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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?

htd
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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 $$ \begin{align} {\rm E}[N_t^2-t\mid\mathcal{F}_s']&={\rm E}[M_{t+t_0}^2+M_{t_0}^2-2M_{t+t_0}M_{t_0}-t\mid\mathcal{F}_{s+t_0}]\\ &={\rm E}[M_{t+t_0}^2-(t+t_0)+M_{t_0}^2-2M_{t+t_0}M_{t_0}-t_0\mid\mathcal{F}_{s+t_0}]\\ &={\rm E}[M_{t+t_0}^2-(t+t_0)\mid\mathcal{F}_{s+t_0}]+M_{t_0}^2-2M_{t_0}{\rm E}[M_{t+t_0}\mid\mathcal{F}_{s+t_0}]-t_0\\ &=M_{s+t_0}^2-(s+t_0)+M_{t_0}^2-2M_{t_0}M_{s+t_0}-t_0\\ &=(M_{s+t_0}-M_{t_0})^2-s\\ &=N_s^2-s. \end{align} $$

Stefan Hansen
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  • Wouldn't you want to use $E[(M_S-M_T)^2|\mathcal{F_S}]=E[M_S^2-M_T^2|\mathcal{F_S}]$ to get rid of the mixed term? - otherwise I cant see why that should behave nicely. Is this by the way the argument why for square integrable martingales the mixed term in the polarization (for the quadratic variation) is 0? – htd Jun 14 '13 at 09:51
  • Henrik: I expanded it a bit - hopefully it's correct :). Not sure what mixed term you're referring to though. – Stefan Hansen Jun 14 '13 at 09:57
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    We have a nice proposition saying that $E[N_t^2|F_s']=E[M_{t_0+t}^2-M_{t_0}^2|F_s]$ (so by mixed term I mean $M_{t_0}M_{t+t_0}$) which makes it a bit simpler. But I'm perfectly fine with the argument why t works (now). It just bothered me that I could find it from the identity above - but that is ok aswell now after realizing that $<M_{t_0+t}M_{t_0}>=t_0$. Thanks! – htd Jun 14 '13 at 10:14