# Jacobian of Inverse Gaussian Transformation in Schwarz & Samanta (1991)

In the sample size $$n=2$$ case when transforming $$\{x_1, x_2\}$$ to $$\{\bar{x}, s\}$$ (where $$X_1, X_2 \overset{iid}{\sim} IG(\mu, \lambda)$$, $$\bar{X}=\frac{\sum_i^2 X_i}{n}$$, and $$S=\sum_i^2 (\frac{1}{X_i}-\frac{1}{\bar{X}}$$), Schwarz and Samanta (1991) write:

I understand the motivation for restricting the transformation to this half-plane, but I'm at a loss for how to find this Jacobian myself. I understand how to find Jacobians for one-to-one transformations, but I'd appreciate a push in the right direction for cases like this.

For simplicity write $$x=x_1$$ and $$y=x_2,$$ so that

$$2\bar x = x+y,\quad s = \frac{1}{x} + \frac{1}{y} - \frac{4}{x+y}.\tag{*}$$

Their Jacobian $$J(\bar x, s)$$ can be computed by comparing the differential elements in the two coordinate systems

$$\mathrm{d} \bar x \wedge \mathrm{d} s = \frac{1}{2}(\mathrm{d} x + \mathrm{d} y) \wedge \left(-\frac{\mathrm{d} x}{x^2} - \frac{\mathrm{d} y}{y^2} + \frac{4(\mathrm{d} x + \mathrm{d} y)}{(x+y)^2}\right)=\frac{1}{2} \left(\frac{1}{x^2}-\frac{1}{y^2}\right)\mathrm{d} x\wedge \mathrm{d} y$$

(These differential forms and the wedge product are described at Construction of Dirichlet distribution with Gamma distribution.)

It remains only to divide both sides by $$(1/x^2 - 1/y^2)/2$$ and then re-express the left hand side entirely in terms of $$\bar x$$ and $$s.$$ Elementary algebra (see below for details) gives

$$\frac{1}{x^2}-\frac{1}{y^2} = \frac{s^{1/2}(2 + s \bar{x})^{3/2}}{\bar{x}^{3/2}},$$

whence

$$\mathrm{d} x\wedge \mathrm{d} y = \frac{1}{\frac{1}{2} \left(\frac{1}{x^2}-\frac{1}{y^2}\right)}\mathrm{d} \bar{x} \wedge \mathrm{d} s =\frac{2\bar{x}^{3/2}}{\sqrt{s}(2 + s \bar{x})^{3/2}}\,\mathrm{d} \bar x \wedge \mathrm{d} s = J(\bar x, s) \,\mathrm{d} \bar x \wedge \mathrm{d} s.$$

Thus, when converting an integral in terms of $$(x,y)$$ to one in terms of $$(\bar x, s)$$, the differential element on the left side must be replaced by the differential element on the right; the absolute value of the factor $$J(\bar x, s)$$ that is introduced is the Jacobian.

Here is one way to do the algebra. Multiplying $$s$$ by $$2\bar x$$ to clear the fraction in $$(*)$$ gives

$$2\bar{x} s = \frac{(y-x)^2}{xy}.$$

Adding $$4$$ to that produces

$$2\bar{x} s + 4 = \frac{(x+y)^2}{xy} = \frac{(2\bar x)^2}{xy},$$

which upon dividing both sides by $$(2\bar x)^2$$ establishes

$$\frac{1}{xy} = \frac{2\bar{x} s + 4}{(2\bar x)^2}.$$

For $$y\gt x \gt 0$$ we may combine the three preceding results to obtain

$$\left(\frac{1}{x^2} - \frac{1}{y^2}\right)^2 = \frac{(y-x)^2(x+y)^2}{(xy)^4} = (2\bar{x}s)(2\bar{x}s + 4)\left(\frac{2\bar{x} s + 4}{(2\bar x)^2}\right)^2.$$

Collecting identical terms in the fraction and taking square roots gives

$$\frac{1}{x^2} - \frac{1}{y^2} = \left(\frac{s(2\bar{x} s + 4)^3}{(2\bar x)^3}\right)^{1/2}$$

which easily simplifies to the expression in the question.