# 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: $$$$Y = \sum_{j=1}^{N} X_j, \quad N\sim \hbox{Poisson}(\lambda), \, X_j\,\, i.i.d. \sim \eta$$$$ 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 $\epsilon \to 0$. 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
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. 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 $\epsilon>0$, $\nu(\mathbb{R}) \le 1$. 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. Oct 29, 2022 at 18:13

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. 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. 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. 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
Oct 28, 2022 at 20:05
• Losif Pinelis and Christophe Leuridan, this question is more specific math.stackexchange.com/questions/4576471/…
– PSE
Nov 14, 2022 at 21:52

Since the compound Poisson distribution is a special infinitely divisible distribution, here we list some impotant facts from the book: B. V. Gnedenko & A. N. Kolmogorov, Limit distributions for sums of Independent Random Variables, Addison-Wesley Publishing Company(1968)(\$18 p.76-- ).

Theorem 1 In order that the function $$\phi(t)$$ be the characteristic function of an infinitive divisible distribution, it is necessary and sufficient that its logarithm be representable in the form $$\begin{equation*} \log \phi(t)=i\gamma t+\int_{\mathbb{R}}\Big[e^{itu}-1-\frac{itu}{1+u^2}\Big] \frac{1+u^2}{u^2}dG(u), \tag{1} \end{equation*}$$ where $$\gamma$$ is a real constant, $$G(u)$$ is a bounded nondecreasing function, and the integrand at $$u=0$$ is defined by the equaton $$\begin{equation*} \Big[\Big\{e^{itu}-1-\frac{itu}{1+u^2} \Big\}\frac{1+u^2}{u^2}\Big]_{u=0} =-\frac{t^2}{2} \end{equation*}$$ The representation of $$\log \phi(t)$$ by the formula (1) is unique.

The formula (1) is called the Kolmogorov canonical form of IDCF(infinitive divisible characteristic function). In the following we also using IDCF (IDDF, IDRV)($$\gamma, G$$) denote characteristic function (distribution function, random variable, respectively) with characteristic function $$\phi$$ form (1).

Leter, we also use $$G$$ to denote the finite measure generated by $$G$$.

Let $$\begin{gather*} \nu(\{0\})=0, \qquad \nu(B)=\int_{B\setminus \{0\}}\frac{1+u^2}{u^2} G(\mathrm{d}u),\quad B\in\mathscr{B}(\mathbb{R}). \\ \sigma^2=G(\{0\}). \tag{2} \end{gather*}$$ Then $$\nu$$ is a $$\sigma$$-finite measure on $$\mathbb{R}$$ with $$\begin{equation*} \int_{\mathbb{R}}(1\wedge u^2)\,\nu(\mathrm{d}u)<+\infty. \tag{3} \end{equation*}$$ The IDCF in (1) may be denoted as following $$\begin{equation*} \log \phi(t)=i\gamma t-\frac{\sigma^2}{2}t^2+\int_{\mathbb{R}}\Big[e^{itu}-1-\frac{itu}{1+u^2}\Big] \nu(\mathrm{d}u), \tag{4} \end{equation*}$$ It is called Lévy canonical form of IDCF. We will denote the $$\phi$$ in (4) as $$\mathrm{IDCF_L}(\gamma,\sigma^2,\nu$$). This representation of $$\log [\phi(t)]$$ by the formula (4) is unique also.

For IDCF, if $$\begin{equation*} \int_{|u|\le1}|u|\,\nu(\mathrm{d}u)<+\infty \quad \Big(\int_{|u|\le1}\frac{1}{|u|}G(\mathrm{d}u)<\infty\Big), \tag{5} \end{equation*}$$ then (4) may be rewrited as $$\begin{equation*} \log \phi(t)=it\Big(\gamma-\int_{\mathbb{R}}\frac{u}{1+u^2} \nu(\mathrm{d}u)\Big) -\frac{\sigma^2}{2}t^2 + \int_{\mathbb{R}}(e^{itu}-1) \nu(\mathrm{d}u). \tag{6} \end{equation*}$$ Furthermore, if $$\sigma^2=0(G(\{0\})=0)$$ and $$\begin{equation*} \gamma=\int_{\mathbb{R}}\frac{u}{1+u^2}\nu(\mathrm{d}u)\quad \Big(\gamma=\int_{\mathbb{R}}\frac{1}{u}G(\mathrm{d}u)\Big), \tag{7} \end{equation*}$$ then (6) may be rewrited as $$\begin{equation*} \log \phi(t)=\int_{\mathbb{R}}(e^{itu}-1)\nu(\mathrm{d}u). \tag{8} \end{equation*}$$

