Check whether the following process is martingale with respect to Brownian filtration.

Question

Is $X(t) = t^2W(t)-2\int_{0}^{t} sW(s) ds$ a martingale? I was thinking to use Ito integral to prove this but couldn't prove the term left on other side is a martingale.

Answer 1

$E((t+u)^{2}W(t+u)-2\int_0^{t+u} sW(s)\, ds|\mathcal F_t)=E((t+u)^{2}(W(t+u)-W(t))|\mathcal F_t)+(t+u)^{2}W(t)-2\int_0^{t}sW(s) \, ds-2E(\int_t^{t+u}sW(s)\, ds|\mathcal F_t)$. The first term is $0$. So all that remains is to evaluate the last term. Write $E(\int_t^{t+u}sW(s)\, ds|\mathcal F_t)$ as $E(\int_t^{t+u}s(W(s)-W(t))\, ds|\mathcal F_t)+W(t) \int_t^{t+u}s ds=0+W(t)(u^{2}+2tu)/2$. You should end up with $$$E((t+u)^{2}W(t+u)-2\int_0^{t+u} sW(s)\, ds|\mathcal F_t)= t^{2}W(t)-2\int_0^{t} sW(s)ds$.