# 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 "$(\sigma_t)\in L^1(B_t)$"? – John Dawkins Jun 1 at 18:18
• $L^1(B_t)$ is the set of predictable processes that satisfy dominated convergence in probability w.r.t $(B_t)$ – W. Volante Jun 1 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? – John Dawkins Jun 1 at 18:47
• You can check out the definition here: almostsuremath.com/2010/01/04/extending-the-stochastic-integral (definition 1) – W. Volante Jun 1 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/… – TheBridge Jun 2 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.