I know that for discrete time martingales, one can show that if they are predictable, they have to be constant. Now for continuous time martingales, that is not true I suppose, since for example a Brownian motion is predictable as it is continuous (correct?). So I was wondering what the relationship between continuous time martingales and predictability is. I could not come up with an example of a non-predictable martingale. Are all continuous time martingales predictable? If not, what would be a counterexample?

  • $\begingroup$ You should define what it means for a martingale (case 1: discrete; case 2: continuous) to be predictable. $\endgroup$ Feb 21, 2021 at 12:00
  • $\begingroup$ Discrete: If $\{X_t\}_{t=0,...,T}$ is a martingale in a filtered probability space $(\Omega,\mathfrak{A},\mathfrak{F},\mathbb{P})$, it is predictable if every $X_t$ is $F(t-1)$-measurable. $\endgroup$
    – joinijo
    Feb 21, 2021 at 12:08
  • $\begingroup$ Continuous: Same space as above. $\mathcal{P}$ is defined as the $\sigma$-field which is generated by all adapted, left-continuous processes $\{X(t)\}_{t\in[0,T]}$ (left-continuous meaning that each trajectory for a fixed $\omega \in \Omega$ is left-continuous). Then a process is called predictable if it is $\mathcal{P}$-measurable. $\endgroup$
    – joinijo
    Feb 21, 2021 at 12:10
  • $\begingroup$ To be more precise, in the continuous case the filtration is w.r.t to the time index set $\{0,...,T\}$, in the discrete case it is w.r.t. $[0,T]$. We even assumed that the continuous time filtration would coincide with the augmented filtration of a Brownian motion, I'm not sure if that makes a difference. $\endgroup$
    – joinijo
    Feb 21, 2021 at 12:17

1 Answer 1


The following may be near what you are after: If $M$ is a (cadlag) continuous time martingale, then ${}^pM$ coincides with $M_-$ almost surely. Here ${}^pM$, the predictable projection of $M$, is characterized as the predictable process such that $E[M_T\mid\mathcal F_{T-}] = {}^pM_{T}$ for all bounded predictable stopping times $T$; and $M_-$ is the left limit process $t\mapsto M_{t-}$. In particular, this shows that $M$ is predictable if and only if the paths of $M$ are continuous.

  • $\begingroup$ I guess I still know too little about martingales to follow everything you're saying :P but essentially, you're saying that continuous time martingales need not be path-continuous in general, hence also not predictable? But why is '$M$ predictable if and only if the paths of $M$ are continuous'? Wouldn't it be sufficient for them to be left-continuous? $\endgroup$
    – joinijo
    Feb 21, 2021 at 20:11
  • $\begingroup$ The cryptic "cadlag" was to indicated tat I was restricting myself to right-continuous martingales, with left limits. $\endgroup$ Feb 22, 2021 at 17:17

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