Given a uniformly integrable discrete martingale $M_n$ on prob. space $(\Omega, \mathcal{F}, \mathbb{P})$, and a.s. finite stopping times $T$ and $S$ with $T\geq S$. Show that $E[M_T|\mathcal{F}_S] = M_S$.

I show that $M_n \rightarrow M$ in $\mathcal{L}^1$ and a.s. by uniform integrability. Therefore $M_T = E[M|\mathcal{F}_T]$ and $E[M_T|\mathcal{F}_S] = E[E[M|\mathcal{F}_T]|\mathcal{F}_S] = E[M | \mathcal{F}_S] = M_S$, however it seems there is something missing here.

  • $\begingroup$ what is missing? i dont follow. $\endgroup$
    – Lost1
    Feb 14, 2014 at 1:04
  • $\begingroup$ I was told that I cannot just conclude that $E[M|\mathcal{F}_S] = M_S$ as this applies only to bounded stopping times, and that to proceed requires proving that the partial averaging property holds here as well. That is, $E[M|\mathcal{F}_n] = M_n$ for fixed $n$ does not really apply here. How do I show this? $\endgroup$
    – J Carter
    Feb 14, 2014 at 1:30

1 Answer 1


Ok, so to show that $E[M |F_T] = M_T$ we have that letting $B \in \mathcal{F}_T$ and $A_i = B \cap \{T = i\}$ that $E[M_T 1_B] = \sum_i E[M_i 1_{A_i}] = \sum_i E[\, E[M | \mathcal{F}_i] 1_{A_i}] = \sum_i E[M 1_{A_i}]= E[M1_B]$. Now, it is clear that $E[M|F_T] = M_T$ and the above works fine.


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