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This is one of the problems from Cinlar's 2011 book - "Probability and Stochastics" (Chapter VI, page 262, exercise 2.36) : Let $N$ be a Poisson random measure on $R^{+}$, defined by

$N(\omega, B) = \int_{\mathbb{R}^{+}\times \mathbb{R}^{+}}M(\omega, dt, dz)I_{B}(t)I_{[0,R_{t}(\omega)]}(z)\text{, }B\in\mathcal{B}(R^{+})\text{, }$

where $M(\omega,dt,dz)$ is a Poisson random measure on $\mathbb{R}^{+}\times \mathbb{R}^{+}$, with intensity measure being equal to the 2-dimensional Lebesgue measure, and $R = \{R_{t}\}_{t\geq 0}$ is a positive, bounded, left-continuous stochastic process, independent of $M$.

The aim is to show that

$\mathbb{E}e^{-Nf}$ = $\mathbb{E}\text{exp_}\int_{\mathbb{R}^{+}}$$dtR_{t}(1-e^{-f(t)})\text{, }f\geq0,\text{ Borel}$,

where $Nf$ denotes the integral of $f$ with respect to the measure $N(\omega,\cdot)$ and exp_ is the shorthand for the function $e^{-x}$.

My idea was to start with $f(t) = cI_{(a,b]}(t)$ and then use linearity and monotone convergence to prove the general case, but even with this simplification I have no idea how to use the independence of $R$ and $M$, or , in fact, how to even compute the expectation of


Any ideas?

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Use double expectation, first conditional on R (i.e., R={Rt}t≥0) and note that the conditional distribution of M given R is the same as that of M. The inner expectation (conditional on R) is the Laplace transorm of the Poisson M(dt,dz) (with mean=dtdz); hence the inner expectation is exp_∫∫[1-f(t)I{0≤z≤Rt}]dtdz. The latter integrand [1-f(t)I{0≤z≤Rt}]=[1-f(t)]I{0≤z≤Rt}, hence exp_∫∫[1-f(t)I{0≤z≤Rt}]dtdz=exp_∫∫[1-f(t)]Rtdt. Take this w.r.t. the outer expectation, you are done. Good luck.

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