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PDF and CDF

  1. Feb 6, 2009 #1
    A question which I amnot able to do....please help:
    Find the PDF of Y if X is a Gaussian PDF:
    fx(x) = (1e-x^2/2)/(2pi)^1/2 ; -infnity<x<+infinity

    Express your answer in terms of CDF of X gven by

    Fi(x) = Integral -infnty to + infnity((1e-x^2/2)/(2pi)^1/2)

    b) Sketch both the PDF, fY(y) and CDF, FY(y) for randon variable Y
     
  2. jcsd
  3. Feb 6, 2009 #2
    You haven't given a relationship between X and Y have you?
     
  4. Feb 6, 2009 #3
    Sorry, it is Y = X^2
     
  5. Feb 6, 2009 #4

    HallsofIvy

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    Then just replace [itex]X^2[/itex] in the formulas by Y!
    [tex]fx(Y) = e^{-Y/2}/(2\pi)^{1/2}[/tex]
    [tex]0\le Y< \infty[/tex]

    [tex]Fi(x) = \int_{-\infty}^\infty (e^{-Y/2}/(2\pi)^{1/2})dy[/tex]
     
  6. Feb 6, 2009 #5
    but if Y = modX
     
  7. Feb 6, 2009 #6
    Also, to find PDF of Y of a Gaussian PDF, do I need to find mean and variance???
     
  8. Feb 6, 2009 #7
    xx yyy
     
  9. Feb 6, 2009 #8

    HallsofIvy

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    I don't know what you mean by "modX" or by "find PDF of Y of a Gaussian PDF".
     
  10. Feb 6, 2009 #9
    I mean
    Y = |X|
     
  11. Feb 6, 2009 #10
    you told me how to do it for Y = X^2 but in another part, how to do it for
    Y = |X|
     
  12. Feb 6, 2009 #11

    statdad

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    Try this. For the distribution function of [tex] X [/tex] write

    [tex]
    F(x) = \Pr(X \le x) = \Phi(x)
    [/tex]

    Here [tex] \Phi(x) [/tex] is the usual notation for the CDF of the standard Gaussian.
    I'll use [tex] G(y) [/tex] as the CDF for your new random variable.

    [tex]
    \begin{align*}
    G(y) & = \Pr(Y \le y) = P(X^2 \le y) \\
    & =\Pr(-\sqrt{y} \le X \le \sqrt{y}) \\
    & = \Phi(\sqrt{y}) - \Phi(-\sqrt{y})
    \end{align*}
    [/tex]

    Since the standard Gaussian is symmetric around 0,

    [tex]
    \Phi(-a) = 1 - \Phi(a)
    [/tex]

    for any number [tex] a [/tex]. From the place where I left off:

    [tex]
    \begin{align*}
    G(y) &= \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) = \Phi(\sqrt{y}) - (1- \Phi(\sqrt{y}))\\
    & = 2\Phi(\sqrt{y}) - 1
    \end{align*}
    [/tex]

    Now use these facts:

    * The density of [tex] Y [/tex] is the derivative of [tex] G(y) [/tex]
    * You need to use the chain rule when you take the derivative of [tex] \Phi(\sqrt{y})[/tex]
    * The derivative of [tex] \Phi(x) [/tex] is the density of the standard Gaussian
    * The random variable [tex] Y [/tex] is defined on [tex] (0, \infty) [/tex]
    * The distribution of [tex] Y [/tex] is one you should be able to recognize

    Edited to add:
    the method for your second question is similar:
    [tex]
    \Pr(|X| \le y) = \Pr(-y \le X \le y)
    [/tex]

    go from here. (I have a second [tex] \le [/tex] between X and y above, but it isn't showing.)
     
  13. Feb 6, 2009 #12
    thanks!!!!!!!!!!
     
  14. Feb 8, 2009 #13
    Any idea how to sketch the PDF, fY(y) and CDF, FY(y) for randon variable Y for this problem?

    Thank you.
     
  15. Feb 8, 2009 #14

    statdad

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    once you have the expression for the density - graph it as any other function.
    You'll need a technology aid to graph the CDF.
     
  16. Feb 9, 2009 #15
    When finding the density of Y by taking the derivative of G(y) as described earlier, we want to take the derivative of G(y) in respect to y correct? Or is it in respect to phi?

    Thank you for all the help
     
  17. Feb 9, 2009 #16

    statdad

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    [tex]
    g(y) = \frac{dG}{dy}
    [/tex]
     
  18. Feb 9, 2009 #17
    Thanks
     
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