= Solution
Work with the <filtration> relative to which the given <Brownian motion> or <Markov jump process> has its stated properties. All <martingale> assertions below concern every finite time interval; no infinite-horizon <uniform integrability> is claimed.
First put $C=\sup_x|\sigma(x)|$. The solution has the representation $X_t=\int_0^t\sigma(X_s)dB_s$, and the <Itô isometry> gives
$$
\mathbb EX_t^2=\mathbb E\int_0^t\sigma(X_s)^2ds\le C^2t.
$$
Thus $X$ is a square-integrable <martingale>, not just a <local martingale>. Applying the <Itô formula> to its square cancels the finite-variation term and gives
$$
Z_t=2\int_0^tX_s\sigma(X_s)dB_s.
$$
The expected squared integrand integral is bounded by
$$
4\mathbb E\int_0^tX_s^2\sigma(X_s)^2ds\le4C^2\int_0^tC^2s\,ds=2C^4t^2.
$$
Therefore \b[both $X$ and $Z$ are true square-integrable <martingales>]. This is the <bounded diffusion coefficient gives square martingales> criterion. Only bounded measurability of the coefficient and existence of the stipulated solution have been used.
For the jump process, the <Lévy kernel of a Markov jump process> uses displacements: a mark $y$ sends the state $x$ to $x+y$. Let $\Lambda=\sup_x\lambda(x)$ and $V=\sup_xv(x)$. The bounded total rate prevents explosion; the number of jumps on a finite interval is dominated by a rate-$\Lambda$ <Poisson process>. Let $N(ds,dy)$ record the jump displacements. Its <predictable compensator of a jump measure> is
$$
\nu(ds,dy)=K(X_{s-},dy)ds.
$$
The <Cauchy-Schwarz inequality> gives $\int |y|K(x,dy)\le\sqrt{\lambda(x)v(x)}$, so the zero first moment is absolutely meaningful. Since that moment vanishes,
$$
X_t=\int_0^t\int_{\mathbb R}y\,(N-\nu)(ds,dy),\qquad \mathbb EX_t^2=\mathbb E\int_0^tv(X_{s-})ds\le Vt.
$$
The compensated integral is a square-integrable <martingale>: its isometry follows first for bounded marks from conditional compensation and then by truncation in the squared-integrand <norm>.
The jump version of the square <Itô formula> is $X_t^2=2\int_0^tX_{s-}dX_s+\sum_{s\le t}(\Delta X_s)^2$. Hence
$$
Z_t=2\int_0^tX_{s-}dX_s+\int_0^t\int_{\mathbb R}y^2(N-\nu)(ds,dy).
$$
The first term is square <integrable> because
$$
4\mathbb E\int_0^tX_{s-}^2v(X_{s-})ds\le2V^2t^2.
$$
The second is an <integrable> <martingale>: the expected absolute uncompensated jump-square sum is at most $Vt$, and the compensator identity, multiplied by any bounded event known at an earlier time, makes its increments conditionally centered. Values $X_s$ and $X_{s-}$ agree for Lebesgue-almost every time, so the displayed compensation is exactly the one defining $Z$. Thus \b[$X$ and $Z$ are true <martingales> in the jump case as well]. A fourth moment of the jumps is unnecessary; in particular we have not asserted that this $Z$ must be square <integrable>. This is <centered finite-rate jump kernel gives square martingales>.
Back to article page
