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1

Note that $\{W_t \mid t \geq 0\}$ is a martingale. Then, for $0<p<q<r$, \begin{align*} E(W_pW_qW_r) &= E\Big( E(W_pW_qW_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q E(W_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q^2\Big)\\ &=E\Big(W_p(W_q-W_p+W_p)^2\Big)\\ &=E\Big(W_p(W_q-W_p)^2+W_p^3+2W_p^2(W_q-W_p) \Big)\\ ...


2

\begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qquad -3t(W_t-W_s)-3tW_s \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3\Big) + W_s^3+3W_sE\Big((W_t-W_s)^2\Big)\\ &\qquad \qquad + 3W_s^2 ...


5

The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows: $$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$ where you'll need to find $C$ such that the equality holds (i.e. $C=W_s^3 + ...


6

We know that $(\tilde{W}_t) := (-W_t)$ is also a Wiener process so $$ E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r] $$ and that implies that $E[W_pW_qW_r] = 0$.


2

I think you are on the right track here. You made a sign error in the first line, unfortunately: $$E[W_p W_q W_r] = E[W_r W_p^2 + W_pW_q^2 - W_qW_p^2]=\\ E[(W_r-W_q)W_p^2]+E[W_pW_q^2]= E[W_pW_q^2] $$ The first term is $0$ by independence (as $p<\text{min}(r,q)$ and the square does not affect independence). To take care of the second term we do the ...


1

You can use that $f(t,W_t)\in C^2$ is Martingale iff:$$\partial_t f+\frac{1}{2}\partial_{WW}f= 0$$ We get:$$\partial_t f=-3W_t$$$$\partial_{WW}f=6W_t$$ Finally: $$-3W_t+3W_t= 0$$ q.e.d. The proof of theorem follows by writing out $f(t,W_t)$ via Ito formula. Proof of theorem:



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