How to prove that the expectation of a random vector lies in the convex hull of its support?

Question

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.

Answer 1

Hint: if it isn't in the closed convex hull, it can be separated from it by a linear functional.

Answer 2

This answer is merely an elaborate version of Robert's Israel sketch:

We shall use the fact that the closed convex hull of $G$ is the intersection of all closed half-spaces containing $G$.

Let $H$ be a closed half-space containing $G$, and write $H=\{ x \in \mathbb R^2 \, | \,\langle x,v \rangle \ge a\}$ for some $a \in \mathbb R$ and $v=(v_1,v_2) \in \mathbb R^2$.

$H \supseteq G=\{(x,F(x))\,|\, x\ge0\}$ implies that $(t,F(t))\in H$ for every $t \ge 0$, or

$$ tv_1+F(t)v_2 \ge a \,\,\,\,\,\text{for every} \, \, \, t \ge 0. \tag{1}$$

Now, recall $Y:X \to [0,\infty)$ is a nonnegative random random variable on a probability space $X$. Putting $t=Y(x)$ in $(1)$, we see that $$ Y(x)v_1+(F\circ Y)(x)v_2 \ge a \,\,\,\,\,\text{for every} \, \, \, x \in X. \tag{2} $$

Integrating this, we get $$ E(Y)v_1+E(F\circ Y)v_2 \ge a $$ or $(E(Y),E(F\circ Y)) \in H$ which is what we wanted.