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What is this well known probability distribution?

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data
    I have a probability density function:
    f[itex]_{X}[/itex](x) = θk[itex]^{θ}[/itex]x[itex]^{-θ-1}[/itex] when x > k and 0 everywhere else
    And:
    Y = θlog(X/k)

    Find the probability density function for Y. What well known density function is this?

    2. Relevant equations
    I dont really know where to start...


    3. The attempt at a solution
    I first tried solving for x and:
    X = ke[itex]^{Y/θ}[/itex]
    Plugging this into f(x):
    f[itex]_{Y}[/itex](y) = θk[itex]^{θ}[/itex](ke[itex]^{y/θ}[/itex])[itex]^{-θ-1}[/itex]

    I also tried going the other way around:
    f[itex]_{Y}[/itex](x) = θlog(θk[itex]^{θ-1}[/itex]x[itex]^{-θ-1}[/itex])

    But, I dont really know what to do and where to start.. ...
     
  2. jcsd
  3. Jun 4, 2012 #2

    vela

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    You can't simply rewrite fX(x) in terms of Y the way you did. It's a little more complicated. The correct relationship is
    $$f_X(x)\,dx = f_Y(y)\,dy$$ where I've assumed both dx and dy are positive. The idea here is that interval (x, x+dx) corresponds to the interval (y, y+dy) so the probability that X is in the interval (x, x+dx), which is given by the lefthand side of the equation, should be equal to the probability Y is in the interval (y, y+dy), which is given by the righthand side of the equation. Solving for fY(y), you get
    $$f_Y(y) = \frac{f_X(x)}{|dy/dx|}.$$ The absolute value is necessary to cover the case where dx>0 corresponds to dy<0. That is, when y decreases as x increases.
     
  4. Jun 4, 2012 #3
    Actually, I had to use the fundamental theorem of calculus:

    [itex]\frac{d}{dy}[/itex]∫[itex]^{y}_{0}[/itex]f(x)dx = f(y)

    Therefore;
    [itex]\frac{d}{dy}[/itex]∫[itex]^{ke^(Y/θ)}_{0}[/itex]θk[itex]^{θ}[/itex]x[itex]^{-(θ+1)}[/itex]dx
    = [itex]\frac{d}{dy}[/itex]-e[itex]^{-y}[/itex]
    = e[itex]^{-y}[/itex]
    And, this is the exponential distribution..

    Or... Am I wrong again?
     
  5. Jun 4, 2012 #4

    Ray Vickson

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    No, you are now correct.

    RGV
     
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