I am trying to solve the following exercise, which should be easy, but I am not managing to, despite different trials. I have a martingale $X_n$, and I know that it is an increasing process, i.e. $X_n\leq X_{n+1}, \forall n$ and I should prove that $X_n=X_0$ a.s. And then show the same for a martingale s.t. Its square is still a martingale.
I tried using the definition of conditional expectation and showing that $E(X_0U) =E(X_{n+1}U)$, for every $U \mathcal{F}_n$-measurable and bounded but nothing. I also tried proceeding by inequalities using various properties of conditional expectation.
Any ideas?
If $\{X_n\}$ is a martingale then $EX_n=EX_0$ for all $n$. If $X_n \leq X_{n+1}$ then $E(X_{n+1}-X_n)=EX_0-EX_0=0$. If a non-negative random variable has mean 0 the it is 0 almost surely. Hence $X_{n+1}-X_n=0$ Second part: consider $$(E(X_{n+1}-X_n)|X_1,X_2,...,X_n))^{2}=(E(X_{n+1}|X_1,X_2,...,X_n))^{2}-2X_nE(X_{n+1}|X_1,X_2,...,X_n))+X_n^{2}=(E(X_{n+1}|X_1,X_2,...,X_n))^{2}-2X_n^{2}+X_n^{2}$$ by martingale property. Now $$(E(X_{n+1}|X_1,X_2,...,X_n))^{2} \leq (E(X_{n+1}^{2}|X_1,X_2,...,X_n))=X_n^{2}$$ since $\{X_n^{2}\}$ is also a martingale. Combining all these we get $$(E(X_{n+1}-X_n)|X_1,X_2,...,X_n))^{2}=0$$. Taking expectation we get $E(X_{n+1}-X_n)^{2}=0$ so $X_{n+1}=X_n$ almost surely for each $n$.