1.1. Definitions and notation

Let $\mathcal{B}(\mathbb{R})$ (respectively $\mathcal{B}((0,\infty))$ ) denote the $\sigma$ -algebra on the field of real numbers $\mathbb{R}$ (respectively $(0,\infty)$ ) generated by the class of open sets. We write $\|\cdot\|$ for the supremum/uniform norm. A point measure is an integer-valued measure on $\mathbb{R}$ and is finite on all compact sets of $\mathbb{R}$ . Let $\mathfrak{N}(\mathbb{R})$ and $\mathfrak{M}(\mathbb{R})$ denote respectively the space of finite point measures and finite Borel measures on $\mathbb{R}$ equipped with the topology of weak convergence. Let $\mathbf{0}$ denote the null measure on $\mathbb{R}$ . We write $\mathfrak{N}^{\circ}(\mathbb{R})\,:\!=\,\mathfrak{N}(\mathbb{R}) \backslash\{\mathbf{0}\}$ and $\mathfrak{M}^{\circ}(\mathbb{R})\,:\!=\,\mathfrak{M}(\mathbb{R}) \backslash\{\mathbf{0}\}$ . Let $\mathcal{B}(\mathfrak{N}(\mathbb{R}))$ , $\mathcal{B}(\mathfrak{N}^{\circ}(\mathbb{R}))$ , $\mathcal{B}(\mathfrak{M}(\mathbb{R}))$ , and $\mathcal{B}(\mathfrak{M}^{\circ}(\mathbb{R}))$ be the Borel $\sigma$ -algebra generated on $\mathfrak{N}(\mathbb{R})$ , $\mathfrak{N}^{\circ}(\mathbb{R})$ , $\mathfrak{M}(\mathbb{R})$ , and $\mathfrak{M}^{\circ}(\mathbb{R})$ by this weak topology, respectively. We use $\mu(h)$ to denote the integral of a function h with respect to a measure $\mu$ . The symbol ${\stackrel{\text{w}}{\longrightarrow}}$ stands for weak convergence throughout this paper.

3.1. Conditional reduced GW processes

Recall that $N=(N_n\colon n\in\mathbb{N})$ is a GW process with offspring distribution p. Let $N_{k,n}$ be the number of particles in the kth generation having at least 1 offspring at time n, then $(N_{k,n}\colon 0\leq k\leq n)$ is called the reduced GW process. Let $\hat{N}_{k,n}$ be a random variable with law

\begin{align*} \mathbb{P}_{\delta_{x}}(\hat{N}_{k,n}\in\cdot\ )=\mathbb{P}_{\delta_{x}}(N_{k,n}\in\cdot\mid N_n>0).\end{align*}

We write $\hat{N}_n\,:\!=\,\hat{N}_{n,n}$ by convention. We say the process $(\hat{N}_{k,n}\colon 0\leq k\leq n)$ is the conditional reduced GW process.

Based on the results of reduced GW processes established in [Reference Fleischmann and Siegmund-Schultze12], we can get the following properties of the conditional reduced GW process.

Proposition 3. The process $(\hat{N}_{k,n}\colon 0\leq k\leq n)$ is a time-inhomogeneous GW process. Suppose that $\hat{p}(k,n)\,:\!=\,\{\hat{p}_l(k,n)\}_{l\in\mathbb{N}}$ is the offspring distribution of a particle of generation $k\ (0\leq k\leq n-1)$ , then its generating function is given by

(11) \begin{align} \hat{f}_{k,n}(s)\,:\!=\,\mathbb{E}[s^{\hat{N}_{k+1,n}} \mid \hat{N}_{k,n}=1] =\frac{f(f_{n-k-1}(0)+s(1-f_{n-k-1}(0)))-f_{n-k}(0)} {1-f_{n-k}(0)}. \end{align}

By differentiating both sides of (11), we get the following proposition.

Proposition 4 Suppose that the offspring distribution p has finite moments of all orders. For $r\in\mathbb{N}_+$ , let $f^{(r)}$ and $\hat{f}_{k,n}^{(r)}$ denote respectively the rth derivative of f and $\hat{f}_{k,n}$ . Then we have

\begin{align*}\hat{f}_{k,n}^{(r)}(1)=\frac{(1-f_{n-k-1}(0))^{r}}{1-f_{n-k}(0)}f^{(r)}(1).\end{align*}

For $j\in\mathbb{N}_+$ , we denote by $\hat{m}_j(k,n)$ the jth moment of $\hat{p}(k,n)$ . It is easy to calculate the first and second moments:

(12) \begin{equation} \hat{m}_1(k,n)=m \frac{1-f_{n-k-1}(0)}{1-f_{n-k}(0)},\end{equation}

(13) \begin{equation} \hat{m}_2(k,n)=\frac{(1-f_{n-k-1}(0))^2}{1-f_{n-k}(0)} f^{(2)}(1)+m \frac{1-f_{n-k-1}(0)}{1-f_{n-k}(0)}. \end{equation}

3.2. Conditional reduced branching Markov chains

Recall that under $\mathbb{P}_{\delta_x}$ , $\mathbb{T}$ is a GW tree with offspring distribution p, starting from one particle. By the definition in [Reference Fleischmann and Siegmund-Schultze12], we can get a reduced GW tree $\mathbb{T}_n$ (rooted at $\varnothing$ ) by removing all branches of $\mathbb{T}$ that do not extend to generation n. Let $\hat{\mathbb{T}}_n$ be a random tree rooted at $\varnothing$ with law

\begin{align*} \mathbb{P}_{\delta_{x}}(\hat{\mathbb{T}}_n\in\cdot\ )=\mathbb{P}_{\delta_{x}}(\mathbb{T}_n\in\cdot\mid N_n>0).\end{align*}

For any $0\leq k\leq n$ , let

\begin{align*}\hat{\mathbb{G}}_{k,n}\,:\!=\,\{u\in\hat{\mathbb{T}}_n\colon |u|=k\}\end{align*}

be the set of individuals in generation k. The conditional reduced branching Markov chain $(V(u)\colon u\in\hat{\mathbb{G}}_{k,n},0\leq k\leq n)$ under $\mathbb{P}_{\delta_x}$ is characterized by the following properties.

1. (i) It starts with a single particle $\varnothing$ born at time 0 and position x, which forms the 0th generation, i.e. $\hat{\mathbb{G}}_{0,n}=\{\varnothing\}$ .

2. (ii) For each $0\leq k\leq n-1$ , each particle $u\in\hat{\mathbb{G}}_{k,n}$ (after one unit time) splits independently of the other particles of the kth generation into a random number of children with law $\hat{p}(k,n)$ . Conditional on $V(u)=y$ , these children are born at positions that are given by independent copies of the transition probability $P(y,\cdot)$ .

Let $\hat{Z}_{k,n}({\cdot})$ denote the counting measure of particles of generation k in the above system. Then $(\hat{Z}_{k,n}\colon 0\leq k\leq n)$ is an $\mathfrak{N}(\mathbb{R})$ -valued Markov process and

(14) \begin{align} \mathbb{P}_{\delta_x} (\hat{Z}_{k,n}\in\cdot\ )=\mathbb{P}_{\delta_{x}}(Z_{k,n}\in\cdot\mid N_n>0).\end{align}

Let $\hat{Z}_{k,n}(B)$ denote the number of particles in generation k located in B. In particular, $\hat{N}_{k,n}=\hat{Z}_{k,n}(\mathbb{R})$ , which forms the conditional reduced process described in Section 3.1.

We make the convention that $\hat{Z}_n\,:\!=\,\hat{Z}_{n,n}$ . It is easy to see that

\begin{align*} \mathbb{P}_{\delta_{x}}(\hat{Z}_n\in\cdot\ )=\mathbb{P}_{\delta_{x}}(Z_n\in\cdot\mid N_n>0).\end{align*}

Let $\hat{Z_n}(h)\,:\!=\,\int_{\mathbb{R}} h(x) \hat{Z}_n(\mathrm{d}x)$ , then the PGFL $\hat{G}_{n}$ of the point process $\hat{Z}_{n}$ under $\mathbb{P}_{\delta_{x}}$ is given by

\begin{align*} \hat{G}_{n}[h](x)=\mathbb{E}_{\delta_{x}}[\mathrm{e}^{\hat{Z}_{n}(\!\log h)}]=\mathbb{E}_{\delta_{x}}[\mathrm{e}^{{Z}_{n}(\!\log h)}\mid N_n>0],\quad h\in\mathcal{V}_0(\mathbb{R}).\end{align*}

Therefore, we only need to consider the moments of $\hat{Z}_n(h)$ , and then obtain the convergence of $\hat{G}_{n}$ , which completes the first step.

4.1. The spinal decomposition theorem and the many-to-few formula

For any fixed $n\in\mathbb{N}$ and $r\in\mathbb{N}_+$ , we now attach r additional random distinguished lines of descent $w^1,\ldots,w^r$ to the conditional reduced branching Markov chain $(V(u)\colon u\in\hat{\mathbb{G}}_{k,n},0\leq k\leq n)$ , which are called spines. For any $x\in\mathbb{R}$ , let us introduce the following system, which is referred to as the conditional reduced branching Markov chain starting from one particle located at x with r spines $w^1,\ldots,w^r$ .

1. (i) At time 0, there is one particle at position x carrying r marks $1,\ldots,r$ , which forms the 0th generation.

2. (ii) For each $1\leq i\leq r$ , we regard the line of descent carrying mark i as the spine $w^i$ . Note that $w^i=(w_1^i,w_2^i,\ldots)$ , where $w_k^i$ denotes the particle carrying mark i at time k. We write $\mathfrak{X}_k^i$ for its position, i.e. $\mathfrak{X}_k^i=V(w_k^i)$ .

3. (iii) A particle u in generation k carrying j marks (after one unit time) splits independently of all others into a random number of children according to the jth sized-biased distribution $\{{l^j \hat{p}_l(k,n)}/{\hat{m}_j(k,n)}\}_{l\in\mathbb{N}}$ . Conditional on $V(u)=y$ , these children are born at positions that are given by independent copies of the law $P(y,\cdot)$ . Given that l particles are born at such a branching event, each of the j marks chooses a particle to follow independently and uniformly at random among these l variables.

4. (iv) Particles that carry no marks at time k have children according to the offspring distribution $\{\hat{p}_l(k,n)\}_{l\in\mathbb{N}}$ of a particle of generation k of conditional reduced GW processes and the one-step transition probability P, just as under $\mathbb{P}_{\delta_x}$ .

5. (v) The children of particles in generation k and their positions (including the marked particles) form the $(k+1)$ th generation.

We denote by $\mathbb{Q}_{\delta_x}^{[r]}$ the law of this new system. In other words, under $\mathbb{Q}_{\delta_x}^{[r]}$ , spine particles give birth to children according to the size-biased distribution, but the n-step transition probability of the spine Markov process $(\mathfrak{X}_k^i\colon k\geq0)$ will be unchanged, which is still $P_n$ . The number of children depends on how many marks the spine particle carries, but the motion does not.

We next give the constructions of the new probability measure $\mathbb{Q}_{\delta_x}^{[r]}$ used in the theory. For $u,v\in\mathcal{U}$ , recall that $|v|$ denotes the generation of v, $\overleftarrow{v}$ is the parent of $v(\neq\varnothing)$ and $v\prec u$ implies that v is an ancestor of u. For fixed $u^1,\ldots,u^r\in\mathcal{U}$ , let

\begin{gather*} {\textrm{skel}}_{u^1,\ldots,u^r}(k)=\{v\in\mathcal{U}\colon |v|\leq k,\text{there exists } 1\leq j\leq r \text{ with } v\prec u^j \}, \\ D_{u^1,\ldots,u^r}(v)=\#\{1\leq j\leq r\colon v\prec u^j\},\end{gather*}

where $\#B$ denotes the cardinality of the set B. For fixed $n\in\mathbb{N}$ , let $(\hat{\mathcal{F}}_{k,n})_{0\leq k\leq n}$ be the natural filtration of the conditional reduced branching Markov chain. For any $r\in\mathbb{N}_+$ , we define an $\hat{\mathcal{F}}_{k,n}$ -adapted process $(W^{[r]}(k,n)\colon 0\leq k\leq n)$ by

(15) \begin{align}W^{[r]}(k,n)\,:\!=\,\sum_{u^1,\ldots,u^r\in\hat{\mathbb{G}}_{k,n}}\prod_{v\in{\textrm{skel}}_{u^1,\ldots,u^r}(k)\backslash\{\varnothing\}}\Big(\hat{m}_{D_{u^1,\ldots,u^r} (\overleftarrow{v})}(|\overleftarrow{v}|,n)\Big)^{-1}.\end{align}

Repeating arguments of [Reference Harris and Roberts19, pp.236–237] and taking the martingale $\zeta\equiv1$ , one only needs to replace respectively the offspring distribution and the motion distribution of the particle u of generation k with $\{\hat{p}_l(k,n)\}_{l\in\mathbb{N}}$ and $P(V(u),\cdot)$ . It then follows that $(W^{[r]}(k,n)\colon 0\leq k\leq n)$ is a martingale with respect to the filtration $\hat{\mathcal{F}}_{k,n}$ under $\mathbb{P}_{\delta_x}$ . Furthermore, repeating the arguments of martingales and changes of measures in [Reference Harris and Roberts19, Section 5.6] and the proof of [Reference Harris and Roberts19, Lemma 6], we get the following lemma in a similar way. The following lemma establishes the link between $\mathbb{P}_{\delta_x}$ and $\mathbb{Q}^{[r]}_{\delta_x}$ , and calculates the probability that the r marks are carried by particles $u^1,\ldots,u^{r}$ at time k.

