Let $\pi_n$ denotes a refining sequence of partitions of a finite closed interval (refining means $\pi_n\subset\pi_{n+1}).$ And we denote $\pi_n B = \sum_{t_i\in \pi_n}(B_{t_{i+1}}-B_{t_i})^2$, where $B$ is a standard Brownian Motion.
The book I am reading (Protter p.18) says that it is straightforward to show $E(\pi_n B | \pi_{n+1} B, \pi_{n+2} B, \dots)=\pi_{n+1}B$. But actually I am having trouble proving this. Could anyone help?
Thanks!
Without loss of generality, we can assume $\pi_{n+1} \backslash \pi_n = \{s\}$ for some $s \in I$ (otherwise we add the missing partitions), i.e. $\pi_{n+1} = \pi_n \cup \{s\}$. Let $$t_j := \sup\{t \in \pi_n; t<s\}$$ Then $$\pi_{n} B- \pi_{n+1} B = (B_{t_{j+1}}-B_{t_j})^2 - (B_{t_{j+1}}-B_s)^2+(B_s-B_{t_j})^2 = 2 (B_{t_{j+1}}-B_s) \cdot (B_s-B_{t_j})$$ Now let $$\mathcal{G}_s := \sigma(B_r, r \leq s; (B_v-B_u)^2, v,u \geq s\}$$ We have $\sigma(\pi_{n+1}B, \pi_{n+2} B,\ldots) \subseteq \mathcal{G}_s$ and $$\mathbb{E}(B_t-B_s \mid \mathcal{G}_s)=0 \tag{1}$$ for all $t \geq s$, thus $$\mathbb{E}(\pi_{n} B- \pi_{n+1} B \mid \pi_{n+1} B, \pi_{n+2}B,\ldots) = \mathbb{E}(\mathbb{E}(\pi_{n} B- \pi_{n+1} B \mid \mathcal{G}_s) \mid \pi_{n+1} B, \pi_{n+2}B,\ldots) = 2\mathbb{E}((B_{s}-B_{t_j}) \cdot \underbrace{\mathbb{E}(B_{t_{j+1}}-B_s \mid \mathcal{G}_s)}_{\stackrel{(1)}{=}0} \mid \pi_{n+1} B, \pi_{n+2} B,\ldots)=0$$ by tower property.
Equation $(1)$ is a consequence of the fact that $(-B_t)_{t \geq 0}$ is also a Brownian motion. Note that sets of the form $$F := \bigcap_{j=1}^m [B_{r_j} \in C_j] \cap \bigcap_{k=1}^n [(B_{u_k}-B_{v_k})^2 \in D_k]$$ where $r_j \leq s, u_k,v_k \geq s, C_j, D_k \in \mathcal{B}(\mathbb{R})$ are a $\cap$-stable generator of $\mathcal{G}_s$, so it suffices to prove $$\int_F (B_t-B_s) \, d\mathbb{P}=0$$
Since $\sigma(B_r; r \leq s)$ and $\sigma((B_u-B_v)^2,u,v \geq s) \subseteq \sigma(B_u-B_s; u \geq s)$ are independent, we have $$\begin{align} \int_F B_t-B_s \, d\mathbb{P} &= \left( \int (B_t-B_s) \cdot \prod_{k=1}^n 1_{D_k}((B_{v_k}-B_{u_k})^2) \, d\mathbb{P} \right) \cdot \left( \int \prod_{j=1}^m 1_{C_j}(B_{r_j}) \, d\mathbb{P} \right) \\ &= \left( \int (W_t-W_s) \cdot \prod_{k=1}^n 1_{D_k}((W_{v_k}-W_{u_k})^2) \, d\mathbb{P} \right) \cdot \left( \int \prod_{j=1}^m 1_{C_j}(B_{r_j}) \, d\mathbb{P} \right)\\ &= \left( \int (B_s-B_t) \cdot \prod_{k=1}^n 1_{D_k}((B_{v_k}-B_{u_k})^2) \, d\mathbb{P} \right) \cdot \left( \int \prod_{j=1}^m 1_{C_j}(B_{r_j}) \, d\mathbb{P} \right) \\ &= \int_F (B_s-B_t) \, d\mathbb{P} \end{align}$$ where $W_t := -B_t$ is the reflected Brownian motion. (Note that the second equality holds since $(W_t)_t$ and $(B_t)_t$ are both Brownian motions, so they have in particular the same finite-dimensional distributions.) Thus $$2\int_F (B_t-B_s) \, d\mathbb{P}=0$$
This proof is found for instance in René L. Schilling, Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes, p. 142f.