# On hitting time of Brownian motion and Ito's lemma

I have two possibly related questions. Let $\tau:=\min\{t\geq0:B_t=1\}$, where $B_t$ is a standard Brownian motion.

1. I am supposed to derive the fact that $\mathbf{E}\tau=\infty$ by applying some properties of stochastic integrals to $\int^{\infty}_0\mathbf{1}_{[0,\tau]}(t)dB_t$. I have tried this: $$\mathbf{E}\tau=\mathbf{E}\int^{\tau}_0dt=\mathbf{E}\int^{\infty}_0\mathbf{1}_{[0,\tau]}(t)dt=\mathbf{E}\left(\int^{\infty}_0\mathbf{1}_{[0,\tau]}(t)dB_t\right)^2$$ And I'm stuck (not even sure if going the right way).

2. Is Ito's lemma false with stopping times as upper limits of the integrals? For example, is $\int^{\tau}_0dB_t=B_{\tau}$ false? How can I see that?

EDIT: A discussion on possible proofs of $\mathbf{E}\tau=\infty$ is presented in the answers here. I am just curious about the particular suggestion.

And the stochastic integral with a hitting time as an upper limit is defined as follows $$\int^{\tau}_0dB_t:=\int^\infty_0\mathbf{1}_{[0,\tau]}(t)dB_t:=L^2-\lim_{N\to\infty}\int^N_0\mathbf{1}_{[0,\tau]}(t)dB_t$$ Then I guess the question 2. would be equivalent to asking if Ito's lemma is true for integrals with infinite upper limit.

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 I think the equation you have with $\tau \wedge n$ instead of $\tau$ is valid. Perhaps you could take a limit as $n \rightarrow \infty$ and argue that if $\mathbf{E}[\tau]$ were finite you could use dominated convergence to get a contradiction. – ShawnD Mar 24 '12 at 20:28 Wald's identity will be helpful. This theorem, applied to the Brownian motion, states that if a stopping time $\tau$ satisfies $\mathbb{E}\tau < \infty$, then $\mathbb{E}(B_{\tau}) = 0$. This is because $B_t$ is a martingales null at 0. Now note that if $\mathbb{E}\tau < \infty$, then $\tau$ is finite a.s., hence $B_{\tau} = 1$ a.s., which contradicts the wald identity $\mathbb{E} B_{\tau} = 0$. – sos440 Mar 25 '12 at 1:18 Curious about... what, exactly? – Did Mar 25 '12 at 21:03

Here is a solution to (1). Not sure if it strictly falls under the requirements of the problem, but it doesn't use anything too fancy.

Let $\tau_r$ be the first time Brownian motion hits either $1$ or $-r$ and $f(x)=x^2$. Since the infinitesimal generator of Brownian motion (restricted to smooth functions) is $1/2 \Delta$, we have by Dynkin's formula that

$$\mathbf{E}[f(B_{\tau_r})]=\mathbf{E}[B_{\tau_r}^2]=f(0)+\mathbf{E}\left[ \int_0^{\tau_r} 1 ds \right]=\mathbf{E}[\tau_r] .$$

Using the Gambler's ruin estimate, we have $$\mathbf{E}[B_{\tau_r}^2]=\frac{r^2}{r+1}+\frac{r}{r+1}.$$ Letting $r \rightarrow \infty$, we see that $\mathbf{E}[\tau_r] \rightarrow \infty$ as $r \rightarrow \infty$. It follows that $\mathbf{E}[\tau]=\infty$.

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This is a good approach. More generally, you could look at the standard proofs for hitting times of the interval $(a,b)$ and then just let $a\to-\infty$ in the arguments. – EMS Mar 24 '12 at 19:35
Thanks for an alternative proof. I believe that something can be done in the direction mentioned in the problem, though. – nokiddn Mar 24 '12 at 19:48
Simpler: since $\tau$ is finite almost surely, $B_\tau=1$ almost surely, hence $E(B_\tau)=1\ne0= E(B_0)$, hence $\tau$ cannot be integrable. – Did Mar 24 '12 at 23:59
@DidierPiau This was my first thought, but I'd be curious to see a precise statement of the theorem you're invoking. In the discrete case you normally need the expectation of the stopping time and increments of the process to be finite to make optional sampling valid and the statement of optional sampling in the continuous case in the one book on my shelf (Karatzas and Schreve) doesn't immediately apply. – ShawnD Mar 25 '12 at 20:23
Assume $\tau$ is integrable. Then $E(B_{t\wedge\tau}^2)=E(t\wedge\tau)\leqslant E(\tau)$, which proves that $(B_{t\wedge\tau})_{t\geqslant0}$ is uniformly integrable. Since $\tau$ is almost surely finite, $B_{t\wedge\tau}\to B_{\tau}=1$ almost surely. But $E(B_{t\wedge\tau})=0$ for every $t$ while $E(B_\tau)=1$. Contradiction. (Kind of rediscovering Wald's identity.) – Did Mar 25 '12 at 21:00
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Hi a direct and analytic proof consists in using the density of $\tau$ and to calclulate $E[\tau]$ directly. $\tau$'s density is given by :

$f(t)_\tau=\frac{1}{\sqrt{2\pi.t^3}}e^{-\frac{1}{2.t}}$ (many books give this e.g. Karlin Taylor, "A First course in Stochastic Processes")

From this you can show your assertion.

