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:

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