Discussion
Let \(Y_{1}, Y_{2} \ldots\) be a sequence of independent random variables with mean zero and let \(\mathbf{F}_{n} = \sigma(Y_{1}, \ldots, Y_{n})\).
Define \(X_{n} = \sum_{k=1}^{n} Y_{k}\); \(W_{1} = Y_{1}\) and \(W_{n+1} = W_{n} + Y_{n+1}W_{n}^{2}\) and \(T_{n} = \sum_{k=1}^{n} (Y_{k} + 1)\). Then :
Answer: A
The answer is both \(\{X_{n}\}\) and \(\{W_{n}\}\) are \(\mathbf{F}_{n}\)-martingales.
\(X_{n+1} = X_{n} + Y_{n+1}\), and \(Y_{n+1}\) is independent of \(\mathbf{F}_{n}\) with mean 0, so \(E[X_{n+1} \mid \mathbf{F}_{n}] = X_{n}\).
\(W_{n}\) is \(\mathbf{F}_{n}\)-measurable, so
\[ \begin{aligned} E[W_{n+1} \mid \mathbf{F}_{n}] &= W_{n} + W_{n}^{2}E[Y_{n+1}] \\ &= W_{n} \end{aligned} \]
(assuming the variables are integrable). But \(T_{n} = X_{n} + n\), so \(E[T_{n+1} \mid \mathbf{F}_{n}] = T_{n} + 1\): it is a submartingale, not a martingale. That rules out the options including \(T_{n}\), and 'only \(X_{n}\)' misses \(W_{n}\).
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