[Math] On hitting time of Brownian motion and Ito’s lemma

brownian motionstochastic-analysisstochastic-integralsstochastic-processes

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.

Best Answer

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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