How to minimize $\Phi(c_0):=c_0^2\int_0^\infty\text E\left[g(c_0,X)\left|\int_0^tf(X_s)-\int f\:{\rm d}\mu\:{\rm d}s\right|^2\right]\:{\rm d}t$?

Question

Let

• $E$ be a $\mathbb R$-Banach space;
• $\lambda$ be a $\sigma$-finite measure on $E$;
• $\mu$ be a probability measure on $(E,\mathcal E)$ with density $p$ with respect to $\lambda$;
• $q$ be a probability density on $E$ with respect to $\lambda$;
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
• $(X_t)_{t\ge0}$ be an $E$-valued right-continuous Markov process on $(\Omega,\mathcal A,\operatorname P)$ with invariant distribution $\mu$;
• $c_0>0$ and $$c:=c_0\frac qp;$$
• $$A_t:=-\int_0^tc(X_s)\:{\rm d}s$$ and $$M_t:=e^{-A_t}$$ for $t\ge0$.

I want to choose $c_0>0$ such that $$\Phi(c_0):=c_0^2\int_0^\infty\operatorname E\left[M_tc(X_t)\left|\int_0^tf(X_s)-\int f\:{\rm d}\mu\:{\rm d}s\right|^2\right]\:{\rm d}t$$ is as small as possible.

Simply letting $c_0\to0+$ doesn't seem to be the solution, since (contrary to the other terms) $M_t\to0+$ as $c_0\to\infty$.

I thought using the Lagrange multiplier theorem should be the solution, but this seems to give a rather complicated expression. If real minimization of $\Phi$ is too complicated, I would also be interested in a choice which makes $\Phi$ sufficiently small.

Answer 1

Analysis of Φ(c₀)

Analysis of Φ(c₀)

We aim to find the minimum of the function Φ(c₀) by differentiating with respect to c₀ and setting the result to zero.

From the provided information:

\[ \Phi(c_0) = c_0^2 + \int_0^\infty E \left[ M_t c(X_t) \int_0^t f(X_s) ds - \int_0^t f(X_s) ds \right]^2 dt \]

Step 1: Differentiation of c₀²

\[ \frac{d}{dc_0} c_0^2 = 2c_0 \]

Step 2: Differentiate the terms inside the integral

a) Differentiation of M_t

we have:

\[ \frac{dM_t}{dc_0} = \frac{d}{dc_0} e^{-\int_0^t c(X_s) ds} \]

Utilizing the chain rule and the relation \:

\[ \frac{dM_t}{dc_0} = M_t \left( \int_0^t \frac{q(X_s)}{p(X_s) \times c_0^2} ds \right) \]

b) Differentiation of c(X_t)

From the relation \(c = \frac{q}{p \times c_0}\):

\[ \frac{dc(X_t)}{dc_0} = -\frac{q(X_t)}{p(X_t) \times c_0^2} \]

Step 3: Incorporate these results into the derivative of the integral term

Let's denote the function inside the expectation as \(G(c_0)\). Using the chain rule:

\[ \frac{dG}{dc_0} = 2 E[g(X_t, c_0)] \times E\left[ \frac{\partial g(X_t, c_0)}{\partial c_0} \right] \]

Substituting in the differentiated terms from step 2:

\[ \frac{dG}{dc_0} = 2 E \left[ M_t c(X_t) \int_0^t f(X_s) ds - \int_0^t f(X_s) ds \right] \times E\left[ \left( \int_0^t \frac{q(X_s)}{p(X_s) \times c_0^2} ds \right) \left( -\frac{q(X_t)}{p(X_t) \times c_0^2} \right) \right] \]

Step 4: Assemble the entire expression

\[ \frac{d\Phi(c_0)}{dc_0} = 2c_0 + \int_0^\infty \frac{dG}{dc_0} dt \]

To find the minimum, set the above to zero and solve. The solutions will provide potential minimum points, which can be further evaluated to determine the actual minimum.