# How can I use transformation properties to obtain the distribution of $h(\mathbf{s})$?

Let that $$\mathbf{s}=(s_1,s_2) \sim Unif(S)$$, where $$S$$ is some spatial area. Suppose $$y=h(\mathbf{s})=1-[exp(exp(\beta_0+\beta_1(\mathbf{s}-\mathbf{x})^T(\mathbf{s}-\mathbf{x})))]^{-1}$$. We have that $$\frac{\partial h}{\partial \mathbf{s}}=2\beta_1e^{\beta_0+\beta_1\cdot Tr[(\mathbf{s}-\mathbf{x})^T(\mathbf{s}-\mathbf{x})]}exp[-(\beta_0+\beta_1\cdot Tr[(\mathbf{s}-\mathbf{x})^T(\mathbf{s}-\mathbf{x})])](\mathbf{s}-\mathbf{x})$$. How can I obtain the distribution for $$h(\mathbf{s})$$ analytically?

My idea:

I was thinking of using Theorem 2.1.5 in Casella and Berger: $$f_y(y)=f_\mathbf{s}(h^{-1}(y))|\frac{d}{d\mathbf{s}}h^{-1}(y)|$$

However, I'm not sure if this theorem extends to the multivariate case that I have.

• Given that $S$ is only some vague "spatial area," could you explain what it might mean to write the distribution of $h(s)$ "analytically"? – whuber Jul 27 at 20:19
• The expression will be nasty even in that case, because level sets of $y$ are circles (concentric around $x$) while the boundaries are portions of lines. That makes this an unappetizing question. Sometimes the original statistical problem will suggest useful approximations. Could you provide some background? – whuber Jul 27 at 21:38
• How far is $x$ from $S$? The problem of approximating $h$ becomes three separate ones depending on whether $x$ is far from $S$ (compared to its diameter), near $S$, or within $S.$ – whuber Jul 27 at 22:11
• You will need to evaluate the integral numerically. – whuber Jul 28 at 13:44
• Clearly the integral depends on the details of the boundary of $S:$ in that sense it's obviously messy except, perhaps, when the boundary is a circle centered at $x.$ It remains possible that an analytical solution exists for the unit square case, but I doubt anyone would want to work it out. Your best bet would be to apply Mathematica to the problem to see if it can produce anything for a specific, extremely simple $x$ (such as the center of the square or one of its corners). If it can't even make progress in that case, the problem is hopeless. – whuber Jul 28 at 15:15