Lemma 6. (The spinal decomposition theorem.) (i) The derivative of $\mathbb{Q}^{[r]}_{\delta_x}$ with respect to $\mathbb{P}_{\delta_x}$ on $\hat{\mathcal{F}}_{k,n}$ is $W^{[r]}(k,n)$ defined by (15), that is,

\begin{align*} \bigg.\frac{\mathrm{d} \mathbb{Q}_{\delta_x}^{[r]}}{\mathrm{d} \mathbb{P}_{\delta_x}}\bigg|_{\hat{\mathcal{F}}_{k,n}}=W^{[r]}(k,n).\end{align*}

(ii) For any $0\leq k\leq n$ and any $u^1,\ldots,u^{r}\in\hat{\mathbb{G}}_{k,n}$ , we have

\begin{align*} & \mathbb{Q}_{\delta_x}^{[r]}\big(w_k^1=u^1,\ldots,w_k^r=u^r \mid \hat{\mathcal{F}}_{k,n}\big) \\ &\quad = \frac{1}{W^{[r]}(k,n)} \prod_{v\in{\textrm{skel}}_{u^1,\ldots,u^r}(k)\backslash\{\varnothing\}}\Big(\hat{m}_{D_{u^1,\ldots,u^r}(\overleftarrow{v})}(|\overleftarrow{v}|,n)\Big)^{-1}.\end{align*}

Let us introduce the many-to-few formula for the conditional reduced branching Markov chain. Let $\hat{\mathcal{G}}_{k,n}^{[r]}$ be the filtration containing all information about the conditional reduced branching Markov chain (including the r spines) up to time k. It follows from [Reference Harris and Roberts19] that if F is measurable with respect to $\hat{\mathcal{G}}_{k,n}^{[r]}$ then it can be expressed as

\begin{align*}F=\sum_{u^1,\ldots,u^r\in\hat{\mathbb{G}}_{k,n}}F(u^1,\ldots,u^r)\mathbf{1}_{\big\{w_k^1=u^1,\ldots,w_k^r=u^r\big\}},\end{align*}

where, for each $u^1,\ldots,u^r\in\hat{\mathbb{G}}_{k,n}$ , the random variable $F(u^1,\ldots,u^r)$ is $\hat{\mathcal{F}}_{k,n}$ measurable. For the sake of simplicity, we still write $\mathbb{Q}_{\delta_x}^{[r]}$ for its associated expectation in this paper.

Lemma 7. (The many-to-few formula.) For $r\in\mathbb{N}_+$ and $0\leq k\leq n$ , if F is measurable with respect to $\hat{\mathcal{G}}_{k,n}^{[r]}$ , then

(16) \begin{align} &\mathbb{E}_{\delta_x}\bigg[\sum_{u^1,\ldots,u^r\in\hat{\mathbb{G}}_{k,n}}F(u^1,\ldots,u^r)\bigg] \nonumber\\ &\quad= \mathbb{Q}_{\delta_x}^{[r]}\bigg[F\big(w_k^1,\ldots,w_k^r\big)\prod_{v\in\text{skel}(k)\backslash\{\varnothing\}}\hat{m}_{D(\overleftarrow{v})}(|\overleftarrow{v}|,n)\bigg],\end{align}

where skel(k) denotes the set of particles that carry at least one mark up to time k, and D(v) is the total number of marks carried by the particle v.

4.2. Moments

In this section we obtain some results for the moments of $\hat{Z}_n(h)$ . Here and in the sequel, we make the convention that $\sum_{j=l}^k=0$ and $\prod_{j=l}^k=1$ for any $k<l$ . Let $\hat{\mathbb{G}}_{n}\,:\!=\,\hat{\mathbb{G}}_{n,n}$ for simplicity. It was proven in [Reference Joffe27] that when $0<m<1$ , the sequence $(1-f_n(s))/(m^n(1-s))$ is monotonically decreasing in n and

(17) \begin{align}\lim_{n\rightarrow\infty}\frac{1-f_n(s)}{m^n(1-s)}=\varphi(s)\in[0,1],\quad \text{ $\varphi(s)$ is a nondecreasing function}.\end{align}

Furthermore, $\varphi(0)>0$ if and only if $\sum_{j}(j\log j)p_j<\infty$ ; see, e.g., [Reference Heathcote, Seneta and Vere-Jones20, Theorem 2].

Lemma 8. (The first moment.) Suppose that condition (C1) holds. For any $h\in\mathcal{B}_b^+(\mathbb{R})$ , we have

(18) \begin{align} \mathbb{E}_{\delta_x}[\hat{Z}_n(h)]=\frac{m^n}{1-f_n(0)}P_nh(x).\end{align}

Proof. Note that $F(w_n^1)=h\big(V\big(w_n^1\big)\big)=h(\mathfrak{X}_n^1)$ implies that $F(u^1)=h(V(u^1))$ . Then by applying (16) for $r=1,k=n$ and using (12), we obtain

\begin{align*}\mathbb{E}_{\delta_x}\big[\hat{Z}_n(h)\big]=\mathbb{E}_{\delta_x}\bigg[\sum_{u^1 \in \hat{\mathbb{G}}_{n}} h\big(V(u^1)\big)\bigg]=\frac{m^n}{1-f_n(0)}\mathbb{Q}^{[1]}_{\delta_x}\big[h\big(\mathfrak{X}_n^1\big)\big],\end{align*}

where $(\mathfrak{X}_n^1\colon n\geq0)$ is the spine Markov process whose n-step transition probability is $P_n$ . Now the desired result follows.

In view of (17), then an application of Lemma 8 gives the following result.

Corollary 3. (The limit of the first moment.) If conditions (C1) and (C3) hold, we have

(19) \begin{align} \lim_{n\rightarrow\infty}\mathbb{E}_{\delta_x}[\hat{Z}_n(h)]=\frac{\widetilde{\pi}(h)}{\varphi(0)},\quad h\in\mathcal{B}_b^+(\mathbb{R}).\end{align}

Lemma 9. (The second moment.) Suppose that condition (C1) holds. For any $h\in\mathcal{B}_b^+(\mathbb{R})$ , we have

(20) \begin{align} \mathbb{E}_{\delta_x}\big[(\hat{Z}_n(h))^2\big]=&\frac{m^n}{1-f_n(0)}P_nh^2(x)\nonumber\\ &+\frac{m^nf^{(2)}(1)}{m(1-f_n(0))}\sum_{j=0}^{n-1}m^j \int_{\mathbb{R}}(P_{j+1}h(y))^2 P_{n-j-1}(x,\mathrm{d} y).\end{align}

Proof. If $F(w_n^1,w_n^2)=h(V(w_n^1))h(V(w_n^2))=h(\mathfrak{X}_n^1)h(\mathfrak{X}_n^2)$ then

\begin{align*}F(u^1,u^2)=h(V(u^1))h(V(u^2)).\end{align*}

We denote by $T_1^{[2]}$ the first split time at which marks 1 and 2 are carried by different particles. Then applying (16) for $r=2$ and $k=n$ we obtain

(21) \begin{align}& \mathbb{E}_{\delta_x}\big[(\hat{Z}_n(h))^2\big] \nonumber\\&\quad= \mathbb{E}_{\delta_x}\Bigg[\sum_{u^1,u^2\in\hat{\mathbb{G}}_{n}} h\big(V(u^1))h(V(u^2)\big) \Bigg] \nonumber\\&\quad= \mathbb{Q}_{\delta_x}^{[2]}\Bigg[\mathbf{1}_{\{T_1^{[2]}>n\}} \big(h(\mathfrak{X}_{n}^{1})\big)^2\prod_{i=0}^{n-1}\hat{m}_2(i,n)\Bigg] \nonumber\\&\hskip29pt+\sum_{j=1}^{n}\mathbb{Q}_{\delta_x}^{[2]}\Bigg[h(\mathfrak{X}_{j-1,n}^{1}) h(\mathfrak{X}_{j-1,n}^{2})\mathbf{1}_{\{T_1^{[2]}=j\}}\prod_{i=0}^{j-1} \hat{m}_2(i,n)\prod_{i=j}^{n-1}(\hat{m}_1(i,n))^2\Bigg],\end{align}

where $(\mathfrak{X}_{n}^{1}\colon n\geq0)$ is the spine Markov process carrying two marks, $(\mathfrak{X}_{j-1,n}^{1}\colon n\geq0)$ and $(\mathfrak{X}_{j-1,n}^{2}\colon n\geq0)$ are two spine Markov processes having the same trajectory before time j and splitting at time j.

Under $\mathbb{Q}_{\delta_x}^{[2]}$ , the initial particle carrying marks 1 and 2 splits at time 1 into l particles with probability $l^2 \hat{p}_l(0,n)/\hat{m}_2(0,n)$ . At such a branching event, the two marks follow the same particle with probability $1/l$ . Thus,

\begin{align*}\mathbb{Q}_{\delta_x}^{[2]}\big(T_1^{[2]}>1\big)=\frac{\hat{m}_1(0,n)}{\hat{m}_2(0,n)}.\end{align*}

Similarly, under $\mathbb{Q}_{\delta_x}^{[2]}$ the particle in generation i carrying two marks splits at time $i+1$ into l particles with law $l^2 \hat{p}_l(i,n)/\hat{m}_2(i,n)$ , and then the two marks follow the same particle with probability $1/l$ . Hence,

(22) \begin{align}\mathbb{Q}_{\delta_x}^{[2]}\big(T_1^{[2]}>j\big)=\prod_{i=0}^{j-1} \frac{\hat{m}_1(i,n)}{\hat{m}_2(i,n)},\quad j=1,2,\ldots,n.\end{align}

Then, for any integer $1\leq j\leq n$ ,

(23) \begin{align}\mathbb{Q}_{\delta_x}^{[2]}\big(T_1^{[2]}=j\big)&=\mathbb{Q}_{\delta_x}^{[2]}\big(T_1^{[2]}>j-1\big)-\mathbb{Q}_{\delta_x}^{[2]}\big(T_1^{[2]}>j\big)\nonumber\\&=\prod_{i=0}^{j-2}\frac{\hat{m}_1(i,n)}{\hat{m}_2(i,n)}\bigg(1-\frac{\hat{m}_1(j-1,n)}{\hat{m}_2(j-1,n)}\bigg),\end{align}

with $\prod_{i=0}^{-1}=1$ by convention. From (12), (13), (22), and (23) it is easy to see that

\begin{align*}\prod_{i=0}^{n-1} \hat{m}_2(i,n) \mathbb{Q}_{\delta_x}^{[2]}\big(T_1^{[2]}>n\big)=\frac{m^n}{1-f_n(0)}\end{align*}

and

\begin{align*}\prod_{i=0}^{j-1}\hat{m}_2(i,n) \prod_{i=j}^{n-1}(\hat{m}_1(i,n))^2 \mathbb{Q}_{\delta_x}^{[2]}\big(T_1^{[2]}=j\big)=\frac{m^{2 n-1}}{1-f_n(0)} f^{(2)}(1) m^{-j}.\end{align*}

Since the reproduction and motion of an individual are independent, we obtain

\begin{align*}\mathbb{E}_{\delta_x}\big[(\hat{Z}_n(h))^2\big]=&\frac{m^n}{1-f_n(0)} \mathbb{Q}_{\delta_x}^{[2]}\big[\big(h\big(\mathfrak{X}_{n}^{1}\big)\big)^2\big]\\&+\frac{m^{2n-1}f^{(2)}(1)}{1-f_n(0)} \sum_{j=1}^n m^{-j}\mathbb{Q}_{\delta_x}^{[2]}\big[h\big(\mathfrak{X}_{j-1,n}^1\big)h\big(\mathfrak{X}_{j-1,n}^2\big)\big].\end{align*}

According to the properties of spine Markov processes $\mathfrak{X}_{n}^{1}$ , $\mathfrak{X}_{j-1,n}^1$ , and $\mathfrak{X}_{j-1,n}^2$ , we have

\begin{align*}& \mathbb{E}_{\delta_x}\big[(\hat{Z}_n(h))^2\big]\\&\quad= \frac{m^n}{1-f_n(0)}P_nh^2(x)+\frac{m^{2 n-1}}{1-f_n(0)} f^{(2)}(1) \sum_{j=1}^n m^{-j} \int_{\mathbb{R}}(P_{n-j+1}h(y))^2 P_{j-1}(x, \mathrm{d} y)\\&\quad= \frac{m^n}{1-f_n(0)}P_nh^2(x)+\frac{m^nf^{(2)}(1)}{m(1-f_n(0))} \sum_{j=0}^{n-1} m^j \int_{\mathbb{R}}(P_{j+1}h(y))^2 P_{n-j-1}(x, \mathrm{d} y).\end{align*}

This gives the desired result.

Corollary 4. (The limit of the second moment.) If conditions (C1) and (C3) hold, for any $h\in\mathcal{B}_b^+(\mathbb{R})$ , we have

(24) \begin{align}\lim_{n\rightarrow\infty} \mathbb{E}_{\delta_x}\big[(\hat{Z}_n(h))^2\big]=\frac{\widetilde{\pi}(h^2)}{\varphi(0)}+\frac{f^{(2)}(1)}{m\varphi(0)} \sum_{j=0}^{\infty}m^j \int_{\mathbb{R}}(P_{j+1}h(y))^2 \widetilde{\pi}(\mathrm{d} y).\end{align}

Proof. By (17), it is not hard to see that the first term on the right-hand side of (20) converges to $\widetilde{\pi}(h^2)/\varphi(0)<\infty$ as $n\rightarrow\infty$ under conditions (C1) and (C3). Let

\begin{align*}a_{j,n}\,:\!=\,\mathbf{1}_{\{0\leq j\leq n-1\}}m^j \int_{\mathbb{R}}(P_{j+1}h(y))^2 P_{n-j-1}(x, \mathrm{d} y),\end{align*}

then the second term on the right-hand side of (20) is equal to

(25) \begin{align}\frac{m^nf^{(2)}(1)}{m(1-f_n(0))} \sum_{j=0}^{\infty}a_{j,n}.\end{align}

Under conditions (C1) and (C3), it is easy to see that $\sum_{j}m^j<\infty$ and

\begin{align*}0\leq a_{j,n}\leq \|h\|^2m^j,\quad \lim_{n\rightarrow\infty}a_{j,n}=m^j \int_{\mathbb{R}}(P_{j+1}h(y))^2 \widetilde{\pi}(\mathrm{d} y).\end{align*}

Then by dominated convergence we have

\begin{align*}\lim_{n\rightarrow\infty}\sum_{j=0}^{\infty}a_{j,n}=\sum_{j=0}^{\infty}m^j \int_{\mathbb{R}}(P_{j+1}h(y))^2 \widetilde{\pi}(\mathrm{d} y).\end{align*}

This combined with (17) and (25) gives the second term on the right-hand side of (24).

