Why do Jacobian transformations in probability densities require a reciprocal?

[IniquiTrance]

Why is it that if you have:

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

Then:

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

While when doing variable transformations in calculus, you have:

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

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!

 

[Stephen Tashi]

While when doing variable transformations in calculus, you have:

du dv = \left|J(h_1(u,v),h_2(u,v))\right| dx dyWe should check that.

We could begin by looking at a simpler case.

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

As I relate this to the notation in your question, x = h_1(u) = 2u.
| J(h_1(u)]| = 2.

 

[IniquiTrance]

Thank you.