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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.

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 $$\int_{\omega\in\Omega}|W_{\tau(\omega)}(\omega)|\,\mathrm d\mathbb P(\omega)=+\infty.$$

Any hints and comments would be much appreciated.

2 Answers 2

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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 $E[W_\tau^2] = E[\tau]$.

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    First, please accept my apologies for the belated reply. Thank you very much for your answer, this is very nice. In fact, this question came into my mind while I was consulting a proof of Skorohod’s embedding theorem, so it is not a coincidence you thought it was relevant. One minor thing: I think the strict inequality $t>1$ in the definition of $\tau$ should be weak: $t\geq 1$. This is because if the inequality is weak, then $\tau=1$ is easily seen to be the same event as $W_1=g(W_1)$ by continuity. However, if you insist on the strict inequality, [continued below] – 2017-02-26
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    [continued] then $W_1=g(W_1)$ does _not_ imply that $\tau=1$: it may happen that $W_t$ just hits $g(W_1)$ at $t=1$, and it does not hit it again for a long time (or ever). The point is that, intuitively, the event $\tau=1$ should not depend (even on zero-probability sets) on what values the motion takes on for $t>1$, since a stopping time is not supposed to “look into the future.” But if the inequality is strict, then the event $\tau=1$ is determined by values of the Brownian motion just a little bit _later_ than $t=1$; the knowledge of $W_1$ alone is not sufficient to decide whether $\tau=1$. – 2017-02-26
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    @triple_sec: Strictly speaking you are right. However, it doesn't really make any difference. It's common to use a definition of stopping time that allows an infinitesimal peek into the future; i.e. the event $\tau = 1$ need only be $\mathcal{F}_{1+}$-measurable. It is also commonly assumed that the filtration is complete, so you can do whatever you want on events of probability 0. And there are many ways to see that the event you are worried about has probability 0. – 2017-02-27
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    I was not aware that some versions of the definition of stopping times would allow instantaneous forward-looking. And, indeed, the probability of the event affected by this minor point vanishes (or at least the affected sample points sit within a zero-probability set if the underlying space is not complete), so that the ultimate distribution is unaffected. Thank you for the clarification and, again, for the brilliant answer! – 2017-03-03
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    Actually, the "instantaneous forward-looking" is usually built into the filtration - it's common to assume the filtration is right continuous, so that $\mathcal{F}_t = \mathcal{F}_{t+}$. This and completeness are called the "usual conditions". – 2017-03-03
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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.

For each $n\in\mathbb N$, define \begin{align*} p_n\equiv&\;\mathbb P\big(\{\omega\in\Omega\,|\,n-1\leq|W_1(\omega)|

Now, define $$\tau(\omega)\equiv t_n\quad\text{if $n-1\leq|W_1(\omega)|

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)|