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?
1 Answer
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".
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
$$(D_{h}I_{f}(0),v)=E\left[f(X_{t})\int_0^t v(s) \, dW\right]=\int_{0}^{t}E\left[f'(X_{r})\right]v(r) dr.$$
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$\begingroup$ Thank you very much! In terms of Frechet derivative I'm asking just about the derivative at zero, therefore the equation simplifies to $(D_hI_f(0),v)=E[f(X_t)\cdot \int_0^t v(s)dW]$. So the remaining question is about the integral kernel of this linear functional. $\endgroup$– VashCommented Jul 19, 2023 at 17:51
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$\begingroup$ do you just want existence eg. math.stackexchange.com/questions/4185344/…? Or do you want to explicitly try to compute it which I think requires an explicit $f$? $\endgroup$ Commented Jul 19, 2023 at 18:02
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$\begingroup$ I want to compute the kernel explicitly. I don't think it requires an explicit f. For example, in the theorem 3.10 (arxiv.org/pdf/1708.09744.pdf) the kernel is computed from a Fokker-Planck perspective. I hope that the expression for the kernel in the form $R(s,t) = E[f(X_t(w))K(W_s(w))])$ (K is sume functional, w - realization of white noise path) is easier to derive, at least non-rigorously. $\endgroup$– VashCommented Jul 19, 2023 at 19:02