# How can we show that $\left(X^{(1)}_t\right)_{t\in[0,\:\tau_1)}$ and $\left(X^{(2)}_t\right)_{t\in[0,\:\tau_2)}$ are independent here?

Let

• $$(\Omega_i,\mathcal A_i)$$ be a measurable space;
• $$(E_i,\mathcal E_i)$$ be a measurable space with $$E_1\cap E_2=\emptyset$$ and $$(E,\mathcal E):=(E_1\cup E_2,E_1\vee E_2)$$;
• $$\Delta_i\not\in E_i$$ and $$(E_i^\ast,\mathcal E_i^\ast)$$ denote the one-point extension of $$(E_i,\mathcal E_i)$$ by $$\Delta_i$$
• $$\left(X^{(i)}_t\right)_{t\ge0}$$ be an $$(E_i^\ast,\mathcal E_i^\ast)$$-valued process on $$(\Omega_i,\mathcal A_i)$$ and $$\tau_i:=\inf\left\{t\ge0:X^{(i)}_t=\Delta_i\right\};$$
• $$T_1$$ be a Markov kernel with source $$(\Omega_1,\mathcal A_1)$$ and target $$(E_1,\mathcal E_2)$$
• $$p_i$$ denote the projection from $$\Omega:=\Omega_1\times\Omega_2$$ onto the $$i$$th coordinate and $$\tilde X^{(i)}:=X^{(i)}\circ p_i$$
• $$\operatorname P^{(i)}_{x_i}$$ be a probability measure on $$(\Omega_i,\mathcal A_i)$$ with $$\operatorname P_{x_i}^{(i)}\left[X^{(i)}_0=x_i\right]=1$$ for $$x_i\in E_i$$
• $$\pi_i(x_i,\;\cdot\;):=\operatorname P_{x_i}^{(i)}$$ for $$x_i\in E_i$$
• $$[\Delta_i]\in\Omega_i$$ with $$X^{(i)}[\Delta_i]\equiv\Delta_i$$
• $$\tilde\tau_i:=\tau\circ p_i$$

Now define $$\operatorname P_{x_1}:=\operatorname P^{(1)}_{x_1}\otimes(T_1\pi_2)$$ for $$x_1\in E_1$$ and $$\operatorname P_{x_2}:=\delta_{\left[\Delta_1\right]}\otimes\operatorname P^{(2)}_{x_2}$$ for $$x_2\in E_2$$.

I would like to show that if $$f:E\to\mathbb R$$ is bounded and $$\mathcal E$$-measurable, then $$\int_0^{\tilde\tau_1}f\left(\tilde X^{(1)}_t\right)\:{\rm d}t\tag1$$ and $$\int_0^{\tilde\tau_2}f\left(\tilde X^{(2)}_t\right)\:{\rm d}t\tag2$$ are $$\operatorname P_{x_1}$$-independent for all $$x_1\in E_1$$.

Further assumptions on $$T_1$$ are obviously necessary. I would like to assume that $$T_1(\;\cdot\;,B_2)$$ is $$\mathcal F_{\tau_1-}$$-measurable for all $$B_2\in\mathcal E_2$$. This should be enough.

However, I don't know how to proceed. Maybe we can show the stronger claim that $$\left(X^{(1)}_t\right)_{t\in[0,\:\tau_1)}$$ and $$\left(X^{(2)}_t\right)_{t\in[0,\:\tau_2)}$$ are $$\operatorname P_{x_1}$$-independent. But I struggle to show the latter ... It might be useful to note that $$\operatorname E_{x_1}\left[g\left(\tilde X^{(2)}_0\right)\mid\mathcal F_{\tilde\tau_1-}\right]=T_1\left.f\right|_{E_2}\circ p_1\tag3$$ for all bounded $$\mathcal E$$-measurable $$g:E\to\mathbb R$$. In particular, if $$T_1(\omega_1,\;\cdot\;)=\mu_1\left(X^{(1)}_{\tau_1-}(\omega_1),\;\cdot\;\right)\;\;\;\text{for all }\omega_1\in\Omega_1$$ (assuming $$\mathcal E_i$$ is the Borel $$\sigma$$-algebra of a topology on $$E_i$$ and $$X^{(1)}$$ is left-regular with respect to that topology), then $$\operatorname E_{x_1}\left[g(\tilde X^{(2)}_0)\mid\tilde X^{(1)}_{\tilde\tau_1-}\right]=(\mu_1\left.f\right|_{E_2})\left(\tilde X^{(1)}_{\tilde\tau_1-}\right)\tag4.$$ So, my interpretation is that $$\tilde X^{(2)}$$ and $$\tilde X^{(1)}$$ are conditionally independent given $$\tilde X^{(1)}_{\tilde\tau_1-}$$ (or maybe better $$\mathcal F_{\tilde\tau_1-}$$)?

Note that it is easy to show that $$\tilde X^{(2)}$$ and $$\tilde X^{(1)}$$ are actually $$\operatorname P_{x_1}$$-independent for all $$x_1\in E_1$$, when $$T_1$$ does not depend on the first argument (i.e. is simply a measure).