For random variable $$X$$, if its characteristic function $$\begin{equation*} \phi_X(u)=\mathsf{E}[e^{iuX}] = \exp\Big[\int_{\mathbb{R}}(e^{itu}-1)\nu(\mathrm{d}u)\Big], \quad \nu(\mathbb{R})<+\infty, \end{equation*}$$ then call $$X$$ is a compound Poisson distributed random variable(CPRV), $$\phi_X(u)$$ is a CPCF(compound Poisson c.f.), denote it by $$\mathrm{CPCF}(\nu)$$. Hence $$\begin{equation*} \mathrm{CPCF}(\nu)=\mathrm{IDCF_L}\Big(\int_{\mathbb{R}}\frac{u}{1+u^2} \nu(\mathrm{d}u), 0,\nu \Big). \tag{9} \end{equation*}$$

In summary, $$\begin{gather*} \mathrm{IDCF}(\gamma, G)\stackrel{\text{by }(2)}{\Longleftrightarrow} \mathrm{IDCF}_L(\gamma,\sigma^2,\nu) + (3). \\ \mathrm{IDCF}_L(\gamma,0,\nu) +\{(\nu(\mathbb{R})<+\infty) + (7)\} \Longleftrightarrow \mathrm{CPCF}(\nu). \end{gather*}$$

Now we are ready to discuss the convergence of sequence of IDCF.

Suppose a sequence $$\phi_n=\mathrm{IDCF}(\gamma_n, G_n)$$ converge to a limit CF $$\phi$$, then $$\phi$$ is a $$\mathrm{IDCF}(\gamma, G)$$.

Theorem 2 For $$\phi_n=\mathrm{IDCF}(\gamma_n, G_n)\to \phi = \mathrm{IDCF}(\gamma, G)$$ as $$n\to\infty$$, it is necessary and sufficient that, \begin{align*} (1) & \lim_{n\to\infty}\int_{\mathbb{R}}f(u)\,G_n(du)= \int_{\mathbb{R}}f(u)\,G(du), \quad \forall f\in C_b(\mathbb{R}) \tag{10}\\ (2) & \lim_{n\to\infty}\gamma_n=\gamma, \end{align*} where $$C_b(\mathbb{R})=\{f:f \text{ is continuous and bounded on \mathbb{R} } \}$$ (cf. Kolmogorov's book p.87 Theorem 19.1).

From this theorem, easy to derive the following result.

Theorem 3 For $$\phi_n=\mathrm{IDCF_L}(\gamma_n, \sigma^2_n, \nu_n) \to \phi=\mathrm{IDCF_L}(\gamma, \sigma^2, \nu)$$ as $$n\to\infty$$, it is necessary and sufficient that, \begin{align*} (1) & \lim_{n\to\infty}\int_{\mathbb{R}}\frac{f(u)u^2}{1+u^2}\,\nu_n(du)= \int_{\mathbb{R}}\,\frac{f(u)u^2}{1+u^2}\nu(du), \quad \forall f\in C_b(\mathbb{R}). \tag{11}\\ (2) &\; \lim_{\epsilon \downarrow 0} \varlimsup_{n\to\infty} \Big[\sigma^2_n+\int_{|u|\le\epsilon} u^2 \nu_n(du) \Big]\\ &\qquad =\lim_{\epsilon \downarrow 0} \varliminf_{n \to\infty} \Big[\sigma^2_n+ \int_{|u|\le\epsilon} u^2\nu_n(du) \Big] =\sigma^2,\\ (3) &\; \lim_{n\to\infty}\gamma_n=\gamma. \end{align*}

Remark 1 The convergence in (11) is stronger than following one: $$\begin{equation*} \lim_{n\to\infty}\int_{\mathbb{R}} f(u)\,\nu_n(\mathrm{d}u) = \int_{\mathbb{R}} f(u)\,\nu(\mathrm{d}u), \quad \forall f\in C_{\#}, \tag{12} \end{equation*}$$ where $$C_{\#}$$ is the class of continuous and bounded functions vanishing on a neighborhood of 0.

Therefore, the convergence of CPCF could be derived as follows.

