Expectation of indicator of the brownian motion inside an interval

Question

Suppose $W_t$ is your usual brownian motion and that you have the following process:

$$ \theta_t = \int_0^t \mathbb{1}_{\alpha \leq W_s \leq \beta} ds $$

How can I calculate the expectation $E[\theta_T]$?

Answer 1

You may apply the Fubini theorem. Its hypotheses are satisfied. So you get:

$$E[\theta_T] = \int_0^T P[W_s \in [\alpha,\beta]] ds = \int_0^T \int_\alpha^\beta \frac{1}{\sqrt{2\pi s}} \exp(-\frac{x^2}{2s}) ~dx ~ds. $$

Etc.

Answer 2

Actually, we can have a semi-analytical formula of the expected value of $\theta$. Thanks to Fubini, we have: $t>0$

\begin{align} E[\theta_t] &= \int_0^t E[1_{\alpha \leq W_s \leq \beta}]ds \\ &= \int_0^t P(\alpha \leq W_s \leq \beta)ds \\ &= \int_0^t \Phi\left(\frac{\beta}{\sqrt{s}}\right)-\Phi\left(\frac{\alpha}{\sqrt{s}}\right)ds \end{align} Where $\Phi$ is cumulative distribution function of standard Gaussian variable. By means of multiple integration by parts, we have \begin{equation} \int_0^t \Phi\left(\frac{x}{\sqrt{s}}\right)ds = (2x^2 + t)\Phi\left(\frac{x}{\sqrt{t}}\right) + 2x\left(\frac{\sqrt{t}e^{-x^2/t}}{\sqrt{\pi}} - x\right) \end{equation}