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.

In the course of proving Wald's second identity $E(B^2_T)=E(T)$, where $(B_t)_{t\geq0}$ is the Brownian motion and $T$ is a stopping time with $E(T)<\infty$, I got stuck with the following problem. The notation used is $T \wedge n = \min(T,n)$.

I already have $$E(T)=E(\lim_{n \to \infty} T \wedge n)\\ =\lim_{n \to \infty} E(T \wedge n)\\ =\lim_{n \to \infty} E(B^2_{T\wedge n}).$$ by monotone convergence and the optional stopping theorem.

Furthermore, by the Lemma of Fatou $$E(B^2_T)=E(\lim_{n \to \infty} B^2_{T\wedge n})\\ \leq \liminf_{n \to \infty} E(B^2_{T\wedge n})\\ = E(T) < \infty.$$

And now I am stuck with the other direction. I tried to use the dominant convergence theorem to exchange the limits in $E(\lim_{n \to \infty} B^2_{T\wedge n})=\lim_{n \to \infty}E(B^2_{T\wedge n})$, but I can't find a suitable integrable dominating function for $B^2_{T\wedge n}$.

Doob's inequality for stopping times yields $$E(\sup_{t \geq 0} B^2_{t \wedge T\wedge n})\leq 4 E(B^2_{T\wedge n}) \leq 4 E(T) <\infty,$$ but what I need is $E(\sup_{n \in \mathrm{N}} B^2_{T\wedge n})<\infty$.

share|improve this question
I just did the random walk version for my stochastic processes class, and what I told them, which may be true for all I know, is that since the $ \sum X_iI_{(T>i-1)}$ is a series of uncorrelated random variables, clearly convergent in $\mathbb L^2$ provided $E(T)\; (= \sum P(T>i-1)) < \infty$ so nothing can go wrong. This is a stochastic integration approach to Wald, and the continuous version would show that $E(B_{T_n} - B_{T_m})^2 \le E(T_n - T_m)$ , making it cauchy in $\mathbb L^2$ –  mike Apr 25 '12 at 21:23
@ Mycroft : You can have a look at Proposition 5 there :math.tau.ac.il/~peledron/Teaching/RW_and_BM_2011/scribe12.pdf Best Regards –  TheBridge Apr 26 '12 at 8:11
I think you can also work with $\int_0^{\infty}Y(t)dB_t$ where $Y(t)=1_{ t \in [0,T]}(t)$ and use Ito isometry. –  Kolmo Apr 27 '12 at 17:09
@TheBridge I thought I might be able to circumvent the strong Markov property apparatus and find a more elegant solution. –  hpschrei Apr 27 '12 at 19:22
@Kolmo Thanks, I will have a look at it. –  hpschrei Apr 27 '12 at 19:22

2 Answers 2

up vote 1 down vote accepted

After leaving the problem for a while, I found the rather obvious solution on reinspection.

The integrable function dominating $B^2_{T \wedge n}$ that I was looking for is $\sup_{t \geq 0} B^2_{T \wedge t}$.

We have $B^2_{T \wedge n} \leq \sup_{t \geq 0} B^2_{T \wedge t} \forall n \in \mathrm{N}$ and by Doob's inequality $E(\sup_{t \geq 0} B^2_{T \wedge t}) \leq 4 E(B^2_T) < \infty$.

share|improve this answer
Thanks. I was trying to solve the exact same problem and this hint helped me along! –  Abhijit May 11 '13 at 23:25

I would just like to point out an easier way to do this. Define $M_n = B_{T \wedge n}$ and note that $M$ is a martingale which is bounded in $L^2$. Namely $E[M_n^2] = E[T\wedge n] \leq E[T]<\infty.$ Since $M$ is a martingale bounded in $L^2$ there exists $M_\infty$ such that $M_n \to M_\infty$ a.s. and in $L^2$. Since $M_n \to B_{T}$ as well, we have that $M_\infty = B_{T}$ a.s. In particular $E[B_T] = \lim E[M_n] = 0$ and $E[B_T^2] = \lim E[M_n^2] = T$ which proves both of Wald's identities.

share|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.