Let $(X_t)$ and $(Y_t)$ be two stochastic processes taking values in $E$ and $F$ respectively. Consider the sigma algebra $(\mathcal{F}_t)$ generated by $Z_t=( X_t,B_t,Y_t)$ where $B_t$ is the Brownian motion. Prove that showing $(X_t)$ is an $(\mathcal{F}_t)$ martingale is equivalent to showing for fixed $0\le s<t\le T$ $$E( h_s X_t)=E(h_s X_s)$$ for all $h_s$ that is $\mathcal {F}_s$ measurable. And as $(\mathcal{F}_s)$ is generated by $(X_r,B_r,Y_r)_{0\le r\le s}$, it is sufficient to show for all $0 \le t_1<\cdots t_n\le s$, and for all continuous functions $f$ on $\Gamma^n$ where $ \Gamma = E \times \mathbb{R}^d \times F$, one has $$E[ f(Z_{t_1}, \cdots, Z_{t_n})X_t] =E[ f(Z_{t_1}, \cdots, Z_{t_n})X_s] $$
Proving $(X_t)$ is a martingale is showing $E(1_A X_t)= E(1_A X_s)$ for all $A \in \mathcal{F}_s$. And as any $h_s$ which is $\mathcal{F}_s$ measurable can be approximated by indicator functions we have $E( h_s X_t)=E(h_s X_s)$ by passing to the limit. How can I prove the other part consisting of continuous function?
Best Answer
Since $f$ is continuous, we know that $f(Z_{t_1}, ..., Z_{t_n})$ is $\cal{F}_s$-measurable. Thus the only thing we need to show is that for $X$ to be a martingale it is sufficient to have \begin{equation}\label{eqn:answer} \mathbb{E}[f(Z_{t_1}, ..., Z_{t_n}) X_t] = \mathbb{E}[f(Z_{t_1}, ..., Z_{t_n}) X_s]. \end{equation} By approximating indicator functions and using dominated convergence you can show that the above implies $\mathbb{E}[\chi_{E} X_t] = \mathbb{E}[\chi_{E} X_s]$ for all $$E \in \bigcup\limits_{n \geq 1} \bigcup\limits_{0 \leq t_1 \leq ... \leq t_n \leq s} \sigma(Z_{t_1}, ..., Z_{t_n}) =: \mathcal{A}.$$ Note that $\mathcal{A}$ is closed under intersections, so it is a $\pi$-system. Now let $\mathcal{M}$ denote the collection of sets $E$ for which $\mathbb{E}[\chi_{E} X_t] = \mathbb{E}[\chi_{E} X_s]$ so that in particular $\mathcal{A} \subseteq \mathcal{M}$. Using the MCT you can show that $\mathcal{M}$ is closed under countable monotone unions, and therefore is a monotone class. By the monotone class theorem we have $\sigma(\mathcal{A}) \subseteq \mathcal{M}$. Since $\sigma(\mathcal{A}) = \mathcal{F}_s$ this proves the claim.