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Homework Help: Function of random variable

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Let X and Y be two independent random variables each exponentially distributed with parameter 1. Define a new random variable:

    [tex]z = \frac{x}{{x + y}}[/tex]

    Find the PDF of Z


    2. Relevant equations



    3. The attempt at a solution
    [tex]\begin{array}{l}
    {F_Z}(z) = P(Z < z) = P\left( {\frac{x}{{x + y}} < z} \right) = P\left( {x \le \frac{{zy}}{{1 - z}}} \right) \\
    {F_Z}(z) = \int\limits_0^\infty {\int_0^{\frac{{zy}}{{1 - z}}} {{f_{xy}}(x,y)dxdy} } \\
    {f_{xy}}(x,y) = {f_x}(x){f_y}(y) \\
    {F_Z}(z) = \int\limits_0^\infty {\int_0^{\frac{{zy}}{{1 - z}}} {{f_x}(x){f_y}(y)dxdy = } } \int\limits_0^\infty {\int_0^{\frac{{zy}}{{1 - z}}} {{e^{ - x}}{e^{ - y}}dxdy} } = \int\limits_0^\infty {{e^{ - y}}\left[ {\int_0^{\frac{{zy}}{{1 - z}}} {{e^{ - x}}dx} } \right]} dy \\
    {F_Z}(z) = \int\limits_0^\infty {{e^{ - y}}\left[ { - {e^{ - \frac{{zy}}{{1 - z}}}} + 1} \right]} dy = \int\limits_0^\infty { - {e^{ - y}}{e^{ - \frac{{zy}}{{1 - z}}}} + {e^{ - y}}} dy = \int\limits_0^\infty { - {e^{ - \frac{{y(1 - z) - zy}}{{1 - z}}}} + {e^{ - y}}} dy \\
    {F_Z}(z) = \int\limits_0^\infty { - {e^{ - \frac{y}{{1 - z}}}} + {e^{ - y}}} dy = (1 - z){e^{ - \frac{y}{{1 - z}}}}|_0^\infty - {e^{ - y}}|_0^\infty = z \\
    \end{array}[/tex]
    Now i know that if i take the derivative of this i will get the "pdf" but its obviously wrong. Any thoughts?
     
  2. jcsd
  3. Mar 14, 2010 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This step is invalid if [itex]z > 1[/itex]. (The inequality gets reversed in that case.)

    But you could instead write

    [tex]P\left( \frac{x}{x+y} < z \right) = P\left(y > \frac{x(1-z)}{z}\right) = 1 - P\left(y \leq \frac{x(1-z)}{z}\right)[/tex]

    I'm not sure if that will be any more helpful, but at least it's correct.

    I wonder if it would be helpful to work with the reciprocal:

    [tex]\frac{1}{z} = \frac{x + y}{x} = 1 + \frac{y}{x}[/tex]

    It shouldn't be hard to work out the pdf of

    [tex]\frac{y}{x}[/tex]

    as it is the quotient of two independent random variables. Adding 1 just shifts the pdf to the right by 1. Then do you know how to find the pdf of the reciprocal of a random variable with known pdf?
     
    Last edited: Mar 14, 2010
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