/Box–Muller_transform

Above describes a method to get a pair of independent normal random variables from two independent random numbers.

 

[rabbed]

https://en.wikipedia.org/wiki/Box–Muller_transform

Above describes a method to get a pair of independent normal random variables from two independent random numbers.

Exactly what I was after. Thank you!

 

[rabbed]

My question then is - when (0 < u < 1) has a uniform distribution, does a radius of sqrt(-2*log(u)) also give a uniform distribution of points over the circle area just like a radius of r*sqrt(u) does? How does that work?

 

[mathman]

Both of these formulas

My question then is - when (0 < u < 1) has a uniform distribution, does a radius of sqrt(-2*log(u)) also give a uniform distribution of points over the circle area just like a radius of r*sqrt(u) does? How does that work?

are for distributions in one dimension. How do they relate to a unit circle?

 

[rabbed]

It's the radial part from the center to a point inside the circle.
The other part (which is the same for both cases) is the angular part = 2*pi*v with (0 < v < 1) coming from a uniform distribution.

Btw, I'm pretty sure that sqrt(-2*log(u)) should be changed to sqrt(-2*ln(u)) since we are dealing with powers of e.

 

[mathman]

r*sqrt(u) gives points between 0 and r, sqrt(-2ln(u)) gives points between 0 and infinity. The first gives points uniform on the unit disc, the second gives (two) normally distributed points.

 

[rabbed]

But according to the link I posted, the normal distribution method can be used to give uniform points on the unit disk also.

I just realized something, the 'basic' method gives uniform points on the unit disk by:
P = sqrt(U)*e^(i*2*pi*V)
using
U_PDF(u) = 1 (0 < u < 1)
V_PDF(v) = 1 (0 < v < 1)

while the other method must do it by:
P = sqrt(-2ln(K))*e^(i*2*pi*V)
using
K = e^(-0.5*k) (0 < k < 1)
V_PDF(v) = 1 (0 < v < 1)

..which will give the same result. Correct?

But what is K, the inverse CDF of some PDF? It should be possible to work backwards to get the normal distribution..