 Solution

ID: past-exam-of-the-stochastic-differential-equations-and-brownian-motion-in-complex-analysis-course-of-the-university-of-oxford/2008/a/3/b/solution

For

f (z) = sin z

f^{'} (z) = cos z

and

f^{''} (z) = - sin z

d sin (B_{t}) = cos (B_{t}) d B_{t} - \frac{1}{2} sin (B_{t}) d t .

Since

B_{0} = 0

, this is exactly

d X_{t} = cos B_{t} d B_{t} - \frac{1}{2} X_{t} d t, X_{0} = 0.

The integrand retains

cos B_{t}

; replacing it by a fixed sign of

1 - X_{t}^{2}

would lose information about the phase of

B_{t}

 New to topics? Read the docs here!