Stopping independent random variables

Question

Let $\{X_n\}$ be independent (and identically distributed if necessary) integrable random variables. Consider the stopping time, adapted to the natural filtration as usual, $\tau$ (that is $\{\tau=n\}\in\mathcal{F}_n$ for all $n\in\mathbb{N}$). Denote the class of stopping times that do not stop before $n$ as $T_n=\{\tau\:\text{stopping time}: \tau\geq n\}$.

Is it true that, as intuitively seems reasonable, for all $n$, $\operatorname{ess\, sup} \mathbb{E}_{\mathcal{F}_n}X_\tau=\sup \mathbb{E}X_\tau$ for the essential supremum and supremum running over all $\tau\in T_{n+1}$? If not, can we show at least that they are deterministic numbers $v_n=\operatorname{ess\, sup} \mathbb{E}_{\mathcal{F}_n}X_\tau$?

The intuition comes from the following reasoning: since $\tau$ does not stop by time $n$, the information given by the history is irrelevant for the future reward. However I cannot formalise this quite well, since the random variables $\mathbb{1}_{\tau=i}$ for all $i\geq n+1$ involved in the conditional expectation above are, I believe, not necessarily independent of $\mathcal{F}_n$, they are only, as far as I know, only $\mathcal{F}_i$-measurable, and although $\mathcal{F}_n\subset\mathcal{F}_{n+1}\subset\ldots$, I do not think this translates directly into independence, just because $\mathcal{F}_{n+1}=\sigma(\mathcal{F}_n,\sigma(X_{n+1}))$ and $\sigma(X_{n+1})$ is independent of $\mathcal{F}_n$. Thanks for any help.

Answer 1

I assume the $X_n$ are iid and that there exists $C\in\Bbb R$ such that $P[X_1\le C]=1$.(The situation in which $P[X_1 >C]>0$ for all real $C$ is treated in a similar fashion.)

Fix $n$. Fix $\epsilon>0$ and let $B=[C-\epsilon,C]$. Then $P[X_k\in B]>0$ (and doesn't depend on $k$) and so the stopping time $$ \tau:=\inf\{k\ge n+1: X_k\in B\} $$ is a.s. finite and an element of $T_n$. Clearly $\Bbb E[X_\tau]\ge C-\epsilon$. It follows that $\sup_{\tau\in T_n}\Bbb E[X_\tau]=C$.

On the other hand $\Bbb E_{\mathcal F_n}[X_\tau]\le C$ a.s. because $\Bbb P[X_k\le C,\forall k]=1$. In the reverse direction fix $n$ and define $\tau$ as before. Notice that $\tau$ and (more importantly) $X_\tau$ are independent of $\mathcal F_n$. Therefore $\Bbb E_{\mathcal F_n}[X_\tau]=\Bbb E[X_\tau]\ge C-\epsilon$, a.s.. It follows that $$ \operatorname{ess\, sup}_{\tau\in T_n}\Bbb E_{\mathcal F_n}[X_\tau]=\sup_{\tau\in T_n}\Bbb E[X_\tau]=C, $$ for all $n$.

Answer 2

When $X_1$ is essentially unbounded above, given any real $L$ we have $\Bbb P[X_1\ge L]>0$. Now define $\tau:=\inf\{k\ge n+1: X_k \ge L\}$. Then $\tau$ is a.s. finite and $\Bbb E[X_\tau]\ge L$. It follows that $\sup_{\tau\in T_n}\Bbb E[X_\tau]=+\infty$. Because the $\tau$ just constructed is independent of $\mathcal F_n$, we even have $\Bbb E_{\mathcal F_n}[X_\tau]\ge L$, so the reverse inequality follows as before.