Radially distributed Cartesian coordinates

In summary, the conversation focuses on creating a cloud of points with a spherically symmetric distribution by using a sphere formula with a uniform direction distribution and a radius function. The goal is to calculate the probability density function (PDF) of the radius variable, R, which is dependent on the three random variables X, Y, and Z. The conversation also discusses the use of the Jacobian and the Box-Muller transformation to find the PDF of R. However, there is confusion regarding the notation and equations used in a document referenced in the conversation.
  • #1
rabbed
243
3
Ok, so randomizing three random variables, X, Y and Z, each from a standard normal distribution, then plotting these in an ordinary cartesian coordinate system gets me a spherically symmetric cloud of points.

Now I want to create this cloud having the same probability distribution but by using a sphere formula having a uniform direction distribution and a radius of some function f(R) where R has some unknown distribution:
X = f(R)*sqrt(1-V^2)*sin(O)
Y = f(R)*V
Z = -f(R)*sqrt(1-V^2)*cos(O)
|J| = | d(XYZ) / d(OVR) | = f(r)^2*f'(r)
O_PDF(o) = 1/(2*pi) where (0 < o < 2*pi)
V_PDF(v) = 1/2 where (-1 < v < 1)
R_PDF(r) = ?

So I want to calculate R_PDF(r). I know that R^2 = X^2 + Y^2 + Z^2 so I use the CDF method to first
find the distribution of XS = X^2 knowing that X_PDF(x) = 1/sqrt(2*pi)*e^(-x^2/2).

This gets me XS_PDF(xs) = 1/sqrt(2*pi*xs)*e^(-xs/2) which is the same for YS_PDF and ZS_PDF

Then I want to calculate RS_PDF(rs) from RS = XS + YS + ZS, knowing XS_PDF(xs), YS_PDF(ys) and ZS_PDF(zs):

RS_PDF(rs) =
RS_CDF'(rs) =
( P(RS < rs) )' =
( P(XS + YS + ZS < rs) )' = ?

What to do here?
 
Last edited:
Physics news on Phys.org
  • #2
  • #3
Hi rabbed:

If you assume that the x, y, and z variable Gaussian distributions all have a zero mean as well as the same standard deviation, then what is the joint probability density p(x,y,z) at a point [x,y,z]? What do you get if you then convert p(x,y,z) to spherical coordinates, p(r,θ,φ)?

Hope this helps.

Regards,
Buzz
 
  • #4
Hi Buzz

XYZ_PDF(x,y,z) =
X_PDF(x) * Y_PDF(y) * Z_PDF(z) =
1/sqrt(2*pi)*e^(-x^2/2) * 1/sqrt(2*pi)*e^(-y^2/2) * 1/sqrt(2*pi)*e^(-z^2/2) =
1/(2*pi)^(3/2)*e^(-(x^2+y^2+z^2)/2)

Substituting with these:
x = r*cos(a)*sin(o)
y = r*sin(a)
z = -r*cos(a)*cos(o)
Gives:
XYZ_PDF(x,y,z) =
1/(2*pi)^(3/2)*e^(-((r*cos(a)*sin(o))^2+(r*sin(a))^2+(-r*cos(a)*cos(o))^2)/2) =
1/(2*pi)^(3/2)*e^(-r^2/2)

Subtituting with these:
X = f(R)*sqrt(1-V^2)*sin(O)
Y = f(R)*V
Z = -f(R)*sqrt(1-V^2)*cos(O)
Gives:
XYZ_PDF(x,y,z) =
1/(2*pi)^(3/2)*e^(-((f(R)*sqrt(1-V^2)*sin(O))^2+(f(R)*V)^2+(-f(R)*sqrt(1-V^2)*cos(O))^2)/2) =
1/(2*pi)^(3/2)*e^(-f(R)^2/2)

So then I have
X = f(R)*sqrt(1-V^2)*sin(O)
Y = f(R)*V
Z = -f(R)*sqrt(1-V^2)*cos(O)
|J| = | d(XYZ) / d(OVR) | = f(r)^2*f'(r)
O_PDF(o) = 1/(2*pi) where (0 < o < 2*pi)
V_PDF(v) = 1/2 where (-1 < v < 1)
R_PDF(r) = 1/(2*pi)^(3/2)*e^(-f(r)^2/2)

But what is f(r)?

Is it now true that O_PDF(o) * V_PDF(v) * R_PDF(r) / |J| = X_PDF(x) * Y_PDF(y) * Z_PDF(z) or are the coordinates of either (O, V, R) or (X, Y, Z) statistically dependent?
 
Last edited:
  • #5
rabbed said:
1/(2*pi)^(3/2)*e^(-r^2/2)
Hi rabbed:

Why is this not the radially symmetric distribution you want?

Regards,
Buzz
 
  • #6
Hi Buzz

It probably is. But after learning about the jacobian and how the area/volume-element for a circle/sphere depends on the radius I'm a bit suspicious about radial distributions.. :)
I guess since I'm just substituting coordinates, it's ok and the first step I should make, so XYZ_PDF(r,a,o) = XYZ_PDF(r,v,o) = 1/(2*pi)^(3/2)*e^(-r^2/2)?
But for finding out this f(R) and formulas to visualize the cloud radially, i would need to continue as I did, right? Because of the Jacobian..
I think my last question is relevant in doing this? Or is it not possible to create such formulas?

