Laplace functional for Poisson Process: $E[e^{-\sum_{n=1}^{\infty}f(W_n)}]= e^{-\lambda\int_0^{\infty}(1-e^{-f(t)})dt}$

Question

Let $W_n$ be the $n$ waiting time of a Poisson process. Proof that $$E[e^{-\sum_{n=1}^{\infty}f(W_n)}]= e^{-\lambda\int_0^{\infty}(1-e^{-f(t)})dt}$$ for a measurable postive function $f$

I really don´t know where to start. The first thing that I tried is: $$E[e^{-\sum_{n=1}^{\infty}f(W_n)}]=E[\prod_{n=1}^{\infty}e^{-f(W_n)}]$$ but I don´t know if the last expression is equivalent to $$\prod_{n=1}^{\infty}E[e^{-f(W_n)}]$$

I would really appreciate any hint or suggestion.

Answer 1

The arrival times $W_1,W_2,\ldots$ are not independent, so you cannot move the expectation past the product.

The key is to start small, with $f$ corresponding to something where $\sum f(W_n)$ has a familiar distribution. For example, if $f$ is the indicator of an interval, then $\sum f(W_n)$ gives the number of arrivals in that interval, and you can easily compute $E[e^{ -\sum f(W_n)}]$. You then verify the result for increasingly large families of $f$.

Here is the outline of the proof:

1. Show the result when $f(t):=\alpha I_A(t)$ is a positive multiple of the indicator of set $A$.

2. Show the result when $f(t):=\sum_{i=1}^k \alpha_i I_{A_i}(t)$ is a sum of functions of type (1), where the sets $A_1, A_2,\ldots, A_k$ are disjoint.

3. Show the result for arbitrary nonnegative $f$ by considering a monotone limit of functions of type (2).

To show (1), note that $\sum_{n=1}^\infty f(W_n)$ is $\alpha$ times $N_A$, the number of arrivals in set $A$. Since the Poisson process has rate $\lambda$, we know $N_A$ has Poisson distribution with mean $\lambda m(A)$ where $m(A)$ is the Lebesgue measure of set $A$. So compute $$ E \exp (-\sum f(W_n))=E \exp(-\alpha N_A) = \exp [-(1-e^{-\alpha})\lambda m(A)]. $$ The trick is to notice that $$(1-e^{-\alpha})m(A)=\int_A(1-e^{-\alpha})\,dt=\int_0^\infty(1-e^{-\alpha})I_A(t)\,dt=\int_0^\infty (1-e^{-\alpha I_A(t)})\,dt. $$

Answer 2

For $x \in \mathbb{R}$ define

$$u(x) := \mathbb{E}\exp \left(- \sum_{n=1}^{\infty} f(x+W_n) \right).$$

The waiting times $W_n$ can be written in the form $W_n = \sum_{j=1}^n \sigma_j$ where each $\sigma_j$ is exponentially distributed with parameter $\lambda$ and $(\sigma_j)_{j \geq 1}$ are independent. Using the tower property of conditional expectation, we find that

\begin{align*} u(x) &= \mathbb{E}\exp \left(- \sum_{n=1}^{\infty} f(x+\sigma_1+\ldots+\sigma_n) \right) \\ &= \mathbb{E} \left[ \mathbb{E}\left( \exp \left(- \sum_{n=1}^{\infty} f(x+\sigma_1+\ldots+\sigma_n) \right) \mid \sigma_1 \right) \right] \\ &= \int_{(0,\infty)} \exp(-f(x+s)) \mathbb{E}\exp\left(- \sum_{n=2}^{\infty} f(x+s+\sigma_2+\ldots+\sigma_n) \right) \, d\mathbb{P}_{\sigma_1}(s) \\ &= \int_{(0,\infty)} \exp(-f(x+s)) \mathbb{E}\exp\left(- \sum_{n=2}^{\infty} f(x+s+\sigma_2+\ldots+\sigma_n) \right) \lambda e^{-\lambda s} \, ds. \tag{1} \end{align*}

Since the random variables $(\sigma_j)_j$ are independent and identically distributed, we have

\begin{align*} &\mathbb{E}\exp\left(- \sum_{n=2}^{\infty} f(x+s+\sigma_2+\ldots+\sigma_n) \right) \\ &= \mathbb{E}\exp\left(- \sum_{n=1}^{\infty} f(x+s+\sigma_1+\ldots+\sigma_n) \right)\\ &= u(x+s),\end{align*}

and so $(1)$ gives

$$u(x) = \int_{(0,\infty)} \exp(-f(s+x)) u(s+x) \lambda e^{-\lambda s} \, ds.$$

By a change of variables, $t=s+x$, we get

$$u(x) =\lambda e^{\lambda x} \int_x^{\infty} \exp(-f(t)) u(t) e^{-\lambda t} \, dt,$$

i.e.

$$(e^{-\lambda x} u(x)) = \int_{x}^{\infty} (e^{-\lambda t} u(t)) \gamma(t) \, dt\tag{2}$$

for $\gamma(t) := e^{-f(t)}$. Assume for the moment that $f$ has compact support, then $u$ satisfies $\lim_{|x| \to \infty} u(x)=0$, and the integral equation $(2)$ has a unique solution. It is straight-forward to verify that this (unique) solution equals

$$u(x) = \exp \left(-\lambda \int_0^{\infty} (1-e^{-f(t+x)}) \, dt \right).$$

For $x=0$ this proves the assertion. To extend the identity for general $f$ (i.e. not necessarily having a compact support), we can use a standard truncation argument.