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The Setup

Suppose I have a stochastic process $f(Z_t)$ where $Z_t$ solve the $d$-dimensional SDE $$ dZ_t = \mu(t,Z_t)dt + \sigma(t,Z_t)dW_t $$ and $f$ is a smooth function.


My Question

Is there a notion of time-derivative "$d_t$" of the process $f(Z_t)$ which satisfies:

  1. Some sort of chain rule like $$ d_tf(Z_t) = \partial_t f(Z_t) d_t(Z_t), $$ where $\partial_t$ is the usual derivative wrt $t$.
  2. If $Z_t$ is deterministic (ie: $\sigma(t,Z_t)=0$) and $\mu(t,z)$ is $C^1$ in $t$ then $$d_t=\partial_t,$$ ie: $d_t$ reduces to the usual derivative when $f(Z_t)$ is a smooth function of $t$.
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    $\begingroup$ Just Multiplication $\endgroup$ – AIM_BLB Sep 14 '16 at 14:37
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    $\begingroup$ I clarified it. Sorry for the mixup. $\endgroup$ – AIM_BLB Sep 14 '16 at 16:16
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    $\begingroup$ Well.. what would the derivative of $Z_t$ be? I was thinking of using Malliavin's construction here but I'm not sure if the Malliavin Derivative of a deterministic function is it's regular derivative.... $\endgroup$ – AIM_BLB Sep 14 '16 at 18:55
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    $\begingroup$ The Malliavin derivative won't play the role of a time derivative, I don't think. If you view the process $Z_t$ as a function $Z(\omega, t)$, then the Malliavin derivative is in some sense the derivative with respect to $\omega$, not $t$. I think the Malliavin derivative of a deterministic function is going to be 0. $\endgroup$ – Nate Eldredge Sep 22 '16 at 14:19
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The Malliavin derivative satisfies the requisite chain rule. However, if $Z_t$ is no longer a function of the Wiener process, then the Malliavin derivative of $f(Z_t)$ is identically zero.

To see this, we briefly recall the nice link between the Girsanov theorem and Malliavin calculus. Let $\epsilon>0$ be a perturbation parameter and let $\phi \in L^2([0,T]; \mathbb{R}^d)$. Recall that Girsanov theorem says that $Z_t$ (given by the OP) on the probability space $(\Omega, \mathcal{F}, P)$ is a weak solution to $$ dZ_t = \mu(t,Z_t) dt + \sigma(t,Z_t) \left( d W_t + \epsilon \phi(t) dt \right) $$ on the probability space $(\Omega, \mathcal{F}, \tilde P)$ where $$ \frac{d \tilde P}{dP}= \exp\left( -\frac{\epsilon^2}{2} \int_0^T | \phi(s)|^2 ds + \epsilon \int_0^T \phi(s) dW(s) \right) $$ More to the point, the variations taken in the Malliavin derivative are precisely of this form: the Wiener process is varied arbitrarily in the direction of the deterministic $\phi(\cdot)$. From this link we immediately see that if $\sigma=0$, then these variations have no effect on $f(Z_t)$.

Reference

Nualart, D. (2006). The Malliavin calculus and related topics. Springer.

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I think the chain rule $d[f(Z_t)]=f'(Z_t)\circ dZ_t$ is valid when the product $\circ$ is defined as in Stratonovich stochastic integral (while the SDE uses Ito's). Note that $g(Z_t)\circ dt$ doesn't differ from the ordinary product $g(Z_t)\ dt$, but $g(W_t)\circ dW_t$ is Ito's $g(W_t)\ dW_t$ plus $\frac12 g'(W_t)\ dt$.

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    $\begingroup$ Sorry, I just didn't manage to understand how $d_tf(Z_t) = \partial_t f(Z_t) d_t(Z_t)$ could qualify as 'some sort of chain rule' in any reasonable sense... $\endgroup$ – Jean Duchon Sep 22 '16 at 15:21

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