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Probability density function of another function

  1. Jun 5, 2014 #1
    Hi guys, I'm working through a problem right now and would like to pick your brains on some stuff.

    I have an function: $$ f(r,\phi)= -\frac{1}{3} -cos(2\phi)(\frac{a^2}{r^2}) \hspace{0.5cm} for \hspace{0.5cm} a<r<b $$. I'm working in radial coordinates so r is the distance from a center and ##\phi## is the angle about that center. Given a particular ## a , b ## which you can think of as the inner and outter rings on an annulus, I can plot this function in Matlab and create a histogram of the values ##f##.

    https://www.dropbox.com/s/k1cipzotenl2del/PF.png

    My question is whether I can arrive at this distribution analytically?

    Thanks in advance for any comments!
     
    Last edited: Jun 5, 2014
  2. jcsd
  3. Jun 5, 2014 #2

    FactChecker

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    Maybe I'm missing something. Is f supposed to be a PDF? It's negative some places so it can't be a PDF. Can you explain what histogram and distribution you are talking about.
     
  4. Jun 5, 2014 #3
    Hi FactChecker,

    f is not a PDF. f is my function. I input ##r## (distance from the center of annulus) and ##a## (inner radius of the annulus). ##b## is the outter radius. When ##r## is between ##a,b##, the values in that region of space is described by ##f(r,\phi)##. That is what the map on the left is showing.

    I am using Matlab so everything is discrete. That is, I simply gather up all the elements with their associated values from inside that annulus and made a histogram of that list of data.

    That histogram is the probability distribution. I want to derive that analytically. It is analogous to say, finding the probability density function of a sine wave.

    I know what the general recipe is. Start with your function. Generally the inverse of the function is the CDF. And you can take the derivative to get to the PDF. I'm wondering if that holds true here as well.
     
  5. Jun 5, 2014 #4

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    I'm still trying to figure out what the random variable is. Here are a couple of my guesses that might help you understand why I am confused.

    1) Maybe a point is randomly chosen with uniform distribution in the domain of f and it's polar coordinates are used to calculate the associated value of f. That would make the values of f a random variable with a PDF. But then are a and b fixed? If so, what values of a and b gave the "histogram" you show?

    2) Values of r, theta, a, and b, are randomly chosen and f is calculated. But I don't know how r, a, and b and chosen or what their distribution is. Are they uniformly distributed over some intervals?
     
  6. Jun 5, 2014 #5
    First, allow me to apologize for being unclear. It was not my intent. I realize what you mean.

    (1) Yes, a and b are fixed. In this case, a is 10.4 and b is 16 (ignored the labels on the graph, those represent the size of my matrix, not the actual values.)

    (2) The random variables are ##r## and ##\phi##. ##r## is uniformly distributed between a and b while ##\phi## is uniformly distributed between ##0## and ##2\pi##
     
  7. Jun 6, 2014 #6

    FactChecker

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    Thanks for the clarification. Yes, the same idea of the inverse is correct except that each value of f can have several pre-images whose PDF values should be summed. A formal mathematical treatment would talk about the probabilities of pre-images of intervals in the range of f. I don't know if it is practical to get a closed-form PDF equation for this problem.

    On a related issue: Knowing the PDF is not as useful as knowing the CDF since the only positive probabilities are for a range of values of f. The CDF gives those probabilities directly while the PDF requires an integration. I would suggest putting your histogram data into a CDF form if you decide to use that data directly.
     
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