# Is the limit of compound Poisson random variables a compound Poisson r.v.?

Let $$Y$$ be an infinitely divisible (I.D.) random variable.

Let $$\nu$$ be any measure not necessarily finite: $$\nu(\mathbb R)\leq \infty$$. Suppose that $$Y \sim (0, \nu,0)_0$$ according to the notation on Page 39, Equation (8.7), from Sato, Ken-Iti, Lévy processes and infinitely divisible distributions, ZBL0973.60001.

That is, the Lévy-Khintchine representation of the characteristic function is given by: $$$$\label{I}\tag{I} \varphi_Y(z) = \exp\left\{ \int_{\mathbb R} [e^{izx} - 1] \, d\nu(x) \right\}$$$$

$$\underline{Remark\,\,1:}$$

Note that if $$\nu(\mathbb R)< \infty$$, we can set $$\lambda:=\nu(\mathbb R)$$ and writte: $$\varphi_Y(z) = \exp\left\{\lambda \int_\mathbb R [e^{izx} - 1] \, d\eta(x) \right\}, \quad d\eta(x):= d\nu(x)/\lambda$$ In this case, we have that $$\eta$$ is a probability measure ($$\eta(\mathbb R)=1$$) and $$Y$$ is a compound Poisson random variable $$Y \sim CP(\lambda, \eta)$$( See page 18, Equation (4.1) from the Sato's book (reference above) or Updated remark 2 below.)

However, in general, we don't have $$\nu(\mathbb R)< \infty$$. For example, see this question and this other question, where we have infinite mass around zero.

So my question is: Given a sequence $$(X_n)_{n \in \mathbb N}$$ of I.D. random variables with $$\varphi_{X_n}(z) = \int_{\mathbb R} [e^{izx}-1 ] \, d\nu_n(x), \quad \nu_n(\mathbb R)< \infty$$ By $$\underline{Remark\,\,1}$$ above, we have that $$X_n \sim CP(\lambda_n, \eta_n)$$ where $$\lambda_n = \nu_n(\mathbb R)$$ and $$d\eta_n(x):= d\nu_n(x)/\lambda_n$$. Now, supppose that $$$$\label{II}\tag{II} X_n \Longrightarrow Y, \quad (n\to \infty)$$$$ where $$Y \sim (0, \nu,0)_0$$ has characterization given by (\ref{I}).

So, in what situations ( assumptions about $$\eta_n$$ and $$\lambda_n$$ ) the convergence given in (\ref{II}) implies that $$Y$$, with characterization given by (\ref{I}), is in fact a compound Poisson random variable? Or in other sufficient way, when we have $$\nu(\mathbb R)< \infty$$?.

One trivial case is when $$(X_n)$$ has the same distribution. I.e. $$\eta_n = \eta$$ and $$\lambda_n = \lambda$$ for all $$n$$. So we exclude this case.

My intuition tells me that it is not possible, given (\ref{II}), to have $$\nu(\mathbb R)< \infty$$.

Help

Updated remarks

$$1.-$$ Using the theorem 8.7, page 41, from the Sato's book (reference above)), we have that if $$f \in C_\#$$ (bounded continuous function vanishing on a neighborhood of $$0$$), then

$$\lim_{n \to \infty} \int_{\mathbb R} f(x) \underbrace{\lambda_n \eta_n (dx)}_{= \nu_n (dx)} = \int_{\mathbb R} f(x) \nu (dx)$$

So, for any $$\epsilon>0$$, taking the indicator function $$f_\epsilon(x) = \chi_{|x|>\epsilon}(x)= 1$$ if $$|x|>\epsilon$$ and $$0$$ other wise, we have

$$\eta_n( E_\epsilon ) \to \nu( E_\epsilon ), \quad E_\epsilon = \{x: |x|>\epsilon\}, \quad (n \to \infty)$$

Could this be a useful way?

After the comment from Christophe Leuridan, we have to take a convenient continuous function and not the indicator function, because it is not continuous.

$$2.-$$ After the comment from Christophe Leuridan, I think it is necessary to specify that, in general, given a probability measure $$\eta$$, $$Y \sim CP(\lambda, \eta)$$ means that: $$$$Y = \sum_{j=1}^N X_j, \quad N\sim \hbox{Poisson}(\lambda), \, X_j\,\, \text{i.i.d.} \sim \eta$$$$ For more details, see this. Moreover, the characteristic function is:

$$\varphi_Y(z) = \exp\left\{\lambda \int_\mathbb R [e^{izx} - 1] \, d\eta(x) \right\} = \exp\left\{ \lambda \int_\mathbb R e^{izx} \, d\eta(x) - 1 \right\}= \exp\left\{ \lambda [\varphi_\eta(z) - 1] \right\}$$

$$3.-$$ I had posted this result above, but I don't know if it will help much. However, I will register it here. We know thar every random variable $$Y$$ is infinitely divisible if and only if there is a sequence $$(X_n)_{n \in \mathbb N}$$ of compound Poisson random variables such that, in weak limits: $$$$X_n \Longrightarrow Y, \quad (n\to \infty)$$$$ C.f. Theorem 16.5, page 333, from Klenke, Achim, Probability theory. A comprehensive course. ZBL1295.60001.

