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A Radially distributed Cartesian coordinates

  1. Apr 27, 2016 #1
    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. jcsd
  3. Apr 27, 2016 #2
  4. Apr 27, 2016 #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
     
  5. Apr 27, 2016 #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: Apr 28, 2016
  6. Apr 28, 2016 #5
    Hi rabbed:

    Why is this not the radially symmetric distribution you want?

    Regards,
    Buzz
     
  7. Apr 28, 2016 #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
     
  8. Apr 28, 2016 #7
    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

     
  9. Apr 29, 2016 #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:
    Can you answer that?
     
  10. Apr 29, 2016 #9
    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
  11. Apr 29, 2016 #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 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
  12. Apr 30, 2016 #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?
     
  13. May 4, 2016 #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: May 4, 2016
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