Before giving the rth moment of $\hat{Z}_n(h)$ , we introduce some notation. For $r_1,l_1,\ldots,$ $r_k,l_k,r\in\mathbb{N}_+$ , let

(26) \begin{align}C_r(r_1,l_1,\ldots,r_k,l_k)\,:\!=\,\frac{r !}{(r_{1} !)^{l_1} \cdots(r_{k} !)^{l_k}} \cdot \frac{1}{l_{1} ! \ldots l_{k} !}.\end{align}

When $l_1 r_1+\cdots+l_k r_k=r$ and $l_1+\cdots+l_k=i$ , $C_r(r_1,l_1,\ldots,r_k,l_k)$ is the number of partitions of dividing r different elements into i indistinguishable blocks, where there are $l_1$ blocks with $r_1$ elements, $\ldots,l_k$ blocks with $r_k$ elements. For $i\geq2$ , let the sum $\sum_{r_\cdot,l_\cdot,k}$ be over all sequences of positive integers satisfying

(27) \begin{equation}\left\{\begin{array}{l}l_1 r_1+\cdots+l_k r_k=r,\\l_1+\cdots+l_k=i,\\0\lt r_1 \lt \cdots \lt r_k\leq r-i+1,\\1\leq k\leq i.\end{array}\right.\end{equation}

Let $M_h(0,x,r)\,:\!=\,Ph^{r}(x)$ and define $M_h(k_1,x,r)$ for $k_1\in\mathbb{N}$ inductively, i.e.

\begin{align*}M_h(k_1,x,r)\,:\!=\,P_{k_1+1}h^{r}(x)+ \sum_{i=2}^{r}\frac{f^{(i)}(1)}{m}&\sum_{r_\cdot,l_\cdot,k}C_{r}(r_1,l_1,\ldots,r_k,l_k)\sum_{k_2=0}^{k_1-1}m^{(i-1)k_2} \\&\times\int_{\mathbb{R}}\Bigg[\prod_{j=1}^{k}\big(M_h(k_2,y,r_j)\big)^{l_j}\Bigg]P_{k_1-k_2}(x,\mathrm{d}y),\end{align*}

with the convention that $\sum_{i=2}^r=0$ for $r\leq 1$ . The following result shows that the rth moment can be represented as a combination of the moments of order lower than r.

Lemma 10 (The moment.) Suppose that condition (C1) holds. For any $r\in\mathbb{N}_+$ and any $h\in\mathcal{B}_b^+(\mathbb{R})$ , we have

(28) \begin{align}\mathbb{E}_{\delta_x}[(\hat{Z}_n(h))^r]=\Lambda_{r,1}(n,h)+\Lambda_{r,2}(n,h),\end{align}

where $\Lambda_{r,1}(n,h)=({m^n}/{(1-f_n(0))})P_nh^r(x) $ and

\begin{align*}\Lambda_{r,2}(n,h)=\sum_{i=2}^{r}\frac{m^nf^{(i)}(1)}{m(1-f_n(0))}\sum_{r_\cdot,l_\cdot,k} C_r&(r_1,l_1,\ldots,r_k,l_k)\sum_{k_1=0}^{n-1}m^{(i-1)k_1}\\&\times\int_{\mathbb{R}}\Bigg[\prod_{j=1}^{k}(M_h(k_1,y,r_j))^{l_j}\Bigg]P_{n-k_1-1}(x,\mathrm{d}y).\end{align*}

In particular, we have

(29) \begin{align} \int_{\mathbb{R}}\mathbb{E}_{\delta_y}[(\hat{Z}_n(h))^r]P(x,\mathrm{d}y)=\frac{m^n}{1-f_n(0)}M_h(n,x,r).\end{align}

Proof. We shall give the proof by induction in $r\geq1$ . For $r=1$ , the result follows from (18). For $r=i=2$ , it is easy to see that $(r_1,l_1,k)=(1,2,1)$ is the unique positive integer solution of (27). Note that

\begin{align*}C_2(1,2)=1\quad\text{and}\quad M_h(k_1,y,1)=P_{k_1+1}h(y).\end{align*}

These combined with (20) yield that the result holds for $r=2$ . Now suppose that (28) holds for $r-1\geq 1$ . It is clearly sufficient to consider the rth moment. Let

\begin{align*}F\big(w_n^1,\ldots,w_n^r\big)=h\big(V\big(w_n^1\big)\big)\cdots h\big(V\big(w_n^r\big)\big)=h\big(V\big(\mathfrak{X}_n^1\big)\big)\cdots h\big(V\big(\mathfrak{X}_n^r\big)\big),\end{align*}

then

\begin{align*}F(u^1,\ldots,u^r)=h(V(u^1))\cdots h(V(u^r)).\end{align*}

Hence, by applying (16) for $k=n$ , we have

(30) \begin{align}\mathbb{E}_{\delta_x}[(\hat{Z}_n(h))^r]&=\mathbb{E}_{\delta_x}\bigg[\sum_{u^1,\ldots,u^r\in\hat{\mathbb{G}}_{n}}F(u^1,\ldots,u^r)\bigg] \nonumber\\&=\mathbb{Q}_{\delta_x}^{[r]}\bigg[F\big(w_n^1,\ldots,w_n^r\big)\prod_{v\in\text{{skel}}(n)\backslash\{\varnothing\}}\hat{m}_{D(\overleftarrow{v})}(|\overleftarrow{v}|,n)\bigg].\end{align}

We next express the product in the expectation of (30) according to the number of marks carried by the particles. Let $T_1^{[r]}$ be the first time at which r marks are carried by different particles. Then the event $\{T_1^{[r]}>n\}$ implies that all r marks are carried by the same particle up to time n and

\begin{align*}\big\{T_1^{[r]}\leq n\big\}=\bigcup_{i=2}^{r}&\{\text{the first time at which the}\ r\ \text{marks are carried by}\ i\ \text{different}\\&\text{particles is less than or equal to}\ n \}.\end{align*}

Furthermore, we are interested in how many different descendants carry these r markers and how many markers are carried by each descendant at the first split time. Suppose that $\{k,r_1,l_1,\ldots,r_k,l_k\}$ is a sequence of positive integers satisfying (27). Let $T^{[r]}_1(r_1,l_1,\ldots,r_k,l_k)$ be the first time at which r marks are carried by i different particles, of which there are $l_1$ particles carrying $r_1$ marks, $\ldots, l_k$ particles carrying $r_k$ marks. Therefore,

\begin{align*}\big\{T_1^{[r]}\leq n\big\}=\bigcup_{i=2}^{r} \bigcup_{r_\cdot,l_\cdot,k} \bigcup_{k_1=1}^{n} \big\{ T_1^{[r]}(r_1,l_1,\ldots,r_k,l_k)=k_1 \big\}.\end{align*}

Hence, we have

(31) \begin{align}\mathbb{E}_{\delta_x}[(\hat{Z}_n(h))^r]=&\mathbb{Q}_{\delta_x}^{[r]}\Bigg[\big(h\big(\mathfrak{X}_{n}^{1}\big)\big)^r\mathbf{1}_{\{T_1^{[r]}>n\}}\prod_{j=0}^{n-1}\hat{m}_r(j,n)\Bigg] \nonumber\\&+\sum_{i=2}^{r}\sum_{r_\cdot,l_\cdot,k}\sum_{k_1=1}^{n}\mathbb{Q}_{\delta_x}^{[r]}\bigg[h\big(\mathfrak{X}_{k_1-1,n}^{1}\big)\cdots h\big(\mathfrak{X}_{k_1-1,n}^{r}\big)\nonumber\\&\hskip24pt\times\mathbf{1}_{\{T_1^{[r]}(r_1,l_1,\ldots,r_k,l_k)=k_1\}}\prod_{v\in\text{{skel}}(n)\backslash\{\varnothing\}}\hat{m}_{D(\overleftarrow{v})}(|\overleftarrow{v}|,n)\bigg],\end{align}

where $(\mathfrak{X}_{n}^{1}\colon n\geq0)$ is the spine Markov process carrying r marks, and $(\mathfrak{X}_{k_1-1,n}^{1}\colon n\geq0),\ldots,$ $(\mathfrak{X}_{k_1-1,n}^{r}\colon n\geq0)$ are spine Markov processes characterized by the following properties.

1. (P1) These r spine Markov processes have the same trajectory up to time $k_1-1$ and split into i independent different trajectories at time $k_1$ .

2. (P2) At time $k_1$ , these r marks are carried by i different particles, of which there are $l_1$ particles carrying $r_1$ marks, $\ldots, l_k$ particles carrying $r_k$ marks.

3. (P3) From time $k_1$ to time n, the trajectories of spine Markov processes carrying $r_j$ marks are equal in law to those in the many-to- $r_j$ formula for the branching process $(\hat{Z}_{j,n}^{\theta^{k_1}}\colon 0\leq j\leq n-k_1)$ , where $\hat{Z}_{j,n}^{\theta^{k_1}}$ is defined as follows.

Suppose that $\{\hat{N}^{\theta^{k_1}}(u)\}$ is a family of independent and identically distributed integer-valued random variables with the same law $\hat{p}(k_1+|u|,n)$ . Let $\hat{\mathbb{T}}_n^{\theta^{k_1}}$ be the shift tree of $\hat{\mathbb{T}}_n$ with defining elements $\{\hat{N}^{\theta^{k_1}}(u)\}$ :

\begin{align*}\hat{\mathbb{T}}_n^{\theta^{k_1}}\,:\!=\,\bigcup_{j=0}^{n-k_1} \hat{\mathbb{G}}_{j,n}^{\theta^{k_1}} \quad\text{with}\quad\hat{\mathbb{G}}_{0,n}^{\theta^{k_1}}\,:\!=\,\{\varnothing\}\text{ and} \, \hat{\mathbb{G}}_{j+1,n}^{\theta^{k_1}}\,:\!=\,\Big\{ui\colon u\in\hat{\mathbb{G}}_{j,n}^{\theta^{k_1}},1\leq i\leq \hat{N}^{\theta^{k_1}}(u)\Big\}.\end{align*}

Then we can get a branching Markov chain $(V(u)\colon u\in\hat{\mathbb{G}}^{\theta^{k_1}}_{j,n},0\leq j\leq n-k_1)$ by endowing each u with the position V(u) in the same way as that in the construction of the branching Markov chain starting from one particle, as outlined in Section 2.1. The corresponding point process is defined by

\begin{align*}\hat{Z}_{j,n}^{\theta^{k_1}}(B)\,:\!=\,\sum_{u\in\hat{\mathbb{G}}_{j,n}^{\theta^{k_1}}} \delta_{V(u)}(B),\quad 0\leq j\leq n-k_1,\, B\in\mathcal{B}(\mathbb{R}).\end{align*}

We write $\hat{Z}_{n-k_1}^{\theta^{k_1}}\,:\!=\,\hat{Z}_{n-k_1,n}^{\theta^{k_1}}$ for simplicity.

For the first term on the right-hand side of (31), by arguments similar to those used for (22) we get

(32) \begin{align}\mathbb{Q}_{\delta_x}^{[r]}\big(T_1^{[r]}>n\big)=\prod_{j=0}^{n-1} \frac{\hat{m}_1(j,n)}{\hat{m}_r(j,n)}.\end{align}

It then follows from (12) that

\begin{align*}\mathbb{Q}_{\delta_x}\Bigg[\big(h(\mathfrak{X}_{n}^{1})\big)^r\mathbf{1}_{\{T_1^{[r]}>n\}}\prod_{j=0}^{n-1}\hat{m}_r(j,n)\Bigg]=\frac{m^n}{1-f_n(0)}P_nh^r(x),\end{align*}

which is equal to $\Lambda_{r,1}(n,h)$ . For the second term on the right-hand side of (31), note that

\begin{align*}\{T_1^{[r]}(r_1,l_1,&\ldots,r_k,l_k)=k_1\}=\{r\ \text{marks follow the same particle until time $k_1-1$;} \\&\text{at time $k_1$ they are carried by}\ i\ \text{different particles, of which there are $l_1$} \\&\text{particles carrying $r_1$ marks, $\ldots, l_k$ particles carrying $r_k$ marks} \}.\end{align*}

Under $\mathbb{Q}_{\delta_x}^{[r]}$ the particle in generation $k_1$ carrying r marks splits at time $k_1+1$ into l particles with probability $l^r\hat{p}_l(k_1,n)/\hat{m}_r(k_1,n)$ . At such a branching event, the r marks are carried by i different particles, of which there are $l_1$ particles carrying $r_1$ marks, $\ldots, l_k$ particles carrying $r_k$ marks with probability

\begin{align*}C_r(r_1,l_1,\ldots,r_k,l_k)\cdot\frac{l(l-1)\cdots(l-i+1)}{l^r},\end{align*}

where $C_r(r_1,l_1,\ldots,r_k,l_k)$ is defined by (26). By the above comments and (32), we have

\begin{align*}& \mathbb{Q}_{\delta_x}^{[r]}\big(T_1^{[r]}(r_1,l_1,\ldots,r_k,l_k)=k_1\big)\nonumber\\&\quad= C_r(r_1,l_1,\ldots,r_k,l_k)\Bigg[\sum_{l\geq1}\frac{l^r\hat{p}_l(k_1-1,n)} {\hat{m}_{r}(k_1-1,n)}\times\frac{l(l-1)\cdots(l-i+1)}{l^r}\Bigg] \prod_{j=0}^{k_1-2}\frac{\hat{m}_{1}(j,n)}{\hat{m}_{r}(j,n)}\nonumber\\&\quad= C_r(r_1,l_1,\ldots,r_k,l_k)\frac{\hat{f}_{k_1-1,n}^{(i)}(1)}{\hat{m}_{r}(k_1-1,n)} \prod_{j=0}^{k_1-2}\frac{\hat{m}_{1}(j,n)}{\hat{m}_{r}(j,n)}\nonumber\\&\quad= C_r(r_1,l_1,\ldots,r_k,l_k)\frac{f^{(i)}(1)}{\hat{m}_{r}(k_1-1,n)} \frac{(1-f_{n-k_1}(0))^i}{1-f_{n-k_1+1}(0)}\prod_{j=0}^{k_1-2} \frac{\hat{m}_{1}(j,n)}{\hat{m}_{r}(j,n)},\end{align*}

