I want to understand an approximation of a compound Poisson distribution in this paper.

First, let's set the environment. Consider $\mathcal{P}$ the class of distributions of real-valued and strictly stationary processes with expectation zero and finite variance. According to this topic, $\mathcal{P}$ is closed with respect to the Mallows metric $d$ (see the topic for a formal definition of $d$). Abusing the notation, we write $X \in \mathcal{P}$ to say that the law of $X$ is in $\mathcal{P}$.

Given $N \sim \hbox{Poisson}(\lambda)$. Let $\xi = (\xi_t)_{t \in \mathbb Z}$ be an ergodic process in $\mathcal{P}$ and $\xi_1,\xi_2, \xi_3,... \overset{iid}{\sim} \xi$ sequence of stochastic process independent of $N$. Define the compound Poisson stochastic process $Y = (Y_t)_{t \in \mathbb Z}$ in $\mathcal{P}$: \begin{equation} Y_t = \sum_{j=1}^N \xi_{t;j}, \quad N \sim \hbox{Poisson}(\lambda) \end{equation} (warning: $Y$ is not the classic Compound Poisson process)

This is a particular case with $(W_t)_{t \in \mathbb{Z}}\equiv 0$, $k=1$ of the Step 2 of the proof of item (ii) of Theorem 1 from the same paper cited above (The statement of the theorem is on page 454 and its proof on page 465). According to the same paper, we can approximate $Y$ by the following sequence of linear processes $(X^{(n)} , n \geq 1)$ (See equation 5.11 on the paper ): \begin{equation} X^{(n)}_t = \sum_{j =1 }^n \bar{\xi}_j U_{t - j ;n} \end{equation} where $(\bar{\xi}_j)_{j\in \mathbb Z}$ is a fixed realization of $\xi = (\xi_t)_{t \in \mathbb Z}$ and $(U_{t;n})_{t \in \mathbb Z} \overset{iid}{\sim} \hbox{Bernoulli}(\lambda/n)$ independent of $\xi$. The approximation or the convergence is with respect to the Mallows metric $d$. The following characterization is useful: $d(X^{(n)},Y) \to 0,\,(n \to \infty)$ is equivalent to:

  1. $X_{t_1,...,t_m}^{(n)}\implies Y_{t_1,...,t_m}\, (n \to \infty)$ for all $t_1,...,t_m \in \mathbb{Z}$ and all $m \in \mathbb{N}$. This is a convergence in distributions, and the method of characteristic functions can be used.

  2. $E[|X^{(n)}_{t}|^2] \to E[|Y_{t}|^2], (n \to \infty)$ for any $t$.

For me, the proof is strange and confusing because $(\bar{\xi}_j)_{j\in \mathbb Z}$ sometimes is fixed realization, and sometimes it is treated as random. This causes conflict with items 1 and 2 above. For example:

To demonstrate the first item, the convergence of the characteristic functions $\varphi^{(n)}(s) \to \varphi_Y(s)$ has to be point-wise, and the paper delivers a convergence in probability. Still in this case, I tried to adapt this answer, but I don't have $(\bar{\xi}_j)_{j\in \mathbb Z}$ iid. So it seems that necessarily I have to first assume $(\bar{\xi}_j)_{j\in \mathbb Z}$ as a fixed realization and then treat it as random. Very strange!

For the second item, the convergence of expectations would have to be a convergence of real numbers, but the paper once again delivers a convergence in probability, first treating $(\bar{\xi}_j)_{j\in \mathbb Z}$ as fixed and then as random. See equation (5.13) on the paper .

Any clarification?

  • $\begingroup$ I do not understand the definition of $Y_t$. What does $\xi_{t:j}$ mean? $\endgroup$ Dec 20, 2022 at 8:00
  • $\begingroup$ First note that a fixed realisation of a random variable is a random variable itself as a constant function. Second, I think that you should see every $\bar{\Xi_j}$ as concatenation of a random variable with a measurable function (the evaluation $ev_\omega: X \mapsto X(\omega)$. Then, the i.i.d. follows (at least under weak conditions) $\endgroup$ Dec 20, 2022 at 9:01
  • $\begingroup$ @ChristopheLeuridan $\xi_{t;j}$ is the $t$-coordinate of the stochastic process $\xi_j=(\xi_{t;j})_{t \in \mathbb Z}$. Note that the $\xi_j$'s are copies of $\xi$. And each of them is a stochastic process. $\endgroup$
    – Fam
    Dec 20, 2022 at 15:07
  • $\begingroup$ @user7427029 I didn't quite understand your second point. I'm finding it strange that the convergence in the paper of the two items above are convergence in probability, when it shouldn't be in probability. $\endgroup$
    – Fam
    Dec 20, 2022 at 15:09

1 Answer 1


The paper seems to be written rather carelessly. In particular, it is indeed unhelpful to denote a random object and its realizations by the same symbol, leaving the job of figuring out which is which here or there to the reader.

Further, it is clear that any sequence (say, the zero sequence) can be a realization of a random sequence $\xi$ (if e.g. $\xi$ is the sequence of iid normal random variables).

So, the result should be true, not for all realizations of $\xi$, but for almost all of them (excluding a set of realizations of probability $0$).

Indeed, the central point in the proof seems to be the use of the ergodic theorem in the middle of page 466 of the paper, where the convergence must of course be in the almost sure sense (but this is not specified on the paper).

  • $\begingroup$ An important point is that the paper states that $d(X^{(n)},Y)\to 0$. For this to be true, the paper itself proposes to show items 1 and 2, placed in my question. But, like you said, the convergence must be in the almost sure sense (or in probability). I don't know if the convergences in probability sense of items 1 and 2 are enough to show in fact that $d(X^{(n)},Y)\to 0$. $\endgroup$
    – Fam
    Dec 22, 2022 at 23:16
  • 1
    $\begingroup$ @Fam : I don't see any mentioning of the convergence in probability in the paper. Overall, as I wrote, it appears that the authors of the paper left too much work to the reader. Apparently, one has to go over the entire long proof (or maybe even over the entire long paper) and rewrite significant portions of it, to get a real proof. $\endgroup$ Dec 23, 2022 at 0:54
  • $\begingroup$ Yes, you're right. I've read most of the article and it's really quite confusing. The idea is to approximate $Y$ by a sequence of linear processes, but such processes seem to have random coefficients ($\xi$ coming from $Y$), although the author starts from a fixed realization of $\xi$. It looks as if convergence is conditional on a realization of $\xi$. Anyway, thanks for your comment! $\endgroup$
    – Fam
    Dec 23, 2022 at 5:57
  • $\begingroup$ Can you help me? mathoverflow.net/questions/438114/… $\endgroup$
    – Fam
    Jan 9 at 4:26

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