Let \(X_t\) be an Ito processes with

\begin{align*}

dX_t&=f_tdt + g_tdW_t

\end{align*}

and \(B_t\) be a second (possibly correlated with \(W\) ) Brownian

motion. We define the Stratanovich integral \(\int X_t \circ dB_t\) by

\begin{align*}

\int_0^T X_t \circ dB_t = \int_0^T X_t dB_t + \frac12 \int_0^T \;d\langle X, B \rangle_t

\end{align*}

Recall that if \(B_t=W_t\) then \(d\langle B, W \rangle_t =dt\) and it is zero if they are independent. Use this definition to calculate:

- \(\int_0^t B_t \circ dB_t\) (Explain why this agrees with the answer you obtained here).
- Let \(F\) be a smooth function. Find equation satisfied by \(Y_t=F(B_t)\) written in terms of Stratanovich integrals. (Use Ito’s formula to find the equation for \(dY_t\) in terms of Ito integrals and then use the above definition to rewrite the Ito integrals as Stratanovich integrals“\(\circ dB_t\)”.) How does this compare to classical calculus ?
- (Integration by parts) Let \(Z_t\) be a second Ito process satisfying

\begin{align*}

dZ_t&=b_tdt + \sigma_tdW_t\;.

\end{align*}

Calculate \(d(X_t Z_t)\) using Ito’s formula and then write it in terms of Stratanovich integrals. Why is this part of the problem labeled integration by parts ? (Write the integral form of the expression you derived for \(d(X_t Z_t)\) in the two cases. What are the differences ?)