Quadratic Brownian Motion If $B(t)$ is standard Brownian Motion then can we say that $B^2(t) -t $ is a martingale?? Given the following theorem:
If $$ \max_{1<k<n} (t_k-t_{k-1}) \to 0$$ as $n \to \infty$ then
$$\sum_1^n |B(t_k)-B(t_{k-1})|^2 \to T$$ a.s
Many thanks  
 A: The theorem you mention shows that the quadratic variation of a Brownian motion on $[0,T]$ equals $T$. This implies in particular that $(B_t^2-t)_{t \geq 0}$ is a martingale. But actually, this argumentation is overkill since the martingale property follows readily from the properties of a Brownian motion:
Let $s \leq t$. Obviously,
$$B_t^2 = ((B_t-B_s)+B_s)^2 = (B_t-B_s)^2 + 2 B_s \cdot (B_t-B_s) + B_s^2$$
Using that $B_t-B_s$ is independent of $\mathcal{F}_s$, $B_s$ is $\mathcal{F}_s$-measurable and $\mathbb{E}(B_r)=0$ for any $r \geq 0$, we find
$$\begin{align*} \mathbb{E}(B_t^2-t \mid \mathcal{F}_s) &= \mathbb{E}((B_t-B_s)^2 \mid \mathcal{F}_s) + 2 \mathbb{E}(B_s \cdot (B_t-B_s) \mid \mathcal{F}_s) + \mathbb{E}(B_s^2 \mid \mathcal{F}_s) -t \\
&= \mathbb{E}((B_t-B_s)^2) + 2B_s \cdot \mathbb{E}(B_t-B_s) + B_s^2-t \end{align*}$$
By the stationarity of the increments, $B_t-B_s \sim B_{t-s}$. From $\mathbb{E}(B_{t-s}^2)=t-s$ and $\mathbb{E}(B_{t-s})=0$ we conclude
$$\mathbb{E}(B_t^2-t \mid \mathcal{F}_s) = (t-s)+B_s^2-t = B_s^2-s$$
This shows that $(B_t^2-t)_{t \geq 0}$ is a martingale.
