14
$\begingroup$

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 $E(\sup_{n \in \mathrm{N}} B^2_{T\wedge n})<\infty$.

$\endgroup$
7
  • $\begingroup$ 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 $\mathbb L^2$ $\endgroup$
    – mike
    Apr 25, 2012 at 21:23
  • $\begingroup$ @ Mycroft : You can have a look at Proposition 5 there :math.tau.ac.il/~peledron/Teaching/RW_and_BM_2011/scribe12.pdf Best Regards $\endgroup$
    – TheBridge
    Apr 26, 2012 at 8:11
  • $\begingroup$ 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. $\endgroup$
    – Kolmo
    Apr 27, 2012 at 17:09
  • $\begingroup$ @TheBridge I thought I might be able to circumvent the strong Markov property apparatus and find a more elegant solution. $\endgroup$
    – Hans-Peter Schrei
    Apr 27, 2012 at 19:22
  • $\begingroup$ @Kolmo Thanks, I will have a look at it. $\endgroup$
    – Hans-Peter Schrei
    Apr 27, 2012 at 19:22

2 Answers 2

Reset to default
8
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $E(\sup_{t \geq 0} B^2_{T \wedge t}) \leq 4 E(B^2_T) < \infty$.

$\endgroup$
3

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .