Let $(\mathcal{F}_n)_{n \in \mathbb{N}}$ be a filtration and $(\mathcal{H}_n)_{n \in \mathbb{N}}$ be a reversed filtration. We consider a sequence $(X_n)_{n \in \mathbb{N}}$ of square integrable random variables such that $(X_n,\mathcal{F}_n)_{n \in \mathbb{N}}$ and $(X_n^2-n,\mathcal{F}_n)_{n \in \mathbb{N}}$ are martingales, and $(\frac{1}{n}X_n,\mathcal{H}_n)_{n \in \mathbb{N}^*}$ and $(\frac{1}{n^2}(X_n^2-n),\mathcal{H}_n)_{n \in \mathbb{N}^*}$ are reversed martingales.
- Prove that there exists $c \in \mathbb{R}$ such that for all $n \in \mathbb{N^*},\frac{1}{n}E[X_{n}^2]=c.$
- Prove that for all $n \in \mathbb{N},X_n \in L^4.$
For 1. we know that $E[X_nX_{n+1}]=E[X_nE[X_{n+1}|\mathcal{F}_n]]=E[X_n^2],$ on the other hand $E[X_nX_{n+1}]=E[X_{n+1}E[X_n|\mathcal{H}_{n+1}]]=nE[X_{n+1}E[\frac{1}{n}X_n|\mathcal{H}_{n+1}]=\frac{n}{n+1}E[X_{n+1}^2].$ Therefore $\frac{1}{n}E[X_n^2]=\frac{1}{n+1}E[X_{n+1}^2].$
How can we prove 2.?