where the last equality comes from Proposition 4. Note that

\begin{align*}& \mathbb{Q}_{\delta_x}^{[r]}\Bigg[h\big(\mathfrak{X}_{k_1-1,n}^{1}\big)\cdots h\big(\mathfrak{X}_{k_1-1,n}^{r}\big)\prod_{v\in\text{{skel}}(n) \backslash\{\varnothing\}}\hat{m}_{D(\overleftarrow{v})}(|\overleftarrow{v}|,n)\Bigg]\\&\quad= \prod_{j=0}^{k_1-1}\hat{m}_r(j,n) \int_{\mathbb{R}} \mathbb{Q}_{\delta_x}^{[r]}\bigg[ h\big(\mathfrak{X}_{k_1-1,n}^{1}\big)\cdots h\big(\mathfrak{X}_{k_1-1,n}^{r}\big)\\&\hskip98pt \times\prod_{v\in\text{{skel}}(n)\backslash{\textrm{skel}}(k_1)} \hat{m}_{D(\overleftarrow{v})}(|\overleftarrow{v}|,n) \Big|\mathfrak{X}_{k_1-1,k_1-1}^{1}=\widetilde{y} \bigg] P_{k_1-1}(x,\mathrm{d}\widetilde{y})\\&\quad= \prod_{j=0}^{k_1-1}\hat{m}_r(j,n) \int_{\mathbb{R}} \Bigg[ \prod_{j=1}^{k}\Big(\int_{\mathbb{R}} \mathbb{E}_{\delta_y}\Big[\Big(\hat{Z}_{n-k_1}^{\theta^{ k_1}}(h)\Big)^{r_{j}}\Big]P(\widetilde{y},\mathrm{d}y)\Big)^{l_{j}} \Bigg] P_{k_1-1}(x,\mathrm{d}\widetilde{y}),\end{align*}

where the last equality follows from properties (P1) and (P3) of the spine Markov processes. Recall that the reproduction and motion of a particle are independent, then it is easy to infer that the second term on the right-hand side of (31) is equal to

\begin{align*} \sum_{i=2}^{r}\sum_{r_\cdot,l_\cdot,k}\sum_{k_1=1}^{n}&\mathbb{Q}_{\delta_x}^{[r]}\big(T_1^{[r]}(r_1,l_1,\ldots,r_k,l_k)=k_1\big)\Bigg[\prod_{j=0}^{k_1-1}\hat{m}_r(j,n)\Bigg] \\&\times\int_{\mathbb{R}}\Bigg[\prod_{j=1}^{k}\bigg(\int_{\mathbb{R}} \mathbb{E}_{\delta_y}\Big[\Big(\hat{Z}_{n-k_1}^{\theta^{ k_1}}(h)\Big)^{r_{j}}\Big]P(\widetilde{y},\mathrm{d}y)\bigg)^{l_{j}} \Bigg]P_{k_1-1}(x,\mathrm{d}\widetilde{y}).\end{align*}

By the definition of $\hat{Z}_{n-k_1}^{\theta^{k_1}}$ and $M_h(k_1,x,r)$ , then we use the induction hypothesis to get

\begin{align*}\int_{\mathbb{R}}\mathbb{E}_{\delta_y}\Big[\Big(\hat{Z}_{n-k_1}^{\theta^{k_1}}(h)\Big)^{r_{j}}\Big]P(\widetilde{y},\mathrm{d}y) =\frac{m^{n-k_1}}{1-f_{n-k_1}(0)}M_h\big(n-k_1,\widetilde{y},r_j\big),\quad 1\leq k_1\leq n.\end{align*}

In view of (12), we obtain

\begin{align*}& \frac{(1-f_{n-k_1}(0))^i}{1-f_{n-k_1+1}(0)}\Bigg[\prod_{j=0}^{k_1-2} \hat{m}_1(j,n)\Bigg]\Bigg[\prod_{j=1}^{k}\bigg(\frac{m^{n-k_1}}{1-f_{n-k_1}(0)}\bigg)^{l_j}\Bigg]\\&\quad= \frac{m^n}{m(1-f_n(0))} m^{(i-1)(n-k_1)}.\end{align*}

Combining the four preceding equalities gives that the second term on the right-hand side of (31) is $\Lambda_{r,2}(n,h)$ , which completes the proof.

We now give some characterizations of the Yaglom limit for GW processes, which play a very important role in studying the limit of the rth moment. Recall that $\nu_{\text{min}}$ is the Yaglom limit defined in Section 1. Let $c_1^{-1}=\varlimsup_{n\rightarrow\infty}(p_n)^{{1}/{n}}$ , then condition (C2) implies that $c_1>1$ and

\begin{align*}\sum_{n\in\mathbb{N}}p_n|s|^n \lt \infty\quad\text{for all }|s|\lt c_1.\end{align*}

Hence, the generating function $f(s)=\sum_{n\in\mathbb{N}} p_n s^n$ is well defined for $|s|<c_1$ .

Proof. (i) Since $m>0$ , the generating function f(s) is strictly increasing on $s\in[0,c_1)$ . Again, note that $m<1$ and $f(1)=1$ , then there is some $c_2 > 1$ such that $1 < f(s) < s$ for $s \in (1, c_2)$ . Iterating this inequality one can see that

\begin{align*}1 \lt f_n(s) \lt f_{n-1}(s) \lt \cdots \lt f(s) \lt s \lt c_2\quad\text{for all } n\geq 1.\end{align*}

Thus, $f_n(s) \downarrow c_{\infty}\in[1,c_2)$ . Since f is a continuous function, we may take limits through the equation $f_{n+1}(s)=f(f_n(s))$ and conclude that $c_{\infty}=f(c_{\infty})$ . Hence, $c_{\infty}=1$ follows from the fact that $c_{\infty}\lt c_2$ and that there are no roots of $s=f(s)$ in $(1,c_2)$ .

(ii) Note that

\begin{align*}\mathbb{E}[\mathrm{e}^{c_3\hat{N}_n}]=\mathbb{E}[\mathrm{e}^{c_3 N_n}\mid N_n\gt 0]=1+\frac{f_n(\mathrm{e}^{c_3})-1}{1-f_n(0)},\end{align*}

then it suffices to show that there exists $\widetilde{c}_3>1$ so that the sequence $(f_n(\widetilde{c}_3)-1)/(1-f_n(0))$ converges. For $s\in(1,c_2)$ , by Taylor’s formula of f at $s=1$ , we have

(33) \begin{align}\frac{f(s)-1}{s-1}=m+\epsilon(s),\end{align}

where

(34) \begin{align}\epsilon(s)\,:\!=\,\frac{f^{(2)}(1+\beta(s-1))}{2}(s-1), \quad \beta\in(0,1),\end{align}

Note that $\beta$ depends on s and we drop the script s when there is no confusion (a similar convention has been used in $\beta_1$ and $k_0$ below). One can easily check that $\epsilon(s)$ is monotonically increasing for $s\in(1,c_2)$ and $\lim_{s\downarrow1}\epsilon(s)=0$ . Replacing s by $f_{k-1}(s)$ in (33) yields

\begin{align*}\frac{f_k(s)-1}{f_{k-1}(s)-1}=m\Big[1+\frac{\epsilon(f_{k-1}(s))}{m}\Big],\quad s\in(1,c_2).\end{align*}

Taking the product on the both side of the above equality for $k=1,\ldots,n$ , we obtain

(35) \begin{align}\frac{f_n(s)-1}{s-1}=m^n \prod_{k=0}^{n-1}\Big[1+\frac{\epsilon(f_k(s))}{m}\Big],\quad s\in(1,c_2).\end{align}

It is clear that the sequence $(f_n(s)-1)/(m^n(s-1))$ is monotonically increasing in n and, thus, converges to a nondecreasing function $\widetilde{\varphi}(s)$ . By (34), we see that $\widetilde{\varphi}(s)<\infty$ if and only if

\begin{align*}\sum_{k=0}^{\infty} f^{(2)}(1+\beta(f_k(s)-1))(f_k(s)-1)\lt \infty.\end{align*}

Given a constant $0\lt \varepsilon\lt c_2-1$ satisfying $f^{(1)}(1+\varepsilon)\lt 1$ . For fixed $s\in(1,c_2)$ , then by Proposition 5(i), there exists $k_0\geq 1$ such that $1\lt f_k(s)\lt 1+\varepsilon$ for $k\geq k_0$ . It then follows that

\begin{align*}f^{(2)}(1+\beta(f_{k+k_0}(s)-1)) \leq f^{(2)}(1+\varepsilon),\quad k\geq0,\end{align*}

as we note that $f^{(3)}(s)\geq0$ for $s\in(1,c_2)$ . By the mean-value theorem, there exists a constant $\beta_1\in(0,1)$ such that

\begin{align*}f_{k+k_0}(s)-1=&\ f^{(1)}(1+\beta_1(f_{k+k_0-1}(s)-1)) \times(f_{k+k_0-1}(s)-1)\\\leq&\ f^{(1)}(1+\varepsilon) \times(f_{k+k_0-1}(s)-1)\\\leq&\ \varepsilon[f^{(1)}(1+\varepsilon)]^k.\end{align*}

Combining the three preceding inequalities gives $\widetilde{\varphi}(s)<\infty$ for $s\in(1,c_2)$ . Thus, it follows by (17) that under condition (C1),

\begin{align*}\lim_{n\rightarrow\infty}\frac{f_n(s)-1}{1-f_n(0)}=\frac{(s-1)\widetilde{\varphi}(s)}{\varphi(0)}<\infty,\quad s\in(1,c_2),\end{align*}

which implies the desired result.

(iii) It is easy to check that $\psi(t)<\infty$ for $t\leq 0$ . If $t>0$ , let $c_3>0$ be chosen as in Proposition 5(ii). By Fatou’s Lemma and Skorokhod’s theorem we get

\begin{align*}\psi(c_3)\leq\lim_{n\rightarrow\infty}\mathbb{E}[\mathrm{e}^{c_3\hat{N}_n}].\end{align*}

Then Proposition 5(iii) follows from Proposition 5(ii).

We continue to study the asymptotic behavior of the rth moment.

Corollary 5. (The limit of the moment.) Suppose that conditions (C1)–(C3) hold. For any $r\in\mathbb{N}_+$ and any $h\in\mathcal{B}_b^+(\mathbb{R})$ , then

(36) \begin{align} \lim_{n\rightarrow\infty}\mathbb{E}_{\delta_x}[(\hat{Z}_n(h))^r]=\Lambda_{r,1}(h)+\Lambda_{r,2}(h),\end{align}

where $\Lambda_{r,1}(h)={\widetilde{\pi}(h^r)}/{\varphi(0)}$ and

\begin{align*}\Lambda_{r,2}(h)=\sum_{i=2}^r \frac{f^{(i)}(1)}{m\varphi(0)}\sum_{r_\cdot,l_\cdot,k} &C_r(r_1,l_1,\ldots,r_k,l_k)\sum_{k_1=0}^{\infty}m^{(i-1)k_1}\\&\times\int_{\mathbb{R}}\Bigg[\prod_{j=1}^{k}(M_h(k_1,y,r_j))^{l_j}\Bigg]\widetilde{\pi}(\mathrm{d}y).\end{align*}

Moreover, let $\Lambda_r(h)\,:\!=\,\Lambda_{r,1}(h)+\Lambda_{r,2}(h)$ , then we have $\Lambda_r(h)<\infty$ .

Proof. For $r\in\mathbb{N}_+$ and $h\in\mathcal{B}_b^+(\mathbb{R})$ , suppose that $\Lambda_{r,1}(n,h)$ and $\Lambda_{r,2}(n,h)$ are defined as in Lemma 10. From (17) it is easy to show that $\lim_{n\rightarrow\infty}\Lambda_{r,1}(n,h)=\Lambda_{r,1}(h)<\infty$ under condition (C3). We next consider the limit of $\Lambda_{r,2}(n,h)$ . Let

\begin{align*}a_{i,k_1,n}(h)\,:\!=\,\mathbf{1}_{\{0\leq k_1\leq n-1\}}m^{(i-1)k_1} \int_{\mathbb{R}}\Bigg[\prod_{j=1}^{k}(M_h(k_1,y,r_j))^{l_j}\Bigg] P_{n-k_1-1}(x, \mathrm{d} y).\end{align*}

Then we have

(37) \begin{align}\lim_{n\rightarrow\infty}\Lambda_{r,2}(n,h)=&\ \frac{1}{\varphi(0)}\lim_{n\rightarrow\infty}\sum_{i=2}^{r}\frac{f^{(i)}(1)}{m}\sum_{r_\cdot,l_\cdot,k} C_r(r_1,l_1,\ldots,r_k,l_k)\sum_{k_1=0}^{\infty}a_{i,k_1,n}(h)\nonumber\\=&\ \sum_{i=2}^{r}\frac{f^{(i)}(1)}{m\varphi(0)}\sum_{r_\cdot,l_\cdot,k} C_r(r_1,l_1,\ldots,r_k,l_k)\lim_{n\rightarrow\infty}\sum_{k_1=0}^{\infty}a_{i,k_1,n}(h).\end{align}

From (29) we see that

\begin{align*}0\leq M_h(k_1,y,r_j)\leq\|h\|^{r_j} \mathbb{E}\big[\big(\hat{N}_{k_1}\big)^{r_j}\big]\frac{1-f_{k_1}(0)}{m^{k_1}}.\end{align*}

It then follows that

\begin{align*}0\leq a_{i,k_1,n}(h)\leq\|h\|^{r} m^{(i-1)k_1}\bigg(\frac{1-f_{k_1}(0)}{m^{k_1}}\bigg)^{i}\Bigg[\prod_{j=1}^{k} \big(\mathbb{E}\big[\big(\hat{N}_{k_1}\big)^{r_j}\big]\big)^{l_j}\Bigg].\end{align*}

Under conditions (C1) and (C2), from Proposition 5(ii) we have $\lim_{n\rightarrow\infty}\mathbb{E}[(\hat{N}_n)^r]<\infty$ for ${r\geq1}$ , and then $\sup_{n}\mathbb{E}[(\,\hat{\!N}_n)^r]\,:\!=\,d_r<\infty$ . In view of (17), we also have $(1-f_n(0))/m^n\leq 1$ . Thus,

(38) \begin{align}0\leq a_{i,k_1,n}(h)\leq\|h\|^{r} m^{(i-1)k_1} \Bigg[\prod_{j=1}^{k}d_{r_j}^{l_j}\Bigg].\end{align}

Note that $\sum_{k_1}m^{(i-1)k_1}<\infty$ for $2\leq i\leq r$ and

\begin{align*}\lim_{n\rightarrow\infty}a_{i,k_1,n}(h)=m^{(i-1)k_1} \int_{\mathbb{R}}\Bigg[\prod_{j=1}^k (M_h(k_1,y,r_j))^{l_j}\Bigg] \widetilde{\pi}(\mathrm{d} y)\,:\!=\,a_{i,k_1,\infty}(h)\quad\text{(under {(C3)})}.\end{align*}

Then by dominated convergence we show that $\sum_{k_1}a_{i,k_1,n}(h)$ converges to $\sum_{k_1}a_{i,k_1,\infty}(h)$ as $n\rightarrow\infty$ . This combined with (37) implies that $\lim_{n\rightarrow\infty}\Lambda_{r,2}(n,h)=\Lambda_{r,2}(h)$ . Hence, (36) holds. In addition, by (37) and (38) we see that

\begin{align*}\Lambda_{r,2}(h)\leq\frac{\|h\|^{r}}{m\varphi(0)}\sum_{i=2}^{r}\frac{f^{(i)}(1)}{1-m^{i-1}}\sum_{r_\cdot,l_\cdot,k} C_r(r_1,l_1,\ldots,r_k,l_k) \Bigg[\prod_{j=1}^{k}d_{r_j}^{l_j}\Bigg]<\infty.\end{align*}

Therefore, $\Lambda_r(h)=\Lambda_{r,1}(h)+\Lambda_{r,2}(h)<\infty$ for any $h\in\mathcal{B}_b^+(\mathbb{R})$ .

