0
$\begingroup$

Edited after comment by Ofer Zeitouni

I have a sequence of discrete time stochastic processes $\big((S_n(i))_{i \geq 1}\big)_{n\geq 1}$ such that for every $n$, $i$,

\begin{equation}S_n(i)=\sum_{j=1}^i Y_n(j)\end{equation} where $Y_n(j)$ are non negative integer valued random variables.

I know that

\begin{equation}\lim_{n \to \infty} \mathbb E\Big[\sum_{j=1}^\infty Y_n(j)\mathbf 1_{Y_n(j)\geq 2}\Big]=0,\end{equation} that is, that the expected contribution to the process from non-unitary jumps converges to $0$, and exists a function $\lambda (t): [0,\infty)\mapsto (0,\infty)$ and a sequence $a_n$ with $a_n \to \infty$ such that, for every $k\in \mathbb N$, and every sequence of arrays $(j_{1,n},j_{2,n},...,j_{k,n})_{n \geq 1}$ such that $(j_{1,n}/a_n,j_{2,n}/a_n,...,j_{k,n}/a_n)_{n \geq 1}\to (t_1,t_2,...,t_k) \in \mathbb [0,\infty)^k$,

\begin{equation}\lim_{n \to \infty}\mathbb E[Y_n(j_{1,n})Y_n(j_{2,n})\cdots Y_n(j_{k,n})]a_n^{k}= \lambda(t_1)\lambda(t_2)\cdots \lambda (t_k).\end{equation}

Can I conclude that the continuous time process $(S_n(\lfloor a_nt\rfloor ))_{t \geq 0}$ converges in distribution to an inhomogeneous Poisson point process with intensity $\lambda (t)$ (even in some quite weak topology, like for finite dimensional distributions)?

$\endgroup$
  • 1
    $\begingroup$ No. Note that your assumption is satisfied if $Y_n(1)=1$ deterministically, which forces $S_n(a_nt)>0$ a.s, which contradicts Poisson behavior. $\endgroup$ – ofer zeitouni Feb 11 at 21:39
  • $\begingroup$ you are right, my assumptions do not take into account lower order corrections on the times. I will try to reformulate the question in order to consider them $\endgroup$ – LopiJ Feb 12 at 10:38
  • $\begingroup$ Unfortunately no: your form of the independence is too weak. For a counter example you can think of $p$ points uniformly on $[1,N]\cap \mathbb{N}$ and $Y_i$ the random variables that is equal to 1 if there is a point in $i$. $\endgroup$ – RaphaelB4 Feb 12 at 14:32
  • $\begingroup$ that would not satisfy the hypothesis , since for $k>p$ the expectation in the last equation would always be $0$ $\endgroup$ – LopiJ Feb 12 at 15:08
  • $\begingroup$ Still not OK. What happens if $j_{1,n}=j_{2,n}$? The assumption then contradicts Cauchy Schwarz :-) $\endgroup$ – ofer zeitouni Feb 12 at 21:35

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.