I'm trying to figure out how the value of a path integral depends on the initial condition (starting point) of the path.
Part1:
Consider the state variable $x(t)$ that is governed by an ODE $$\dot{x}(t) = f(x(t))$$ with some initial condition. This defines our path in time.
Let's say I have a path integral, for example a cost-to-go $C(x_t,t)$ that evaluates how much cost will be accumulated until a final time $t_f$ for a path starting at $x_t$ at time $t$ and governed by the ODE above: $$ C(x_t,t) = \int_t^{t_f} g(x(\tau)) \; d\tau \qquad \text{given: } x(t)=x_t, \; \dot{x}(t) = f(x(t))$$
Now my question: What is ${\partial C(x_t,t) \over \partial x_t} $, i.e., the sensitivity of the cost-to-go w.r.t. the starting state?
I'm having troubles figuring this out because I don't know how to formalize the dependence of future states on $x_t$.
Part2:
How does the answer change when we consider a stochastic path (random walk) instead $$dx = f(x,t) dt + \sigma(x,t) dw$$ where $dw$ is standard Brownian motion and the cost-to-go is now an expectation over all paths that start at $x_t$ at time $t$: $$ C(x_t,t) = {\mathbb E}_{x_t} \left[ \int_t^{t_f} g(x(\tau)) \; d\tau \right] \qquad \text{given: } x(t)=x_t $$
I've tried to apply some Ito calculus rules here but I'm getting stuck again...
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