Martingale and Stopping Time with Finite Expectation

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.

• to see the DCT argument wont work in general try taking $B$ to be a Brownian motion and $\tau = \inf \{t>0 : B_t = a \}$ Commented Feb 10, 2018 at 20:39
• To apply DCT for the particular example you are interested in, note that $|B_{t \wedge \tau}| \leq \max\{|a|,|b|\}$.
– saz
Commented Feb 10, 2018 at 20:44
• I see. The argument is that $\tau$ is finite a.s. because $B_t$ is unbounded and continuous and $E(B_\tau) = a \neq 0 = E(B_0)$, given that $a \neq 0$, of course. Commented Feb 10, 2018 at 20:46
• Ah that's right, thank you @saz! Commented Feb 10, 2018 at 20:48

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.
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.