Theorem 4 For $$\phi_n=\mathrm{CPCF}(\nu_n) \to \mathrm{CPCF}(\nu)$$ as $$n\to\infty$$, it is necessary and sufficient that, \begin{align*} (1) & \lim_{n\to\infty}\int_{\mathbb{R}}\frac{f(u)u^2}{1+u^2}\,\nu_n(\mathrm{d}u)= \int_{\mathbb{R}}\,\frac{f(u)u^2}{1+u^2}\nu(\mathrm{d}u), \quad \forall f\in C_b(\mathbb{R}). \tag{13} \\ (2) &\; \lim_{\epsilon \downarrow 0} \varlimsup_{n\to\infty} \Big[\int_{|u|\le\epsilon} u^2 \nu_n(\mathrm{d}u) \Big] =\lim_{\epsilon \downarrow 0} \varliminf_{n \to\infty} \Big[ \int_{|u|\le\epsilon} u^2\nu_n(\mathrm{d}u) \Big] =0. \tag{14}\\ (3) &\lim_{n\to\infty}\int_{\mathbb{R}}\frac{u}{1+u^2}\nu_n(\mathrm{d}u) = \int_{\mathbb{R}}\frac{u}{1+u^2} \nu(\mathrm{d}u). \tag{15} \end{align*}

Remark 2 From (13), easy to get following: $$\begin{equation*} \varliminf_{n \to\infty}\nu_n(\mathbb{R})\ge \nu(\mathbb{R}). \end{equation*}$$

Theorem 5 For $$\phi_n=\mathrm{CPCF}(\nu_n) \to \mathrm{CPCF}(\nu)$$ as $$n\to\infty$$, the following conditions are sufficient, \begin{align*} (1) & \lim_{n\to\infty}\int_{\mathbb{R}}f(u)\,\nu_n(\mathrm{d}u)= \int_{\mathbb{R}}\,f(u)\nu(\mathrm{d}u), \quad \forall f\in C_b(\mathbb{R}), \tag{16} \\ (2) &\; \lim_{\epsilon \downarrow 0} \varlimsup_{n\to\infty} \Big[\int_{|u|\le\epsilon} u^2 \nu_n(\mathrm{d}u) \Big] =\lim_{\epsilon \downarrow 0} \varliminf_{n \to\infty} \Big[\int_{|u|\le\epsilon} u^2\nu_n(\mathrm{d}u) \Big] =0. \end{align*}

In the following, we use $$\eta_{\{x\}}$$ denote the unit measure concentrateed at single point $$x$$, i.e., $$\begin{equation*} \eta_{\{x\}}(A)=I_{A}(x), \quad A\in\mathscr{B}(\mathbb{R}), \quad x\in\mathbb{R}. \end{equation*}$$ Denote $$\lambda$$ the Lebesgue measure on $$\mathbb{R}$$. And denote formally $$\begin{equation*} \frac{\mathrm{d}\eta_{x}}{\mathrm{d}\lambda}(u)=\delta_{\{x\}}(u). \end{equation*}$$

Example 1 Let $$$$\nu_n=n\delta_{\{1/\sqrt{n}\}}, \qquad \phi_n(t)=\mathrm{CPCF}(\nu_n).$$$$ Then $$\phi_n(t)\to \nu=\exp(-t^2/2)$$, $$\nu$$ isn't a CPCF. This means (14) is necessary for $$\mathrm{CPCF}(\nu_n)$$ converge to CPCF.

Example 2 Suppose $$\begin{equation*} \frac{\mathrm{d}\nu_n}{\mathrm{d}\lambda}(u)=\frac{(1+(-1)^n)(1+u^2)}{u^3} I_{[\mathrm{e}/n,2\mathrm{e}/n]}(u) + \delta_{\{-1\}}(u) , \quad \nu=\eta_{\{-1\}}. \end{equation*}$$ In this time, \begin{align*} \nu_n(\mathbb{R})-\nu(\mathbb{R}) & =\int_{\mathbb{R}}\frac{u^2}{1+u^2}\nu_n(\mathrm{d}u)- \int_{\mathbb{R}}\frac{u^2}{1+u^2}\nu(\mathrm{d}u)\\ & =[1+(-1)^n]\int_{\mathrm{e}/n}^{2\mathrm{e}/n} \frac1{u}\mathrm{d}u =[1+(-1)^n], \end{align*} hence (13) is not hold. But if $$f\in C_{\#}$$ and $$f(u)1_{\{|u|<\epsilon\}}=0$$, then \begin{align*} &\Big|\int_{\mathbb{R}}f(u)\nu_n(\mathrm{d}u)- \int_{\mathbb{R}}f(u)\nu(\mathrm{d}u)\Big|\\ &\quad =\Big|\int_{\mathbb{R}}f(u)\nu_n(\mathrm{d}u)-f(-1)\Big|=0,\quad \text{as } n>2\mathrm{e}\epsilon^{-1}. \end{align*}

Therfore, (12) is true. This example also interpret (12) is insufficient for $$\mathrm{CPCF}(\nu_n) \to \mathrm{CPCF}(\nu)$$.

