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For a semimartingale $X$, we want to solve the SDE $dZ_t=Z_{t-}dX_t$. I was able to prove that $$ Z_t:=\exp{(X_t-\frac{1}{2}\langle X\rangle_t^c)}\prod_{0<s\le t}(1+\Delta X_s)\exp{(-\Delta X_s)} $$

is the solution if I can show that $Z_s=Z_{s-}(1+\Delta X_s)$ and $Z_{s-}\Delta K_s=Z_{s-}\Delta X_s$, where $K_s:=X_t-\frac{1}{2}\langle X\rangle_t^c$. Can someone tell me, why these equations are true?

Thanks in advance!

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In general we have that $Z_t=Z_{t-}+\Delta Z_t$, because $\Delta Z_t=Z_t-Z_{t-}$. Now since $$ Z_t=\int_0^t Z_{s-}\mathrm{d} X_s $$ we have that $\Delta Z_t = Z_{t-}\Delta X_t$ and hence $Z_t=Z_{t-}(1+\Delta X_t)$. For the second equality it suffices to show that $\Delta K_t=\Delta X_t$. This is true because $\langle X\rangle_t^c$ is continuous and hence $$ \Delta K_t=\Delta X_t+\frac{1}{2} \Delta\langle X\rangle_t^c=\Delta X_t. $$

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  • $\begingroup$ How can I see from $Z_t=\int_0^t Z_{s-}dX_s$ that $\Delta Z_t=Z_{t-}\Delta X_t$? $\endgroup$
    – Mathfreak
    Commented May 6, 2016 at 10:23

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