The following question arise from the proof of Levy's theorem in Richard Bass - Stochastic processes (can be seen via Google books, its on page 77).
So we have $(M_t)_{t\geq 0}$ a continuous local martingale, $M_0=0$ adapted to $\{\mathcal{F}_t\}$ s.t. $<M>_t = t$.
We let $t_0>0$ and define $N_t=M_{t_0+t}-M_{t_0}$ and now its routine (Bass claims) to show $<N>_t=t$.
Had the quadratic variation been linear - then no problem, but by my objection is that we only know the mixed variation is a biliniar form so that $$<N>_t=<M_{t_0+t}-M_{t_0},M_{t_0+t}-M_{t_0}>=<M_{t_0+t}>+<M_{t_0}>-2<M_{t_0},M_{t_0+t}>$$ which I cant see is the sought.
On the other hand, for square integrable martingales we have that $E[(M_S-M_T)^2|\mathcal{F_S}]=E[M_S^2-M_T^2|\mathcal{F_S}]$ (his proposition 9.6 p 56) so in this case the quadratic variation would be linear (one can just check it works) - but how is this not in conflict with the above bilinearity?
By definition, we just need to check that $(N_t^2-t)_{t\geq 0}$ is a local martingale with respect to the filtration $(\mathcal{F}_t')_{t\geq 0}$, where $\mathcal{F}_t'=\mathcal{F}_{t+t_0}$ using that $(M_t^2-t)_{t\geq 0}$ and $(M_t)_{t\geq 0}$ are local martingales with respect to $(\mathcal{F}_t)_{t\geq 0}$.
A localization argument shows that we can treat the problem for true martingales, and thus have to show that $(N_t^2-t)$ is a true $(\mathcal{F}_t')_{t\geq 0}$-martingale. To that end, let $0\leq s\leq t$ be given and then $$ \begin{align} {\rm E}[N_t^2-t\mid\mathcal{F}_s']&={\rm E}[M_{t+t_0}^2+M_{t_0}^2-2M_{t+t_0}M_{t_0}-t\mid\mathcal{F}_{s+t_0}]\\ &={\rm E}[M_{t+t_0}^2-(t+t_0)+M_{t_0}^2-2M_{t+t_0}M_{t_0}-t_0\mid\mathcal{F}_{s+t_0}]\\ &={\rm E}[M_{t+t_0}^2-(t+t_0)\mid\mathcal{F}_{s+t_0}]+M_{t_0}^2-2M_{t_0}{\rm E}[M_{t+t_0}\mid\mathcal{F}_{s+t_0}]-t_0\\ &=M_{s+t_0}^2-(s+t_0)+M_{t_0}^2-2M_{t_0}M_{s+t_0}-t_0\\ &=(M_{s+t_0}-M_{t_0})^2-s\\ &=N_s^2-s. \end{align} $$