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.


2 Answers 2


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).

  • $\begingroup$ Can you provide a proof or a hint for that result? Thanks in advance! $\endgroup$
    – Feng
    Commented Apr 2, 2020 at 9:03
  • $\begingroup$ (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 $n\to \infty$. $\endgroup$
    – user140541
    Commented Apr 2, 2020 at 15:28
  • $\begingroup$ @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 $\mathbb{E}(W_T^2) \neq \mathbb{E}(T)$.) $\endgroup$
    – saz
    Commented Apr 3, 2020 at 18:40
  • $\begingroup$ @saz In your example $W_T^2=a^2$ and $\mathsf{E}T=a^2$, aren't they? $\endgroup$
    – user140541
    Commented Apr 3, 2020 at 18:49
  • 1
    $\begingroup$ 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 $E[W_T^2] < \infty = E[T]$. $\endgroup$
    – Jason
    Commented Apr 3, 2020 at 19:12

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