How can I show that for a sequence of independent martingales $(X_{i}^{j})_{i,j}$ we have $E[X_{i+1}^{j}\lvert \mathcal G_{i}]=X_{i}^{j}$

Question

How can I show that for a sequence of independent martingales $(X_{i}^{j})_{i\in \mathbb N,j\in [m]}$ with respect to the natural filtrations $\mathbb F^{j}, j \in [m]$, that is

$\mathbb F^{j}=(\mathcal{F}_{i}^{j}:i \in \mathbb N)$ and $\mathcal{F}_{i}^{j}=\sigma(X_{h}^{j}: h \in [i]) $.

For notation purposes $[i]:=\{1,...,i\}$.

Now define a new filtration $\mathbb G = (\mathcal{G}_{i}:i \in \mathbb N)$ and $\mathcal{G}_{i}=\sigma(X_{h}^{j}: h \in [i], j \in [m]) $

Question:

How can I show that for $j \in [m]$:

$E[X_{i+1}^{j}\lvert \mathcal G_{i}]=X_{i}^{j}$ a.s.?

Does the Dynkin Lemma help?

Answer 1

Hint: use the following property of conditional expectation, see e.g. Section 9.7 in Williams' Probability with Martingales.

If $\mathcal{H}$ and $\mathcal{G}$ are two $\sigma$-fields and $X$ is a random variable such that $\mathcal{H}$ is independent of $\sigma(\sigma(X),\mathcal{G})$, then $$E\left[X\middle| \sigma(\mathcal{G},\mathcal{H})\right] = E[X|\mathcal{G}] \quad \text{a.s.}$$