Rgds
rabbed
 
  • #7
rabbed said:
XYZ_PDF(r,a,o) = XYZ_PDF(r,v,o) = 1/(2*pi)^(3/2)*e^(-r^2/2)?
Hi rabbed:

The density function gives a value for a point in the domain of the distribution. To get the probability of a random point being in a specific volume, you would integrate the density function over the volume.
P = prob {random point is in V} = ∫ p dV .​
For Cartesian coordinates,
P = ∫ p(x,y,z) dx dy dz .​
For polar coordinates,
P = ∫ p(r) r2 dr dφ sin θ dθ .​

I think that is a ll you need.

Regards,
Buzz
 
  • #8
Hi again Buzz

So to find f(R), I need to check out how to treat the error function? Or Box Muller in 3D? I've also seen something about covariance matrices..

I still think this link was interesting, because it will lead me into Gamma and Chi squared, which I've understood are important:
rabbed said:
Maybe the change of variables/PDF method is better..
I found this document: https://www2.stat.duke.edu/courses/Spring11/sta114/lec/114mvnorm.pdf

I don't understand how, on page 2, fY(y) = 2/sqrt(2*pi)*e^(-y/2) / |2*sqrt(y)|
shouldn't it be fY(y) = 1/sqrt(2*pi)*e^(-y/2) / |2*sqrt(y)| since fY(y) = fX(g^-1(y)) / g'(g^-(y)) ?

Can you answer that?
 
  • #9
rabbed said:
Can you answer that?
Hi rabbed:

Sorry, but I can't help you with this. The notation for expressing problems used at the link you gave is different than the notations I learned many decades ago, and I am unable to make sense of it. The equations you quoted also use notations I don't understand. For example, what does fY(y) mean? If you wrote English defintions for all the notations, then perhaps I might be able to answer your question.

Regards,
Buzz
 
Last edited:
  • #10
Hi

Glad to see someone else having issues with notation.
fY(y) should be the wanted PDF of Y where Y is a random variable expressed as a function, g, of the random variable X.
Since Y = g(X) = X^2, the inverse of g gives us X = g^-1(Y) = +/- sqrt(Y)
I thought the PDF of X was 1/sqrt(2*pi)*e^(-x^2/2), but the X~No(0,1) and the sigma x:x^2=y makes me think now that what's being calculated is different from what I was trying to do in the first post of this thread.

Hope you or someone else can explain
 
Last edited:
  • #11
According to this page http://www.gamlss.org/wp-content/uploads/2014/10/distributions.pdf on page 225, NO(0,1) = 1/sqrt(2*pi)*e^(-x^2/2)
So that means maybe X_PDF(x) = 1/sqrt(2*pi)*e^(-x^2/2) in the other document, after all.

Maybe the sigma is summing X_PDF's so that we get:
Y_PDF(y) = (X_PDF(x) + X_PDF(x)) / g'(X) = 2*X_PDF(g^-1(y)) / g'(g^-1(y)) = 2*X_PDF(sqrt(y))) / g'(sqrt(y)) = 2/sqrt(2*pi)*e^(-y/2) / |2*sqrt(y)|
Instead of:
Y_PDF(y) = X_PDF(x) / g'(X) = X_PDF(g^-1(y)) / g'(g^-1(y)) = X_PDF(sqrt(y))) / g'(sqrt(y)) = 1/sqrt(2*pi)*e^(-y/2) / |2*sqrt(y)|
?

It could be tied to the fact that the Gamma function has one special case of parameters, (Ga(1/2,1/2) according to the document), which is why the Box Muller method can sample the normal distribution. I read something about that.

Can someone confirm?
 
  • #12
Hi again

If we skip the side track, is it possible to sample from the radial distribution 1/(2*pi)^(3/2)*e^(-r^2/2) that was derived?

According to this link:
http://mathematics.livejournal.com/1144375.html
I think I should sample the "lower incomplete gamma function", somehow..

Rgds
rabbed
 
Last edited:

1. What are radially distributed Cartesian coordinates?

Radially distributed Cartesian coordinates are a coordinate system used in physics and mathematics to represent points in three-dimensional space. They are based on the Cartesian coordinate system, but instead of using only x, y, and z axes, they also include a radial axis that extends from the origin of the system to represent distance from the center.

2. How are radially distributed Cartesian coordinates different from traditional Cartesian coordinates?

Traditional Cartesian coordinates only use x, y, and z axes to represent points in three-dimensional space. Radially distributed Cartesian coordinates, on the other hand, also include a radial axis to represent distance from the center. This allows for a more comprehensive representation of points in space.

3. What are the advantages of using radially distributed Cartesian coordinates?

Radially distributed Cartesian coordinates offer several advantages, including a more comprehensive representation of points in space, easier visualization of spherical objects, and simplified calculations for certain physical phenomena such as electric and magnetic fields.

4. How are radially distributed Cartesian coordinates used in physics?

In physics, radially distributed Cartesian coordinates are commonly used to describe the position, velocity, and acceleration of objects moving in a circular or spherical path. They are also used to represent the electric and magnetic fields around charged or magnetic objects.

5. Are radially distributed Cartesian coordinates used in any other fields?

Yes, radially distributed Cartesian coordinates are also used in computer graphics, particularly in 3D modeling and animation. They are also used in robotics, particularly in the design and control of robotic arms and manipulators.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
8
Views
867
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
2K
  • General Math
Replies
4
Views
4K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
25
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
13
Views
1K
  • Programming and Computer Science
Replies
2
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
4
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
2K
Back
Top