Anyway, I think that the solution given by ShawnD is more in the spirit of your problem, and essentially is an application of Doob's Optimal Sampling Theorem for martingales.

Best regards

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To answer your second question first, you should be careful to determine exactly how you define a stochastic integral with a stopping time in the limit. But any reasonable definition should result in $\int_0^\tau dB_t = B_\tau$.

Edit: I take it back. Using your definition, $\int_0^\tau dB_t$ does not exist when $\tau = \inf\{t \ge 0 : B_t = 1\}$. For we have $\int_0^N 1_{[0, \tau]} dB_t = B_{\tau \wedge N}$, and this does not converge in $L^2$ as $N \to \infty$ (if it did, it would also converge in $L^1$, but we have $E[B_{\tau \wedge N}]=0$ while $E[B_\tau] = 1$). Of course, it does converge almost surely.

For your main question, here's an argument possibly in the spirit of your question, though it seems too complicated still.

Let $\tau_k = \inf\{t \ge 0 : B_t = k\}$ be the hitting time of $k$. Note that, $0 = \tau_0 \le \tau_1 \le \tau_2 \le \dots$ by continuity of Brownian motion, and by recurrence $\tau_k < \infty$. Let $\sigma_k = \tau_k - \tau_{k-1}$ be the time needed to get from $k-1$ to $k$. By the strong Markov property and translation invariance, the $\sigma_k$ are iid. Thus we have $$E[\tau_k] = E[\sigma_1 + \dots + \sigma_k] = k E[\sigma_1].\quad(1)$$

Now for any $k,n$, we have $E \int_0^\infty |1_{[0, \tau_k \wedge n]}(t)|^2\,dt = E[\tau_k \wedge n] \le n < \infty$, so the Itô isometry gives $$E[\tau_k \wedge n] = E\left[\left(\int_0^\infty 1_{[0, \tau_k \wedge n]}(t)\,dB_t\right)^2\right] = E[B_{\tau_k \wedge n}^2]\quad(2)$$ as in your argument. (Another way to get this identity is to note that $X_t = B_t^2 - t$ is a martingale, hence so is $X_{\tau_k \wedge t}$.) By monotone convergence, $E[\tau_k \wedge n] \uparrow E[\tau_k]$ as $n \to \infty$. And by Fatou's lemma, $\liminf_{n \to \infty} E[B_{\tau_k \wedge n}^2] \ge E[B_{\tau_k}^2] = k^2$. So passing to the limit, we have $$E[\tau_k] \ge k^2.\quad(3)$$

Combining (1) and (3), we see $E[\sigma_1] \ge k$. Since $k$ was arbitrary, we must have $E[\sigma_1] = \infty$, which completes the proof.

Note the essential observation here is that the strong Markov property suggests that $E[\tau_k]$ scales linearly with $k$, as in (1), while (2) suggests quadratic scaling. The only way out of this paradox is to have $E[\tau_k] = \infty$.

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 Why is $\int_0^N 1_{[0, \tau]} dB_t = B_{\tau \wedge N}$ true? Do you use Ito's lemma here? How about the fact that the indicator function is not differentiable? – nokiddn Mar 26 '12 at 18:25 @nokiddn: First, it is the same as $\int_0^N 1_{[0, \tau \wedge N]}dB_t$ since the integrands are the same on the integral $[0,N]$. Next, I think it might be easiest to get directly from the definition of the integral. For a fixed number $n$, divide up the interval $[0,N]$ into $n$ pieces and let $X_t = 1_{[0, i/n]}(t)$ on the event $A_{i,n} = \{i/n \le \tau \wedge N \le (i+1)/n\}$. Now $X_t$ is an elementary predictable process and by definition we get $\int_0^N X_t dB_t = B_{i/n}$ on the event $A_{i,n}$. As $n \to \infty$, we have $X_t \to 1_{[0, \tau \wedge N]}(t)$... – Nate Eldredge Mar 26 '12 at 19:43 @nokiddn: And the integrals converge to the right thing as well. – Nate Eldredge Mar 26 '12 at 19:43