Stratonovich integral

Question

I'm having some troubles to calculate the Stratonovich integral $I(sin)(t)=\int_{0}^{t}\sin{B_{s}}dB_{s}$. I've tried with the limit of $\sum_{j>0}^{t}\sin(B_{\frac{t_{j+1}+t_{j}}{2}})\chi_{[t_{j}+t_{j+1}]}$ but I don't know how to do, Can you please help me?

Answer 1

Via Ito integral and using Ito Lemma (third and penultimate steps):

$$ \int_0^t \sin(B_u)\circ dB_u = \int_0^t \sin(B_u) dB_u +\frac{1}{2}\int_0^t d(\sin(B_u))dB_u$$

$$ = \int_0^t \sin(B_u) dB_u + \frac{1}{2}\int_0^t\left(\sin'(B_u) dB_u +\frac{1}{2}\sin''(B_u) du\right)dB_u $$

$$ = \int_0^t \sin(B_u) dB_u + \frac{1}{2}\int_0^t\sin'(B_u) du $$ $$ = \int_0^t (-\cos)'(B_u) dB_u + \frac{1}{2}\int_0^t(-\cos)''(B_u) du $$

$$ = \int_0^t d(-\cos)(B_u) = (-\cos) (B_t) - (-\cos)( B_0).$$

Answer 2

Almost by definition, $$\int_0^tf'(B_s)\circ\mathrm dB_s=f(B_t)-f(B_0).$$ Sure you cannot find some $f$ wich applies to your setting?