Can anyone help me with the following problem?
Let $W(t), t\geq 0$ be a Brownian motion with filtration:$F(t)$. Let $0\leq s\leq t$.
1- Show that $E\left [ W^{3}(t)\mid F(s) \right ]=W^{3}(s)+3(t-s)W(s)$. where $E\left [ (*)\mid* \right ]$ is the conditional expectation.
2- Show that: $ W^{3}(t)-3\int_{0}^{t}W(u)du$ is a Martingale.
For the First Part: I tried the following: $E\left [ W^{3}(t)\mid F(s) \right ]=E\left [ W^{3}(t)-W^{3}(s)+W^{3}(s) \right ]=E\left [ W^{3}(t)-W^{3}(s) \right ]+E\left [W^{3}(s) \right ]=E\left [ (W(s)-W(t))^{3} +3W(t)W(s)(W(t)-W(s))\right ]+W^{3}(s)=...$
I couldn't move from that point. Can anyone write in detail how I can complete the solution.
For the second part: I did the following $E\left [ W^{3}(t)-3\int_{0}^{t}W(u)du\mid F(s) \right ]=E\left [ W^{3}(t) \mid F(s)\right ]-3E\left [ \int_{0}^{t}W(u)du\mid F(s) \right ]=W^{3}(s)+3(t-s)W(s)-3\int_{0}^{t}E\left [ W(t)\mid F(s) \right ]du$.
Can anyone let me know how to finish my proof?