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Math Help - probability distributions

  1. #1
    Junior Member
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    probability distributions

    Suppose that X ~ U(-2, 2). What is the DENSITY of Y = X^4? Note that if
    Z ~ U(a, b) then it has density f(z) = 1/(b - a) and cdf F(z) = (z - a)/(b - a) for a < z < b.

    So I used change of variables ie.
    f(y^(1/4)) = 1/4*dx/dy = (1/4)*(1/4) * y^(-3/4)

    and this is wrong. any ideas what is the mistake?
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  2. #2
    MHF Contributor matheagle's Avatar
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    Note that this is a two to one mapping, Y is defined on (0,16).
    I'll derive it, but it's just like the squaring of a st normal to get a chi-square.

    F_Y(y)=P(Y\le y)=P(X^4\le y)

    here's the algebra involving the 2-1 mapping

    =P(-y^{1/4}\le X\le y^{1/4} )= F_X(y^{1/4})- F_X(-y^{1/4})

    If you differentiate, which is not necessary, you get the sum of the two solutions TIMES the jacobian.

    f_Y(y)=\Biggl(f_X(y^{1/4})+f_X(-y^{1/4})\Biggr){1\over 4y^{3/4}}

    AND when 0<y<16 we get................

    f_Y(y)=\Biggl({1\over 4}+{1\over 4}\Biggr){1\over 4y^{3/4}}={1\over 8}y^{-3/4}

    If y is outside that region you get zero.
    Last edited by matheagle; December 6th 2009 at 11:12 PM.
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