Quadratic variation of a process

Question

I am given a filtered probability space and $(B_t)_{t \geq 0}$ as a standard $\mathcal{F}_t$ Brownian Motion. In addition, I have the following process: $X = (B_t^2 - t)_{t \geq 0}$. I am asked to show that the quadratic variation (aka sharp brackets) of X is given by: $$\langle X \rangle _ t = \int_0^t4 \cdot B_s^2 \ ds$$ Is there any theorem that can help me working through this? If not, would someone give me a good hint on how to proceed?

Answer 1

For a suitable integrand $Y$, the quadratic variation of the process $I_t := \int_0^t Y_u dB_u, 0 \leq t \leq T$ is the process $\langle I \rangle_t = \int_0^t Y^2_u du, 0 \leq t \leq T.$ (*)

Now, applying Ito's formula with the function $f(x)=x^2 -t,$ we obtain $dX_t = 2B_tdB_t.$ Applying this together with (*) we get the result.

Answer 2

Both

$$\big(B_t^4-6\int_0^tB_s^2\ ds \big)_{t\ge 0}$$

and

$$\big(tB_t^2-\int_0^t(B_s^2+s)\ ds \big)_{t\ge 0}=\big(tB_t^2-\frac{1}{2}t^2-\int_0^tB_s^2\ ds \big)_{t\ge 0}$$

are martingales, whereas the latter one can be found from a markov process with generator of the function $f(t,x)=tx^2$.

From that we find, that

$$\big(B_t^4-6\int_0^tB_s^2\ ds -2\big(tB_t^2-\frac{1}{2}t^2-\int_0^tB_s^2\ ds\big)\big)_{t\ge 0}=\big((B_t^2-t)^2-4\int_0^tB_s^2\ ds\big)_{t\ge 0}$$

is a martingale.