Conditions for quadratic variation

Question

Good evening everyone. We are struggling showing the following:
we are given a filtered probability space (each filtration is complete) and $(B_t)_{t \geq 0}$ as a standard $\mathcal{F}_t$ Brownian Motion. In addition, we have the following process: $X = (B_t^2 - t)_{t \geq 0}$. We are asked to show that the quadratic variation (aka sharp brackets) of X is given by: $$\langle X \rangle _ t = \int_0^t4 \cdot B_s^2 \ ds := A_t$$ We have seen a theorem that guarantees that $A_t$ is indeed the quadratic variation if the following four conditions are met:

1. X is a martingale;
2. $A_t$ is adapted, a.s. continuous, increasing and such that $A_0 = 0$;
3. X is squared integrable
4. $(X_t^2 - A_t)_{t \geq 0}$ is a martingale.

We showed those point in this way (and then get stuck in the last one):

1. X is integrable because sum of two integrable functions (namely $B_t^2$ and $-t$). Then we have to check: $$E[X_t - X_s | \mathcal{F_s}] = 0 \\ E[X_t - X_s] = 0 \\ E[B_t^2 - t - B_s^2 + s] = 0$$ which is true by linearity of expectation and recalling that $E[B_t^2] = t \ \forall t \geq 0$
2. $A_t$ is adapted because it is integral of a continuous and adapted process (our $B_t^2$). It is a.s. continuous because integral of a.s. continuous function. It is increasing because $4 \cdot B_t^2 \geq 0 \ \forall t$ and basically we are just enlarging our interval of integration each subsequent step. $A_0 = 0$ trivially.
3. We checked that $E[X^2_t] < \infty$ which is true because we have: $$E[(B_t^2 - t)^2] = E[B_t^4 - 2B_t^2t + t^2] < \infty$$ because we know that $E[B_t^4] = 3t^2$

Is it quite right what we have done so far?
We are left with showing 4. Are there any hints on how to proceed? May Fubini be useful?

Thanks in advance and we hope this argument here will be useful for other users!!

Answer 1

1. First calculate

\begin{align*} X_{t}^{2} &= (B_{t}^{2}-t)^{2} \\ &= ([(B_{t}-B_{u})+B_{u}]^{2} - t)^{2} \\ &= (B_{t}-B_{u})^{4} + 4B_{u}(B_{t}-B_{u})^{3} + 6B_{u}^{2}(B_{t}-B_{u})^{2} + 4B_{u}^{3}(B_{t}-B_{u}) + B_{u}^{4} \\ &\quad - 2t[(B_{t}-B_{u})^{2} + 2B_{u}(B_{t}-B_{u}) + B_{u}^{2}] + t^{2} \end{align*} and $$X_{u}^{2} = B_{u}^{4}-2uB_{u}^{2}+u^{2}$$

before deducing that

\begin{align*} \mathbb{E}_{u}[X_{t}^{2}] &= 3(t-u)^{2} + 6B_{u}^{2}(t-u) + B_{u}^{4} -2t(t-u) -2tB_{u}^{2} + t^{2} \\ &= B_{u}^{4}-2uB_{u}^{2}+u^{2} + 4(t-u)B_{u}^{2} + 2t^{2}-4ut+2u^{2} \\ &= X_{u}^{2} + 4(t-u)B_{u}^{2} + 2(t-u)^{2}. \end{align*}

2. Next calculate

\begin{align*} A_{t} &= \int_{0}^{t} 4B_{s}^{2}\, ds \\ &= A_{u} + \int_{u}^{t} 4B_{s}^{2}\, ds \\ &= A_{u} + 4\int_{u}^{t} (B_{s}-B_{u})^2 + 2B_{u}(B_{s}-B_{u}) + B_{u}^{2}\, ds, \end{align*}

and take a conditional expectation to get

\begin{align*} \mathbb{E}_{u}[A_{t}] &= A_{u} + 4 \int_{u}^{t} (s-u)+B_{u}^{2}\, ds \\ &= A_{u} + 4B_{u}^{2}(t-u) + 2(t^{2}-u^{2}) - 4u(t-u) \\ &= A_{u} + 4B_{u}^{2}(t-u) + 2(t-u)^{2}. \end{align*}

3. Last but not least, take the difference to get

$$\mathbb{E}_{u}[X_{t}^{2}-A_{t}^{2}] = X_{u}^{2}-A_{u}.$$

NOTE: Obviously (for reference) Itô's lemma is a little faster...

\begin{align*} d(X_{t}^{2}-A_{t}) &= 2X_{t}\, dX_{t} + d[X,X]_{t} - dA_{t} \\ &= 2X_{t}(2B_{t}\, dB_{t}) + 4B_{t}^{2}\, dt - 4B_{t}^{2}\, dt \\ &= 4X_{t}B_{t}\, dB_{t}. \end{align*}