Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

From David William's book Probability with Martingales.

Doob's Optional stopping theorem:

Let $T$ be a stopping time and $X$ be a supermartingale. Then $X_T$ is integrable and $E(X_T)\le E(X_0)$ in each of the following situations

(i) $T$ is bounded (for some $N \in \mathbb{N}$, $T(\omega)\le N$);

(ii) $X$ is bounded (for some $K \in \mathbb{R}^{+}$, $|X(\omega)|\le K$ for every $n$ and every $\omega$) and $T$ is a.s finite.

(There are more conditions but I'm not too worried about them)

My question is that what is the difference between (i) and (ii)? I understand the proof behind them in the sense that I can justify all the steps in the proof but I believe I'm missing some subtlety not required in the proof because intuitively $I$ thought the the conditions placed on T were the same. Proof 1: choose $n=N$, proof 2 uses Dominated Convergence Theorem.

share|improve this question
5  
If $T \sim Geo(p)$, $T$ is a.s. finite but there is no $N$ such that $T(w) \leq N a.s.$ –  madprob Nov 11 '12 at 2:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.