I know that the process $X_t$ is a martingale when:
- $\mathbb{E}[|X_t|]<\infty$ for each $t$,
- $\mathbb{E}[X_t|\mathcal{F_s}] = X_s$ for $s<t$.
Using properties of the conditional expectation we can show that:
$\mathbb{E}[X_s] = \mathbb{E}[X_t]$.
My question is:
Can we use this property ($\mathbb{E}[X_s] = \mathbb{E}[X_t]$) to prove that the process is a martingale?
I noted that in many sources people show that $\mathbb{E}[X_t|\mathcal{F_s}] = X_s$ to prove that the process is a martingale.
I think that these conditions are equivalent and showing the first condition should be easier. Am I right?
Fractional Brownian motion with $H\neq 1/2$ is not a martingale yet $E(B_s)=E(B_t)=0$ for all $s,t$.