2
$\begingroup$

Let $(N_t)_{t \geq 0}$ be a non-homogeneus Poisson process with intensity function $t \mapsto \lambda(t)$. Let $(T_n)_{n \in \mathbb{N}}$ be the corresponding simple point process (arrival times), namely $$N_t = \sum_n \mathbf{1}_{\{0 < T_n \leq t \}}.$$ For a measurable subset $A \subseteq [0, \infty)$ denote by $N(A)$ the number of arrivals in $A$.

Campbell's theorem states that for a measurable function $f\colon [0, \infty) \rightarrow \mathbb{R}$ such that $$\int_{[0, \infty)} \min\{|f(t)|, 1\} \lambda(t)dt <\infty $$ the characteristic function of $$\Sigma := \sum_{i=1}^{N([0, \infty))} f(T_i) = \int_0^\infty f(t)dN(t)$$ is given by $$\varphi_{\Sigma}(r) = \mathbb{E}[e^{ir\Sigma}] = \exp\left\{\int_{[0, \infty)} (e^{irf(t)}-1)\lambda(t)dt\right\}$$ provided that the integral on the right-hand side converges for each $r$. Therefore, we can get all the moments of $\Sigma$.

There is also a paper by Privault "Moments of Poisson stochastic integrals with random integrands" arXiv which gives all moments to a more general problem.

Assuming that the intensity function $\lambda$ is a PDF of a distribution supported on $[0, \infty)$, I would like to find moments (at least first two) of $$\frac{1}{1+N([0, \infty))} \Sigma?$$

I was not able to calculate it, the issue is that we have a product of two correlated random variables. I could not find it in the literature on the Poisson process, there was nothing on "random averages" over a point process. Maybe it is trivial, but I have no idea how to approach it. I would be grateful for any relevant references.

My attempt: Let $f$ be a measurable function.

We obtain \begin{align*} \mathbb{E}\left[ \frac{1}{1+ N([0, \infty))} \Sigma \right] =& \sum_{m \geq 1} \mathbb{E}\left[ \frac{1}{1+ m} \sum_{k=1}^{m} f(T_k) \Big| N([0, \infty))=m)\right]\mathbb{P}(N([0, \infty))=m) \\=& \sum_{m \geq 1} \frac{1}{1+m} \sum_{k=1}^m \mathbb{E}[f(T_k)]\mathbb{P}(N([0, \infty))=m) \\ =& \sum_{m \geq 1} \frac{1}{1+m} \mathbb{P}(N([0, \infty))=m) \sum_{k=1}^m \int_0^{\infty} f(t) p_k(t) dt, \end{align*} where $p_k$ is the pdf of $k$-th point $T_k$.

$\endgroup$
  • 1
    $\begingroup$ Just derive it from the first principles like you did with Campbell's theorem (in which you forgot exponentiation on the RHS). Note that to get $N([0,+\infty))$ finite with probability $1$, you have to assume that $\int_0^\infty\lambda<+\infty$. Condition upon $N([0,+\infty))=k$ and get a series representation for the answer. $\endgroup$ – fedja May 30 '18 at 12:09
  • $\begingroup$ Thanks, I simplified it. It looks almost like $\mathbb{E}[\frac{1}{1+N([0, \infty])}] \int_0^{\infty} f(t) \lambda(t)dt$. Do you agree, now? Or is there a way or further simplification? $\endgroup$ – user83658 May 30 '18 at 20:20
  • $\begingroup$ Almost but not quite. The actual factor is $e^{-1}\sum_{k\ge 1}\frac 1{k+1}\frac 1{(k-1)!}$ while your factor $E[\frac 1{1+N}]$ is $e^{-1}\sum_{k\ge 0}\frac 1{k+1}\frac 1{k!}$. If we agree on that, then everything is correct. $\endgroup$ – fedja May 30 '18 at 22:14
  • $\begingroup$ I cannot work out, why it should be $\frac{1}{(k-1)!}$ instead of $\frac{1}{k!}$. Could you please give another hint, I had this idea with changing the indices, but then I decided that it was wrong. We have to exclude the case when $N=0$ because then there would be no arrivals. $\endgroup$ – user83658 May 31 '18 at 9:39
  • $\begingroup$ It is just $\frac 1{k+1}P(N=k)E]\sum_{j=1}^k f(X_j)]=\frac 1{k+1}e^{-1}\frac 1{k!}k\int f\lambda$ where $E$ is taken with respect to the distribution with pdf $\lambda$. $\endgroup$ – fedja May 31 '18 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy