Let $(M_n)_{n\in\mathbb{N}}$ be a discrete martingale with $M_n \in L^2\ \forall n\in\mathbb{N}$ and $S,T$ bounded stopping times with $S\leq T$ a.s.. Show $$E[(M_T − M_S)^2|\mathcal{F_s}]= E[M_T^2 − M_S^2|\cal{F_s}]\ a.s.$$

I tried $$E[(M_T − M_S)^2|\mathcal{F_s}]=E[(M_T^2 -2M_TM_S + M_S^2)^2|\mathcal{F_s}]= E[M_T^2|\mathcal{F_s}]-2E[M_TM_S|\mathcal{F_s}]+E[M_S^2|\mathcal{F_s}]$$ but it does not seem very useful...

I really don't know what to do now. Thanks for any help!!


1 Answer 1


$$E[(M_T − M_S)^2|\mathcal{F_s}]- E[M_T^2 − M_S^2|\mathcal{F_s}]$$ $$=E[2M_S^{2}-2M_TM_S|\mathcal F_S)$$ $$=2M_SE[M_S-M_T |\mathcal{F_s}]=0$$ by Optional Sampling Theorem.

[$2M_S$ pulls out of the conditional expectation because it is measurable w.r.t. $\mathcal F_S$].

  • $\begingroup$ Thank you! What exacty did you use to obtain $=0$? Our theorem says "$E(M_T|\mathcal{F}_S)=M_{min\{S,T\}}$" for all bounded stopping times $S,T$, so I could use $E(M_T-M_S|\mathcal{F}_S)=E(M_T|\mathcal{F}_S)-E(M_S|\mathcal{F}_S)=M_S-M_S=0$? $\endgroup$
    – Uhmm
    Apr 11 at 12:12
  • 1
    $\begingroup$ @Uhmm Yes, that is correct. $\endgroup$ Apr 11 at 12:13

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