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.





About aquazorcarson

math PhD at Stanford, studying probability
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