Stochastic exponential jump process equality

Question

I've read in a Paper this equality. However i couldnt find a proof for it. $\mathbb{E}\left[\exp\Big\{\int_{[0,{t}]\times E}\ln\left(1+h(s,e)\right))N(dt,de)\Big\}\right]=\mathbb{E}\left[\exp\Big\{\int_{[0,{t}]\times E}h(s,e)\nu(de) dt\Big\}\right]$ N(dt,de) is a poisson random measure and $\nu(de)\otimes dt$ its compensator. Does anybody have a proof or an idea ?

Answer 1

To prove above equality, we use following Dolean's stochastic exponential $\mathscr{E}(Y)$ of semimartingale $Y$(Please refer to J. Jacod, and A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003.p.59, Theorem 1.4.61), \begin{equation*} \mathscr{E}(Y)_t=\exp\Big(Y_t-Y_0-\frac12\langle Y^c\rangle_t\Big)\prod_{s\le t}[(1+\Delta Y_s)e^{-\Delta Y_s}]. \tag{1} \end{equation*}

At first, let \begin{equation*} X_t=\int_{]0,t]\times E}e\,[N(dt,de)-1_{|e|\le1}(e)\nu(de)dt]. \end{equation*} then $X$ is an $E$-valued-Levy process with $X_0=0$.

Next, let \begin{equation*} Y_t=\sum_{s\le t}h(s,\Delta X_s)=\int_{[0,t]\times E} h(s,e)N(ds,de), \tag{2} \end{equation*} where $h>-1$ is a non-random measurable function of $(s,e)$ and satisfies $\nu$-integrable condition, then $Y$ is a process with independent increments, $Y^c=0$, \begin{equation*} \Delta Y_t=h(t,\Delta X_t) \tag{3} \end{equation*} and \begin{equation*} \mathsf{E}[Y_t]=\int_{[0,t]\times E} h(s,e)\,\nu(de)\mathrm{d}s, \tag{4} \end{equation*} which is continuous in $t$, $Y_t-\mathsf{E}[Y_t]$ and $\mathscr{E}(Y-\mathsf{E}[Y])_t$ are martingales and \begin{equation*} \mathsf{E}[\mathscr{E}(Y-\mathsf{E}[Y])_t]=1. \tag{5} \end{equation*} Now, according (1)--(4), \begin{align*} \mathscr{E}(Y-\mathsf{E}[Y])_t &=\exp(Y_t-\mathsf{E}[Y_t])\prod_{s\le t}[(1+\Delta Y_s)e^{-\Delta Y_s}]\\ &=\prod_{s\le t}(1+\Delta Y_s)\exp(-\mathsf{E}[Y_t])\\ &=\exp\Big\{\sum_{s\le t}\ln(1+\Delta Y_s)\Big\}\exp(-\mathsf{E}[Y_t])\\ &=\exp\Big\{\int_{]0,t]\times E}\ln(1+ h(s,e))N(ds,de)\Big\} \exp(-\mathsf{E}[Y_t]). \end{align*} \begin{align*} &\mathsf{E}[\mathscr{E}(Y-\mathsf{E}[Y])_t]\\ &\quad=\mathsf{E}\Big[\exp\Big\{\int_{]0,t]\times E}\ln(1+ h(s,e))N(ds,de)\Big\}\Big] \exp(-\mathsf{E}[Y_t])\\ &\quad=\mathsf{E}\Big[\exp\Big\{\int_{]0,t]\times E}\ln(1+ h(s,e))N(ds,de)\Big\}\Big]\\ &\qquad\qquad \times \exp\Big[-\int_{[0,t]\times E} h(s,e)\,\nu(de)\mathrm{d}s\Big] \end{align*} At last, from above equality and (5), get \begin{align*} &\mathsf{E}\Big[\exp\Big\{\int_{]0,t]\times E}\ln(1+ h(s,e))N(ds,de)\Big\}\Big]\\ &\quad=\exp\Big[\int_{[0,t]\times E} h(s,e)\,\nu(de)\mathrm{d}s\Big]. \end{align*} which is required.