We know that for a submartingle $A(t)$, $A(t)-\langle A\rangle_t$ is a martingale where $\langle A\rangle_t$ is its quadratic variation.
For processes like $W^3(t)$ ($W(t)$ being standard Brownian Motion), It$\hat{ \mathrm{o}}$'s formula gives that
$$d(W^3(t))=3W^2(t)dW(t)+3W(t)dt$$
Then we have $W^3(t)-\int^t_03W(t)dt$ is a martingale.
Can we conclude that
$$\langle W^3(t)\rangle = \int^t_03W(s)ds$$
using the uniqueness of quadratic variation?
Furthermore, how can we calculate $\int^t_03W(t)dt$? It's not an It$\hat{ \mathrm{o}}$ integral.
Moreover, we can check that $W^3(t)-3tW(t)$ is a martingale.
Does it imply $\int^t_03W(s)ds = 3tW(t)$?
Best Answer
The quadratic variation of a continuous local martingale $X$ (say) is the predictable increasing process $V$ such that $X(t)^2-V(t)$ is a local martingale. More generally, the quadratic variation of a continuous semimartingale coincides with the quadratic variation of its martingale part. In your example, this would be $\langle W^3\rangle(t) =9\int_0^t W(s)^4\,ds$.