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LaPlacian joint probability density function.

  1. May 7, 2013 #1
    A joint pdf is given as pxy(x,y)=(1/4)^2 exp[-1/2 (|x| + |y|)] for x and y between minus and plus infinity.

    Find the joint pdf W=XY and Z=Y/X.

    f(w,z)=∫∫f(x,y)=∫∫(1/4)^2*e^(-(|x|+|y|)/2)dxdy -∞<x,y<∞
    Someone told me I can not use Jacobian because of the absolute value. Is that true?
    So far this is what I have but I feel like I am not going anywhere.

    f(w,z)=(1/4)^2∫∫e^(-(|x|+|y|)/2)dxdy
    =(1/4)^2∫∫[e^-|x|/2]*e^-|y|/2]
    =(1/4)^2∫[e^-|x|/2]∫e^-|y|/2]

    =(1/4)^2[∫[e^(-x/2)+∫e^(x/2))] * [∫[e^(-y/2)+∫e^(y/2))] the limits from -∞<x,y<0 and 0<x,y<∞

    =[(1/4)^2 ]*4*[ ∫ [e^(x/2)dx] + ∫ [e^(y/2)dy] ] 0<x,y<∞



    I have problems with transforming the limits for the new functions.
     
    Last edited: May 8, 2013
  2. jcsd
  3. May 8, 2013 #2

    Ray Vickson

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    Of course you can use the Jacobian, but you need to be careful to ensure that the variables therein do not cross zero; that is, |x| is perfectly well differentiable as long as x > 0 or x < 0. So, for positive w and z, what is the WY-region of {w < W < w+dw, z < Z < z+dz}? You ought to be able to find it as the union of two disjoint subregions, and in each subregion you can certainly use a Jacobian.

    Now you need to look at the other cases where w> 0 and z > 0 do not both hold.
     
  4. May 8, 2013 #3
    That is the part where I got stuck.
    What is the best way to solve this problem? should I use Jacobian or should I use the distribution properties?

    But my biggest question and where I need help is with the boundaries.
     
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