Consider a family of stochastic processes $dX^h\_t=(g(X^h\_t)+h(s))\,dt+dW\_t$ and a functional $I\_f:h(s) \rightarrow E[f(X\_t^h)] $. I would like to compute the kernel of the derivative of this functional with respect to $h$ using the Girsanov formula or Malliavin calculus. For example, Gisranov's formula gives
\begin{align}
E[f(X^h\_t)] & = E \left[ f(X\_t) \exp \left( \int\_0^t h(s) \, dW-\frac{1}{2}\int\_0^t h(s)^2 \, ds \right) \right] \\[8pt]
& \sim E\left[f(X\_t) \left(\int\_0^t h(s)\, dW\_s+1\right)\right].
\end{align}
Malliavin calculus also gives the same weight: $\int\_0^t h(s) \, dW\_s$. So the question is, how can I proceed from that? If the response function function $R(s,t)$ defined by $$E\left[f(X\_t) \int h(s) \, dW\_s\right] \sim \int h(s)R(s,t)\, ds + o(h^2),$$ then I would expect that formally $R(s,t) = E\left[f(X\_t)\dot W\_s \right]$. Is there a way to make this statement precise or express $E\left[ f(X\_t) \dot W\_s \right]$ in a more common way?



Linear response for SDE
For the original question of computing the kernel of the derivative of this functional with respect to h, I am more leaning on just regular Frechet derivative (if h is some deterministic function) and using the Girsanov formula to study the difference $I\_{f}(h+\epsilon v)-I\_{f}(h)$. For just the exponentials we have


$$\frac{1}{\epsilon}\exp \left( \int\_0^t h(s) \, dW-\frac{1}{2}\int\_0^t h(s)^2 \, ds \right)\cdot\left[\exp\left( \epsilon\left(\int\_0^t v(s) \, dW-\int\_0^t h(s)v(s) \, ds\right)-\frac{\epsilon^{2}}{2}\int\_0^t v(s)^2 \, ds\right) -1\right]. $$


From here only the first order survives and so in the limit we get


$$\exp \left( \int\_0^t h(s) \, dW-\frac{1}{2}\int\_0^t h(s)^2 \, ds \right)\cdot\left[\int\_0^t v(s) \, dW-\int\_0^t h(s)v(s) \, ds\right]. $$


i.e.


$(D\_{h}I\_{f}(h),v)=E\left[f(X\_{t})\exp \left( \int\_0^t h(s) \, dW-\frac{1}{2}\int\_0^t h(s)^2 \, ds \right)\cdot\left[\int\_0^t v(s) \, dW-\int\_0^t h(s)v(s) \, ds\right]\right].$


From here it is a bit hard to proceed without information on $f$.


Does this work for you or do you want something more?


In terms of getting an expression


$$(D\_{h}I\_{f}(0),v)=\int\_{0}^{t}R(r,t)v(r) dr$$


for some $R(u,t)$ kernel, we get inspiration from the work ["Mathematical foundation of nonequilibrium fluctuation-dissipation theorems for inhomogeneous diffusion processes with unbounded coefficients"](https://arxiv.org/pdf/1708.09744.pdf).


In particular, we apply Ito's formula to $f$


$$f(X\_{t})=f(x\_0)+\int\_{0}^{t}f'(X\_{r})g(X\_{r})dr+\int\_{0}^{t}f'(X\_{r})dW\_{r}+\frac{1}{2}\int\_{0}^{t}f''(X\_{r})dr.$$


By [Ito isometry](https://math.stackexchange.com/questions/3169779/the-expectation-of-the-product-of-two-stochastic-integrals)


$$(D\_{h}I\_{f}(0),v)=E\left[f(X\_{t})\int\_0^t v(s) \, dW\right]=\int\_{0}^{t}E\left[f'(X\_{r})\right]v(r) dr.$$


