# Expected first time that $|B(t)|=1$ for a standard Brownian motion

I want to calculate $$\mathbb{E}[T]$$ where $$T = \inf \{t \geq 0 \mid |B(t)| = 1\}$$ and $$B(t)$$ is a Brownian motion with mean $$0$$. I saw some similar posts but for a one-sided hitting time, and in those cases the expectation is infinite. I tried the following:

$$\mathbb{E}[T] = \int_{0}^\infty s \mathbb{P}[T=s]ds = \int_{0}^\infty \mathbb{P}[T > s]ds.$$ By the reflection principle, $$\mathbb{P}[T>s] = \mathbb{P}[\sup_{t\in [0,s]} |B(t)| <1] = 2 \mathbb{P}[|B(s)| < 1]$$. So this gives $$\mathbb{E}[T]= 2 \int_0^\infty \mathbb{P}[B(s) < 1]ds.$$ Now $$B(s)$$ is distributed as a normal random variable with variance $$\sigma^2 = s$$. I have no idea how to continue, or maybe there is a simpler way?

EDIT: Maybe a nice way can be to view the Brownian motion as (the limit of) the sum of $$\{-1,1\}$$ random variables. But I am also not sure how to prove the result in this case..

So I have come some way since I asked this question, and I think I know roughly how to do it. Here goes:

Let $$T_1$$ be the first time when $$|B(t)|=1$$, and $$T_2$$ be the first time after $$T_1$$ that $$|B(T_2)-B(T_1)| = 1$$, $$T_3$$ the first time afterwards where $$|B(T_3)-B(T_2)| = 1$$, etc.. Let $$X_i$$ correspond to $$B(T_i) - B(T_{i-1})$$, with $$T_{0}=0$$. By independent increments, all these are independent, and $$\mathbb{P}[X_i=1] = \mathbb{P}[X_i=-1]=\frac12$$. Also, $$\mathbb{E}T_1 = \mathbb{E}T_i$$ for all $$i$$. Now, by the LLN,

$$\mathbb{P}[ | T_1 + \ldots + T_n - n \mathbb{E}T_1| > \epsilon n] \to 0.$$ Moreover, $$B(T_1+\ldots + T_n) - B(n \mathbb{E} T_1)$$ is a distributed as $$\mathcal{N}(0, T_1 +\ldots + T_n - n \mathbb{E}T_1)$$. With increasingly high probability (as $$n\to \infty$$), this variance is bounded by $$\epsilon n$$. Now,

\begin{align} \mathbb{P}[ |B(T_1+\ldots T_n) - B(n \mathbb{E} T_1)| > \sqrt{\epsilon n} ] & \leq 2\int_{\sqrt{\epsilon n}}^\infty \frac{1}{\sqrt{2\pi \epsilon n}}e^{-x^2/(2 \epsilon n)}dx\\ &=2 \int_{1}^\infty \frac{1}{\sqrt{2 \pi} } e^{-x^2/2} dx \\ &\leq 0.25 \label{eq:adadssa} \end{align} However, $$B(T_1 + \ldots + T_n) = X_1+\ldots + X_n$$, which by the Central limit theorem converges to a $$\mathcal{N}(0,n)$$ random variable. If it then were the case that $$\alpha := \mathbb{E}T_1 \neq 1$$, then $$B(n \mathbb{E}T_1)$$ is $$\mathcal{N}(0, \alpha n)$$, while $$B(T_1+ \ldots + T_n)$$ is roughly $$\mathcal{N}(0, n)$$. I think this should somehow contradict the fact that $$\mathbb{P}[ |B(T_1+\ldots T_n) - B(n \mathbb{E} T_1)| > \sqrt{\epsilon n} ] < 0.25$$.

We note that: $$B_T \in \{-1,1\}.$$

So:

$$P(B_T=-1) + P(B_T=1) = 1.$$

We also know that $$(B_t^2-t)_{t\geq 0}$$ is a martingale.

Doob's optional stopping theorem gives then:

$$E[B_T^2 - T] = E[B_0^2 -0] =0,$$

from where we have

$$E[T]=E[B_T^2] = (-1)^2P(B_T=-1) + 1^2 P(B_T=1) = 1.$$

• Thank you for your answer, I am afraid this is currently a bit above my pay grade :) We have not covered stopping times or Doobs stopping theorem. Commented Jun 10, 2021 at 9:33