1. Apr 27, 2016

rabbed

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: Apr 27, 2016
2. Apr 27, 2016

rabbed

3. Apr 27, 2016

Buzz Bloom

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. Apr 27, 2016

rabbed

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: Apr 28, 2016
5. Apr 28, 2016

Buzz Bloom

Hi rabbed:

Why is this not the radially symmetric distribution you want?

Regards,
Buzz

6. Apr 28, 2016

rabbed

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. Apr 28, 2016

Buzz Bloom

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. Apr 29, 2016

rabbed

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:

9. Apr 29, 2016

Buzz Bloom

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: Apr 29, 2016
10. Apr 29, 2016

rabbed

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 whats 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: Apr 29, 2016
11. Apr 30, 2016

rabbed

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)|
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. May 4, 2016

rabbed

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?