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

Suppose $$B(t)B(t)$$ is the standard Brownian motion. Does it make sense to compute the total time of upward movement
$$∫t0I(dBτ>0)dτ\int_0^t I(dB_\tau>0)d\tau$$
where $$II$$ 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 $$t2\frac t2$$. 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.

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 $$t>0t>0$$ and a $$δ>0\delta>0$$ such that $$sups∈[t−δ,t]Bs=infs∈[t,t+δ]Bs\sup_{s\in[t-\delta,t]}B_s=\inf_{s\in[t,t+\delta]}B_s$$ is $$00$$.