# Prove that $(B_t)$ is a brownian motion $\iff$ $(B_t)$ and $(B_t^2-t)$ are continuous martingale.

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 ?

• 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 $\mathbb{E}(B_s^2-s) = \mathbb{E}(B_0^2-0)=0$.) – saz Sep 11 at 11:31