Given prob space $(\Omega, \mathscr{F}, P)$ and a Wiener process $(W_t)_{t \geq 0}$, define filtration $\mathscr{F}_t = \sigma(W_u : u \leq t)$
Let $(A_t)_{t \geq 0}$ where $A_t = W_t^3 – 3tW_t$. Show that $E[A_t|\mathscr{F}_s] = A_s$ whenever $s < t$.
I think this all comes down to manipulation since there are martingales somewhere
My attempt:
Splitting up into $E[W_t^3|\mathscr{F}_s] – 3E[tW_t|\mathscr{F}_s]$ doesn't do anything since those guys aren't martingales? So, I tried splitting it up into:
$E[W_t(W_t^2 – 3t)|\mathscr{F}_s]$
$= E[W_t(W_t^2 – t -2 t)|\mathscr{F}_s]$
$= E[W_t(W_t^2 – t) -2 tW_t)|\mathscr{F}_s]$
$= E[W_t(W_t^2 – t)|\mathscr{F}_s] -2E[ tW_t|\mathscr{F}_s]$
$W_t$ is not $\mathscr{F}_s$-measurable, so we can't take that out…
$tW_{1/t}$ is Brownian and thus a martingale, but I don't know about $tW_t$…
$cW_{t/c^2}$ is Brownian and thus a martingale, but I don't think we can set c = t…
Help please?
Best Answer
Writing
$$W_t^3 = ((W_t-W_s)+W_s)^3 = (W_t-W_s)^3+ 3 W_s (W_t-W_s)^2 + 3 W_s^2 (W_t-W_s) + W_s^3$$
we find using the independence of the increments
$$\begin{align*} \mathbb{E}(W_t^3 \mid \mathcal{F}_s) &= \underbrace{\mathbb{E}((W_t-W_s)^3)}_{\mathbb{E}(W_{t-s}^3)=0} + 3W_s \underbrace{\mathbb{E}(W_t-W_s)^2}_{\mathbb{E}(W_{t-s}^2)=t-s} + 3W_s^2 \underbrace{\mathbb{E}(W_t-W_s)}_{0}+W_s^3 \\ &= 3(t-s) W_s + W_s^3. \end{align*}$$
Consequently,
$$\mathbb{E}(A_t \mid \mathcal{F}_s) = \mathbb{E}(W_t^3 - 3t W_t \mid \mathcal{F}_s) = 3(t-s)W_s+ W_s^3 - 3t \underbrace{\mathbb{E}(W_t \mid \mathcal{F}_s)}_{W_s} = A_s.$$
A general remark: If you want to prove a process of this form a martingale, it is always a good idea to write $$W_t = (W_t-W_s)+W_s$$ and separate both terms, since this allows us to use the independence of the increments (which in turn makes calculation of conditional expectations much more easier).