Continuing from my previous question Decomposition of semimartingales,

In his answer, George Lowther mentioned that if $X$ is a local martingale, then $X^d$ and $X^c$ in its decomposition

$X_t - X_0 = X_t^d + X_t^c$

(where $X^c$ is the continuous local martingale part of $X$ and $X^d$ is the purely discontinuous semimartingale part) are in some sense orthogonal to each other.

Q1: In what sense? Can we say $X^d$ and $X^c$ are independent of each other?

Q2: What about the special case where

(1) $X^c$ is locally square integrable with respect to the filtration generated by a standard Brownian motion $W$ and, hence, can be written

$X_t^c - X_0^c = \int_0^t \! H_u \, \mathrm{d}W_u,$

where $H$ is predictable, and

(2) $X_t^d \equiv \sum_{0\leq u\leq t}\Delta X_u.$

Can we say anything about the independence between $H$, $W$, or $X^d$?


1 Answer 1


Q1: The orthogonality term introduced in this context is just a definition, i.e. two local martingales $M$ and $N$ are said to be orthogonal if $MN$ is a local martingale. I guess the term comes from the fact that if $M,N\in\mathcal{H}^2$ ($\mathcal{H}^2$ being the set of square integrable martingales) then $M$ and $N$ are orthogonal in this sense if and only if for every stopping time $\tau$, $M^\tau$ and $N-N_0$ is orthogonal in $\mathcal{H}^2$ (seeing $\mathcal{H}^2$ as a Hilbert space with inner product $(M,N)=E[M_\infty N_\infty]$).

As to whether $X^d$ and $X^c$ are independent, I'm pretty sure that this isn't the case. One can show that if $M,N\in\mathcal{H}^2$ then $M$ and $N$ are orthogonal if and only if $\langle M,N\rangle=0$, where $\langle M,N\rangle$ denotes the predictable quadratic covariation/sharp-bracket process. But $\langle M,N\rangle =0$ does not imply $M$ and $N$ being independent.

  • $\begingroup$ do you know where to find a proof of your second paragraph? $\endgroup$
    – Mathfreak
    Jun 17, 2016 at 9:27
  • $\begingroup$ @Mathfreak: Try Limit Theorems for Stochastic Processes by Jacod & Shiryaev. $\endgroup$ Jun 17, 2016 at 9:36

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