Are all stopped Brownian motions integrable?

Question

Let $(W_t)_{t\in[0,\infty)}$ be a standard Brownian motion and let $\tau$ be a stopping time that is finite almost everywhere, but otherwise arbitrary—it need not be a hitting time, in particular.

I am wondering whether the random variable $\omega\mapsto W_{\tau(\omega)}({\omega})$ is integrable for all such stopping times or whether a counterexample can be constructed.

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Intuitively, the expected value of $|W_t(\omega)|$ for a fixed $t\geq0$ is proportional to $\sqrt{t}$, which tends to make it less likely that $\omega\mapsto W_{\tau(\omega)}({\omega})$ is integrable if one can make $\tau$ take on large values with sufficiently high probabilities. On the other hand, the probability of $\tau$ being “large enough” is small (given that $\tau$ is finite almost everywhere), which tends to favor integrability.

I cannot see whether a counterexample can be constructed making the first of these two effects “win” against the second, yielding $$\int_{\omega\in\Omega}|W_{\tau(\omega)}(\omega)|\,\mathrm d\mathbb P(\omega)=+\infty.$$

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Any hints and comments would be much appreciated.

Answer 1

Using the idea in your answer we can prove something stronger.

For any cumulative distribution function $F$ on $\mathbb{R}$, there is an a.s. finite stopping time such that $W_\tau$ has the distribution $F$.

In particular we can choose an $F$ with infinite first moment.

Recall the following lemma which is fairly standard and not hard to prove:

For any cdf $F$, define $F^{\leftarrow} : (0,1) \to \mathbb{R}$ by $F^{\leftarrow}(y) = \sup\{x : F(x) < y\}$. If $U$ is a random variable with a uniform(0,1) distribution, then $F^{\leftarrow}(U)$ has the distribution $F$.

Note that if $F$ is continuous and strictly increasing then $F^{\leftarrow} = F^{-1}$.

Next let $\Phi$ be the standard normal cdf. If $Z$ has a standard normal distribution, then $\Phi(Z)$ has a uniform(0,1) distribution. In particular this is true for $\Phi(W_1)$. So let $g = F^{\leftarrow} \circ \Phi$; then $g(W_1)$ has the distribution $F$.

Now let $$\tau = \inf\{t \ge 1 : W_t = g(W_1)\}.$$ As in your answer, by the strong Markov property, $X_s = W_{s+1} - W_1$ is a Brownian motion independent of $W_1$. Brownian motion is recurrent, so for any $x$, $X_s$ hits the value $x$ almost surely. Since $\{X_s\}$ is independent of $W_1$, this implies that $X_s$ hits the value $g(W_1) - W_1$ almost surely. This happens when $W_t$ hits the value $g(W_1)$, so $\tau < \infty$ almost surely. And it is clear that $W_\tau = g(W_1)$, so $W_\tau$ has the distribution $F$.

It's worth mentioning here the Skorohod embedding theorem, which says that if $F$ has zero mean and finite variance, then we can find an integrable stopping time $\tau$ such that $W_\tau \sim F$. You can find a proof in Brownian Motion by Mörters and Peres. The assumption of zero mean and finite variance is necessary here, since it follows from the Wald lemmas that if $\tau$ is integrable then $E[W_\tau] = 0$ and $E[W_\tau^2] = E[\tau]$.

Answer 2

Here is a (somewhat contrived) counterexample. The idea is this: wait until the period $t=1$. Then, decide when to stop based solely on the value of $W_1$. In particular, choose for each realization of $W_1(\omega)$ such an extremely late time of eventual stopping that makes the first effect mentioned above “win” against the second, rendering an infinite expected absolute value of the stopped process.

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For each $n\in\mathbb N$, define \begin{align*} p_n\equiv&\;\mathbb P\big(\{\omega\in\Omega\,|\,n-1\leq|W_1(\omega)|<n\}\big),\\ t_n\equiv&\;\frac{\pi}{2p_n^2}+1. \end{align*} Since $W_1$ has the standard normal distribution, the sequence $(p_n)_{n\in\mathbb N}$ of probabilities is strictly positive, strictly decreasing, and converges to $0$. Correspondingly, the sequence $(t_n)_{n\in\mathbb N}$ is well-defined, strictly increasing, always greater than $1$, and diverges to $+\infty$.

Now, define $$\tau(\omega)\equiv t_n\quad\text{if $n-1\leq|W_1(\omega)|<n$.}$$ It is not difficult to verify that $\tau$ is a stopping time that is everywhere finite-valued.

The following argument [which exploits the facts that for any $t>1$, (i) $W_1$ and $W_t-W_1$ are independent; and (ii) $W_t-W_1$ has the normal distribution of mean $0$ and variance $t-1$] establishes that $W_{\tau}-W_1$ is not integrable, which implies that $W_{\tau}$ is not integrable, either: \begin{align*} &\;\int_{\omega\in\Omega}|W_{\tau(\omega)}(\omega)-W_1(\omega)|\,\mathrm d\mathbb P(\omega)=\sum_{n=1}^{\infty}\int_{\{\omega\in\Omega\,|\,\tau(\omega)=t_n\}}|W_{t_n}(\omega)-W_1(\omega)|\,\mathrm d\mathbb P(\omega)\\ =&\;\sum_{n=1}^{\infty}\int_{\{\omega\in\Omega\,|\,n-1\leq|W_1(\omega)|<n\}}|W_{t_n}(\omega)-W_1(\omega)|\,\mathrm d\mathbb P(\omega)\\ =&\;\sum_{n=1}^{\infty}\mathbb P\big(\{\omega\in\Omega\,|\,n-1\leq|W_1(\omega)|<n\}\big)\times\mathbb E(|W_{t_n}-W_1|)=\sum_{n=1}^{\infty}p_n\times\sqrt{\frac{2}{\pi}\times(t_n-1)}\\ =&\;\sum_{n=1}^{\infty}1=+\infty. \end{align*}