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Let $X_t= X_0 + \int_0^t \mu(s,X_s)ds + \int_0^t \sigma(s,X_s)dW_s$ where $\mu$ and $\sigma$ are $C^1$ functions satisfying the usual growth restriction and $W_t$ is a $d$-dimensional Brownian motion. Let $f\in C^2(\mathbb{R}^d;\mathbb{R})$ and let $D_t$ denote the Malliavin derivative at time $t$; then how would one go about computing $$ D_tf(X_t) =? $$

So Far: I know that if $X_t= \int_0^t \sigma(s)dW_s$ then by definition 2.2 in these lectures by D. Nualart $$ D_tf(X_t) = \sigma_t^T\left[\frac{\partial f}{\partial x} (X_t)\right]... $$ but what about the general case?

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    $\begingroup$ The Malliavin derivative of solutions of SDEs like the one for X_t solves an equation whose solution can be written in term of the first variation process. See fabricebaudoin.wordpress.com/2012/11/08/… $\endgroup$ Oct 24, 2019 at 22:48
  • $\begingroup$ Perfect, Thank you Fabrice. Also, I'm a big fan of your rough path theoretic stuff. :) $\endgroup$
    – ABIM
    Oct 25, 2019 at 14:36

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