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Homework Help: Random variable probability problem

  1. Sep 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Continuous random variable X has probability density function defined as

    f(x)= 1/4 , -1<x<3

    =0 , otherwise

    Continuous random variable Y is defined by Y=X^2

    Find G(y), the cummulative distribution function of Y

    2. Relevant equations

    3. The attempt at a solution

    G(y) = P(-sqrt(y)<=X<=sqrt(y))

    What does this mean? G(Y) is defined as P(-sqrt(y)<=X<=sqrt(y)) for 0<=X<=9 ?
  2. jcsd
  3. Sep 7, 2010 #2
    Re: probability

    G(y), the cummulative probability distribution function of Y, is a function that gives the probability that Y is smaller than y.

    P(-sqrt(Y)<=X<=sqrt(Y)) is a way of writing: Probability that X is in between plus and minus the square root of Y (note the use of capital Y here, y and Y are not the same thing).
  4. Sep 7, 2010 #3
    Re: probability

    Thanks Gerben, the next step will be to integrate the pdf?

    ie [tex]G(y)=\int^{\sqrt{y}}_{-\sqrt{y}}\frac{1}{4}dx[/tex] for [tex]0\leq y\leq 9 [/tex]

  5. Sep 7, 2010 #4
    Re: probability

    yes exactly
  6. Sep 10, 2010 #5
    Re: probability

    There will be 3 cases here,

    case 1 : G(y)=0 for y<0

    case 2 : G(y)= integration from -sqrt(y) to sqrt(y) 1/4 dx for 0<=y<=9

    case 3: G(y)=1, for y>9

    Is this the domain? I doubt my domain for case 2 is correct.
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