Martingale and Stopping Time with Finite Expectation

Question

Let $(M_t, \mathcal{F}_t, 0 \leq t < \infty)$ be a martingale. For bounded stopping time $T$, we can deduce from Doob's Optional Sampling that $\mathbb E(M_T)=\mathbb E(M_0)$. Now let $T$ be a stopping time with finite expectation, i.e. $\mathbb E(T)<+\infty$. Can we deduce using $T\wedge n$ and perhaps Lebesgue's Dominated Convergence Theorem that $\mathbb E(M_T)=\mathbb E(M_0)$? If not, is there a sufficient condition for this to be true?

This question arose in studying Brownian motion. For $\tau:=\inf\{t>0:B_0=a,\,B_t=-b\}$ stopping time, we wish to show $\mathbb E(\tau)=ab$. One solution suggests $\mathbb E(B_\tau^2-\tau)=0$ by Doob's Optional Sampling Theorem. I wish to justify this claim with the above.

Answer 1

To see that $E[T] < \infty$ isn't sufficient in general, a discrete-time counterexample is the "doubling martingale". Let $\xi_i$ be iid Rademacher (i.e. taking the values $\pm 1$ with probability 1/2) and let $M_n = \sum_{i=1}^n 2^i \xi_i$, with $M_0 = 0$. (Imagine betting on fair coin flips, where you double your wager on every round.) It's easy to see $M_n$ is a martingale. Let $T = \inf\{i : \xi_i = 1\}$ be the first time that heads is flipped. Then $T$ is a stopping time and we have $M_T = \sum_{i=1}^{T-1} (-1) \cdot 2^i + 2^T = 2$; as soon as you flip a heads, you win back everything you have lost, plus 2 dollars. So the optional sampling theorem fails for $T$. Yet $T$ has a geometric distribution with success probability $1/2$ and one can easily compute that $E[T]=2$.

I haven't checked, but I think one could use similar constructions to show that no "light tails" condition on $T$ could suffice, short of requiring $T$ to actually be bounded.

Answer 2

1) You first need to prove that $E(B_{\tau_{-b,a}})=0$ using Doob's Theorem. In order to do so, observe that $B_{t\wedge \tau_{-b,a}}$ is a bounded martingale. Moreover it can be easily proved that $P(\tau_{-b,a}<\infty)=1$. This is enough to apply Doob's Optional Stopping Theorem (see Revuz Yor's Continuous Martingales and Brownian Motion). Therefore, $$0=E(B_{\tau_{-b,a}})= aP(\tau_{-b,a}=a)-b(1-P(\tau_{-b,a}=a)). $$ This entails that $P(\tau_{-b,a}=a)=\frac{b}{b+a} $.

2) By Ito's formula it follows $$E(B^2_t-t)=0. $$ Applying Doob's Theorem as in step 1): $$ E(B^2_{\tau_{-b,a}})= E(\tau_{-b,a}), $$ but the left hand side equals $$b^2(1-\frac{b}{b+a} )+a^2\frac{b}{b+a} = ab. $$ The result is then proved.