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Find the probability distribution

  1. Oct 1, 2004 #1
    Given the probability distribution function

    [tex]

    f(x) = \frac{2}{9}(x-1), -1<x<2

    [/tex]

    find the pdf of [tex] Y = X^2 [/tex]

    My Solution:

    When x = -1, y = 1 and when x = 2, y = 4, so the range of y is

    [tex]

    1 \leq y \leq 4

    [/tex]

    So to find the pdf of Y = X^2, we need to find the cdf of Y. Since Y is nontrivial over the interval [tex] 1 \leq y \leq 4 [/tex],

    [tex]

    F(y) = P(1 \leq Y \leq y)
    =P(1 \leq x^2 \leq y)

    [/tex]

    I'm at a loss here. I do not know how I should deal with square rooting the x, since [tex] x^2 \leq y [/tex] is rewritten as [tex] -\sqrt{y} \leq x \leq \sqrt{y}[/tex]

    Any suggestions?
     
  2. jcsd
  3. Oct 1, 2004 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What you have is correct: [itex]-\sqrt{y}\leqx\leq\sqrt{y}[/itex]. Now integrate your pdf for x, (2/9)(x-1) from -y to y (be careful about y> 1!).
     
  4. Oct 1, 2004 #3
    err... when you mean about be careful of y>1, do you mean that I have to seperate the integral like this:

    [tex]

    \int^{1}_{-\sqrt{y}} f(x)dx + \int^{\sqrt{y}}_{1} f(x)dx

    [/tex]

    Because I know that y=1 should not be within the range of y (since y is just greater than 1)

    Edit:

    I got a cdf of [tex]-4\sqrt{y} [/tex]. Im sure I did something wrong and it must have something to do with that y>1 precaustion...
     
    Last edited: Oct 1, 2004
  5. Oct 1, 2004 #4
    Hello relinquished! Isn't it that it should be 2/9 (x + 1) as the given?
     
  6. Oct 1, 2004 #5
    well... if the pdf really is [tex] \frac{2}{9}(x+1) [/tex] then I should look back at the book ^^;
     
  7. Oct 2, 2004 #6
    I don't think it is found in the book.
     
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