# Pulling random times out of conditional expectation

Problem
Let $$G$$ be a positive random variable (a random time) that is a.s. finite, $$(X)_{t \geq 0}$$ be a càdlàg process taking values in $$\mathbb{R}^d$$ and $$g$$ is some sufficiently nice real-valued function that takes as input a part of the path of $$X$$ (i.e. $$g( (X_s)_{s \in [a,b]} )$$ is a real valued random variable for fixed $$0 \leq a \leq b \leq \infty$$).

I want to have the following "pull out" property $$\mathbb{E}\left[g( (X_s)_{s \geq G} ) \mid (X_s)_{s \leq G} \right] = \mathbb{E}\left[g( (X_s)_{s \geq t} ) \mid (X_s)_{s \leq t} \right] \mid _{t = G}$$ where $$\mid_{t=G}$$ means evaluating the function of $$t$$ in $$G$$. It is somewhat similar to the question here, but there, one assumes that densities exsist.

Here, $$G$$ is $$\sigma( (X_s)_{s \leq G } )$$-measurable, but is not a $$\sigma( (X_s)_{s \leq t})$$ stopping time (It is unobserved from only observing $$X$$). Also,

$$$$\sigma( (X_{s})_{s \geq t} ) \perp \!\!\! \perp \sigma(G 1_{G \leq t} )\; \mid \; \sigma( (X_{s})_{s \leq t} ) \label{eq:independence}$$$$

My attempt:
By the tower property: $$\mathbb{E}[g( (X_s)_{s \geq G} ) \mid (X_s)_{s \leq G} ] = \mathbb{E}\Big[ \mathbb{E}[ g( (X_s)_{s \geq G} ) \mid (X_s)_{s \leq G}, G ] \mid (X_s)_{s \leq G} \Big]$$ Now by disintegration (Kallenberg 1997, Theorem 5.4) $$\mathbb{E}\Big[ \mathbb{E}[ g( (X_s)_{s \geq G} ) \mid (X_s)_{s \leq G}, G ] \mid (X_s)_{s \leq G} \Big] = \int \mathbb{E}[ g( (X_s)_{s \geq G} ) \mid (X_s)_{s \leq G}, G=t ] \; \mu((X_s)_{s \leq G} , dt)$$ where $$\mu(x,\cdot)$$ is the regular conditional distribution of $$G$$ given $$(X_s)_{s \leq G}=x$$ (it will be a one-point measure, since $$G$$ is $$(X_s)_{s \leq G}$$ measurable).

Now by a little dubious subtitution, we get \begin{align*} \mathbb{E}[ g( (X_s)_{s \geq G} ) \mid (X_s)_{s \leq G}, G=t ] &= \mathbb{E}[ g( (X_s)_{s \geq t} ) \mid (X_s)_{s \leq t}, G=t ] \\ &= \mathbb{E}[ g( (X_s)_{s \geq t} ) \mid (X_s)_{s \leq t}, G 1_{G \leq t}=t ] \\ &= \mathbb{E}[ g( (X_s)_{s \geq t} ) \mid (X_s)_{s \leq t} ] \end{align*} where the last equality followed by the conditional independence.

Finally, using that $$\mu((X_s)_{s \leq G} , dt) = \delta_{G}(dt)$$ (the dirach measure in $$G$$), we get the desired result $$\mathbb{E}[g( (X_s)_{s \geq G} ) \mid (X_s)_{s \leq G} ] = \mathbb{E}[ g( (X_s)_{s \geq t} ) \mid (X_s)_{s \leq t} ] \mid_{t=G}$$

My problem here is that i am aware of the problems and paradoxes that occur when using heuristical arguments such as the above "dubious substitutions" inside regular conditional expectations. So is there a way to formalize the argument above, or give a different argument for the same result? If there is additional regularity required, what is then needed?

This seems related to the difference between the usual Markov property and the strong Markov property (since if the "pull out" property always held, it seems like the Markov property would imply the strong Markov property), but $$G$$ is not a stopping time here, and some additional conditional independence is known. Maybe some of the arguments from there can be used e.g. approximating the random time by a sequence of discrete valued random times?