# Uniform points on a circle

1. Jan 27, 2016

### 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.

2. Jan 27, 2016

### Svein

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

3. Jan 27, 2016

### 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?

4. Jan 27, 2016

### Svein

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.

5. Jan 27, 2016

### 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.

6. Jan 27, 2016

### 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.

Last edited: Jan 27, 2016
7. Jan 27, 2016

### mathman

8. Jan 27, 2016

### rabbed

9. Jan 28, 2016

### 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?

10. Jan 28, 2016

### mathman

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

11. Jan 29, 2016

### 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.

12. Jan 29, 2016

### 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.

13. Jan 30, 2016

### 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..

Last edited: Jan 30, 2016