1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Find the probability distribution

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


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


    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


    1 \leq y \leq 4


    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],


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


    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


    User Avatar
    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:


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


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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook