2
$\begingroup$

Two probability distributions $\mu, \nu$ on $\mathbb R^d$ are said to be increasing in convex order if

$$\int_{\mathbb R^d} |x|\mu(dx) + \int_{\mathbb R^d} |x|\nu(dx)<\infty$$

and

$$\int_{\mathbb R^d} f(x)\mu(dx) \le \int_{\mathbb R^d} f(x)\nu(dx),\quad \mbox{for all convex function } f:\mathbb R^d\to\mathbb R \mbox{ of linear growth}.$$

It is known that for $d=1$, $\mu, \nu$ are increasing in convex order iff

$$\int_{\mathbb R} (x-K)^+\mu(dx) \le \int_{\mathbb R} (x-K)^+\nu(dx),\quad \mbox{for all } K\in\mathbb R.$$

For $d=2$, do we still have a similar class of test functions (instead of all convex functions) ensuring that $\mu, \nu$ are increasing in convex order?

$\endgroup$

0

You must log in to answer this question.

Browse other questions tagged .