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Math Help - Transformation

  1. #1
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    Transformation

    X denotes a random variable with pdf <br />
f(x)=\left\{\begin{array}{cc}\frac{2}{x^2} & 1 < x < 2\\  0 &elsewhere \end{array}\right.
    let Y=X^2, find the cdf of Y and hence the pdf of Y and cov(X,Y).

    I know that to find the pdf from the cdf we differentiate, but I'm not sure how to handle to transformation to  Y = X^2
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  2. #2
    MHF Contributor matheagle's Avatar
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    How can you get the covariance?
    You don't have a joint density

    I guess you get E(XY)-E(X)E(Y)=E(X^3)-E(X)E(X^2)

    This is a one to one mapping with probability one since x>0.

    f_Y(y)=f_X(x)\Biggl|{dx\over dy}\Biggr|={2\over y}\Biggl|{1\over 2\sqrt y}\Biggr|=y^{-3/2}

    where 1<y<4.
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  3. #3
    Super Member Random Variable's Avatar
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    I prefer the method of distribution functions.

     G(y) = P(Y \le y) = P(X^{2}\le y) = P(-\sqrt{y} \le X \le \sqrt{y}) = P(X \le \sqrt{y}) (since X is not defined for any negative numbers)

     = \int^{\sqrt{y}}_{0}\ \frac{2}{x^{2}} \ dx = -\frac{2}{x} \Big|^{\sqrt{y}}_{0}= -\frac{2}{\sqrt{y}}

    then  g(y) = G'(y) = \frac{1}{y^{3/2}} for \ 1 \le y \le 4
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  4. #4
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    Quote Originally Posted by Random Variable View Post
    I prefer the method of distribution functions.

     G(y) = P(Y \le y) = P(X^{2}\le y) = P(-\sqrt{y} \le X \le \sqrt{y}) = P(X \le \sqrt{y}) (since X is not defined for any negative numbers)

     = \int^{\sqrt{y}}_{0}\ \frac{2}{x^{2}} \ dx = -\frac{2}{x} \Big|^{\sqrt{y}}_{0}= -\frac{2}{\sqrt{y}}

    then  g(y) = G'(y) = \frac{1}{y^{3/2}} for \ 1 \le y \le 4
    An antiderivative of \frac{1}{x^2} is \frac{-2}{x^3}
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Moo View Post
    An antiderivative of \frac{1}{x^2} is \frac{-2}{x^3}
    On what planet?
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  6. #6
    Super Member Random Variable's Avatar
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    I made a mistake.

    It should be  P(-\sqrt{y} \le X \le \sqrt{y}) = P(1 \le X \le \sqrt{y}) = \int^{\sqrt{y}}_{1}\frac{2}{x^{2}} dx = -\frac{2}{\sqrt{y}} +2
    Last edited by Random Variable; August 18th 2009 at 04:29 PM.
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  7. #7
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    P(-\sqrt{y} \le X \le \sqrt{y}) = P(1 \le X \le \sqrt{y}) = \int^{\sqrt{y}}_{1}\frac{2}{x^{2}} dx = -\frac{-2}{\sqrt{y}} +2<br />
     f(y) = \frac{-2}{\sqrt{y}}+2
    this is the resulting pdf of Y, with range 1<y<4?

    Also, if my pdf is
    <br />
P(-\sqrt{y} \le X \le \sqrt{y}) = P(1 \le X \le \sqrt{y}) = \int^{\sqrt{y}}_{1}\frac{2}{x^{2}} dx = -\frac{-2}{\sqrt{y}} +2<br />
    the cdf is
    F(x)=\frac{-2}{x}, 1<x<2 and then if Y=X^2 how would I then find the cdf of Y?
    Last edited by bigdoggy; August 18th 2009 at 02:16 PM.
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  8. #8
    MHF Contributor matheagle's Avatar
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    and I thought I solved this, this morning

    A negative cdf?
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  9. #9
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    Quote Originally Posted by matheagle View Post
    and I thought I solved this, this morning
    I can understand the method of distribution functions better....

