Solution

Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 33 / 2 / c / ii / Solution

{title2=$Y_t=e^{W_t}\cos B_t,\quad[Y]_t=\int_0^te^{2W_s}\,ds$}

For independent standard <Brownian motions> $B,W$, <Itô formula> gives $dY=e^W\cos B\,dW-e^W\sin B\,dB$: the two second-order drifts cancel. Independence gives zero cross <quadratic covariation>, so the displayed bracket follows. The process itself is not Brownian because its bracket is random. Its inverse-bracket time change is Brownian motion starting at one, in the time-changed <filtration>; subtracting one gives standard Brownian motion. The <Divergence of a driftless Brownian exponential clock> guarantees that the inverse is finite at every new time.

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 Incoming links: Cosine-exponential Brownian local martingale