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Jacobian Transformations

  1. Aug 11, 2012 #1
    Why is it that if you have:

    [tex] U=g_1 (x, y), \quad V = g_2 (x,y)[/tex]
    [tex] X = h_1 (u,v), \quad Y = h_2 (u,v)[/tex]


    [tex]f_{U,V} (u,v) du dv = f_{X,Y} (h_1(u,v), h_2 (u,v)) \left|J(h_1(u,v),h_2(u,v))\right|^{-1} dxdy[/tex]

    While when doing variable transformations in calculus, you have:

    [tex]du dv = \left|J(h_1(u,v),h_2(u,v))\right| dx dy[/tex]

    without the reciprocal. Why is it that with the probability densities, you take the reciprocal, rather than how it's typically done without the reciprocal?

  2. jcsd
  3. Aug 14, 2012 #2

    Stephen Tashi

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    Science Advisor

    [tex]du dv = \left|J(h_1(u,v),h_2(u,v))\right| dx dy[/tex]

    We should check that.

    We could begin by looking at a simpler case.

    If we consider the integral [itex] \int_{0}^{1} 1 du [/itex] and used the substitution [itex] x = 2u [/itex], we have [itex] du = (1/2) dx [/itex] and the range of [itex] x [/itex] in the integration is [0,2].

    As I relate this to the notation in your question, [itex] x = h_1(u) = 2u [/itex].
    [itex] | J(h_1(u)]| = 2 [/itex].
  4. Aug 14, 2012 #3
    Thank you.
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