1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bivariate Transformation of Random Variables

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Two RVs X1 and X2 are continuous and have joint pdf [itex]
    f_{X_1,X_2}(x_1, x_2) = \begin{cases} x_1+x_2 &\mbox{for } 0 < x_1 < 1; 0 < x_2 < 1
    \\
    0 & \mbox{ } \text{otherwise}. \end{cases} [/itex]

    Find the pdf of [itex]Y = X_1X_2[/itex].


    2. Relevant equations
    I'm using the transformation "shortcut' that says if U = g1(x,y), V = g2(x,y) and h1(u,v) = x, h2(u,v) = y then
    [itex]f_{U,V}(u,v) = f_{X,Y}(h_1(u,v), h_2(u,v) \times |J| [/itex] where |J| is the absolute value of the determinant of the Jacobian of h1 and h2.


    3. The attempt at a solution
    In this setting I let [itex]Y = X_1X_2[/itex] and [itex]Z = X_1[/itex] so [itex]h_1(y,z) = Z[/itex] and [itex]h_2(y,z) = Y/Z[/itex]. Then I calculated [itex]|J| = \frac{1}{Z}[/itex] so plugging in I got that
    [itex]f_{Y,Z}(y,z) = (z + \frac{y}{z})(\frac{1}{z}) = 1 + \frac{y}{z^2}[/itex] where [itex]0 < z < 1, 0 < y < z[/itex].
    So to find the pdf of Y, integrate with respect to z over all possible values of z, but this integral does not converge when evaluated from 0 to 1.

    I've tried other choices for Z, but I've ended up with the same problem.
     
    Last edited: Oct 16, 2013
  2. jcsd
  3. Oct 16, 2013 #2

    statdad

    User Avatar
    Homework Helper

    Note: if both [itex] 0 \le x_1 \le 1 [/itex] and [itex] 0 \le x_2 \le 1 [/itex], how does the product
    [itex] X_1 X_2 [/itex] compare in size to [itex] X_1 [/itex] alone?

    Edited to add: I have not stepped through all of your work to simplify the transformation, simply noted that you have [itex] 0 < z < 1 [/itex] and [itex] 0 < y < 1 [/itex].
     
  4. Oct 16, 2013 #3
    Since [itex]X_1[/itex] and [itex]X_2[/itex] are both smaller than 1, then their product will be smaller than [itex]X_1[/itex] alone, but I'm not seeing how this helps me find the pdf of [itex]Y[/itex].

    Are the limits [itex]0 < y < z[/itex] not correct? It seems that this needs to be true in order for [itex]Y/Z[/itex] to be less than 1?

    Now that I'm looking at it again, is this transformation shortcut even valid in this situation? I'm not sure if my [itex]h_1, h_2[/itex] are workable in this case.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Bivariate Transformation of Random Variables
Loading...