Wald lemma for Brownian motion

Question

In the following question, $B$ refers to Brownian motion.

I am having a problem with a particular part of the solution of the following question (picture below). More precisely, how to prove that $$M(4^{p+1})-M(4^p) \leq |B_{4^{p+1}}-B_{4^p}|?$$

Picture:

We wish to show that when $T$ is a stopping time with $\mathbb{E}\left[T^{1 / 2}\right]<\infty$, Wald's lemma still applies and $\mathbb{E}\left[B_{T}\right]=0$

(1) Define $\tau:=\min \left\{k: 4^{k} \geq T\right\}$. Set $M(t):=\max _{[0, t]} B$ and $X_{k}:=M\left(4^{k}\right)-2^{k+2}$. Show that $\left(X_{k}\right)$ is a supermartingale for the filtration $\left(\mathcal{F}_{4^{k}}\right)_{k}$, and that $\tau$ is a stopping time.

Solution (1) Define $\tau:=\min \left\{k: 4^{k} \geq T\right\}$. Set $M(t):=\max _{[0, t]} B$ and $$ \mathbb{E}\left[X_{k+1}-X_{k} \mid \mathcal{F}_{4^{k}}\right]=\mathbb{E}\left[M\left(4^{k+1}\right)-M\left(4^{k}\right) \mid \mathcal{F}_{4^{k}}\right]-4 \times 2^{k} $$ Since we know that almost surely $M\left(4^{k+1}\right)-M\left(4^{k}\right) \leq\left|B_{4^{k+1}}-B_{4^{k}}\right|$ which is independent of $\mathcal{F}_{4^{k}}$ and distributed like $\left|B_{4^{k+1}-4^{k}}\right|$, then $$ \mathbb{E}\left[X_{k+1}-X_{k} \mid \mathcal{F}_{4^{k}}\right] \leq \mathbb{E}\left[\left|B_{4^{k+1}-4^{k}}\right|\right]-4 \times 2^{k}=\sqrt{3 \times 4^{k}} \mathbb{E}\left[\left|B_{1}\right|\right]-4 \times 2^{k} $$ A simple application of Cauchy-Schwarz or Jensen gives $\mathbb{E}\left[\left|B_{1}\right|\right] \leq \sqrt{\mathbb{E}\left[\left|B_{1}\right|^{2}\right]}=1$, and the expectation above is bounded by 0 .

Answer 1

The proof seems wrong take the event where $B_{4^{p+1}}=B_{4^p}=M(4^p)=2$, and $M(4^{p+1})=4$ then $2=M(4^{p+1})-M(4^p) \not\leq |B_{4^{p+1}}-B_{4^p}|=0$

This event has null proba but you can take intervals of non null proba instead and obtain the claim wrong. For example take the event where, $B_{4^p}\in (1,2)$, $|B_{4^{p+1}}-B_{4^p}|< 1/2$, $|M_{4^{p}}-B_{4^p}|\leq 1$ and $M(4^{p+1}) \in (9,10)$ then
$M(4^{p+1})-M(4^p)>9 -(2+1) =6$ and $|B_{4^{p+1}}-B_{4^p}|\leq 1/2$

All those events combined have non null probability (even though I really don't want to calculate this)

So we don't have almost surely :

$$M(4^{p+1})-M(4^p) \leq |B_{4^{p+1}}-B_{4^p}|$$

Answer 2

I am the original author of the document you lifted this solution from. Indeed the solution is plain wrong but the stochastic domination of $M\left(4^{k+1}\right)-M\left(4^{k}\right)$ by $\left|B_{4^{k+1}-4^{k}}\right|$ independently of $\mathcal{F}_{4^{k}}$ holds, as the reflection principle can show. I thank you and all other posters for your careful reading and will update the document in the near future.

Set $M(t):=\max _{[0, t]} B$.

We have the following mutually exclusive cases.

• Either $M\left(4^{k+1}\right) = M\left(4^{k}\right)$ in which case $M\left(4^{k+1}\right)-M\left(4^{k}\right) = 0$.

• Or $M\left(4^{k+1}\right) > M\left(4^{k}\right)$ meaning $M\left(4^{k+1}\right)= \max_{[4^k,4^{k+1}]} B$. On the other hand, $M\left(4^{k}\right)\geq B_{4^{k}}$. Together, this gives $M\left(4^{k+1}\right) - M\left(4^{k}\right) \leq \max_{[4^k,4^{k+1}]} B - B_{4^{k}}$.

Putting the two cases together implies the certain inequality $$M\left(4^{k+1}\right) - M\left(4^{k}\right) \leq \max_{[4^k,4^{k+1}]} B - B_{4^{k}}$$ (bear in mind the r.h.s. is nonnegative).

Now the r.h.s. is

• Independent from $\mathcal{F}_{4^{k}}$, and distributed like $\max_{[0,4^{k+1}-4^k]} B$ by Markov property.
• Distributed in turn like $\left|B_{4^{k+1}-4^{k}}\right|$ by reflection principle.