My question comes from the book Credit risk modeling: Valuation and hedging by Tomasz R. Bielecki and Marek Rutkowski, Page 137 at the bottom. It is easily described as follows:
Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $\tau:\Omega\rightarrow\mathbb{R}_+$ be a non-negative random variable defined on $(\Omega,\mathcal{F},\mathbb{P})$ with $\mathbb{P}(\tau=0)=0$ and $\mathbb{P}(\tau>t)>0$ for all $t\in\mathbb{R}_+$. Its cdf is $F(t)=\mathbb{P}(\tau\leq t)$. Define its hazard function $\Gamma(t)$ as $$ \Gamma(t)=-\ln (1-F(t)) \quad\text{or}\quad F(t)=1-e^{-\Gamma(t)} $$ Then, both $F(t)$ and $\Gamma(t)$ are right-continuous. Let $H_t=\mathbb{I}_{\{\tau\leq t\}}$ be a jump process and let $L_t=(1-H_t)e^{\Gamma(t)}$.
The question involves applying the product rule to $L_t$ when $\Gamma(t)$ is discontinuous. The book gives the following calculation: $$ \begin{split} L_t & = L_0+\int_{]0,t]}(1-H_u)de^{\Gamma(u)}-\int_{]0,t]}e^{\Gamma(u_{-})}dH_u \\ & = 1 + \int_{]0,t]}e^{\Gamma(u_{-})}((1-H_u)d\Gamma(u)-dH_u)+\sum_{s\leq t,s<\tau}(\Delta e^{\Gamma(s)}-e^{\Gamma(s_{-})}\Delta\Gamma(s)) \end{split} $$ where $\Delta e^{\Gamma(s)}=e^{\Gamma(s)}-e^{\Gamma(s_{-})}$ and $\Delta\Gamma(s)=\Gamma(s)-\Gamma(s_{-})$.
However, I am not sure how the last term is obtained: $\sum_{s\leq t,s<\tau}(\Delta e^{\Gamma(s)}-e^{\Gamma(s_{-})}\Delta\Gamma(s))$. $\Delta e^{\Gamma(s)}$ seems reasonable to me, but where does $-e^{\Gamma(s_{-})}\Delta\Gamma(s)$ come from? And why does it have a negative sign?
Thank you very much in advance!
UPDATE: I think I got it. The first term $\Delta e^{\Gamma(s)}$ takes care of jumps in $e^{\Gamma(u)}$ due to jumps in $\Gamma(u)$. After this, the integral should be over $\Gamma(u)^c$, i.e. the continuous part of $\Gamma$. Hence, in order to write it as an integral over $\Gamma(u)$ instead, jumps in $\Gamma(u)$ should be subtracted, which is done by the second term $e^{\Gamma(s_{-})}\Delta\Gamma(s)$: $$ \begin{split} \int_{]0,t]}(1-H_u)de^{\Gamma(u)} &=\int_{]0,t]}(1-H_u)d(e^{\Gamma(u)})^c+\sum_{s\leq t,s<\tau}\Delta e^{\Gamma(s)} \\ &=\int_{]0,t]}e^{\Gamma(u)}(1-H_u)d\Gamma(u)^c+\sum_{s\leq t,s<\tau}\Delta e^{\Gamma(s)} \\ &=\int_{]0,t]}e^{\Gamma(u_{-})}(1-H_u)d\Gamma(u)+\sum_{s\leq t,s<\tau}(\Delta e^{\Gamma(s)}-e^{\Gamma(s_{-})}\Delta\Gamma(s)) \end{split} $$