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
2026-02-23 15:13:14.1771859594
Continuous Square integrable martingale Quadratic Variation
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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.