I'm reading a proposition in the class of Introduction to Martingale:
If $\left(X_{n}\right)_{n \in \mathbb{Z}_{+}}$ is a submartingale relative to some filtration $\left(\mathcal{F}_{n}\right)_{n \in \mathbb{Z}_{+}}$, then it is also a submartingale with respect to its natural filtration $\left(\mathcal{G}_{n}\right)_{n \in \mathbb{Z}_{+}}$ where $\mathcal{G}_{n}:=\sigma\left(X_{0}, \ldots, X_{n}\right)$, for any $n$.
I'm trying to show $X_n \le \mathbb E [X_{n+1} \mid \mathcal G_n]$. But I can not infer any useful information from the fact that $\left(X_{n}\right)_{n \in \mathbb{Z}_{+}}$ is a submartingale w.r.t $\left(\mathcal{F}_{n}\right)_{n \in \mathbb{Z}_{+}}$. It seems to me that $(\mathcal{G}_{n})$ is not related to $(\mathcal{F}_{n})$.
Could you please elaborate on this proposition? Have I missed something?
Best Answer
Since $(X_n)_{n \in \mathbb{N}_0}$ is a submartingale w.r.t to $(\mathcal{F}_n)_{n \in \mathbb{N}_0}$, the process $(X_n)_{n \in \mathbb{N}_0}$ is adapted to $(\mathcal{F}_n)_{n \in \mathbb{N}_0}$, i.e. $$\mathcal{G}_n :=\sigma(X_0,\ldots,X_n) \subseteq \mathcal{F}_n. \tag{1}$$
By the tower property of conditional expectation, this implies $$\mathbb{E}(X_{n+1} \mid \mathcal{G}_n) = \mathbb{E} \bigg[ \mathbb{E}(X_{n+1} \mid \mathcal{F}_n) \mid \mathcal{G}_n \bigg].$$
From the submartingale property $$ \mathbb{E}(X_{n+1} \mid \mathcal{F}_n) \geq X_n$$ we get
$$\mathbb{E}(X_{n+1} \mid \mathcal{G}_n) \geq \mathbb{E}(X_n \mid \mathcal{G}_n) = X_n,$$
and so $(X_n)_{n \in \mathbb{N}_0}$ is a submartingale w.r.t. $(\mathcal{G}_n)_{n \in \mathbb{N}_0}$.