Discussion
Let \(\{(X_{n}, F_{n})\}_{n=1}^{\infty}\) be a martingale. Let \(\zeta_{n} = \sigma(X_{1}, X_{2}, ......, X_{n})\). Then, \(\{(X_{n}, \zeta_{n})\}_{n=1}^{\infty}\) is a ................ .
Answer: A
\(\{(X_{n}, \zeta_{n})\}\) is a martingale.
Since \(X_{n}\) is \(F_{n}\)-measurable for each \(n\), the natural filtration satisfies \(\zeta_{n} = \sigma(X_{1}, \ldots, X_{n}) \subseteq F_{n}\). By the tower property,
\[ \begin{aligned} E(X_{n+1}|\zeta_{n}) &= E(E(X_{n+1}|F_{n})|\zeta_{n}) \\ &= E(X_{n}|\zeta_{n}) = X_{n} \end{aligned} \]
Integrability is unchanged, so the process is a martingale with respect to its own filtration. Sub- and supermartingales would need an inequality, which does not arise here, and white noise is an uncorrelated sequence, not this structure.
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