Exponential formula for Poisson random measure

Question

We want to prove the following result.

• Let $M$ be a Poisson random measure with intensity measure $\mu$.
• Then for every measurable set $B$ such that $\mu(B) < \infty$ and for all functions $f$ such that $\int_B e^{f(x)} \mu(dx) < \infty$

$$ \mathbb{E} \exp \left\{ \int_B f(x) M(dx) \right\} = \exp \left\{ \int_B (e^{f(x)}-1 ) \mu(dx) \right\}$$

We should condition by the expectation on $\mu(B)$.

From Course Online:

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Definition 1.6 Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, $E\subset \mathbb{R}^k$, and $\rho$ a measure on $(E,\mathcal{E})$. A Poisson random measure on $E$ with intensity $\rho$ is a function with values in $\mathbb{N}$: $$ \begin{array}{ccccc} M : & \Omega \times \mathcal{E} & \to & \mathbb{N} & \text{s.t.} \\ & (\omega, A) & \mapsto & M(\omega, A), & \end{array}$$

1. $\forall \omega \in \Omega$, $M(\omega, \cdot)$ is a Radon measure on $E$, that is, $\forall A \in \mathcal{E}$ measurable and bounded, $M(A) < \infty$ is a r.v. with values in $\mathbb{N}$;
2. $\forall A \in \mathcal{E}$, $M(\cdot,A) = M(A)$ is a Poisson r.v. with parameter $\rho(A)$;
3. for any $A_1, \dots, A_n$ disjointed sets, the r.v. $M(A_1), \dots, M(A_n)$ are independents.

Now let $M$ be a Poisson random measure with intensity measure $\mu$ and let $A$ be a measurable subset s.t. $0 < \mu(A) < +\infty$. Then the following two random measures on the subsets of $A$ have the same distribution conditionally on $M(A)$:

• $M|_A$, the restriction of $M$ to $A$.
• $\widehat{M}\!{}_A$ defined by $\widehat{M}\!{}_A(B) = \sharp\{X_i \in B\}$ for all measurable subsets $B$ of $A$, where $X_i$, $i = 1, \dots, M(A)$ are independent and distributed on $A$ with the law $\frac{\mu(\mathrm{d}x)}{\mu(A)}$.

This implies in particular that

$$ \boxed{\mathbb{E}\exp\left(\int_{A}f(X) \, M(dx)\right) = \exp\left( \int_{A} (e^{f(x)} - 1) \mu(dx) \right)} \tag{1.7}$$

for any function $f$ such that $\int_A e^{f(x)} \mu(dx) < +\infty$. This can be obtained by conditioning the expectation on $\mu(A)$ and by the previous result on $M|_A$ and $\widehat{M}\!{}_A$.

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Q. We have a proof here, could you explain the first line $E(e^{iuX} \mid N=n)$ ?

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Answer 1

• We fix a subset $A$

• For given $\omega$, $M(A)$ is a random variable that follows a Poisson distribution of parameter $( \mu(A))$

• ${M|}_A$ the restriction of $M$ the Poisson random measure on the subset $A$ admits a representation

• For $B \subset A$, ${M|}_A(B)= \# \{ Y_i \in B , i \in [1, \dots N] \}$ where $N$ follows a a Poisson distribution of parameter $( \mu(A))$ and $Y_i$ is distributed according to $P(Y_i \in B) = \dfrac{ \mu(B)}{ \mu{B}}$

• That is to say, we draw first $N$

• Therefore ${M|}_A(B) = \sum_{k=1}^N \mathbb{1}_{Y_i \in B}$

• and $\int f(x) {M|}_A(dx) = \sum_{i=1}^{N} f(Y_i)$

• it explains the proof