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Let $(M_t)_{t\geq 0}$ a continuous Gaussian process that is a martingale with $M_0=0$. Show that $\langle M,M \rangle _t=f(t)$ a.s. where $f$ is a continuous increasing function.

In the particular case of Brownian motion we have $\langle M,M \rangle _t=t$. I have tried to imitate this proof. First we can show that for all $s>0$, $M_{t+s}-M_t$ is independant from $\sigma (M_r, r\leq t)$ since $M$ is a martingale. Then we can remark that $t\to E(M_t^2)=f(t)$ is a continuous increasing function. So we can define a Gaussian measure $G$ on $L^2(\mathbb{R}^+, df)$ such that $G(1_{[0,t]})=M_t$. We can now show as for brownian motion that $$\lim \sum (B_{t_i^n}-B_{t_{i-1}^n})^2=f(t) $$ in $L^2$. Since $(\langle M,M \rangle _t)_{t\geq 0}$ is unique (if we identify indistinguishable processes), we obtain that $\langle M,M \rangle _t=f(t)$ a.s.

I didn't write details, but is it a correct way in order to proof the asserted result ? Is there another method ?

Is the converse true ? Dubins-Schwartz theorem can apply only if $\langle M,M \rangle _\infty =+\infty$ a.s.

Edit : For the first part we can prove directly that $M^2_t-f(t)$ is a martingale. Indeed for t>s, $\mathbb{E}(M_t^2-M_s^2|\mathcal{F}_s)=\mathbb{E}((M_t-M_s)^2|\mathcal{F}_s)=\mathbb{E}((M_t-M_s)^2)=f(t)-f(s). $

Patissot
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  • The fact that $(M_t)t$ is a martingale does not imply that $(M_t){t \geq 0}$ has independent increments. What do you know about the quadratic variation $\langle M \rangle_t$? (E.g.: How is it defined?) – saz Jun 07 '15 at 14:50
  • We can define $\langle M, M \rangle $ as the unique increasing (adapted) process such that $M^2-\langle M, M \rangle $ or the limit in probability of $ \sum (B_{t_i^n}-B_{t_{i-1}^n})^2$.

    The assertion $M_{t+s}-M_t$ is independent from $\sigma (M_r, r\leq t) $ is false ?

     – Patissot Jun 07 '15 at 15:55
  • @Patissot There are martingales where the increments for instance are bigger if the previous increments have been bigger, but this growth is "symmetric". I think $W_t^2-t$ is an example of this. – Ian Jun 07 '15 at 21:07
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    @saz: But one can show that a Gaussian martingale has independent increments. – Nik Quine Apr 28 '16 at 20:57
  • If you define the Gaussian measure $G$ on $L^2(\mathbb{R}+,df)$ such that $G(1{[0,t]})=M_t$ then you will have if s<t $E(M_sM_t)=E(M^2_s)$ and in general this is not true. – foubw Oct 29 '16 at 21:53
  • Have you find a correct solution of the problem? – foubw Oct 29 '16 at 21:56

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