5.1. Proof of Theorem 1

Suppose that conditions (C1)–(C3) hold, then by Corollary 5 we see that

\begin{align*} \Lambda_r(h)=\lim_{n\rightarrow\infty}\mathbb{E}_{\delta_x}[(Z_n(h))^r \mid N_n>0]<\infty,\quad h\in\mathcal{B}_b^+(\mathbb{R}).\end{align*}

The following result is an application of the moment method, which describes the weak convergence of the conditional distribution of $Z_n(h)$ given $N_n>0$ .

Proof. Since $\lim_{n\rightarrow\infty}\mathbb{E}_{\delta_x}[(Z_n(h))^r\mid N_n>0]=\Lambda_r(h)<\infty$ under conditions (C1)–(C3), by [Reference Durrett8, Theorem 3.3.12] it suffices to show that

(39) \begin{align}\varlimsup_{r\rightarrow\infty}\frac{(\Lambda_{2r}(h))^{1/2r}}{2r}<\infty.\end{align}

Clearly, $\mathbb{E}_{\delta_x}[(\hat{Z}_n(h))^r]\leq \|h\|^r\mathbb{E}[(\hat{N}_n)^r]$ implies that $\Lambda_r(h)\leq\|h\|^r\Lambda_r(1)$ . Thus, we only need to prove that

(40) \begin{align}\varlimsup_{r\rightarrow\infty}\frac{(\Lambda_{2r}(1))^{1/2r}}{2r}<\infty.\end{align}

By Proposition 5(ii), we can easily show that $\sup_n\mathbb{E}[(\hat{N}_n)^r]<\infty$ for all $r\geq1$ under conditions (C1) and (C2). Recall that $\hat{N}_n$ converges in distribution to $\eta_1$ as $n\rightarrow\infty$ . Then

\begin{align*}\mathbb{E}[(\eta_1)^r]=\lim_{n \rightarrow \infty}\mathbb{E}[(\hat{N}_n)^r]=\Lambda_r(1).\end{align*}

Hence, by Taylor’s expansion of $\mathrm{e}^x$ and monotone convergence, we see that

\begin{align*}\mathbb{E}[\mathrm{e}^{t\,\eta_1}] = \mathbb{E}\bigg[ \sum_{r=0}^{\infty} \frac{(t \eta_1)^r}{r!} \bigg] = \sum_{r=0}^{\infty} \frac{t^r}{r!} \mathbb{E}[ (\eta_1)^r] = \sum_{r=0}^{\infty} \frac{t^r}{r!} \Lambda_r(1).\end{align*}

From Proposition 5(iii), conditions (C1) and (C2) imply that $\mathbb{E}[\mathrm{e}^{t\eta_1}]<\infty$ for all $t\in(-c_3,c_3)$ , where $c_3>0$ . This means the radius of convergence of $\sum_{r=0}^{\infty} (t^r \Lambda_r(1))/r!$ is greater than 0, i.e.

\begin{align*}\varlimsup_{r\rightarrow\infty} \Big(\frac{\Lambda_r(1)}{r!}\Big)^{1/r} <\infty.\end{align*}

Stirling’s formula (see [Reference Feller10, p.52]) tells us that, as $r\rightarrow\infty$ ,

\begin{align*}r! \sim r^r \mathrm{e}^{-r} \sqrt{2 \pi r}.\end{align*}

Combining the two preceding results gives (40). Thus, (39) holds.

For any $h\in\mathcal{B}_b^+(\mathbb{R})$ , suppose that conditions (C1)–(C3) hold and $F_h$ denotes the distribution of the random variable $\eta_h$ given by Lemma 11. Since $Z_n(h)$ is a $[0,\infty)$ -valued random variable, then by Lemma 11 and the continuity theorem for Laplace transforms we have

(41) \begin{align}\mathbb{E}_{\delta_x}[\mathrm{e}^{-\lambda Z_n(h)}\mid N_n>0]\xrightarrow[n\rightarrow\infty]{}\int_{[0,\infty)}\mathrm{e}^{-\lambda y}\mathrm{d}F_h(y),\quad x\in\mathbb{R},\, \lambda\geq0,\end{align}

where the distribution $F_h$ is uniquely determined by its moments $\Lambda_r(h)$ . In particular,

(42) \begin{align}\mathbb{E}_{\delta_x}[\mathrm{e}^{-Z_n(h)}\mid N_n>0]\xrightarrow[n\rightarrow\infty]{}\int_{[0,\infty)}\mathrm{e}^{-y}\mathrm{d}F_h(y),\quad x\in\mathbb{R}.\end{align}

Lemma 12. Suppose that conditions (C1)–(C3) hold. For any $x\in\mathbb{R}$ , there exists a unique probability measure $\mathbf{Q}_{\textrm{min}}$ on $\mathfrak{N}^{\circ}(\mathbb{R})$ such that

(43) \begin{align} \mathbb{P}_{\delta_x}(Z_n\in\cdot\mid N_n>0)\xrightarrow[n \rightarrow \infty]{\text{w}} \mathbf{Q}_{\textrm{min}}({\cdot}),\end{align}

and its Laplace functional is given by

(44) \begin{align} L_{\mathbf{Q}_{\textrm{min}}} [h] = \int_{[0,\infty)}\mathrm{e}^{-x}\mathrm{d}F_h(x),\quad h\in\mathcal{B}_b^+(\mathbb{R}),\end{align}

where the distribution $F_h$ is uniquely determined by its moments $\Lambda_r(h)$ .

Proof. Assume that conditions (C1)–(C3) hold. For any $x\in\mathbb{R}$ , we shall apply Lemma 5 for $\mathbf{Q}_n({\cdot})=\mathbb{P}_{\delta_x}(Z_n\in\cdot\mid N_n>0)$ to prove the existence of $\mathbf{Q}_{\textrm{min}}$ . By (42), we show that Lemma 5(i) holds and

\begin{align*}L[h] = \int_{[0,\infty)}\mathrm{e}^{-x}\mathrm{d}F_h(x),\quad h\in\mathcal{V}_0(\mathbb{R}),\end{align*}

where $F_h$ is uniquely decided by its moments $\Lambda_r(h)$ . We next prove that Lemma 5(ii) holds. Suppose that the sequence $\{h,h_1,h_2,\ldots\}\subset\mathcal{B}_b^+(\mathbb{R})$ satisfies $h_n\rightarrow h$ and $\sup_n \|h_n\|<\infty$ , then it suffices to show that $L(h_n)\rightarrow L(h)$ . Let $c_4\,:\!=\,\sup_n\|h_n\|<\infty$ . For any $r\in\mathbb{N}_+$ , by Corollary 5, we have

\begin{align*}\Lambda_r(h_n)=\Lambda_{r,1}(h_n)+\Lambda_{r,2}(h_n),\quad \Lambda_r( h)=\Lambda_{r,1}(h)+\Lambda_{r,2}(h).\end{align*}

By dominated convergence it is easy to show that $\Lambda_{r,1}(h_n)\rightarrow\Lambda_{r,1}(h)$ as $n\rightarrow\infty$ . On the other hand, by similar calculations to (38) we obtain

\begin{align*}m^{(i-1)k_1} \Bigg[\prod_{j=1}^k (M_{h_n}(k_1,y,r_j))^{l_j}\Bigg]\leq c_4^r m^{(i-1)k_1} \Bigg[\prod_{j=1}^{k}d_{r_j}^{l_j}\Bigg].\end{align*}

We conclude by dominated convergence that

\begin{align*}\lim_{n\rightarrow\infty}\Lambda_{r,2}(h_n)=\sum_{i=2}^r\frac{f^{(i)}(1)}{m\varphi(0)}\sum_{r_\cdot,l_\cdot,k} C_r&(r_1,l_1,\ldots,r_k,l_k)\sum_{k_1=0}^{\infty}m^{(i-1)k_1}\\&\times\int_{\mathbb{R}}\lim_{n\rightarrow\infty}\Bigg[\prod_{j=1}^k (M_{h_n}(k_1,y,r_j))^{l_j}\Bigg]\widetilde{\pi}(\mathrm{d}y).\end{align*}

Since $\sup_{n}\|h_n\|=c_4<\infty$ , $\mathbb{E}[(\hat{N}_{k_1})^{r_j}]<\infty$ and $h_n \rightarrow h$ , by dominated convergence we have

\begin{align*}\lim_{n\rightarrow\infty} \int_{\mathbb{R}} \mathbb{E}_{\delta_{\widetilde{y}}} \big[ \big(\hat{Z}_{k_1} (h_n)\big)^{r_j} \big] P(y, \mathrm{d}\widetilde{y} ) = \int_{\mathbb{R}} \mathbb{E}_{\delta_{\widetilde{y}}} \big[\big( \hat{Z}_{k_1} (h)\big)^{r_j} \big] P(y, \mathrm{d}\widetilde{y} ).\end{align*}

This combined with (29) implies that

\begin{align*}\lim_{n\rightarrow\infty}M_{h_n}(k_1,y,r_j)= M_{h}(k_1,y,r_j).\end{align*}

Then we get $\lim_{n\rightarrow\infty}\Lambda_{r,2}(h_n)=\Lambda_{r,2}(h)$ from the definition of $\Lambda_{r,2}(h)$ . In conclusion,

(45) \begin{align}\Lambda_r(h_n)\xrightarrow [n\rightarrow\infty]{}\Lambda_r(h),\quad r\in\mathbb{N}.\end{align}

Recall that $\Lambda_r(h_n)$ and $\Lambda_r(h)$ are the rth moment of $F_{h_n}$ and $F_h$ , respectively. Hence, in view of (39) and (45), we find that $F_{h_n}$ converges weakly to $F_h$ by using [Reference Durrett8, Theorem 3.3.12]. Then $L(h_n)\rightarrow L(h)$ follows from the fact that the function $x\mapsto\mathrm{e}^{-x}$ on $x\in[0,\infty)$ is bound continuous. Summing up, by Lemma 5, there exists a unique probability measure $\mathbf{Q}_{\text{min}}$ on $\mathfrak{N}(\mathbb{R})$ such that (43) and (44) hold.

We now prove that $\mathbf{Q}_{\text{min}}$ is the probability measure on $\mathfrak{N}^{\circ}(\mathbb{R})$ . Note that

Then we obtain the desired result.

We shall now give the proof of Theorem 1, which is an application of Lemma 12.

Proof of Theorem 1. Without loss of generality, we can assume that $\mu=\sum_{i=1}^{l}\delta_{x_i}\in\mathfrak{N}^{\circ}(\mathbb{R})$ with $l\geq1, x_i\in\mathbb{R}$ . By the construction of the branching Markov chain starting from $\mu$ , then $Z_n$ is equal in law (under $\mathbb{P}_{\mu}$ ) to $Z_{n}^{(1)}+\cdots+Z_{n}^{(l)}$ , where $Z_{n}^{(1)},\ldots,Z_{n}^{(l)}$ are mutually independent random variables satisfying

\begin{align*}\mathbb{P}_{\mu}\big(Z_n^{(i)}\in\cdot\ \big)=\mathbb{P}_{\delta_{x_i}}(Z_n\in\cdot\ )\quad\text{for}\ i\in\{1,\ldots,l\}.\end{align*}

Let $N_n^{(i)}\,:\!=\,Z_n^{(i)}(\mathbb{R})$ , then $N_n$ is equal in law (under $\mathbb{P}_{\mu}$ ) to $N_{n}^{(1)}+\cdots+N_{n}^{(l)}$ and $N_{n}^{(1)},\ldots,N_{n}^{(l)}$ are independent copies of $N_n$ . Hence, for any $h\in\mathcal{B}_b^+(\mathbb{R})$ ,

(46) \begin{align}&\mathbb{E}_\mu\big[\mathrm{e}^{-Z_n(h)}\mid N_n > 0\big]\nonumber\\&\quad= \frac{\mathbb{E}_\mu\big[\mathrm{e}^{- Z_n(h)}\big]-\mathbb{P}_\mu(N_n=0)}{\mathbb{P}_\mu(N_n> 0)}\nonumber\\&\quad= \frac{\mathbb{E}_\mu\big[\prod_{i=1}^{l} \mathrm{e}^{-Z_n^{(i)}(h)}\big]-\mathbb{P}_\mu\big(\sum_{i=1}^{l} N_n^{(i)}=0\big)}{1-\mathbb{P}_\mu\big(\sum_{i=1}^{l} N_n^{(i)}=0\big)}\nonumber\\&\quad= \frac{\prod_{i=1}^{l}\mathbb{E}_{\delta_{x_i}}\big[ \mathrm{e}^{-Z_n(h)} \big]-(\mathbb{P}(N_n=0))^l}{1-(\mathbb{P}(N_n=0))^l}\nonumber\\&\quad= \frac{\prod_{i=1}^{l}\mathbb{E}_{\delta_{x_i}}\big[ \mathrm{e}^{-Z_n(h)}\big]-\big(\mathbb{E}\big[\mathrm{e}^{-Z_n(h)}\big]\big)^l} {1-(\mathbb{P}(N_n=0))^l}+\frac{\big(\mathbb{E}\big[\mathrm{e}^{-Z_n(h)}\big] \big)^l-(\mathbb{P}(N_n=0))^l}{1-(\mathbb{P}(N_n=0))^l}.\end{align}

