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Function of a random variable

  1. May 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose a random variable X has probability density function(pdf)
    f(x) { 1/3 for [tex]1 \leq x \leq 4[/tex]

    find the density function of [tex] Y= \sqrt{X}[/tex]

    3. The attempt at a solution
    [tex] y=g(x)=\sqrt{x}[/tex]
    so [tex]g^-1(y)=x=y^2[/tex]

    [tex]A= \{ x: 1 \leq x \leq 4 \}[/tex]
    is monotonic onto
    [tex]B= \{y: 1 \leq y \leq 2 \}[/tex]


    [tex]f(y)=fx (g^-1(y). |(g^-1(y))^'[/tex][tex]|[/tex]
    which gives me
    [tex]f(x)=[/tex][tex] \{ 2y/3 [/tex] for [tex]1 \leq y \leq 2 \}[/tex]

    This seems to be an invalid pdf as for y=2 f(y)=1.3333 which is >1
    Can anyone tell me where I went wrong?
  2. jcsd
  3. May 19, 2008 #2


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    Science Advisor
    Homework Helper

    It's not invalid. You did it right. A probability DENSITY can be greater than 1. A probability can't be greater than 1. The integral of f(y)dy from 1 to 2 is one.
    Last edited: May 19, 2008
  4. May 20, 2008 #3
    Thanks, You are the man Dick.
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