Prove that if $ (X_{n}, \mathcal{F}_n)_{n=0}^\infty $ is a supermartingale such as
$ EX_{n} = EX_{0} $ for all $ n $ then $ (X_{n}, \mathcal{F}_n)_{n=0}^\infty $ is a martingale
Is it enough to say that if $ EX_{n} = EX_{0} $ for all $ n $ then $ EX_{n} = const $ for all $ n $ and use a prove of theory saying that that every supermartingale $X_{t}$ for which the map $t\mapsto E(X_{t})$ is constant is already a martingale?
Best Answer
$E(X_{n+1} |\mathcal F_n) \geq X_n$. But $E (E(X_{n+1} |\mathcal F_n) - X_n) =EX_{n=1} -EX_n=EX_0-EX_0=0$. If a non-negative random variable has zero mean it is $0$ almost surely. Hence $E(X_{n+1} |\mathcal F_n) = X_n$.