Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    Then:

    [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?

    Thanks!
     
  2. jcsd
  3. Aug 14, 2012 #2

    Stephen Tashi

    User Avatar
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Jacobian Transformations
Loading...