My question came from M. Hairer's paper "Homogenization of periodic linear degenerate PDEs" (page 2469), I think it could be regarded as a basic exercise of Malliavin calculus thought. I will put is into a nutshell.
Let $\Omega$ and $\bar\Omega$ be two independent copies of $1$-dimensional Wiener space, and denote by $B$ and $\bar B$ the corresponding canonical Wiener processes, $\tau$ is a stopping time defined on $\Omega$. Now define a process $W$ on the product space $\Omega\times\bar\Omega$ by \begin{equation} W_t(\omega,\bar\omega)=\left\{ \begin{split} B_t(\omega) & \quad\text{for } t\le\tau(\omega), \\ B_\tau(\omega)+\bar B_{t-\tau}(\bar\omega) & \quad\text{for } t\ge\tau(\omega). \end{split} \right. \end{equation} Then it is easy to see that $W$ is again a $1$-dimensional Wiener process. Consider the process $X_t$ which is the solution of the SDE $$\tag{SDE} X_t=x+\int_0^tb(X_r)\,dr+\int_0^t\sigma(X_r)\,dW_r. $$ We suppose the coefficients $b$ and $\sigma$ to be smooth enough such that (SDE) has an unique solution and $X_t\in\mathbb D^\infty$.
My question is: How to calculate the Malliavin derivative $\mathcal D_sX_t$ with respect to $\bar B$?
I know I could refer to Theorem 2.2.1 of Nualart's book "Malliavin Calculus And Related Topics", but how to treat the case of Malliavin derivative w.r.t part of samples?
Any comments will be appreciate. Thanks in advance.
Ok, finally I find it out by myself.
From (SDE), \begin{equation} \begin{split} X_t(\omega,\bar\omega)= &\ x+\int_0^tb(X_r)\,dr+\mathbf 1_{[0,\tau(\omega))}(t)\int_0^t\sigma(X_r)\,dB_r(\omega) \\ & +\mathbf 1_{[\tau(\omega),+\infty)}(t)\bigg(\int_0^{\tau(\omega)}\sigma(X_r)\,dB_r(\omega)+\int_{\tau(\omega)}^t\sigma(X_r)\,d\bar B_{r-\tau(\omega)}(\bar\omega)\bigg). \end{split} \end{equation} Thus, \begin{align} \mathcal D_s^{\bar\omega}X_t &= \begin{aligned}[t] \mathbf 1_{[s,+\infty)}(t)\bigg\{ & \int_s^tDb(X_r)\mathcal D_s^{\bar\omega}X_r\,dr+\mathbf 1_{[0,\tau(\omega))}(t)\int_s^tD\sigma(X_r)\mathcal D_s^{\bar\omega}X_r\,dB_r(\omega) \\ & \begin{aligned} +\mathbf 1_{[\tau(\omega),+\infty)}(t) \bigg[ & \sigma(X_s)+\mathbf 1_{[\tau(\omega),+\infty)}(s)\int_s^tD\sigma(X_r)\mathcal D_s^{\bar\omega}X_r\,d\bar B_{r-\tau(\omega)}(\bar\omega) \\ & \begin{aligned} +\mathbf 1_{[0,\tau(\omega))}(s)\bigg( & \int_s^{\tau(\omega)}D\sigma(X_r)\mathcal D_s^{\bar\omega}X_r\,dB_r(\omega) \\ & \begin{aligned} +\int_{\tau(\omega)}^tD\sigma(X_r)\mathcal D_s^{\bar\omega}X_r\,d\bar B_{r-\tau(\omega)}(\bar\omega)\bigg) \bigg]\bigg\} \end{aligned} \end{aligned} \end{aligned} \end{aligned}\\ &= \begin{aligned}[t] \mathbf 1_{[s,+\infty)}(t)\bigg\{ & \int_s^tDb(X_r)\mathcal D_s^{\bar\omega}X_r\,dr+\mathbf 1_{[0,\tau(\omega))}(t)\int_s^tD\sigma(X_r)\mathcal D_s^{\bar\omega}X_r\,dW_r \\ & \begin{aligned} +\mathbf 1_{[\tau(\omega),+\infty)}(t) \bigg[ & \sigma(X_s)+\int_s^tD\sigma(X_r)\mathcal D_s^{\bar\omega}X_r\,dW_r \bigg]\bigg\}. \end{aligned} \end{aligned} \end{align}