Today in class my professor mentioned that a backwards submartingale is equivalent to a supermartingale; however, I don’t quite understand why that’s true.
Let’s say $(X_n, \mathscr{F}_n)_{n \ge 0}$ is a backwards submartingale. Then we have that $E[X_n | \mathscr{F}_{n+1}] \ge X_{n+1}$. Therefore, $E[X_n] = E[E[X_n | \mathscr{F}_{n+1}]] \ge E[X_{n+1}]$ $\forall n$.
However, in order to be a supermartingale, I believe we specifically need that $E[X_{n+1}|\mathscr{F}_{n}] \le X_n$. By the backwards property, we know that $\mathscr{F}_{n+1} \subset \mathscr{F}_n$; therefore, I’m not quite sure if this conditional expectation is well defined? I believe that $E[X|G]$ only works if $G$ is a sub $\sigma-$field of $F$ where the underlying probability space is defined as $(\Omega, F, \mathbb{P} )$.