We shall prove the first term on the right-hand side of (46) tends to 0 as $n\rightarrow\infty$ . Using [Reference Durrett8, Lemma 3.4.3] and the equality

(47) \begin{align}a^l-b^l=(a-b)\big(a^{l-1}+a^{l-2} b+\cdots+b^{l-1}\big),\end{align}

we obtain

\begin{align*}& \frac{\prod_{i=1}^{l}\mathbb{E}_{\delta_{x_i}}\big[ \mathrm{e}^{-Z_n(h)}\big]-\big(\mathbb{E}\big[\mathrm{e}^{-Z_n(h)}\big]\big)^l} {1-(\mathbb{P}(N_n=0))^l}\\&\quad\leq \frac{\sum_{i=1}^{l} \big|\mathbb{E}_{\delta_{x_i}}\big[\mathrm{e}^{-Z_n(h)}\big]-\mathbb{E} \big[\mathrm{e}^{-Z_n(h)}\big]\big|}{(1-\mathbb{P}(N_n=0)) (1+\mathbb{P}(N_n=0)+\cdots+(\mathbb{P}(N_n=0))^{l-1})}\\&\quad\leq \bigg[\sum_{i=0}^{l-1}(\mathbb{P}(N_n=0))^{i}\bigg]^{-1}\sum_{i=1}^{l}\big|\mathbb{E}_{\delta_{x_i}}\big[\mathrm{e}^{-Z_n(h)}\mid N_n>0\big]-\mathbb{E}\big[\mathrm{e}^{-Z_n(h)}\mid N_n>0\big]\big|.\end{align*}

From Lemmas 4 and 12, under conditions (C1)–(C3), there exists a unique probability measure $\mathbf{Q}_{\text{min}}$ on $\mathfrak{N}^{\circ}(\mathbb{R})$ such that

(48) \begin{align}\lim_{n\rightarrow\infty}\mathbb{E}_{\delta_ x}\big[\mathrm{e}^{-Z_n(h)}\mid N_n>0\big]=L_{\mathbf{Q}_{\text{min}}}[h]\quad\text{for all }x\in\mathbb{R}.\end{align}

Then the first term of (46) tends to 0 by the fact that $\lim_{n\rightarrow\infty}\mathbb{P}(N_n=0)=1$ .

For the second term on the right-hand side of (46), by (47) we have

\begin{align*}& \frac{\big(\mathbb{E}\big[\mathrm{e}^{-Z_n(h)}\big]\big)^l-(\mathbb{P}(N_n=0) )^l}{1-(\mathbb{P}(N_n=0))^l}\\&\quad= \frac{\mathbb{E}\big[\mathrm{e}^{-Z_{n}(h)}\big]- \mathbb{P}(N_n=0)}{ 1- \mathbb{P}(N_n=0) } \, \frac{\sum_{i=0}^{l-1}\big(\mathbb{E}\big[\mathrm{e}^{-Z_{n}(h)}\big]\big)^i (\mathbb{P}(N_n=0))^{l-i-1}}{\sum_{i=0}^{l-1}(\mathbb{P}(N_n=0))^i}\\&\quad= \mathbb{E}\big[\mathrm{e}^{ -Z_{n} (h)}\mid N_n>0\big] \frac{\sum_{i=0}^{l-1}\big(\mathbb{E}\big[\mathrm{e}^{-Z_{n}(h)}\big]\big)^i (\mathbb{P}(N_n=0))^{l-i-1}}{\sum_{i=0}^{l-1}(\mathbb{P}(N_n=0))^i}.\end{align*}

Note that both $\mathbb{P}(N_n=0)$ and $\mathbb{E}[\mathrm{e}^{-Z_n(h)}]$ converge to 1 as $n\rightarrow\infty$ . Therefore, by (48), the right-hand side of the above equality tends to $L_{\mathbf{Q}_{\text{min}}}[h]$ .

5.2. Proofs of Corollaries 1 and 2

Proof of Corollary 1. For fixed $l\in\mathbb{N}$ , $\mu\in\mathfrak{N}^{\circ}(\mathbb{R})$ and $h\in\mathcal{B}_b^+(\mathbb{R})$ , we have

(49) \begin{align}& \mathbb{E}_\mu \big[\mathrm{e}^{-Z_n(h)}\mid N_{n+l}>0\big] \nonumber\\&\quad= \frac{\mathbb{P}_\mu(N_n>0)}{\mathbb{P}_\mu(N_{n+l}>0)} \, \frac{\mathbb{E}_\mu \big[\mathrm{e}^{- Z_n(h)} ; N_{n+l}>0 \big]}{\mathbb{P}_\mu(N_n>0)} \nonumber\\&\quad= \frac{\mathbb{P}_\mu(N_n>0)}{\mathbb{P}_\mu(N_{n+l}>0)} \mathbb{E}_\mu \big[\mathrm{e}^{- Z_n(h)} \big(1-f_l(0)^{N_n}\big)\mid N_n>0\big]\quad \text{(by the branching property)} \nonumber\\&\quad= \frac{\mathbb{P}_\mu(N_n>0)}{\mathbb{P}_\mu(N_{n+l}>0)} \big( \mathbb{E}_\mu \big[\mathrm{e}^{- Z_n(h)}\mid N_n>0\big] - \mathbb{E}_\mu \big[\mathrm{e}^{- Z_n(h)} f_l(0)^{N_n}\mid N_n>0\big] \big).\end{align}

From Lemma 4 and Theorem 1, we see that, under conditions (C1)–(C3),

(50) \begin{align}\lim_{n\rightarrow\infty} \mathbb{E}_{\mu} \big[ \mathrm{e}^{-Z_n(h)}\mid N_n>0 \big] = L_{\mathbf{Q}_{\text{min}}}[h],\quad h\in\mathcal{B}_b^+(\mathbb{R}),\end{align}

where $L_{\mathbf{Q}_{\text{min}}}$ is the Laplace functional of $\mathbf{Q}_{\text{min}}$ . Since $f_l(0)\geq p_0>0$ , by replacing h with $h-\log f_l(0)$ in the above equality we obtain

\begin{align*}\lim_{n\rightarrow\infty} \mathbb{E}_{\mu} [ \mathrm{e}^{-Z_n(h)} f_l(0)^{N_n} \mid N_n>0 ] = L_{\mathbf{Q}_{\text{min}}}[h-\log f_l(0)].\end{align*}

Note that $f_n(0)\rightarrow1$ as $n\rightarrow\infty$ implies that ${ (1-f_n(0)^{\mu(\mathbb{R})} )}/{ (1-f_n(0) )}\rightarrow\mu(\mathbb{R})$ , then

\begin{align*}\frac{\mathbb{P}_\mu(N_n>0)}{\mathbb{P}_\mu(N_{n+l}>0)} = \frac{1-f_n(0)^{\mu(\mathbb{R})}}{1-f_n(0)} \, \frac{1-f_n(0)}{1-f_l(f_n(0))} \, \frac{1-f_{n+l}(0)}{1-f_{n+l}(0)^{\mu(\mathbb{R})}} \xrightarrow[n\rightarrow\infty]{} m^{-l}.\end{align*}

Combining the four preceding equalities gives

(51) \begin{align}\lim_{n\rightarrow\infty} \mathbb{E}_\mu \big[\mathrm{e}^{-Z_n(h)}\mid N_{n+l}>0\big] = m^{-l} \big( L_{\mathbf{Q}_{\text{min}}}[h]- L_{\mathbf{Q}_{\text{min}}}[h-\log f_l(0)] \big).\end{align}

Let $h_n,h\in\mathcal{B}_b^+(\mathbb{R})$ satisfy $h_n\rightarrow h$ and $\sup_n\|h_n\|<\infty$ . By dominated convergence it is easy to show that

(52) \begin{align}\lim_{n\rightarrow\infty} L_{\mathbf{Q}_{\text{min}}}[h_n] = L_{\mathbf{Q}_{\text{min}}}[h],\quad \lim_{n\rightarrow\infty} L_{\mathbf{Q}_{\text{min}}}[h_n-\log f_l(0)] = L_{\mathbf{Q}_{\text{min}}}[h-\log f_l(0)].\end{align}

In view of (51) and (52), by Lemma 5, there exists a unique probability measure $\mathbf{Q}_{(l)}$ on $\mathfrak{N}(\mathbb{R})$ such that

\begin{align*}\mathbb{P}_{\mu} (Z_n\in\cdot \mid N_{n+l}>0) \xrightarrow[n\rightarrow\infty]{\text{w}} \mathbf{Q}_{(l)}({\cdot})\end{align*}

and its Laplace functional satisfies $L_{\mathbf{Q}_{(l)}} [h] = m^{-l} ( L_{\mathbf{Q}_{\text{min}}}[h]- L_{\mathbf{Q}_{\text{min}}}[h-\log f_l(0)])$ . Note that

(53) \begin{align}\mathbf{Q}_{(l)}(\{ \mu\in\mathfrak{N}(\mathbb{R})\colon \mu=0 \})&= \lim_{\lambda\rightarrow\infty} \int_{\mathfrak{N}(\mathbb{R})} \mathrm{e}^{-\lambda \mu(\mathbb{R})} \mathbf{Q}_{(l)}(\mathrm{d}\mu)\nonumber\\&= \lim_{\lambda\rightarrow\infty} L_{\mathbf{Q}_{(l)}} [\lambda] \nonumber\\&= m^{-l} \lim_{\lambda\rightarrow\infty} ( L_{\mathbf{Q}_{\text{min}}}[\lambda]- L_{\mathbf{Q}_{\text{min}}}[\lambda-\log f_l(0)] ),\end{align}

where for a constant $\lambda$ , we write $L_{\mathbf{Q}_{(l)}} [\lambda]$ for the action of $L_{\mathbf{Q}_{(l)}}$ on a constant function $h(x) \equiv \lambda$ . Since $\mathbf{Q}_{\text{min}}$ is a probability measure on $\mathfrak{N}^{\circ}(\mathbb{R})$ , we have

\begin{align*}&\lim_{\lambda\rightarrow\infty} L_{\mathbf{Q}_{\text{min}}}[\lambda]=\mathbf{Q}_{\text{min}}(\{ \mu\in\mathfrak{N}(\mathbb{R})\colon \mu=0 \})=0\end{align*}

and $\lim_{\lambda\rightarrow\infty} L_{\mathbf{Q}_{\text{min}}}[\lambda-\log f_l(0)]=0$ similarly. These combined with (53) show that $\mathbf{Q}_{(l)}$ is a probability measure on $\mathfrak{N}^{\circ}(\mathbb{R})$ .

Proof of Corollary 2. (i) For any $\mu\in\mathfrak{N}^{\circ}(\mathbb{R})$ and any $h\in\mathcal{B}_b^+(\mathbb{R})$ , in view of (49) we get

(54) \begin{align}& \mathbb{E}_\mu \big[\mathrm{e}^{-Z_n(h)}\mid N_{n+l}>0\big] \nonumber\\&\quad= \frac{\mathbb{P}_\mu(N_n>0)}{\mathbb{P}_\mu(N_{n+l}>0)} \mathbb{E}_\mu \big[\mathrm{e}^{- Z_n(h)} \big(1-f_l(0)^{N_n}\big) \mid N_n>0\big] \nonumber\\&\quad= \frac{1-f_l(0)}{1-f_{n+l}(0)^{\mu(\mathbb{R})}} \mathbb{E}_{\mu} \bigg[ \frac{\mathrm{e}^{- Z_n(h)} (1-f_l(0)^{N_n})}{1-f_l(0)}\mathbf{1}_{\{ N_{n}>0 \}} \bigg].\end{align}

It is easy to check that

(55) \begin{align}\frac{1-f_l(0)}{1-f_{n+l}(0)^{\mu(\mathbb{R})}} = \frac{1-f_l(0)}{1-f_n( f_l(0) )} \, \frac{1-f_{n+l}(0)}{1-f_{n+l}(0)^{\mu(\mathbb{R})}} \xrightarrow[l\rightarrow\infty]{} \frac{1}{m^n \mu(\mathbb{R})}.\end{align}

We first consider the expectation on the right-hand side of (54). By (47) we have

\begin{align*}0\leq \frac{\mathrm{e}^{- Z_n(h)} \big(1-f_l(0)^{N_n}\big)}{1-f_l(0)} \mathbf{1}_{\{ N_{n}>0 \}} = \mathrm{e}^{- Z_n(h)} \big( f_l(0)^{N_n-1}+\cdots+1 \big) \leq N_n.\end{align*}

Since $(1-f_l(0)^{N_n})/(1-f_l(0))\rightarrow N_n$ as $l\rightarrow\infty$ and

(56) \begin{align}\mathbb{E}_{\mu} [N_n] = \mu(\mathbb{R})m^n<\infty,\end{align}

by dominated convergence we obtain

\begin{align*}\lim_{l\rightarrow\infty} \mathbb{E}_{\mu} \bigg[ \frac{\mathrm{e}^{- Z_n(h)} \big(1-f_l(0)^{N_n}\big)}{1-f_l(0)} \mathbf{1}_{\{ N_{n}>0 \}} \bigg] = \mathbb{E}_{\mu} \big. \big[ N_n \mathrm{e}^{- Z_n(h)} \mathbf{1}_{\{ N_{n}>0 \}} \big] = \mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- Z_n(h)} \big].\end{align*}

This together with (54) and (55) yields

(57) \begin{align}\lim_{l\rightarrow\infty} \mathbb{E}_\mu\big[\mathrm{e}^{-Z_n(h)} \mid N_{n+l}>0\big] = \frac{\mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- Z_n(h)} \big]}{m^n \mu(\mathbb{R})} .\end{align}

Let $h_r,h\in\mathcal{B}_b^+(\mathbb{R})$ satisfy $\sup_r\|h_r\|<\infty$ and $\lim_{r\rightarrow\infty}h_r= h$ . Note that $Z_n$ is a finite point process, then by dominated convergence it is easy to show that $Z_n(h_r)$ converges almost surely to $Z_n(h)$ as $r\rightarrow\infty$ . Hence, $\lim_{r\rightarrow\infty}N_n \mathrm{e}^{-Z_n(h_r)}= N_n \mathrm{e}^{-Z_n(h)}$ almost surely. We also note that $0\leq N_n \mathrm{e}^{-Z_n(h_r)} \leq N_n$ . Therefore, by (56) and dominated convergence, we obtain

\begin{align*}\lim_{r\rightarrow\infty} \mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- Z_n(h_r)} \big] = \mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- Z_n(h)} \big].\end{align*}

According to Lemma 5, the above equality combined with (57) implies that there exits a unique probability measure $\mathbf{Q}_{\{n\}}$ on $\mathfrak{N}(\mathbb{R})$ such that

(58) \begin{align}\mathbb{P}_{\mu} (Z_n\in\cdot \mid N_{n+l}>0) \xrightarrow[l\rightarrow\infty]{\text{w}} \mathbf{Q}_{\{n\}}({\cdot})\end{align}

and its Laplace functional is given by

(59) \begin{align}L_{\mathbf{Q}_{\{n\}}} [h] = \big(m^n \mu(\mathbb{R})\big)^{-1} \mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- Z_n(h)} \big],\quad h\in\mathcal{B}_b^+(\mathbb{R}).\end{align}

By dominated convergence it is simple to get

\begin{align*}\mathbf{Q}_{\{n\}}(\{ \mu\in\mathfrak{N}(\mathbb{R})\colon \mu=0 \})= \lim_{\lambda\rightarrow\infty} L_{\mathbf{Q}_{\{n\}}} [\lambda]= \lim_{\lambda\rightarrow\infty} (m^n \mu(\mathbb{R}))^{-1} \mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- \lambda N_n} \big]=0.\end{align*}

Summing up, there is a unique probability measure $\mathbf{Q}_{\{n\}}$ on $\mathfrak{N}^{\circ}(\mathbb{R})$ satisfying (58) and (59).