Example 3 Suppose \begin{align*} \frac{\mathrm{d}\nu_n}{\mathrm{d}\lambda}(u)&=\frac{(1+(-1)^n)(1+u^2)}{u^2} [I_{[-2\mathrm{e}/n,-\mathrm{e}/n]}(u)+I_{[\mathrm{e}/n,2\mathrm{e}/n]}(u)] \\ &\quad +\delta_{\{-1\}}(u),\\ \nu&=\eta_{\{-1\}}. \end{align*} For $$f\in C_b(\mathbb{R}), |f|\le M$$, as $$n\to\infty$$ \begin{align*} &\Big|\int_{\mathbb{R}}\frac{f(u)u^2}{1+u^2}\nu_n(\mathrm{d}u)- \int_{\mathbb{R}}\frac{f(u)u^2}{1+u^2}\nu(\mathrm{d}u)\Big| \\ &\qquad \le 2\Big[\int_{-2\mathrm{e}/n}^{-\mathrm{e}/n}|f(u)|\mathrm{d}u + \int_{\mathrm{e}/n}^{2\mathrm{e}/n}|f(u)|\mathrm{d}u \Big] \le \frac{12M}{n}\to 0. \\ &\Big|\int_{\mathbb{R}}\nu_n(\mathrm{d}u) - \int_{\mathbb{R}}\nu(\mathrm{d}u)\Big|\\ &\quad = (1+(-1)^n) \Big[ \int_{\mathrm{-2e}/n}^{\mathrm{-e}/n} \frac{1+u^2}{u^2}\mathrm{d}u + \int_{\mathrm{e}/n}^{2\mathrm{e}/n} \frac{1+u^2}{u^2}\mathrm{d}u \Big]\\ &\quad \ge \frac{(1+(-1)^n)n}{\mathrm{e}}\nrightarrow 0 . \end{align*}
Hence (13) is true, but (16) doesn't hold for $$f=1$$. It also interpret that (16) is strictly stronger than (13). Meanwhile, \begin{align*} &(1+(-1)^n)\Big[\int_{\mathbb{R}}\frac{u}{1+u^2}\nu_n(\mathrm{d}u)- \int_{\mathbb{R}}\frac{u}{1+u^2}\nu(\mathrm{d}u)\Big] \\ &\qquad = (1+(-1)^n)\Big[\int_{-2\mathrm{e}/n}^{-\mathrm{e}/n}\frac{1}{u}\mathrm{d}u + \int_{\mathrm{e}/n}^{2\mathrm{e}/n}\frac{1}{u}\mathrm{d}u \Big] =0, \end{align*} so (15) is true and $$\phi_n=\mathrm{CPCF}(\nu_n) \to \mathrm{CPCF}(\nu)$$ as $$n\to\infty$$, it also interpret that (16) is unnecessary.

Example 4 Suppose $$\begin{equation*} \frac{\mathrm{d}\nu_n}{\mathrm{d}\lambda}(u)=\frac{1}{u^2} I_{[n,\mathrm{e}n]}(u) + \delta_{\{-1\}}(u) , \quad \nu=\eta_{\{-1\}}. \end{equation*}$$ For $$f\in C_b(\mathbb{R}), |f|\le M$$, as $$n\to\infty$$, \begin{align*} \Big|\int_{\mathbb{R}}f(u)\nu_n(\mathrm{d}u)-f(-1)\Big| =\Big|\int_{n}^{\mathrm{e}n}f(u)\frac{1}{u^2}\mathrm{d}u\Big| \le \frac{M}{n} \to 0. \end{align*} Hence (16) holds. Now suppose $$X_n=\mathrm{CPRV}(\nu_n), X=\mathrm{CPRV}(\nu)$$, then $$X_n\stackrel{df}{\longrightarrow}X$$, \begin{align*} \mathsf{E}[X_n] &=\int_{n}^{\mathrm{e}n}\frac{u}{u^2}\mathrm{d}u-1 = (\ln u)\Big|_{u=n}^{\mathrm{e}n} -1=0,\\ \mathsf{E}[X] &= -1. \end{align*}