If $(X_t)$ is Markov, are $f(X_t)$ and $\int_0^tc(X_s)\:{\rm d}s$ uncorrelaed?

Question

Let

• $(E,\mathcal E)$ be a measurable space
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A,\operatorname P)$ such that $(X_r)_{r\in[0,\:s]}$ is conditionally independent of $X_t$ given $X_s$ for all $t\ge s\ge0$ and $X:\Omega\times[0,\infty)\to E$ is $(\mathcal A\otimes\mathcal B([0,\infty)),\mathcal E)$-measurable
• $c:E\to[0,\infty)$ be $\mathcal E$-measurable and $$A_t:=\int_0^tc(X_s)\:{\rm d}s\;\;\;\text{for }t\ge0.$$

Are we able to show that $$\operatorname E[A_tf(X_t)]=\operatorname E[A_t]\operatorname E[f(X_t)]\tag1$$ for all $t\ge0$ and bounded $\mathcal E$-measurable $f:E\to\mathbb R$?

The reason whiy I believe the claim could be true is that $\{t\}$ is a Lebesgue null set and hence $A_t$ does effectively depend only on $(X_s)_{s\in(0,\:t)}$ for all $t\ge0$.

Answer 1

This is not true. Let $X_t = 0$ for $t < 1$ and at $t=1$, flip a coin such that $\mathbb{P}(X_1 = 1) = \mathbb{P}(X_1 = -1) = 1/2$. Then let $X_t = X_1$ for $ t \geq 1$. This defines a Markov process. Set $f(x) := x$ for $x \in [-1,1]$ and $f(x) = 0$ for $|x|>1$, and define $c(x) := 1_{[0,\infty]}(x)$. Then, with $t=2$, we get $$ 1/2 = \mathbb{E}[A_2 f(X_2)] \neq \mathbb{E}[A_2]\mathbb{E}[f(X_2)] = 0, $$ since $\mathbb{E}[f(X_2)] = \mathbb{E}[X_1] = 0$.