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Let $X= (X_1,\dots, X_d)$ be a fixed vector of random variables on the space $(\Omega, \mathcal{F}, \mathbb{P})$. Consider the following set.

\begin{equation}\label{main12} C= \{x\in \mathbb{R}^d ~|~ h(x)\leq \mathop{\sup}_{Q\in \mathcal{D}}\mathbb{E}_{Q}(h(X)),\qquad \forall h\in \mathcal{A}\}, \end{equation}

where $\mathbb{E}_{Q}$ is the expectation with respect to a measure $Q$ (integration in measure theory language), $\mathcal{D}$ is a set of probability measures and $\mathcal{A} = \{h~|~ h: \mathbb{R}^d \rightarrow \mathbb{R},\qquad h \hbox{ is a non-linear convex function}\}$.

Can we characterize the set $C$? Or, at least can we say something about how to approximate the elements of $C$?

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Suppose $x\in C$. Then we can make $h_n$ closer and closer to the projection $g$ given by $g(x)=x_i$, and $\hat h_n$ closer and closer to $\hat g(x)=-x_i$, so upon taking limits, $x_i\le \sup\mathbb E_Q(X_i)$ and $-x_i\le\sup \mathbb E_Q(-X_i)=-\inf \mathbb E_Q(X_i)$.

(May need to assume $\mathcal D$ is nice enough that the sup's are actually max's here.)

Then by Jensen's inequality, $$ \{E_Q(X): Q\in \mathcal D\}\subseteq C\subseteq \{x: x_i\in [\inf_Q\mathbb E_Q(X_i), \sup_Q\mathbb E_Q(X_i)]\quad\forall i\}. $$

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  • $\begingroup$ You're welcome, it is a kind of approximation of $C$... $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Mar 19, 2014 at 0:53
  • $\begingroup$ Can we change the class $\mathcal{A}$ of functions with another class like $\mathcal{A_0}$ of functions such that we could get almost the same set as $C$ or some set which gives you a good approximation of $C$? Some class,$\mathcal{A_0},$ for which we have a nice representation... $\endgroup$
    – Math123
    Commented Mar 19, 2014 at 15:11
  • $\begingroup$ I don't know... what's the motivation? $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Mar 19, 2014 at 15:55
  • $\begingroup$ What we can say about the border of $C$, $\partial C$? $\endgroup$
    – Math123
    Commented Mar 21, 2014 at 17:49
  • $\begingroup$ I'm not sure, maybe ask that as a new question... $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Mar 21, 2014 at 17:51

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