explicit characterization of the stochastic integrand

Let $V$ be a cadlag positive supermartingale with the following decomposition:

$$V_t=V_0+\int_0^tH_sdX_s-K_t$$

where $X$ is a cadlag local martingale and $K$ is an adapted increasing process with $K_0=0$. By Theorem 4.52 in "Limit Theorems for Stochastic Processes", one has

$$[V,X]_t=\langle V^c,X^c \rangle_t+\sum_{s\le t}\Delta V_s\Delta X_x$$

where $[V,X]$ denotes the quadratic co-variation, $V^c$ and $X^c$ denote respectively the continuous local martingale part of $V$ and $X$ and $\Delta V_t=V_t-V_{t-}$ and $\Delta X_t=X_t-X_{t-}$.

Could some help me calculate explicitly $[V,X]_t$, $\langle V^c,X^c \rangle_t$ and $\Delta V_s\Delta X_s$ only by $H$ and $X$? (not involving $K$!) Thanks a lot!

• Or more precisely, how to express $H$ by $V$ and $X$? Commented Apr 2, 2014 at 20:12
• Related question: mathoverflow.net/questions/106808/… Commented Apr 3, 2014 at 0:36

I think jumps of $V$ are given by $\Delta V _t = H_t \Delta X _t - \Delta K _t$, which allows us to compute $\Delta V _t \Delta X _t$; also, because $K$ is of bounded variation,
$$\langle V^c,X^c \rangle_t = \langle [\int_0^t H_sdX_s ]^c ,X^c \rangle_t = \int_0^t H_sd \langle X , X \rangle ^c _s.$$
• Thanks for the reply. I am sorry that my question is not so clear. In fact I would a expression without K. Formally, $V_t=V_0+\int_0^tH_s1_{\{\Delta X_s= 0\}}dX_s^c+\sum_{s\le t}H_s1_{\{\Delta X_s\neq 0\}}\Delta X_s-K_t$. Thus we have formally $H_t1_{\{\Delta X_t=0\}}=\frac{d<V^c,X^c>_t}{d<X^c,X^c>_t}$, which is given by $V^c$ and $X^c$ Commented Apr 3, 2014 at 12:25
• But I would also like to know how to calculate $H_t1_{\Delta X_t\neq 0}$? Commented Apr 3, 2014 at 12:28
• Hm, I think one can not express $\Delta V_s\Delta X_s$ only by $H$ and $X$ (without $K$), since they do not determine $V$ uniquely. Commented Apr 3, 2014 at 13:28
• In fact by classical Doob-Meyer decomposition, there exists a cadlag martingale $M$ and a predictable increasing process $K'$ s.t. $V_t=V_0+M_t-K'_t$, which is really like the previous decomposition. If in addition we know $K$ is predictable, thus by unicity we have $K'=K$ and thus $K$ is uniquely determined by $V$. Commented Apr 3, 2014 at 14:56
• But I don't know how to obtain explicitly $H_t1_{\Delta X_t\neq 0}$ even when $K$ is predictable.... Commented Apr 3, 2014 at 14:57