0
$\begingroup$

This question pertains to the theory of Hadamard/elementwise functions of multivariate r.v.s/random vectors, which is unfortunately not a very popular topic:

References for the theory of Hadamard functions and compositions of random vectors

Let ${\mathbf{f}} = \left( {{f_1},...,{f_n}} \right)$ and ${\mathbf{g}} = \left( {{g_1},...,{g_n}} \right)$ be two real random vectors of dimension $n$ . Both of them have an improper, degenerate multivariate Gaussian regularization (prior) probability distribution

$p\left( {\left. {\mathbf{f}} \right|{\varepsilon _f}} \right) \propto {\varepsilon _f}^n{e^{ - \frac{{{\varepsilon _f}^2}}{2}{{\mathbf{f}}^{\mathbf{T}}}{\mathbf{Rf}}}}$ and $p\left( {\left. {\mathbf{g}} \right|{\varepsilon _g}} \right) \propto {\varepsilon _g}^n{e^{ - \frac{{{\varepsilon _g}^2}}{2}{{\mathbf{g}}^{\mathbf{T}}}{\mathbf{Rg}}}}$

where ${\varepsilon _f}$ and ${\varepsilon _g}$ are regularization hyperparameters and ${\mathbf{R}}$ is a singular regularization matrix of dimension $n$ (e.g. of rank deficiency $2$ ).

Let ${\mathbf{h}} = {\mathbf{f}} \oslash {\mathbf{g}} \triangleq \left( {{f_1}/{g_1},...,{f_n}/{g_n}} \right)$ be their Hadamard, elementwise ratio.

What’s the (possibly degenerate) probability distribution of ${\mathbf{h}}$?

Should I first call the disintegration theorem in order to fall on the non-degenerate case? (that would not be natural because the regularization priors are naturally, intrinsically degenerate and that’s perfectly OK as long as the posteriors are not).

Could the degenerate case be simpler than the non-degenerate one? (which I don’t know, a kind of multivariate Cauchy-like distribution?)

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.