Let $Y$ be a nonnegative random random variable on some probability space, and Let $F:[0,\infty) \to \mathbb R$ be continuous function.
How to prove that the "vector expectation" $(E(Y),E(F(Y)))$ lies
in the closed convex hull of $G=\{(x,F(x))\,|\, x\ge0\} \subseteq \mathbb R^2$?
Edit:
Robert's answer here uses the hyperplane separation theorem. I wonder whether a more constructive approach is possible-say by expressing the integral as a limit of averages. Here is a possible approach:
Since $Y \ge 0$, there exist an increasing sequence of simple functions $Y_n$ converging to $Y$;
Write
$$ Y_n=\sum_i a_n^i \chi_{E_n^i}, \, \,\,\, (E_n^i)_i\text{ are disjoint}. $$
Then $ E(Y_n)=\sum_i a_n^i \mu(E_n^i), E(F(Y_n))=\sum_i F(a_n^i) \mu(E_n^i)$.
Thus
$$
\big( E(Y_n),E(F(Y_n))\big)=\sum_i \mu(E_n^i) \big( a_n^i,F(a_n^i) \big) \in \text{conv}(G)
$$
is a convex combination of elements of $G$, for every $n$.
Since $Y_n$ is increasing, the monotone convergence theorem implies that
$$
E(Y)=\lim_{n \to \infty} E(Y_n).
$$
If we could make $E(F(Y))=\lim_{n \to \infty} E(F(Y_n))$ as well, then we would realize
$$
(E(Y),E(F(Y)))=\lim_{n \to \infty} (E(Y_n),E(F(Y_n)))
$$
as a limit of elements in $\text{conv}(G)$, thus it lies in $\overline{\text{conv}(G)}$ as required.
$E(F(Y))=\lim_{n \to \infty} E(F(Y_n))$ holds whenever $F$ is monotonic for instance, but I am not sure if we can arrange it in general.
Best Answer
Hint: if it isn't in the closed convex hull, it can be separated from it by a linear functional.