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Change of variable

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Let x,y be iid and [tex]x, y \sim U(0,1) [/tex] (uniform on the open set (0,1)) and let [tex] z = xy^2. [/tex]
    Find the density of z.


    2. Relevant equations



    3. The attempt at a solution

    [itex]P(z \leq w) = P(xy^2 \leq w) = P(- \sqrt{\frac{w}{x}} \leq y \leq \sqrt{\frac{w}{x}}) = \int^{ \sqrt{\frac{w}{x}}}_{ -\sqrt{\frac{w}{x}}} dy = 2 \sqrt{\frac{w}{x}}
    [/itex]

    Is this right. Seems like I am missing something, not sure.

    Thanks.
     
    Last edited: Dec 11, 2012
  2. jcsd
  3. Dec 11, 2012 #2

    Ray Vickson

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    No. Y cannot be < 0, so you cannot have the region ## - \sqrt{w/x} \leq y < 0##. Anyway, you need an answer that contains w only, so you still need to get rid of the 'x'.
     
  4. Dec 11, 2012 #3

    pasmith

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    [itex]P(z \leq w)[/itex] should be a function of w only, so something's wrong.

    Fix x. Then [itex]P(y^2 \leq w/x) = P(0 \leq y \leq \sqrt{w/x}) = \min(1,\sqrt{w/x})[/itex] since [itex]0 \leq y \leq 1[/itex]. Then
    [tex]
    P(z \leq w) = P(xy^2 \leq w) = \int_0^1 \min\left(1,\sqrt{\frac wx}\right)\,\mathrm{d}x
    = \int_0^w 1\,\mathrm{d}x + \int_w^1 \sqrt{\frac wx}\,\mathrm{d}x
    [/tex]
     
  5. Dec 11, 2012 #4
    Makes sense it should be a function of w only. I do not understand though how the integral with the minimum is broken up into the two integrals at the end. Any insight?

    Thanks for the help.
     
  6. Dec 11, 2012 #5

    Ray Vickson

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    Look at the two cases x > w and x < w.
     
  7. Dec 11, 2012 #6
    The first integral ranges from ##x=0## to ##x=w##, so the minimum is equal to 1. In the second integral, you have ##x\geq w##, and so the minimum is the square-root expression.
     
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