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Math Help - density function question..

  1. #1
    MHF Contributor
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    density function question..

    Y\sim U(-2\pi,2\pi)
    find the density function of z=tan(Y)
    ?

    i had a similar question

    X~U(0,1)
    find the density function of W=a+bx
    the solution is
    W~U(a,a+b)

    how to solve the first question ??
    Last edited by CaptainBlack; February 2nd 2010 at 03:21 AM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    READ below, I think the orginal density is WRONG.

    USE f_Z(z)=f_Y(y)\left| {dy\over dz}\right| which is easy to derive.

    It's the derivatives of the CDFs, it's just calc 1.

    So, f_Y(y)={1\over 4\pi} on (-2\pi, 2\pi)

    and y=\arctan z giving you {dy\over dz}={1\over 1+z^2}

    Hence f_Z(z)={1\over  4\pi(1+z^2)}

    I THINK there is a problem here, we should be getting a Cauchy.

    I think the original density should be a uniform on (-\pi/2, \pi/2) because tangent isn't defined at \pi/2
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  3. #3
    MHF Contributor matheagle's Avatar
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    The second one is obvious.

    f_X(x)=1 on 0<x<1.

    If you let W=a+bX then W exists on (a,a+b), by letting X=0 and X=1.
    Next, I will show that W is uniform on that interval.

    USE f_W(w)=f_X(x)\left| {dx\over dw}\right|

    With w=a+bx we have {dw\over dx}=b

    So, USE f_W(w)=(1)\left| {1\over b}\right|={1\over b} IF b>0.

    which is a uniform density.
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