Question: Prove $X_t = {W_t}^3 – 3\int^{t}_{0} W_s \, ds$ is a martingale using definition, that is, for any $0\leq s <t, $ we have
$$\mathbb{E}(X_t|\mathcal{F}_s) = X_s$$
where $\mathcal{F}_s$ is the filtration generated by $X_s.$
I can solve the question by showing that the SDE satisfied by $X_t$ has no drift term, and thus $X_t$ is a martingale.
But I do not know how to show using definition of martingale.
Best Answer
$$X_t-X_s=(W_t-W_s+W_s)^3-W_s^3-3 \int_{s}^{t}{(W_u-W_s+W_s)du}$$
$$=\left[(W_t-W_s)^3+3(W_t-W_s)^2W_s+3(W_t-W_s)W_s^2+W_s^3\right]-W_s^3-3 \int_{s}^{t}{(W_u-W_s)du}-3W_s(t-s)$$
Taking the expectation, using the fact the independence between $W_u-W_s$ and $\mathcal{F_s}$ with $s\leq u \leq t$
$$E[(W_t-W_s)^3| \mathcal{F_s}]=E[(W_t-W_s)^3]=0$$ $$E[3(W_t-W_s)^2W_s| \mathcal{F_s}]=3W_sE[(W_t-W_s)^2]=3W_s(t-s)$$ $$E[3(W_t-W_s)W_s^2| \mathcal{F_s}]=3W_s^2E[(W_t-W_s)]=0$$ $$E\left[3 \int_{s}^{t}{(W_u-W_s)du}| \mathcal{F_s}\right]=3 \int_{s}^{t}{E\left[(W_u-W_s)| \mathcal{F_s}\right]du}=0$$
Therefore $$E[(X_t-X_s)| \mathcal{F_s}]=3W_s(t-s)-3W_s(t-s)=0$$
or $$E[X_t| \mathcal{F_s}]=X_s$$