Let $M\_t$ and $N\_t$ be two purely discontinuous martingales such that $[M]\_t=[N]\_t $ almost surely. Can one conclude that $M$ and $N$ have the same law?



Is there any analogous to Levy characterization theorem for purely discontinuous martingales?
No. If $N\_t = -M\_t$, then $[M]\_t = [N]\_t$, but $N\_t$ and $M\_t$ may have different law (for example if $N\_t$ is a Poisson process with drift).


A less trivial example: take two independent Poisson processes $X\_t$, $Y\_t$ and take $N\_t = X\_t + Y\_t - 2 t$, $M\_t = X\_t - Y\_t$. Then $[M]\_t = [N]\_t = X\_t + Y\_t$.


In general: $[M]\_t = [N]\_t$ if and only if $M\_t$ and $N\_t$ have jumps of equal magnitude, i.e. $|\Delta M\_t| = |\Delta N\_t|$ for all $t$. Their signs may differ in a completely arbitrary way.


Remark: Such martingales are mutually *differentially subordinate*; $M\_t$ is differentially subordinate to $N\_t$ if $|\Delta N\_t| \leqslant |\Delta M\_t|$ (when both martingales are of pure-jump type). This concept was recently used to find bounds for some Fourier multipliers; see, for example, [an article by Bañuelos and Bogdan](https://arxiv.org/abs/math/0609432).


