Let $X_1, \dotsc, X_n \sim F$, where $F$ is a distribution function with support in $[0,1]$.
For $t \in [0,1]$, define the sigma-algebra:
$$ \mathscr{F}_t = \sigma(1_{\{X_i \leq s\}}\;,\; 1 \geq s \geq t, i=1,\dotsc,n) $$
Similarly, let:
$$ \mathscr{G}_t = \sigma(1_{\{X_i < s\}}\;,\; 1 \geq s \geq t, i=1,\dotsc,n) $$
Both of these families define filtrations on $[0,1]$ (time direction reversed).
My question is the following:
Assume we now have a stopping rule defined based on the order statistics $X_{(1)}, \dotsc, X_{(n)}$, where we set $X_{(0)} = 0$, and the following borel-measurable functions:
$f: \mathbb R \to \{0,1\}$
$g_k: \mathbb R^{k} \to \{0, 1, \dotsc, n-k\}$
Rule:
- Start at $t = X_{(n)}$.
- Inductive step: (step $k$, $t = X_{(n-k+1)}$) If $f(t) = 1$ or $t=0$, then stop at $t$. Else set $\hat{n} = g_k(X_{(n-k+1)}, \dotsc, X_{(n)})$ and $t = X_{(\hat{n})}$. Continue with the next step.
Is this rule a stopping time in regards to the above filtrations? Intuitively it seems to be that way, since whether we stop on $t$ depends only on values $\geq t$.