Is $(B_t^2 - t^2)_{t\geq 0}$ a local martingale?

Question

Let $B$ be a standard $\mathbb{R}$-valued Brownian motion. It is very easy to show that $(B_t^2 - t^2)_{t\geq 0}$ is not a martingale by checking the martingale condition. First, note that $(B_t^2 - t)_{t\geq 0}$ is a martingale. Using the martingale condition of $(B_t^2 - t)_{t\geq 0}$ yields: \begin{align*} &\mathbb{E}\left[B_t^2 - t^2\vert\mathcal{F}_s\right] = \mathbb{E}\left[B_t^2 - t\vert\mathcal{F}_s\right] + t - t^2 = B_s^2 - s + t - t^2 \overset{!}{=} B_s^2 - s^2\\ &\Leftrightarrow t - t^2 = s - s^2 \end{align*} The last equation is not always true (e.g. take $s=1$ and $t=2$) and thus $(B_t^2 - t^2)_{t\geq 0}$ is not a martingale. However, I have no clue how to investigate if $(B_t^2 - t^2)_{t\geq 0}$ is a local martingale.

Answer 1

Note that the process $(B_t^2 - t)_{t\ge 0}$ is a martingale. This implies (by Doob's Theorem) that for any finite stopping time $\tau$, the stopped process $X_t = (B_{t \wedge \tau}^2 - t\wedge \tau)_{t \ge 0}$ is a martingale.

In order for $(B_t^2 - t^2)$ to be a local martingale, we would in particular need the existence of stopping time $\tau$ such that $(B_{t \wedge \tau}^2 - (t \wedge \tau)^2 )_{t \ge 0}$ is a martingale. But if this process is a martingale, then for any $t \ge 0$, $$ 0 = \mathbb E[B_{t \wedge \tau}^2 - (t \wedge \tau)^2] = \mathbb E[t \wedge \tau - (t \wedge \tau)^2].$$ Since $s - s^2 \ge 0$ for $s \in (0,1)$ we see the latter for $t \in (0,1)$ implies $t \wedge \tau = (t \wedge \tau)^2$ almost surely, i.e., $\tau = 0$ almost surely due to $t \in (0,1)$.

This shows that there cannot exist a sequence of stopping times $T_n \to +\infty$ such that $(B_{t \wedge T_n}^2 - (t \wedge T_n)^2)_{t \ge 0}$ are martingales for any $n$.

Answer 2

If $B^2_t-t^2$ were a local martingale the so too would $t-t^2=(B^2-t)-(B^2-t^2)$ be a (continuous) local martingale, but also a process of finite variation. Such a process must be constant in time, forcing $t-t^2=0$ for all $t>0$, a contradiction. Therefore $B^2_t-t^2$ is no local martingale.