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


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!


1 Answer 1


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,

$$\langle V^c,X^c \rangle_t = \langle [\int_0^t H_sdX_s ]^c ,X^c \rangle_t = \int_0^t H_sd \langle X , X \rangle ^c _s. $$

  • $\begingroup$ 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 $X^c$ $\endgroup$
    – CodeGolf
    Apr 3, 2014 at 12:25
  • $\begingroup$ But I would also like to know how to calculate $H_t1_{\Delta X_t\neq 0}$? $\endgroup$
    – CodeGolf
    Apr 3, 2014 at 12:28
  • 1
    $\begingroup$ 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. $\endgroup$
    – Viktor B
    Apr 3, 2014 at 13:28
  • $\begingroup$ 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 $V$. $\endgroup$
    – CodeGolf
    Apr 3, 2014 at 14:56
  • $\begingroup$ But I don't know how to obtain explicitly $H_t1_{\Delta X_t\neq 0}$ even when $K$ is predictable.... $\endgroup$
    – CodeGolf
    Apr 3, 2014 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.