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)?

  • 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

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