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

Suppose B(t) is the standard Brownian motion. Does it make sense to compute the total time of upward movement
t0I(dBτ>0)dτ
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 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

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 δ>0 such that sups[tδ,t]Bs=infs[t,t+δ]Bs 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)].

Attribution
Source : Link , Question Author : Hans , Answer Author : John Dawkins

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