How to compute the time of upward movement of a Brownian motion

Question

Suppose $B(t)$ is the standard Brownian motion. Does it make sense to compute the total time of upward movement $$\int_0^t I(dB_\tau>0)d\tau$$ where $I$ is the characteristic function? What is a rigorous way to define it if it does make sense? Intuition dictates that the expectation of the above integral should be $\frac t2$. I am aware of Doob's upcrossing lemma. Perhaps we can use local time and Tanaka's formula. But I do not have a way to define my integral in similar terms.

Answer 1

One answer to your question, in the negative, is this: With probability 1 the Brownian motion has no times of increase; that is, the probability that there exists a time $t>0$ and a $\delta>0$ such that $\sup_{s\in[t-\delta,t]}B_s=\inf_{s\in[t,t+\delta]}B_s$ is $0$. This result if due to Dvoretzky, Erdos, & Kakutani ["Nonincrease everywhere of the Brownian motion process", (1961)]. A wonderful and quite brief proof was found thirty years later by K. Burdzy ["On nonincrease of Brownian motion" (1990)].

Answer 2

for almost any $\omega$, $ \tau \mapsto I_{B_\tau > 0}$ is a measurable function (since $B$ is continuous). You can therefore define and calculate $$\int_0^tI_{B_\tau>0}d\tau$$ for almost any $\omega$.

However, it is not constantly equal to $\frac t2$: why would it be?

What we can say is that its expected value is $\frac t2$, indeed $$ \mathbb E\Bigg[\int_0^tI_{B_\tau>0}d\tau\Bigg] = \int_0^t\mathbb E\Big[I_{B_\tau>0}\Big]d\tau = \int_0^t\frac12d\tau = \frac t2.$$