0
$\begingroup$

If I have $X,Y\stackrel{\text{iid}}\sim \text{Cauchy}$ so $$ f_X(x) = \frac{1}{\pi(1+x^2)} $$ and identically for $f_Y$ then I should have $$ f_{XY}(x,y) = f_X(x) f_Y(y) = \frac{1}{\pi^2 (1+x^2)(1+y^2)}. $$

But the pdf of a central multivariate $t$ distribution with one degree of freedom is $$ f_t(x; \Sigma, p) = \frac{\Gamma\left(\frac{p+1}2\right)}{\pi^{(p+1)/2}|\Sigma|^{1/2}} \left[1 + x^T\Sigma^{-1}x\right]^{-(p+1)/2} $$ (according to wikipedia) and these two are definitely not the same no matter my choice for $\Sigma$. Why is this? Why doesn't the multivariate $t$ include the case with two independent Cauchy RVs? Why does the multivariate $t$ force some amount of dependence?

$\endgroup$
  • $\begingroup$ According to the same article, it is an extension of this (see the notes on independency there). $\endgroup$ – metamorphy Oct 3 '18 at 15:20
  • $\begingroup$ @metamorphy i did see that, but I guess my question is more about why. Maybe there's no satisfying answer, but I find this to be a very surprising property. It seems very weird to me that a "generalization" would lose the ability to model independent RVs $\endgroup$ – alfalfa Oct 3 '18 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.