For $s<t$, write $X_t=X_s+Y$, where the increment $Y=X_t-X_s$ is independent of the <natural filtration> at time $s$, has mean zero, and has variance $(t-s)\sigma^2$. Hence
$$
\mathbb E[X_t^2\mid\mathcal F_s]
=X_s^2+\mathbb E[Y^2]
=X_s^2+(t-s)\sigma^2.
$$
It follows that $M_t=X_t^2-t\sigma^2$ is a martingale, as asserted by the <centered square-integrable Lévy martingale> identity.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 201 / 6 / c / Solution

{title2=$\mathcal F_t^X$}
{wiki=Filtration_(probability_theory)}

The natural filtration of a stochastic process $X$ records its history:
$$
\mathcal F_t^X=\sigma(X_s:0\leq s\leq t),
$$
possibly completed and made right-continuous when the usual conditions are required.


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