I have a question regarding the definition of 'Markov process' given by Shreeve, book II (continuous models).
It says (see below) this: "Assume that for all $0 \leq s \leq t \leq T$ and for every nonnegative Borel-measurable function $f$, there is another Borel-measurable function $g$ such that $\mathbb E [ f(X(t)) \mid \mathcal F (s) ] = g(X(s)$.
I get very confused from his comments in the Remark below it. It says there that $f$ is "permitted to depend on $t$" and $g$ depend on $s$.
I do understand the Definition as stated, but I do not understand the Remark following it.
What does this mean? Why is this necessary? If we pick some times $s,t$ with $0 \leq s \leq t \leq T$ then these times are fixed--It's unclear to me what a function "depending on $t$" means then.
Can someone clarify this definition? I do not understand what it means.
Together with this, I also do not understand why we can take $g$ to be the same symbol as $f$.
In general, I just don't understand what is meant here. Can someone explain? It would be highly appreciated if someone could give a more precise statement/formulation of what Shreve means here. Thanks a lot.
I too find this Remark 2.3.7. a bit hard to follow probably even slightly incorrect, in the sense that Shreve should have written that "and the function $g$ will depend on $\color{red}{t}$" (this might be a typo).
Regardless if $f$ depends on $t$ or not, a way of writing the Markov property is $$ \mathbb E\big[f\big(X_t)\,\big|\,{\cal F}_s\big]=\mathbb E\big[f(X_t)\,\big|\,X_s\big]\,. $$ From the RHS we see that the function $g$ should always depend on $t\,$ (and $X_s\,$).
When $f$ depends on $t$ also we use the fact that $(t,X_t)$ is trivially a Markov process (with the same filtration as $X\,$). Then $$ \mathbb E\big[f\big(t,X_t)\,\big|\,{\cal F}_s\big]=\mathbb E\big[f(t,X_t)\,\big|\,X_s\big]\,. $$ The function $g$ on the RHS now depends on $t$ and $X_s$ again but not on $s$ because that is not random: $\sigma(s,X_s)=\sigma(X_s)\,.$
Simple example: Brownian motion (or any other martingale): $$ \mathbb E[t\,W_t\,|\,{\cal F}_s]=\color{red}{t}\,W_s=g(\color{red}t,W_s)\,. $$