Let's consider a function $\Psi\in \mathcal{C}_B(\mathcal{C}[t,T])$ continuous and bounded
\begin{align*}
\Psi \colon \mathcal{C}[t,T] \longrightarrow [0,+\infty)
\end{align*}
Then my question is: Is the following equality true?
\begin{align*}
\mathbb{E} \left [ \Psi\left(S_s, s\in [t,T]\right) \big| \mathcal {F}_t \right ] = \mathbb{E} \left [ \Psi\left(S_s, s\in [t,T]\right) \big| S_t \right ]
\end{align*}
where $S$ is the unique strong solution of a standar SDE with Lipschitz and sublinear coefficients $$dS_s=b(s,S_s)ds+\sigma(s,S_s)dW_s$$ with $S_t=\bar{s}$, $t\leq s \leq T$.
We obviously know $S$ is a Markov process by standard theory, so we have the standard Markov property:
let $s>t$
\begin{align*}
\mathbb{E} \left [ \Phi(S_s) \big| \mathcal {F}_t \right ] = \mathbb{E} \left [ \Phi(S_s) \big| S_t \right ]
\end{align*}
where $\mathcal{F}_t$ is the filtration with respect to $W$ is a brownian motion.
In other word the question is if this Markov property is preserved for a path-dependent process $\Psi=\Psi\left(S_s, s\in [t,T] \right)$ ?
Thanks everybody for your help.
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$\begingroup$ What is $Y_t$ here? $\endgroup$– Iosif PinelisCommented Oct 20, 2020 at 13:55
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$\begingroup$ sorry my mistake, $Y$ was something that shouldn't have been there, just deleted (the sde was a two dimensional sde but nothing changes from this point of view). @Pinelis $\endgroup$– defex95Commented Oct 20, 2020 at 14:17
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$\begingroup$ I also think you need to fix your notations. As it is presented now, the right-hand side of the equality in question depends on $\bar s$, whereas the left-and side does not. $\endgroup$– Iosif PinelisCommented Oct 21, 2020 at 0:35
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$\begingroup$ $\bar{s}$ was the deterministic value of $S$ at time $t$ under the filtration $\mathcal{F}_t$. Ok now it should be ok. $\endgroup$– defex95Commented Oct 21, 2020 at 10:56
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