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

Question

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: \begin{equation} \label{I}\tag{I} \varphi_Y(z) = \exp\left\{ \int_{\mathbb R} [e^{izx} - 1] \, d\nu(x) \right\} \end{equation}

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

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 \begin{equation}\label{II}\tag{II} X_n \Longrightarrow Y, \quad (n\to \infty) \end{equation} 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: \begin{equation} Y = \sum_{j=1}^N X_j, \quad N\sim \hbox{Poisson}(\lambda), \, X_j\,\, \text{i.i.d.} \sim \eta \end{equation} 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: \begin{equation}X_n \Longrightarrow Y, \quad (n\to \infty) \end{equation} C.f. Theorem 16.5, page 333, from Klenke, Achim, Probability theory. A comprehensive course. ZBL1295.60001.

Answer 1

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 $\nu(\mathbb{R}) \le 1$.