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 $X^d$?