We next consider the asymptotic behavior of $\mathbf{Q}_{\{n\}}$ as $n\rightarrow\infty$ . When h is replaced by $\lambda_1+\lambda_2 h$ in (50), we see that under conditions (C1)–(C3), for any $\lambda_1,\lambda_2\geq0$ ,

\begin{align*}\lim_{n \rightarrow \infty} \mathbb{E}_\mu [\mathrm{e}^{-\lambda_1 N_n-\lambda_2 Z_n(h)}\mid N_n>0] = L_{\mathbf{Q}_{\text{min}}} [\lambda_1+\lambda_2 h]= \int_{\mathfrak{N}^{\circ}(\mathbb{R})} \mathrm{e}^{-\nu(\lambda_1+\lambda_2 h)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu).\end{align*}

Suppose that $\{({\varkappa},{\varkappa}_h);\mathbb{P}\}$ is a random vector with the Laplace transform

(60) \begin{align}\mathbb{E} [ \mathrm{e}^{-\lambda_1 \varkappa - \lambda_2 \varkappa_h} ] = \int_{\mathfrak{N}^{\circ}(\mathbb{R})} \mathrm{e}^{-\nu(\lambda_1+\lambda_2 h)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu),\quad \lambda_1,\lambda_2\geq0.\end{align}

It then follows that

\begin{align*}\mathbb{P}_{\mu} ( ({N}_n, {Z}_n(h))\in\cdot \mid N_n>0) \xrightarrow[n\rightarrow\infty]{{\text{w}}} \mathbb{P} ( ({\varkappa},{\varkappa}_h) \in \cdot ).\end{align*}

Hence, $\varkappa$ is equal in law to the Yaglom limit of the GW process. Note that $\lim_{n\rightarrow\infty} \mathbb{E}_{\mu}[N_n \mid N_n>0] = 1/\varphi(0) <\infty$ under condition (C1). Together with the bound $0\leq N_n \mathrm{e}^{- Z_n(h)}\leq N_n$ , we apply Skorokhod’s theorem and generalized dominated convergence (see [Reference Folland15, Chapter 2, Exercise 20]) to obtain

\begin{align*}\lim_{n \rightarrow \infty} \mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- Z_n(h)} \mid N_n>0 \big] = \mathbb{E} \big[ \varkappa \mathrm{e}^{\varkappa_h} \big].\end{align*}

By differentiating (60) with respect to $\lambda_1$ at $\lambda_1=0, \lambda_2=1$ and using dominated convergence, we obtain

\begin{align*}\mathbb{E} [ \varkappa \mathrm{e}^{\varkappa_h} ] = \int_{\mathfrak{N}^{\circ}(\mathbb{R})} \nu(\mathbb{R}) \mathrm{e}^{-\nu(h)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu).\end{align*}

In summary, we have

\begin{align*}\lim_{n \rightarrow \infty} \mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- Z_n(h)} \mid N_n>0 \big] = \int_{\mathfrak{N}^{\circ}(\mathbb{R})} \nu(\mathbb{R}) \mathrm{e}^{-\nu(h)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu).\end{align*}

Under condition (C1), (17) implies that $m^{-n}\mathbb{P}_{\mu}(N_n>0)\rightarrow \mu(\mathbb{R})\varphi(0)$ as $n\rightarrow\infty$ . Consequently, letting $n\rightarrow\infty$ in (59) we see that

(61) \begin{align}\lim_{n\rightarrow\infty} L_{\mathbf{Q}_{\{n\}}} [h]&= \lim_{n\rightarrow\infty} \frac{\mathbb{P}_{\mu}(N_n>0)}{m^n \mu(\mathbb{R})} \mathbb{E}_{\mu} \big[ N_n \mathrm{e}^{- Z_n(h)} \mid N_n>0 \big] \nonumber \\&=\varphi(0)\int_{\mathfrak{N}^{\circ}(\mathbb{R})} \nu(\mathbb{R}) \mathrm{e}^{-\nu(h)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu).\end{align}

Suppose that the sequence $\{h,h_1,h_2,\ldots\}\subset\mathcal{B}_b^+(\mathbb{R})$ satisfies $h_n\rightarrow h$ and $\sup_n\|h_n\|<\infty$ . Then, for all $\nu\in\mathfrak{N}(\mathbb{R})$ , we have $\lim_{n\rightarrow\infty}\nu(h_n)=\nu(h)$ . On the other hand, by Remark 11 we see that under conditions (C1) and (C2), $\mathbb{E}[\eta_1] = 1/\varphi(0)$ . By applying Theorem 1 and the uniqueness of the Yaglom limit of the GW process, it follows that

(62) \begin{align}\int_{\mathfrak{N}^{\circ}(\mathbb{R})} \nu(\mathbb{R}) \mathbf{Q}_{\text{min}}(\mathrm{d} \nu)=\mathbb{E}[\eta_1] =\frac{1}{\varphi(0)}<\infty.\end{align}

Therefore, by dominated convergence we obtain

(63) \begin{align}\lim_{n\rightarrow\infty} \int_{\mathfrak{N}^{\circ}(\mathbb{R})} \nu(\mathbb{R}) \mathrm{e}^{-\nu(h_n)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu) = \int_{\mathfrak{N}^{\circ}(\mathbb{R})} \nu(\mathbb{R}) \mathrm{e}^{-\nu(h)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu).\end{align}

Using (61), (63), and Lemma 5 it is simple to deduce that there exits a unique probability measure $\mathbf{Q}_{\infty}$ on $\mathfrak{N}^{\circ}(\mathbb{R})$ such that $\mathbf{Q}_{\{n\}}\stackrel{\text{w}}{\longrightarrow}\mathbf{Q}_{\infty}$ as $n\rightarrow\infty$ and

\begin{align*}L_{\mathbf{Q}_{\infty}} [h] = \varphi(0)\int_{\mathfrak{N}^{\circ}(\mathbb{R})} \nu(\mathbb{R}) \mathrm{e}^{-\nu(h)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu),\quad h\in\mathcal{B}_b^+(\mathbb{R}).\end{align*}

Then Corollary 2(i) follows from (58).

(ii) We assume that conditions (C1)–(C3) hold. Suppose that $\mathbf{Q}_{(l)}$ is given in Corollary 1 and its Laplace functional $L_{\mathbf{Q}_{(l)}}$ is defined by (7). Then

\begin{align*}\lim_{l\rightarrow\infty} L_{\mathbf{Q}_{(l)}} [h]&= \lim_{l\rightarrow\infty} \frac{ L_{\mathbf{Q}_{\text{min}}}[h]- L_{\mathbf{Q}_{\text{min}}}[h-\log f_l(0)] }{m^{l} }\\&= \lim_{l\rightarrow\infty} \int_{\mathfrak{N}^{\circ}(\mathbb{R})} \mathrm{e}^{-\nu(h)} \frac{1-f_l(0)^{\nu(\mathbb{R})}}{m^l} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu).\end{align*}

Note that

\begin{gather*} \frac{1-f_l(0)^{\nu(\mathbb{R})}}{m^l} = \frac{1-f_l(0)^{\nu(\mathbb{R})}}{1-f_l(0)} \, \frac{1-f_l(0)}{m^l} \xrightarrow[l\rightarrow\infty]{} \nu(\mathbb{R}) \varphi(0), \\ \frac{1-f_l(0)^{\nu(\mathbb{R})}}{m^l} = \frac{1-f_l(0)^{\nu(\mathbb{R})}}{1-f_l(0)} \, \frac{1-f_l(0)}{m^l} \leq \nu(\mathbb{R}).\end{gather*}

In view of (62) and the above results, by dominated convergence we get

\begin{align*}\lim_{l\rightarrow\infty} L_{\mathbf{Q}_{(l)}} [h] = \varphi(0)\int_{\mathfrak{N}^{\circ}(\mathbb{R})} \nu(\mathbb{R}) \mathrm{e}^{-\nu(h)} \mathbf{Q}_{\text{min}}(\mathrm{d} \nu) = L_{\mathbf{Q}_{\infty}} [h],\quad h\in\mathcal{B}_b^+(\mathbb{R}),\end{align*}

which implies Corollary 2(ii).

6.1. Proof of Proposition 2

Proof of Proposition 2. Assume that $\mathbf{Q}$ is a probability measure on $\mathfrak{N}^{\circ}(\mathbb{R})$ . Then by Lemma 2, $\mathbf{Q}$ is the QSD of the process Z if and only if its PGFL $G_{\mathbf{Q}}$ satisfies

(67) \begin{align}(\mathbf{Q}\mathbb{E})\big[\mathrm{e}^{Z_n(\!\log h)}\mid N_n>0\big]=G_{\mathbf{Q}}[h],\quad h\in\mathcal{V}_0(\mathbb{R}).\end{align}

Hence, we only need to prove that (9) is equivalent to (67).

If (9) holds then taking $h\equiv1$ in (9) yields the fact that $G_{\mathbf{Q}}[f(0)]=1-m^{\alpha}$ . For $n\in\mathbb{N}$ and $h\in\mathcal{V}_0(\mathbb{R})$ , replacing h by $G_{n-1}[h]$ in (9), we can then conclude by Proposition 1 that

\begin{align*}G_{\mathbf{Q}} [ G_n[h]({\cdot}) ] = m^{\alpha} G_{\mathbf{Q}}[G_{n-1}[h]({\cdot})]+G_{\mathbf{Q}}[f(0)].\end{align*}

By iterating this equality it is easy to see that

(68) \begin{align}G_{\mathbf{Q}} [ G_n[h]({\cdot}) ] =m^{\alpha n} G_{\mathbf{Q}}[h]+1-m^{\alpha n} .\end{align}

Therefore, by a combination of (65) and (68), we obtain

(69) \begin{align}(\mathbf{Q} \mathbb{P}) (N_n=0) &= \lim_{\lambda\rightarrow\infty} G_{\mathbf{Q}}[ f_n(\mathrm{e}^{ -\lambda}) ] \nonumber\\&= m^{\alpha n} \lim_{\lambda\rightarrow\infty} G_{\mathbf{Q}}[\mathrm{e}^{ -\lambda}]+1-m^{\alpha n} \nonumber\\&= 1-m^{\alpha n}\in(0,1).\end{align}

Hence,

which implies that (67) holds.

Suppose that a probability measure $\mathbf{Q}$ on $\mathfrak{N}^{\circ}(\mathbb{R})$ is a QSD of the process Z, then

\begin{align*}(\mathbf{Q} \mathbb{E}) \big[ \mathrm{e}^{Z_n(\!\log h)} \big] -(\mathbf{Q} \mathbb{P}) (N_n=0) = (\mathbf{Q} \mathbb{P}) (N_n>0) G_{\mathbf{Q}}[h].\end{align*}

In view of (64) and (65) we get

(70) \begin{align}G_{\mathbf{Q}}[G_n[h] ({\cdot}) ]-G_{\mathbf{Q}}[f_n(0)]=(\mathbf{Q} \mathbb{P}) (N_n>0) G_{\mathbf{Q}}[h].\end{align}

By a proof similar to that of [Reference Méléard and Villemonais36, Proposition 2], there exists $\alpha\in(0,\infty)$ such that $(\mathbf{Q} \mathbb{P} )(N_n>0)=m^{\alpha n}$ . Then (9) follows by taking $n=1$ in (70).

From the proof of Proposition 2 we immediately get the following result.

Corollary 6. For any $\alpha\in(0,\infty)$ , the probability measure $\mathbf{Q}_{\alpha}$ on $\mathfrak{N}^{\circ}(\mathbb{R})$ is a QSD of Z with eigenvalue $m^{\alpha}$ if and only if its PGFL $G_{\mathbf{Q}_{\alpha}}$ satisfies the following equation:

(71) \begin{align}G_{\mathbf{Q}_{\alpha}}[G_n[h]({\cdot})]-G_{\mathbf{Q}_{\alpha}} [f_n(0)]=m^{\alpha n} G_{\mathbf{Q}_{\alpha}}[h ],\quad n\in\mathbb{N},\, h\in\mathcal{V}_0(\mathbb{R}).\end{align}

The relationship between the Yaglom limit and the QSD of Z is described as follows.

