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Math Help - uniform distribution / pdf

  1. #1
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    uniform distribution / pdf

    Let X have the uniform distribution U(-1,3). How do you find the p.d.f. of Y=X^{2}?
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  2. #2
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    Quote Originally Posted by laser View Post
    Let X have the uniform distribution U(-1,3). How do you find the p.d.f. of Y=X^{2}?
    Calculate the cdf of Y and then differentiate it to get the pdf.
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    MHF Contributor matheagle's Avatar
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    You need to be careful.
    This one to one on 1<x<3 and two to one -1<x<1.

    USE f_Y(y)= \biggl(f_X(-\sqrt y) +f_X(\sqrt y)\biggr)\biggl({1\over 2\sqrt y}\biggr).

    Now f_X(x)={1\over 4} on -1<x<3.

    Next notice that y's range is 0 to 9.
    WHEN 0<y<1, we have a 2-1 mapping and both
    f_X(\sqrt y)=f_X(-\sqrt y)=1/4.
    So, for 0<y<1, we have

    f_Y(y)= \biggl( {1\over 4} +{1\over 4} \biggr)\biggl({1\over 2\sqrt y}\biggr)= {1\over 4\sqrt y}.

    But when 1<y<9, we have a 1-1 mapping (with probability one)
    So, f_X(\sqrt y)=1/4 while f_X(-\sqrt y)=0.

    Thus for 1<y<9, we have

    f_Y(y)= \biggl( 0 +{1\over 4} \biggr)\biggl({1\over 2\sqrt y}\biggr)= {1\over 8\sqrt y}.

    AND zero everwhere else.
    Last edited by matheagle; April 23rd 2009 at 10:33 PM.
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