Let $(\Omega, \mathfrak A, P)$ be probability space and $\mathcal F$ a filtration. Moreover, let $\mathcal Q$ be the space of probability measures $Q$ and $\mathcal T:=\mathcal T_0:= \{\tau |\tau \text{ a stopping time with } \tau \le T\}\text{ also }\mathcal T_t:=\{\tau \in\mathcal T|\tau \ge t \} \text{ where } t \le T \in \mathbb R$.
I would like to know why for $H_t \in \mathscr L^1(Q)$ the following is obvious:
$$\text{ess inf}_{Q\in \mathcal Q} \text{ ess sup}_{\tau \in \mathcal T_t}E_Q[H_\tau|\mathcal F_t] \ge \text{ess sup}_{\tau \in \mathcal T_t} \text{ ess inf}_{Q\in \mathcal Q}E_Q[H_\tau|\mathcal F_t].$$
My attempt is the following: It is true that
$$\begin{align*} \text{ess sup}_{\tau \in \mathcal T_t}E_Q[H_\tau|\mathcal F_t] &\ge E_Q[H_\tau|\mathcal F_t] \\ \Rightarrow \text{ ess inf}_{Q\in \mathcal Q} \ \text{ess sup}_{\tau \in \mathcal T_t}E_Q[H_\tau|\mathcal F_t] &\ge \text{ ess inf}_{Q\in \mathcal Q}E_Q[H_\tau|\mathcal F_t] \\ \Rightarrow \text{ess sup}_{\tau \in \mathcal T_t}\left(\text{ ess inf}_{Q\in \mathcal Q} \ \text{ess sup}_{\tau \in \mathcal T_t}E_Q[H_\tau|\mathcal F_t]\right) &\ge \text{ess sup}_{\tau \in \mathcal T_t}\text{ ess inf}_{Q\in \mathcal Q}E_Q[H_\tau|\mathcal F_t]. \tag 1 \end{align*}$$ Since we took the essential supremum of $\tau$ on the left side in $(1)$ we can ignore the uttered essential supremum in $(1)$ and get $$\text{ess inf}_{Q\in \mathcal Q} \text{ ess sup}_{\tau \in \mathcal T_t}E_Q[H_\tau|\mathcal F_t] \ge \text{ess sup}_{\tau \in \mathcal T_t} \text{ ess inf}_{Q\in \mathcal Q}E_Q[H_\tau|\mathcal F_t].$$
Is this correct?
Any help is appreciated!
We write $Y=:\text{ess sup}_{i \in I}X_i$.
Now the "index" is the set of all probability measures. Hope this makes it clear.
– scholar Nov 22 '22 at 22:37