# 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 $E(M_T) = E(M_0)$. Now let $T$ be a stopping time with finite expectation, i.e. $E(T) < \infty$. Can we deduce using $T\wedge n$ and perhaps Lebesgue Dominated Convergence Theorem that $E(M_T) = E(M_0)$? If not, is there a sufficient condition for this to be true?

This question arose in studying Brownian motion. For $\tau := \inf\{t > 0 : B_t = a, B_t = -b\}$ stopping time, we wish to show $E(\tau) = ab$. One solution suggests $E(B_\tau^2 - \tau) = 0$ 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 $\tau = \inf \{t>0 : B_t = a \}$ – Rhys Steele Feb 10 '18 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 Feb 10 '18 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. – James Yang Feb 10 '18 at 20:46
• Ah that's right, thank you @saz! – James Yang Feb 10 '18 at 20:48

## 1 Answer

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.