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I would like to prove that $(B_t)$ is a brownian motion $\iff$ $(B_t)$ and $(B_t^2-t)$ are continuous martingale. I did the implication, but I have difficulties for the converse. Continuity is fine. How can I prove that $\mathbb E[B_t]=0$ and $\mathbb E[B_tB_s]=t\wedge s$ and how can I prove that $(B_t)$ is a Gaussian process ?


For $\mathbb E[B_t]=0$ I did as follow $$\mathbb E[B_t]=\mathbb E[\mathbb E[B_t\mid \mathcal F_0]]=\mathbb E[B_0]=0.$$ Q1) Does it works ? I know that if $(M_t)$ is a martingale $\mathbb E[M_t\mid \mathcal F_s]=M_s$ for all $t>s>0$, but does it works for $s=0$ ?


For $\mathbb E[B_tB_s]=t\wedge s$ I tried as follow : Let $t>s$. $$\mathbb E[B_tB_s]=\mathbb E[(B_t-B_s)B_s]+\mathbb E[B_s^2]=\mathbb E[B_s\mathbb E[B_t-B_s\mid \mathcal F_s]]+\mathbb E[B_s^2]=\mathbb E[B_s^2]=\mathbb E[B_s^2-s]+s.$$

Now, I can imagine that $\mathbb E[B_s^2-s]=0$ but I can't prove it. I tried as $$\mathbb E[B_s^2-s]=\mathbb E[\mathbb E[B_s^2-s\mid \mathcal F_0]]=\mathbb E[B_0^2]=0,$$ but I'm not sure if this is possible.

Q2) Does it work ?

Q3) How can I prove that it's a gaussian process ?

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    $\begingroup$ Searching for "Lévy's characterization of Brownian motion" will lead you to the answer... (and regarding the expectation of $B_s^2-s$: Note that martingales have constant expectation, hence $\mathbb{E}(B_s^2-s) = \mathbb{E}(B_0^2-0)=0$.) $\endgroup$ – saz Sep 11 at 11:31

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