# If $M_t= M_0 + \int_0^t \sigma_s dB_s$ and $E[M_1 | M_0]=M_0$, can we say that $(M_t)$ is a martingale?

If $$(B_t)$$ is a standard Brownian motion and $$M_t= M_0 + \int_0^t \sigma_s dB_s$$ for some process $$(\sigma_t) \in L^1(B_t)$$ and $$E[M_1 | M_0]=M_0$$, can we say that $$(M_t)$$ is a martingale on $$[0,1]$$ ?

Basically, does $$E[M_1 | M_0]=M_0$$ implies $$E[M_t | \mathcal F_s ]=M_s$$ for $$s in $$[0,1]$$ and $$\mathcal F_s$$ the filtration generated by $$(M_s)$$ ?

• What is the precise meaning of "$(\sigma_t)\in L^1(B_t)$"? Jun 1, 2021 at 18:18
• $L^1(B_t)$ is the set of predictable processes that satisfy dominated convergence in probability w.r.t $(B_t)$ Jun 1, 2021 at 18:34
• But what does "dominated convergence in probability w.r.t $(B_t)$" mean? I presume you require $\int_0^1\sigma_s^2 ds<\infty$ a.s., but what more does $(\sigma_t)$ need to satisfy? Jun 1, 2021 at 18:47
• You can check out the definition here: almostsuremath.com/2010/01/04/extending-the-stochastic-integral (definition 1) Jun 1, 2021 at 18:57
• As you cite almost sure looking in this blog provides the answer to your question, and as you may know only local martingale property is acheived, for more on the martingale necessary and sufficient condition have a look here : math.stackexchange.com/questions/38908/… Jun 2, 2021 at 5:07

According to a result of R.M. Dudley [Wiener functionals as Itô integrals, Ann. Probability 5 (1977), no. 1, 140–141], given an $$\mathcal F_1$$-measurable random variable $$X$$, there is a predictable $$(H_t)$$ with $$\int_0^t H_s^2 ds<\infty$$ such that $$X=\int_0^1 H_s dB_s$$, a.s. Adapting this in the obvious way, one has predictable processes $$(H_t)$$ and $$(K_t)$$ such that (i) $$H_t=0$$ for $$1/2, (ii) $$K_t=0$$ for $$0\le t\le 1/2$$, and (iii) $$\int_0^1 H_s dB_s =1 =\int_0^1 K_s dB_s$$, a.s. The difference $$\sigma_t:=H_t-K_t$$ is then non-trivial but $$\int_0^1 \sigma_s dB_s =0$$, a.s. Take $$M_0=0$$ and you have a counterexample: $$M_t=\int_0^t \sigma_s dB_s$$ is a continuous local martingale, but not a martingale.