# Stable distribution on sphere

1. Apr 20, 2012

### IttyBittyBit

The normal distribution has the property that if X and Y are i.i.d. and are both normally distributed, then X + Y will also be normally distributed.

My question is about directional statistics. Consider probability distributions on the unit circle, where all points can be parametrized by angle. The wrapped normal distribution also has the property that if X and Y are angles that are i.i.d., their sum will be distributed according to the wrapped normal distribution. Note that the von Mises distribution does not have this property, but since the von Mises distribution is fairly close in shape to the wrapped normal distribution (and is often used in place of it), some confuse the two.

Here is my question: what is the extension of this idea to the sphere? Is it possible to come up with a distribution P, and a parametrization of points on the sphere, in such a way that if X and Y are the parametrizations of two random variables that are i.i.d. and distributed according to P, X+Y will also be distributed according to P?

2. Apr 20, 2012

### chiro

Hey IttyBittyBit.

For the surface of a sphere you have the identity x^2 + y^2 + z^2 = r^2 where r > 0.

Now you can parameterize a sphere in two dimensions using some parameterization. The intuitive reason for this is that r is constant and if you have any pair including (x,y), (x,z), or (y,z) then you can find the other parameter. The only thing is however that a proper parameterization will usually be in trig form and you supply an angle to generate (x,y,z) instead of doing the implicit form where you will usually get two solutions.

According to: http://www.math.hmc.edu/~gu/curves_and_surfaces/surfaces/sphere.html, the parameterization of a sphere is:

x[u,v] = a cos cos[v]
y[u,v] = a cos[v] sin
z[u,v] = a sin[v]

where a is the radius.

Now u and v are completely independent and for a sphere, they would be considered uniform if you want the density of points to be equal across all values of the angles, but there is no reason why you can't use a different distribution, but it would make sense to make sure your values are in the range [0,2π) even though the trig functions are periodic meaning that you could use all of R, but this will just complicate analysis.

Now in terms of general distributions, you are correct that not all distributions have the property that if X, Y is i.i.d P distribution then X + Y <> P for at least some P.

The easiest one is the uniform distribution, which would be a good assumption to use for a normal sphere. U1 + U2 <> a Uniform distribution.

There are two ways that are often used to check this: the first way is through Moment Generation Functions and the Second is through the convolution theorem for finding the sum of independent random variables (don't need to be identical).

If M1 = MGF(X), M2 = MGF(Y), then MGF(X+Y) = M1M2. If MGF(X+Y) is not in form for MGF of X and Y (in terms of its structure: the parameters will of course be different), then you have shown that the distribution is not the same form and you're done.

You can also do a convolution between X and Y and if the PDF isn't in the same form, then you're done here as well.

If you want to do this for yourself, Take f(x) = 1/(2π) for [0,2π) and calculate the convolution of X and Y and you'll see that you get a triangle distribution which is clearly not a uniform distribution.

3. Apr 20, 2012

### IttyBittyBit

Thanks for the reply. You got the general gist of my question, but i'm specifically looking for a non-uniform distribution with this property. I.e. one whose mean is not the center of the sphere.

I'm aware of the methods using moment-generating functions and the convolution theorem. However, you first need to have a candidate function to apply those methods to. My problem right now is that I have no candidate function with the required properties. I have checked the Kent and von Mises-Fisher distribution and neither have this property.

4. Apr 20, 2012

### chiro

What you could do is start with some function that has roots at 0 and 2pi and also that it's area under the curve is 1 (i.e. a valid PDF).

Then using MGF's you can show the conditions of a PDF of polynomial of such degree when multiplied give the same form.

You don't need to make the distribution symmetric if you don't want to: a quadratic function will be symmetric, but higher order polynomials don't necessarily have to be symmetric which means that you skew the mean in any direction if you want this.

5. Apr 20, 2012

### IttyBittyBit

Alright, assuming I did this, how would I extend a function on the real interval [0,2pi] to a function on the sphere, maintaining the property of being stable?

6. Apr 20, 2012

### chiro

Do you mean in terms of X,Y,Z? What specific random variable do you want to calculate if it is in terms of X,Y,Z where these are the co-ordinates in R^3?

7. Apr 20, 2012

### IttyBittyBit

The specific parametrization is not important.

Perhaps my question would become clearer if phrased in terms of 3-d rotations instead of points on the sphere, and the CLT instead of stability.

Let's say you have an arbitrary probability distribution on 3d rotations (i.e. rotations in the familiar 3d space). Let's say you pick a large number of random variables from this distribution X1, X2, X3, etc. By the notation X1+X2 I mean "rotate by X2 then rotate by X1". What would be the probability distribution of X1+X2+X3+... ?

8. Apr 20, 2012

### chiro

Given this information, I think you could go about it in a few ways.

You could think about the rotation in matrix form and form the distribution that way. So in other words instead of thinking in terms of the angles, you think in terms of the actual rotation involving the sin and cos functions of the angles themselves.

If you have the matrix form of a rotation then you can just treat the rotations as matrix multiplication and you can then derive properties of your distribution for the sums of random variables.

So with this what you can do is start with two rotations and expand the matrix definition of these rotations which will give you a messy representation, Then try and link that back to the original matrix definition for one rotation by looking at what variables you can use as a function of the angles u,v (as expressed above) and once you get a transform that is the same for one matrix as for the composition of two matrices that gives the same PDF, then you're done. This will help you solve the general problem and by solving this you will know the exact constraints required that the distribution has to lie in.

A general rotation on a 3D vector is given by the angle-axis rotation matrix which is 3x3 and it is pretty messy. You could compose x,y,z (in terms of axis) rotations but you risk what is known as gimbal lock so I would recommend axis-angle representation instead.

So then you have one matrix corresponding to one rotation (call it M1) and another matrix corresponding to two rotations (call it M2) and then you have to figure out a transformation of u and v for the appropriate cells that have the same structure. I am just re-emphasizing what I've said above here for clarification.

Once you have done this, you have solved the general problem because if you have found the distribution that is the same for X vs Y + Z, then it's a simple induction argument for any number of rotation compositions.

I would look at the trig functions and look at the trig-identities to get M1 and M2 in the same 'form' as a start. Once you get these in the same 'form' then look at the structure of the arguments in this form and that will tell you what your distribution for the angles u and b should be.

9. Apr 21, 2012

### IttyBittyBit

The matrix method is easy to do for 2d rotations, but quickly falls apart for 3d ones because you can't separate the angles from each other. At any rate, I just want to know if it is a known distribution and if work has been done on it. Given that it's such a simple question to pose, I'm sure people have considered it before but I couldn't find the answer anywhere.

Anyway, thanks for the help, if you don't know the answer to this question that's fine.

10. Apr 21, 2012

### chiro

I've never worked on the problem myself and I don't know the answer, but I'm just outlining what I would do personally to tackle the problem.

I wish you all the best in solving it.

11. Apr 21, 2012

### chiro

Also just before I forget, you could also look at the quaternion expressions of rotations which are a hell of a lot easier to work with. They are a lot simpler (I think if I recall correctly) so that might be a good avenue to look into.

12. Apr 21, 2012