Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The distribution of ratio of two uniform variables

  1. Jun 18, 2005 #1
    Let X ~ U(0,1), Y ~U(0,1), and independent from each other. To calculate the density of U=Y/X, let V=X, then:
    [tex]f_{U,V}(u,v)=f_{X,Y}(v,uv)|v|[/tex] by change of variables.
    [tex]f_{U}(u)=\int_{0}^{1}{f_{X,Y}(v,uv)|v|dv}=\int_{0}^{1}{vdv}={1\over 2}, 0<u<\infty[/tex], which is not integrated to 1.
    Where I am wrong?
    gim :cry:
    Last edited: Jun 18, 2005
  2. jcsd
  3. Jun 18, 2005 #2
    you only integrated with respect to fu(u). Now you have to integrate with respect to fv(v). Or you could have just used a double integral to start with...
  4. Jun 19, 2005 #3
    Actually the marginal distribution of U, namely the distribution of the ratio of two uniform variables, is the only thing that I am interested. To be more clear:
    Now the question is my result 1/2 is not a reasonable density since it's not integrated to 1.
    gim :bugeye:
  5. May 4, 2010 #4
    Question says Let X and Y be independent random variables with join cumulative distribution function (CDF) F subscript X,Y of (x,y)= P (X</= x, Y</=y). Show that the CDF Fz(z) of the random variable Z=min (X,Y) can be computed via
    Fz(z)= Fx(z) + Fy(z) - Fx(z) . Fy(z) = 1 - (1 - Fx (z)) . (1- Fy (z))

    Please reply to this asap. I need to submit this answer by Friday. Thanks!
  6. May 4, 2010 #5


    User Avatar
    Homework Helper

    Don't jump into the thread of another. What have you done so far?
  7. May 4, 2010 #6
    Well, I'm new here and had problems starting a new thread.

    I looked at PDF of an exponential function which is (1 - Fx (x)) and also since its also given its independent, we know it can be split into Fx (x) . Fy (y) but I can't put these things together.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook