# Wald’s identity for Brownian motion with $E[\sqrt T]<\infty$.

It's the Exercise 3.3.35 of Karatzas and Shereve: Brownian Motion and Stochastic Calculus on page 168.

Let $$W=\{W_t,\mathscr{F}_t; 0\leq t<\infty\}$$ be a standard, one-dimensional Brownian motion, and let $$T$$ be a stopping time of $$\{\mathscr{F}_t\}$$ with $$E[\sqrt T]<\infty$$. Prove that $$E[W_T]=0, E[W_T^2]=E[T].$$

For each $$t>0$$, we have $$E[W_{T\wedge t}]=0, E[W_{T\wedge t}^2]=E[T\wedge t].$$ It suffices to show that $$W_{T\wedge t}$$ converges to $$W_T$$ as $$t\to\infty$$ in $$L^2$$ and thus in $$L^1$$. If $$E[T]<\infty$$, this post gives a proof. But here we only have $$E[\sqrt T]<\infty$$. By the Burkholder-Davis-Gundy inequality, $$E[\sup_{0\leq s\leq T}|W_s|]\leq CE[\langle W\rangle_T^{1/2}]=CE[\sqrt T]<\infty,$$ hence $$W_{T\wedge t}$$ converges to $$W_T$$ as $$t\to\infty$$ in $$L^1$$ and now the first identity follows.

As for the $$L^2$$ convergence, I have no idea.

Any help would be appreciated.

Since you have shown $$W_{T\wedge t} \to W_T$$ in $$L^1$$, one has $$W_{T\wedge t} = E[W_T \, | \, \mathscr F_t]$$, and so by Jensen's inequality,

$$W_{T\wedge t}^2 \le E[W_T^2 \, | \, \mathscr F_t].$$

Taking expectation and letting $$t\to\infty$$, one sees

$$\limsup_{t\to\infty} E[W_{T\wedge t}^2] \le E[W_T^2].$$

Applying Fatou's lemma gives the complementary inequality, so $$E[W_{T\wedge t}^2] \to E[W_T^2]$$ as $$t\to\infty$$. Since $$E[T\wedge t] \to E[T]$$ by the monotone convergence theorem, the result follows.

If $$\mathsf{E}T<\infty$$, then the second identity holds (as in the linked question). On the other hand, when $$\mathsf{E}\sqrt{T}<\infty$$ and $$\mathsf{E}T=\infty$$, we have $$\mathsf{E}W_T^2=\infty$$ (see, e.g., Exercise 2.12 here).

• Can you provide a proof or a hint for that result? Thanks in advance!
– Feng
Apr 2, 2020 at 9:03
• (1) $\mathsf{E}\sqrt{T}<\infty$ implies that $T<\infty$ a.s. (2) Consider $S_n=\inf\{t:|W_t|=n\}$ s.t. $S_n\nearrow\infty$. Then $\mathsf{E}[T\wedge S_n]\le n^2$ and so $\mathsf{E}\!\left[W_{T\wedge S_n}^2\right]=\mathsf{E}[T\wedge S_n]$. Now send $n\to \infty$.
– user140541
Apr 2, 2020 at 15:28
• @d.k.o. How do yo send $n \to \infty$ on the left-hand side? (E.g. for $T=\inf\{t; |W_t|=a\}$ for fixed $a>0$ we have $T<\infty$ a.s. but $\mathbb{E}(W_T^2) \neq \mathbb{E}(T)$.)
– saz
Apr 3, 2020 at 18:40
• @saz In your example $W_T^2=a^2$ and $\mathsf{E}T=a^2$, aren't they?
– user140541
Apr 3, 2020 at 18:49
• Right, but you haven't addressed saz's question - why is it true that $\lim_{n\to\infty}E[W_{T\wedge S_n}^2]=E[W_T^2]$? Moreover, nowhere in your argument do you use $E[\sqrt{T}]<\infty$, you only use $T<\infty$ almost surely. But saz has provided an example where $T<\infty$ almost surely, and yet $E[W_T^2] < \infty = E[T]$. Apr 3, 2020 at 19:12