## a nice optional sampling problem

While learning the proof of Burkholder-Gundy-Davis inequality, I had to review the statement and proof of Doob’s maximal inequality (one of my favorite results in elementary probability). In the proof of the latter, I had to recall the meaning of the stopped filtration $\mathcal{F}_\tau$, as a critical step involves the

$\displaystyle X^*_\tau 1_{X^* \ge K} = K 1_{\tau \le t} \le \mathbb{E}(X_t 1_{\tau \le t} | \mathcal{F}_\tau)$

where $X^*_t$ is the running max of $X_t$, and $\tau = \inf\{t: X_t \ge K\}$ is the first time $X$ exceeds $K$.

The last inequality appears a bit mysterious to me (still). So I looked into optional sampling theorem on wiki. Unfortunately nothing about the stopped filtration $\mathcal{F}_\tau$ is explained there, but I did find an interesting exercise that took me a bit of time:

Suppose S(N) is with probability one either 100 or 0 and that S(0) = 50. Suppose further there is at least a sixty percent probability that the price will at some point dip to below 40 and then subsequently rise to above 60 before time N. Prove that S(n) cannot be a martingale.

Under the excuse of giving my readers a relaxing time, here is how I approached the problem (perhaps the only way): consider the stopping time $\tau = \min\{n: S_n \in \{40,60\}\}$. Then first observe that $\tau \le N$ almost surely, since $S_n$ has to reach either $0$ and or $100$ at $N$ and by continuity it has to go through $40$ or $60$. However $\mathbb{E}_0(S_\tau) \le 40 * 60\% + 60 * 40\% < 50$ by the probability assumption, thus violating optional sampling theorem.

Back to the meaning of the stopped filtration: the definition I recall is $A \in \mathcal{F}_\tau$ iff $A \cap \{\tau < t\} \in \mathcal{F}_t \forall t$. This definition is both opaque and natural. Then I thought maybe in the special case when the filtration $\mathcal{F}_t$ is generated by a single process $X_t$, $\mathcal{F}_\tau$ can be alternatively defined as $\sigma(\{X_{\tau \wedge t}: t \ge 0\})$. This is the subject of a stackexchange post (see ref 3). The conclusion there seems that the alternative definition holds only when $X_t$ is cadlag (continue a droit limite a gauche), or in other words the filtration is progressively measurable. The proof doesn't seem so trivial. I am sure it's true in the discrete case though still need to find time to see the actual proof.

http://almostsure.wordpress.com/2010/04/06/the-burkholder-davis-gundy-inequality/

http://math.stackexchange.com/questions/168566/stopped-filtration-filtration-generated-by-stopped-process