We know that given a continuous square integrable martingale there exists unique (up to indistinguishability) continuous, natural and increasing process which is quadratic variation process of the given martingale. I would like to know the converse i.e., given any continuous, natural and increasing process is it possible to get a martingale whose quadratic variation will be the given process? If not, please provide a counter example. Thanks


1 Answer 1


Yes, just take the Brownian motion with the change of time $A_s$ that's it to define $M_s = B_{A_s}$. This process will have the quadratic variation $A_s$.

The resulting $M$ is a local martingale with respect to the canonical filtration of the process $B$. If you want this to be square integrable you will have to ask $E(A_s) < \infty$ for each $s$. This is a consequence of the Dubins-Shwartz theorem.

  • $\begingroup$ This works only if you know that $A_s$ is a quadratic variation of some process! But my question was whether you can find any such process whose quadratic variation will be the given process. $\endgroup$
    – chandu1729
    Commented Nov 29, 2013 at 18:47
  • $\begingroup$ @chandu1729: No, this works. Just take $B$ to be a Brownian motion independent of $A$. (Actually, it should be added that in general you just get a local martingale, and not necessarily a proper martingale) $\endgroup$ Commented Nov 30, 2013 at 11:53
  • $\begingroup$ @George Lowther: I am not able to establish that $B_{A_s}$ is a martingale adapted to $\mathcal{F}_t$. Can you please give some ideas on it. Thanks. $\endgroup$
    – chandu1729
    Commented Dec 1, 2013 at 9:14

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