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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?

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  • $\begingroup$ One analog of Levy's characterization of Brownian motion is due to S. Watanabe: if $\{N_t, t\ge 0\}$, is a counting process (right continuous pure jump process with jumps of size $+1$ only) with $N_t=0$ and if $N_t-\lambda t$ a martingale (for some $\lambda>0$) then $\{N_t, t\ge 0\}$ is a rate-$\lambda$ Poisson process. $\endgroup$
    – John Dawkins
    Commented Oct 8, 2017 at 6:29

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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.

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