    I've just noticed the negative cdf...sometimes I get caught up in the calculating and don't use common sense!

    Perhaps I should give the question in full....
    the pdf is given above, I then have to find the cdf(by integrating the pdf given) then the transformation Y=X^2 is given and I have to then find the cdf of Y and the pdf of Y, and finally the covariance of (X,Y)....
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  10. #10
    Super Member Random Variable's Avatar
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    Quote Originally Posted by matheagle View Post
    and I thought I solved this, this morning

    A negative cdf?
    I fixed my mistake in a subsequent post.
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  11. #11
    Super Member Random Variable's Avatar
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    Let me do the problem over again from scratch to clear up the confusion.


     G(y) = P(Y \le y) = P( X^{2} \le y) = P( -\sqrt{y} \le X \le \sqrt{y})

    But X is not defined for values less than 1

    so  P( -\sqrt{y} \le X \le \sqrt{y}) = P(1 < X \le \sqrt{y})

     = \int^{\sqrt{y}}_{1} \frac{2}{x^{2}} \ dx =  -\frac{2}{x}\Big|^{\sqrt{y}}_{1}= -\frac{2}{\sqrt{y}} +2 for 1 < y < 4

    then  g(y) = G'(y) = \frac{1}{y^{3/2}} for 1 < y < 4
    Last edited by Random Variable; August 18th 2009 at 04:42 PM.
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  12. #12
    MHF Contributor matheagle's Avatar
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    These TWO techniques are the same. IF you differentiate...

     G(y) = P(Y \le y) = P( X^{2} \le y) = P( -\sqrt{y} \le X \le \sqrt{y})

    wrt the variable y (this is not a random variable, this is calc 1) you get

     f_Y(y) = f_X(\sqrt y ){d\over dy}(\sqrt y)={2\over y}\Biggl({1\over 2\sqrt y}\Biggr)=y^{-3/2}

    The general formula, which is in most books is

     f_Y(y) = f_X(\sqrt y )\cdot {d\over dy}(\sqrt y)-f_X(-\sqrt y )\cdot {d\over dy}(-\sqrt y)

    =f_X(\sqrt y ){1\over 2\sqrt y}+f_X(-\sqrt y ){1\over 2\sqrt y}

    but since x>0 we have f_X(-\sqrt y )=0
    Last edited by matheagle; August 18th 2009 at 08:20 PM.
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  13. #13
    Super Member Random Variable's Avatar
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    I prefer to use the distribution function technique over the change of variable technique because I find the former more intuitive (especially when dealing with distributions of two random variables).
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  14. #14
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    Thanks.
    So given the pdf f(x)=\left\{\begin{array}{cc}\frac{2}{x^2} & 1 < x < 2\\  0 &elsewhere \end{array}\right.....


    The replies have helped alot, however I'm trying to find the cdf from this pdf then the transformation Y=X^2 and then find the cdf of Y and so then the pdf of Y....
    1)What would the cdf be as I get a negative answer as I shown a couple posts ago..??
    2)What is the process to obtaining the cdf of Y?
    3)and finally then the pdf of Y?
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  15. #15
    Super Member Random Variable's Avatar
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    Quote Originally Posted by bigdoggy View Post
    Thanks.
    So given the pdf f(x)=\left\{\begin{array}{cc}\frac{2}{x^2} & 1 < x < 2\\ 0 &elsewhere \end{array}\right.....


    The replies have helped alot, however I'm trying to find the cdf from this pdf then the transformation Y=X^2 and then find the cdf of Y and so then the pdf of Y....
    1)What would the cdf be as I get a negative answer as I shown a couple posts ago..??
    2)What is the process to obtaining the cdf of Y?
    3)and finally then the pdf of Y?
    G(y) is the cdf of Y and g(y) = G'(y) is the pdf of Y
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