# Multivariate t vs independent Cauchy random variables

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?

• According to the same article, it is an extension of this (see the notes on independency there). – metamorphy Oct 3 '18 at 15:20
• @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 – alfalfa Oct 3 '18 at 15:31