I was working on this problem:
Use Itô's formula to prove that the following stochastic processes
are martingales:
$$ \begin{equation}
\bullet \quad X_t = e^\frac{t}{2}\cos B_t \\
\bullet \quad X_t = e^\frac{t}{2}\sin B_t \\
\bullet \quad X_t = (B_t + t)e^{-B_t – \frac{t}{2}} \\
\end{equation}$$
I used Ito's lemma to prove that the resulting stochastic differential equation is independent of any drift factor. How does it imply that $X_t$ is a martingale, mathematically. How can you prove it?
$$ \begin{equation}
\bullet \quad dX_t = -e^\frac{t}{2}\sin B_t dB_t \\
\bullet \quad dX_t = e^\frac{t}{2}\cos B_t dB_t \\
\bullet \quad dX_t = e^{-B_t – \frac{t}{2}}(1 – B_t -t)dB_t \\
\end{equation}$$
These are the resulting differential equations, all independent of drift. How can you prove, in general, that an SDE independent of drift, is a Martingale?
Best Answer
This is in general because $dB_t$ is independent of $B_t$ by the definition of the Wiener process/Brownian motion. More specifically, for the unbound parts of the path in the future of time point $s$, $t\ge s$ $\newcommand{\IE}{\Bbb E}$ $$ \IE_s(f(t,B_t)dB_t)=\IE_s(f(t,B_t))\IE_s(dB_t)=0 $$ which taken together with the fixed parts of the path under $\cal F_s$ in the past of time point $s$ implies $$ \IE_s(X_t-X_0)=\IE_s\left(\int_0^t f(u,B_u)dB_u\right)=\int_0^s f(u,B_u)dB_u=X_s-X_0. $$ This is the main property defining a martingale.