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Math Help - joint probability density function, using the Jacobian

  1. #1
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    joint probability density function, using the Jacobian

    if X1 and X2 are independent random variables from the standard normal distribution i have worked out the joint probability density function of X1,X2 to be
    (1/2π) exp[-1(x12 + x22 ) /2 ] (is this correct?)

    then if Y1=X1 and Y2=X2/X1 the joint probability density function of Y1,Y2 is
    (1/2π) exp[-1y12 (1+y22)/2) ] |J|

    i worked out J to be y1

    is that all correct and what does the joint probability density function of Y1, Y2 then become?
    thanks

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  2. #2
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    *where it says x12 and y22 etc this is x1 squared and y2 squared etc, dont no why it came out like that
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  3. #3
    MHF Contributor harish21's Avatar
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    good job.

    Although your notations are really hard to understand, your method looks correct.. You just need to plug in the value of |J| and you have the joint pdf
    Last edited by harish21; November 5th 2010 at 09:45 AM.
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  4. #4
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    i will try and write it out a bit fuller and clearer;

    X1 and X2 are independent random variables from a standard normal distribution
    so the joint p.d.f of X1, X2 = p.d.f X1 x p.d.f X2
    = ( 1/(2pie)^1/2) ) exp[ -1(x1^2)/2 ] ( 1/(2pie)^1/2) ) exp[ -1(x2^2)/2 ]
    = (1/2pie) exp[ -1(x1^2 + x2^2)/2 ]

    If Y1 = X1 and Y2 = X2/X1 then the joint p.d.f of Y1, Y2 is
    = (1/2pie) exp[ -1(y1^2 + (y1y2)^2)/2 ] |J|
    = (1/2pie) exp[ -1(y1^2(1+y2^2))/2 ] |J|

    where J = y1, so joint p.d.f of Y1, Y2 is
    = (y1/2pie) exp[ -1(y1^2(1+y2^2))/2 ]

    is this all correct?
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  5. #5
    MHF Contributor harish21's Avatar
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    Do not forget the SUPPORT of the distribution.

    and,

    Pie: the one you eat.

    Pi: \pi \;=\; 3.14159
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