Suppose $\{X_t\}_{t \geq 0}$ is a Markov process, then we know $\mathbb{E}[X_T \mid \mathcal{F}_t]= \mathbb{E}[X_T \mid X_t]$. However, if we only have the form of the process explicility, without being told whether it is Markovian or not, apart from the definition, do we have other ways to check. This question is related to the following link Markovian Hawkes Process elementary proof and I also agree that literature just mention for a Hawkes process that $\{N_t,\lambda_t\}_{t\geq 0}$ is a Markov process.
I am also wondering that since we know if $\mathbb{E}[g_1(X_T,T)\mid \mathcal{F}_t]=\mathbb{E}[g_2(X_T,T)\mid \mathcal{F}_t]$ for two measurable function $g_1$ and $g_2$, then we have $\mathbb{E}[g_1(X_T,T)]=\mathbb{E}[g_2(X_T,T)]$ by taking one expectation. Then, under what condition, the reverse would be true? Will $\{X_t\}_{t \geq 0}$ being a Markov process helpful here?
Any reference is apprciated. Many thanks in advance
