The task at hand is a simple calculation of the quadratic variation of $(tB_t)_{t\geq0}$, where $B$ is Brownian motion. Technically I have a solution but get a wrong result after many recalculations.
I first apply Ito's time-dependent formula, after checking the integrability conditions for the existence of the stochastic integral:
$\forall t \geq 0: t^2B_t^2= \int_0^t2s^2B_sdB_s + \int_0^t (s^2 + 2sB_s^2)ds \tag{1}$
Now, since $(tB_t)_{t \geq 0} \in L^2$, the quadratic variation process exists and is uniquely determined as the continuous adapted process starting at $0$ for $t=0$ making $(t^2B_t^2 – \langle tB_t \rangle)_{t \geq 0}$ a martingale.
Thus, since $\int_0^t2s^2B_s^2dB_s$ is a martingale due to the integrability conditions, we conclude: the quadratic variation is $\langle \cdot B_\cdot \rangle_t = \int_0^t (s^2 + 2sB_s^2)ds \tag{2}$
However, when we take the expected value we obtain:
$\mathbb{E}[\langle \cdot B_\cdot \rangle_t] = \mathbb{E}[\int_0^t (s^2 + 2sB_s^2)ds] \stackrel{(Tonelli)}{=} \int_0^t s^2 + 2s^2 ds = t^3 \tag{3}$
On the other hand, it is well known that $\langle \cdot B_\cdot \rangle_t = t^3/3$ a.s.
Is there a mistake in the solution?
Best Answer
The problem is that you are wrongly assuming that $tB_t$ is a martingale, but it's not! In fact $E(tB_t| \mathcal F_s)=tB_s$, and hence you cannot expect the Doob-Meyer decomposition theorem to hold in this case.