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Homework Help: Cauchy distribution

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img132.imageshack.us/img132/1/48572399ly5.png [Broken]



    3. The attempt at a solution
    I tried dividing the two pdf's but that isn't right. If you have X normal and Y normal distributed how can you derive the distribution function of X/Y?
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 12, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    Of course you don't divide the pdf's. To get the quotient pdf as a function of Z, you take the dirac delta function [itex]\delta(Z-X/Y)[/itex] and integrate it times the pdf's for X and Y, dXdY.
     
  4. Dec 12, 2008 #3

    statdad

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    Another method (based techniques typically presented early in mathematical stats) is this:

    Define these two variables (one you already have)

    [tex]
    W = \frac{Z_1}{Z_2}, \quad V = Z_2
    [/tex]

    Then

    [tex]
    Z_1 = V \cdot W, \quad Z_2 = V
    [/tex]

    By their definitions both new random variables range over [tex] (-\infty, \infty) [/tex].

    Use the basic ideas for transformation of a joint distribution to get the distribution of [tex] V [/tex] and [tex] W [/tex], then integrate out [tex] V [/tex].
     
  5. Aug 1, 2010 #4
    Hi,

    I'm actually going over some probability problems and I got a bit stuck in this one too.

    If you let:

    W=Z1/Z2 and V=Z2

    Then truly Z1=V*W and Z2=V

    And if you calculate the Jacobian determinant of such transformations you get:

    Jacobian determinant = V (here we take the absolute value when sticking into formula below)

    Therefore:

    f(w,v) =[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP91219bfg00e7b70caif00005aa636h4egb540i1?MSPStoreType=image/gif&s=39&w=148&h=44 [Broken]

    and so all you need to get the probability density function of W is to integrate the joint probability with respect to v as follows:

    First note that: d/dv (e-v2(1+w2)/2) = -v(1+w2)*e-v2(1+w2)/2

    =>[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP243119bff8ch835e1b7i00001639e2c5d96b97gg?MSPStoreType=image/gif&s=39&w=366&h=54 [Broken]

    and here is where I seem to be overlooking something, in order to get f(w) you must evaluate the integral from minus infinity to plus infinity and so I believe you get:

    [PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP237019bff8ch83i380ib0000641a5786ggg043f3?MSPStoreType=image/gif&s=39&w=124&h=43 [Broken]

    Which is just plainly equal to zero, so I must've done something wrong, can anyone spot what was it? I would appreciate if someone did. Thanks.
     
    Last edited by a moderator: May 4, 2017
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