Suppose $(M_t)_{t \in [a,b]}$ is a stochastic process. Denote $(\mathcal{F}_t)_{t \in[a,b]}$ to be the natural filtration generated by the process $(M_t)_{t \in [a,b]}$. Moreover, suppose $(M_t)_{t \in [a,b]}$ is martingale, i.e. $\forall s \le t \in [a,b]$, $$ E [ M_t \mid \mathcal{F}_s ] = E [ M_s ] \qquad a.s.,$$ where $E [ \cdot \mid \mathcal{F}_s ]$ denotes conditional expectation given sub-$\sigma$-algebra $\mathcal{F}_s$.

How can I show that the map $t \mapsto E [ M_t ]$ is continuous?

I am currently reading "Davar Khoshnevisan Multiparameter Resources An Introduction to Random Fields". And at pp. 226 he remarks that the above question is true. But I don't know how to show it. Can someone help?

I am thinking along the line of computing $$E [ M_t ] - E [ M_s ]$$ by using the definition of martingale as defined above, properties of conditional expectation, etc, and then take the limit $s \to t$ to hopefully get a zero as the finally result. (This means continuity.)

Any help is much appreciated!


1 Answer 1


I may be misunderstanding something but it looks to me like the unconditional expectation of a martingale is constant: taking $s=a$ in the definition gives $$ EM_t=EE[M_t|M_a]=EM_a \forall\,t\in[a;b]. $$ And a constant function is clearly continuous...

  • $\begingroup$ @GGYYD Do you agree? $\endgroup$
    – Vossler
    Mar 14, 2016 at 3:08
  • $\begingroup$ XD. Agree! Oh my god. How did I not think of taking conditional expectation againt to the least value $a$. Thanks! I think this solves it. $\endgroup$
    – GGYYD
    Mar 14, 2016 at 6:40

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