Dominated convergence problems with Wald's identity for the Brownian Motion

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 $E(T)&lt\infty$, 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 $$E(B^2_T)=E(\lim_{n \to \infty} B^2_{T\wedge n})\\ \leq \liminf_{n \to \infty} E(B^2_{T\wedge n})\\ = E(T) &lt \infty.$$

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 $$E(\sup_{t \geq 0} B^2_{t \wedge T\wedge n})\leq 4 E(B^2_{T\wedge n}) \leq 4 E(T) &lt\infty,$$ but what I need is $E(\sup_{n \in \mathrm{N}} B^2_{T\wedge n})&lt\infty$.

Solution 1:

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.

Solution 2:

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