Suppose I have a martingale $M(t)$ adapted to some filtration $\mathbb{F}=\{\mathcal{F}_t\}_{t\geq 0}$ and a positive monotonically increasing time change $T(t)=\int_0^tv(s)\mathrm{d}s$ with an $\mathbb{F}$-adapted process $v(s)$ which is not necessarily independent of $M(t)$. Is the time changed process $M(T(t))$ still a martingale? This is true when we have independence of $M(t)$ and $v(t)$ because then we can integrate things out by conditioning. But does it still hold when the processes are not independent? And if so, to which filtration is the time changed martingale adapted? (I allow the filtration to be enlarged).

  • $\begingroup$ Do you assume $v(s) \leq 1$...? Otherwise $M(T(t))$ does, in general, fail to be $\mathcal{F}_t$-measurable $\endgroup$ – saz Aug 17 '18 at 20:08
  • $\begingroup$ No not necessarily, $v(t)$ might even be unbounded. In a paper by Carr and Wu (semanticscholar.org/paper/…) they time change a Doleans-Dade exponential (which is martingale) by an integrated CIR process and claim that this were again a martingale. For this see Theorem 1 in the cited paper. In the proof of the theorem they assume that $T(t)$ is a stopping time for each $t\geq 0$ which would then be a valid argument. $\endgroup$ – lbf_1994 Aug 18 '18 at 2:48
  • $\begingroup$ But later on they apply theorem 1 to the integrated CIR process which obviously does not fullfill the conditions of theorem 1. For this see for example subsection 4.3.1 of that paper. Hence, the proof of the theorem is correct but later on they apply the theorem to special cases where the conditions of the theorem are not met. I'm interested if the results of that examples are still true. $\endgroup$ – lbf_1994 Aug 18 '18 at 2:48
  • $\begingroup$ I also want to note that $M(T(t))$ is allowed to be a martingale to a potentially enlarged filtration which I asumme to be $\mathcal{G}_t:=\sigma\left(M((T(s));0\leq s\leq t\right)$ $\endgroup$ – lbf_1994 Aug 18 '18 at 2:54

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