• A good assumption is that the sequence $(\eta_n(\mathbb{R}))_{n \ge 1}$ is bounded. Without this assumption, you may have non-compound-Poisson distribution at the limit. For example, one can approach $\int_0^\infty \frac{e^{izx-1}}{z^{3/2}}dx$ by $\int_\epsilon^\infty \frac{e^{izx-1}}{z^{3/2}}dx$ as $\epsilon \to 0$. Commented Oct 27, 2022 at 18:19
• Note that if $X_n \sim CP(1, \eta_n)$, then $\eta_n$ is a measure, i.e., $\eta_n(\mathbb R)=1$ for all $n$. c.f. Definition 1.2 , page 4 from cutt.ly/dNjXB6z (this definition is for C.P process, but for r.v. is the same) So $(\eta(\mathbb R))_{n \geq 1}$ is bounded, necessarily.
– PSE
Commented Oct 27, 2022 at 18:54
• Sorry, I am not familiar with the notations, and I confused the probability measure $\eta_n$ with $\lambda_n\eta_n$. If I understand correctly, you assume the total mass $\lambda_n$ to be constant. Commented Oct 27, 2022 at 20:55
• My impression is that your first remark gives the answer. Les $f_\epsilon$ be a continuous (hence not an indicator) function vanishing on $[-\epsilon/2,\epsilon/2]$ and equal to 1 on the complement of $]-\epsilon,\epsilon[$. Then $\nu(]-\epsilon,\epsilon[^c) \le \int f_\epsilon d\nu \le 1$. Since it holds for every $\epsilon>0$, $\nu(\mathbb{R}) \le 1$. Commented Oct 28, 2022 at 7:41
• Do you assume that $(\lambda_n)$ is bounded or not ? If yes, I think that $\nu$ is necessarily bounded. Otherwise, it may be finite or infinite. Commented Oct 29, 2022 at 18:13

## 1 Answer

I hope I did not make a mistake, but I think it works.

The convergence $$\exp\Big(\int_\mathbb{R} (e^{izx}-1)d\eta_n(x) \Big) \to \exp\Big(\int_\mathbb{R} (e^{izx}-1)d\nu(x)\Big)$$ yields the convergence $$\int_\mathbb{R} (e^{izx}-1)d\eta_n(x) \to \int_\mathbb{R} (e^{izx}-1)d\nu(x).$$ I take the real parts and change the signs to have non-negative functions. $$\int_\mathbb{R} (1-\cos(zx))d\eta_n(x) \to \exp \int_\mathbb{R} (1-\cos(zx))d\nu(x).$$ By Fubini's theorem and Fatou's lemma, for every $$T>0$$, $$\begin{eqnarray*} \int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\nu(x) &=& \frac{1}{T}\int_0^T\Big(\int_\mathbb{R} (1-\cos(zx))d\nu(x)\Big)dz \\ &=& \frac{1}{T} \int_0^T \lim_n\Big(\int_\mathbb{R} (1-\cos(zx))d\eta_n(x)\Big)dz \\ &\le& \liminf_n \frac{1}{T}\int_0^T \Big(\int_\mathbb{R} (1-\cos(zx))d\eta_n(x)\Big)dz \\ &=& \liminf_n \int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\eta_n(x) \\ &\le& 1+1/\pi, \end{eqnarray*}$$ since the function sinc is bounded below by $$-1/\pi$$. Applying Fatou's lemma again, $$\begin{eqnarray*} \int_\mathbb{R} 1d\nu(x) &\le& \liminf_{T \to +\infty} \int_\mathbb{R} (1-(Tx)^{-1}\sin(Tx))d\nu(x) \\ &\le& \liminf_{T \to +\infty} 1+1/\pi. \end{eqnarray*}$$ Thus $$\nu$$ is finite. I guess that a refinement of this argument shows that $$\nu(\mathbb{R}) \le 1$$.

• It is a bit unclear to me how you get your second display from the first one, given that $\ln$ has different branches. Commented Oct 27, 2022 at 23:45
• @Iosif Pinellis. You are right. In the second display, both sides are continuous functions of $x$ vanishing at $0$, but the convergence is not necessarily uniform on compact sets. Commented Oct 28, 2022 at 7:45
• @Iosif Pinellis. Yes, you are right. Although both members are continuous functions vanishing at $0$ (the function $x \mapsto \min(|x|,1)$ is assumed to be $\nu$-integrable), it is not se obvious, since we have not necessarily convergence on compact sets. Commented Oct 28, 2022 at 8:06
• Dear, I apologize. I'm actually interested in the case where $\lambda$ is not constant equal to one. I edited the question one more time. I hope it becomes clearer. Sorry.
– PSE
Commented Oct 28, 2022 at 20:05
• Losif Pinelis and Christophe Leuridan, this question is more specific math.stackexchange.com/questions/4576471/…
– PSE
Commented Nov 14, 2022 at 21:52