Proof. By (66) we obtain, for any $x\in\mathbb{R}$ ,

\begin{align*}G_{\mathbf{Q}_{\text{min}}}[h]=&\ \lim _{n \rightarrow \infty} \frac{G_{n+1}[h](x)- f_{n+1}(0) }{1-f_{n+1}(0)} \nonumber \\=&\ \lim _{n \rightarrow \infty} \bigg(\frac{G_{n+1}[h](x)-f_n(0)}{1-f_n(0)}-\frac{f_{n+1}(0)-f_n(0)}{1-f_n(0)}\bigg) \frac{1-f_n(0)}{1-f_{n+1}(0)}.\end{align*}

Note that $f_{n+1}(0)=G_n[f(0)]$ and $(1-f_n(0))/(1-f_{n+1}(0))\rightarrow 1/m$ as $n\rightarrow\infty$ . Therefore,

\begin{align*}G_{\mathbf{Q}_{\text{min}}}[h]=& \frac{1}{m} \lim _{n \rightarrow \infty} \bigg\{ \frac{G_{n+1}[h](x) -f_n(0)}{1-f_n(0)} - \frac{G_n[ f(0) ]-f_n(0)}{1-f_n(0)} \bigg\}\\=& \frac{1}{m} \lim _{n \rightarrow \infty} \bigg\{ \frac{G_n[ G[h]({\cdot})](x) -f_n(0)}{1-f_n(0)} - \frac{G_n[ f(0) ]-f_n(0)}{1-f_n(0)} \bigg\}\\=& \frac{1}{m} ( G_{\mathbf{Q}_{\text{min}}} [ G[h]({\cdot}) ] - G_{\mathbf{Q}_{\text{min}}} [ f(0) ] ),\end{align*}

where the second equality and the last equality follow from Proposition 1 and (66), respectively. Then we get the desired result.

6.2. Proof of Theorem 2

6.2.1. Proof of the necessity part of Theorem 2

For any $\alpha\in(0,\infty)$ , suppose that a probability measure $\mathbf{Q}_{\alpha}$ on $\mathfrak{N}^{\circ}(\mathbb{R})$ is a QSD of Z with eigenvalue $m^{\alpha}$ . We denote by $G_{\mathbf{Q}_{\alpha}}$ its PGFL. Let $q_n\,:\!=\,\mathbb{P}(N_n>0)=1-f_n(0)$ . Then by Corollary 6 we obtain, for any $h\in\mathcal{V}_0(\mathbb{R})$ ,

where the last equality comes from the independence. For any $n\in\mathbb{N}$ , we can define the measure $\gamma_n$ by $\gamma_n(B) = m^{-\alpha n} \mathbf{Q}_\alpha(q_n^{-1} B)$ for $B\in\mathcal{B}(\mathfrak{M}^{\circ}(\mathbb{R}))$ . Then a simple calculation shows that

(72) \begin{align}G_{\mathbf{Q}_{\alpha}} [h]&= \int_{\mathfrak{M}^{\circ}(\mathbb{R})} \bigg[\exp \bigg(q_n^{-1} \int_{\mathbb{R}} \log G_n[h](x) \nu(\mathrm{d} x)\bigg)-(1-q_n)^{q_n^{-1} \nu(\mathbb{R})}\bigg] \gamma_n(\mathrm{d} \nu) \nonumber\\&= \int_{\mathfrak{M}^{\circ}(\mathbb{R})} U_{h,n}(\nu) \widetilde{\gamma}_n(\mathrm{d} \nu),\end{align}

where $U_{h,n}$ is a functional on $\mathfrak{M}^{\circ}(\mathbb{R})$ defined by

(73) \begin{align}U_{h, n}(\nu)\,:\!=\,\frac{ \exp (q_n^{-1} \int_{\mathbb{R}} \log G_n[h](x) \nu(\mathrm{d} x))-(1-q_n)^{q_n^{-1} \nu(\mathbb{R})} }{\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}}\end{align}

and $\widetilde{\gamma}_n$ is a measure on $\mathfrak{M}(\mathbb{R})$ given by

(74) \begin{align}\widetilde{\gamma}_n(\mathrm{d} \nu)\,:\!=\,\left\{\begin{array}{l@{\quad}l} \nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})} \gamma_n(\mathrm{d} \nu) & \text{if}\ \nu \in \mathfrak{M}^{\circ}(\mathbb{R}), \\ 0 & \text{if}\ \nu=\mathbf{0}.\end{array}\right.\end{align}

In the proof of the necessity part of Theorem 2, we use the following three propositions whose proofs are postponed to Section 6.3.

Proposition 6. Assume that conditions (C1)–(C3) hold. Suppose that $U_{h,n}$ is defined by (73). Then, for any $h\in\mathcal{V}_0(\mathbb{R})$ , we have

(75) \begin{align} \lim_{n\rightarrow\infty}U_{h,n}(\nu)= U_{h,\infty} (\nu) \,:\!=\, \frac{\mathrm{e}^{G_{\mathbf{Q}_{\textrm{min}}}[h] \nu(\mathbb{R})}-1}{\nu(\mathbb{R}) } ,\quad \nu\in\mathfrak{M}^{\circ}(\mathbb{R}),\end{align}

where $\mathbf{Q}_{\textrm{min}}$ is the Yaglom limit and its PGFL is $G_{\mathbf{Q}_{\textrm{min}}}$ .

Proposition 7. Suppose that $\sum_{j} (j\log j)p_j<\infty$ . Let $\widetilde{\gamma}_n$ be a measure on $\mathfrak{M}(\mathbb{R})$ defined by (74), then the sequence $\{\widetilde{\gamma}_n\colon n\geq 0\}$ is tight in $\mathfrak{M}(\mathbb{R})$ and

(76) \begin{align} \sup_{n\in\mathbb{N}} \int_{\mathfrak{M}(\mathbb{R})} \frac{ \mathrm{e}^{ \nu(\mathbb{R})} - 1 }{\nu(\mathbb{R}) } \widetilde{\gamma}_n(\mathrm{d} \nu) <\infty.\end{align}

In particular, we get $\sup_{n} {\widetilde{\gamma}_n} (\mathfrak{M}(\mathbb{R}))<\infty$ .

Proposition 8. Assume that conditions (C1), (C2), and (C4) hold. Let $G_n$ be the PGFL of $Z_n$ defined by (3). Suppose that $\mathbf{Q}_{\textrm{min}}$ is the Yaglom limit and $G_{\mathbf{Q}_{\textrm{min}}}$ is its PGFL. Then we have

\begin{align*} \lim_{n\rightarrow\infty} \frac{G_n[h](x) - f_n(0)}{1-f_n(0)} = G_{\mathbf{Q}_{\textrm{min}}} [h]\end{align*}

uniformly in $x\in\mathbb{R}$ for each $h\in\mathcal{V}_0(\mathbb{R})$ .

Remark 12. In Proposition 8, under the strengthened condition (C4), it then follows by using the size-biased measure that the pointwise convergence in Yaglom’s theorem is strengthened to uniform convergence in x. We point out that Proposition 8 is a key step in giving the integral characterization (77) of the PGFL of the QSD with eigenvalue $m^{\alpha}$ .

Proof of the necessity of Theorem 2. Since $\mathbb{R}$ is a Polish space, so is $\mathfrak{M}(\mathbb{R})$ ; see, e.g., [Reference Kallenberg28, Lemmas 4.3 and 4.5]. Then by Proposition 7 and Prokhorov’s theorem, there exists a subsequence $\{\widetilde{\gamma}_{n_k}\}\subset\{\widetilde{\gamma}_{n}\}$ such that $\widetilde{\gamma}_{n_k}$ converges weakly to a finite measure $\widetilde{\gamma}$ on $\mathfrak{M}(\mathbb{R})$ . For any $h\in\mathcal{V}_0(\mathbb{R})$ , we claim that

(77) \begin{align}G_{\mathbf{Q}_{\alpha}}[h] = \int_{\mathfrak{M}^{\circ}(\mathbb{R})} U_{h,\infty} (\nu) \widetilde{\gamma} (\mathrm{d} \nu) + G_{\mathbf{Q}_{\textrm{min}}} [h] \widetilde{\gamma} (\{\mathbf{0}\}),\end{align}

where $U_{h,\infty}$ is defined by (75). Choose $L\geq1$ with $\widetilde{\gamma}(\{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})=L\})=0$ . It then follows from (72) that

(78) \begin{align}G_{\mathbf{Q}_{\alpha}} [h]=&\ \int_{\mathfrak{M}^{\circ}(\mathbb{R})} U_{h,n_k}(\nu) \widetilde{\gamma}_{n_k}(\mathrm{d} \nu) \nonumber \\=&\ \int_{ \{\nu\in\mathfrak{M}^{\circ}(\mathbb{R})\colon \nu(\mathbb{R})\leq L\} } U_{h,n_k}(\nu) \widetilde{\gamma}_{n_k}(\mathrm{d} \nu) + \int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})>L\} } U_{h,n_k}(\nu) \widetilde{\gamma}_{n_k}(\mathrm{d} \nu).\end{align}

Let $\mathcal{L}\,:\!=\,\{h\in\mathcal{V}_0(\mathbb{R})\colon \sup_{x\in\mathbb{R}} h(x)<1\}$ . We first prove that (77) holds for $h\in\mathcal{L}$ through the following three steps.

Step 1. For $h\in\mathcal{L}$ , we shall prove that the second term on the right-hand side of (78) tends to 0 as $k\rightarrow\infty$ first and then $L\rightarrow\infty$ . Note that

(79) \begin{align}G_{n_k}[h] (x) = 1-q_{n_k}\Big(1-\frac{G_{n_k}[h] (x)-(1-q_{n_k})}{q_{n_k}}\Big).\end{align}

Since $\log (1-x)\leq -x$ for $x<1$ , we have

\begin{align*}q_{n_k}^{-1} \log G_{n_k}[h](x) \leq \frac{G_{n_k}[h] (x)-(1-q_{n_k})}{q_{n_k}} -1,\quad x\in\mathbb{R}.\end{align*}

Let $\bar{c}_1\,:\!=\,\sup_{x\in\mathbb{R}} h(x)$ , then $h\in\mathcal{L}$ implies that $0< h(x)\leq \bar{c}_1<1$ for all $x\in\mathbb{R}$ . By dominated convergence, we have

\begin{align*}\lim_{\lambda\downarrow0} G_n[\lambda](x)=\lim_{\lambda\downarrow0}\mathbb{E}_{\delta_x}[\lambda^{N_n}]=\mathbb{P}_{\delta_x}(N_n=0)=f_n(0)=1-q_n.\end{align*}

It then follows that

(80) \begin{align}0\leq \frac{G_{n_k}[h] (x)-(1-q_{n_k})}{q_{n_k}} \leq 1.\end{align}

Furthermore, by (17) it is easy to see that when $\sum(j \log j)p_j<\infty$ , we have

\begin{align*}\frac{G_{n_k}[h] (x)-(1-q_{n_k})}{q_{n_k}} \leq \frac{ f_{n_k}(\bar{c}_1)-f_{n_k}(0) }{ 1-f_{n_k}(0) } \xrightarrow[k\rightarrow\infty]{} 1-\bar{c}_2< 1,\end{align*}

where $\bar{c}_2\,:\!=\,{(1-\bar{c}_1)\varphi(\bar{c}_1)}/{\varphi(0)}>0$ . From the preceding two inequalities one can choose sufficiently large $k\geq1$ to obtain $q_{n_k}^{-1} \log G_{n_k}[h](x) \leq -\bar{c}_2/2$ . On the other hand, for sufficiently large $k\geq1$ , there exists $\underline{c}\in(0,1)$ such that $(1-q_{n_k})^{q_{n_k}^{-1} \nu(\mathbb{R})} \geq \mathrm{e}^{-(1+\underline{c}) \nu(\mathbb{R})}$ . Consequently,

\begin{align*}U_{h,n_k}(\nu) &= \frac{ \exp \big(q_{n_k}^{-1} \int_{\mathbb{R}} \log G_{n_k}[h](x) \nu(\mathrm{d} x)\big)-(1-q_{n_k})^{q_{n_k}^{-1} \nu(\mathbb{R})} }{\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}} \\ &\leq \frac{\mathrm{e}^{-\bar{c}_2\nu(\mathbb{R})/2}-\mathrm{e}^{-(1+\underline{c}) \nu(\mathbb{R})}}{\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}}.\end{align*}

It then follows that

(81) \begin{align}\!\!\!\!\int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})>L\}}\! U_{h,n_k}(\nu) \widetilde{\gamma}_{n_k}(\mathrm{d} \nu)&\,{\leq}\int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})>L\}}\! \frac{\mathrm{e}^{-\bar{c}_2\nu(\mathbb{R})/2}\,{-}\,\mathrm{e}^{-(1+\underline{c}) \nu(\mathbb{R})}}{\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}} \widetilde{\gamma}_{n_k}(\mathrm{d} \nu)\nonumber\\&\,{=}\int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})>L\}}\!\big( \mathrm{e}^{-\bar{c}_2\nu(\mathbb{R})/2}\,{-}\,\mathrm{e}^{-(1+\underline{c}) \nu(\mathbb{R})} \big) \gamma_{n_k}(\mathrm{d} \nu).\end{align}

For any $n\in\mathbb{N}$ , we define two measures $\widetilde{\kappa}_n$ and $\widetilde{\kappa}$ on $[0,\infty)$ by $\widetilde{\kappa}_n(\mathrm{d} x) \,:\!=\, \widetilde{\gamma}_n(\{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})\in\mathrm{d} x\})$ and $\widetilde{\kappa}(\mathrm{d} x) \,:\!=\, \widetilde{\gamma}(\{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})\in\mathrm{d} x\})$ , respectively. Let $\kappa_n$ and $\kappa$ be two measures on $(0,\infty)$ given by $\kappa_n(\mathrm{d} x) \,:\!=\, x^{-1} \mathrm{e}^{x} \widetilde{\kappa}_n |_{(0,\infty)}(\mathrm{d} x)$ and $\kappa(\mathrm{d} x)\,:\!=\,x^{-1} \mathrm{e}^x \widetilde{\kappa}|_{(0, \infty)}(\mathrm{d} x)$ , where the symbol $\widetilde{\kappa}|_{(0, \infty)}$ is the restriction of $\widetilde{\kappa}$ to $(0,\infty)$ . As a consequence, (81) can be written as

(82) \begin{align}\int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})>L\} } U_{h,n_k}(\nu) \widetilde{\gamma}_{n_k}(\mathrm{d} \nu) \leq \int_{(L,\infty)} \big( \mathrm{e}^{-\bar{c}_2 x/2}-\mathrm{e}^{-(1+\underline{c}) x} \big) \kappa_{n_k}(\mathrm{d} x).\end{align}

Note that $\widetilde{\gamma}_{n_k}$ converges weakly to $\widetilde{\gamma}$ as $k\rightarrow\infty$ implies that $\widetilde{\kappa}_{n_k}\xrightarrow[]{\text{w}}\widetilde{\kappa}$ . Recall from (76) that $\sup_n \int_{[0,\infty)} ({(\mathrm{e}^{x}-1)}/{x}) \widetilde{\kappa}_n(\mathrm{d}x)<\infty$ . Hence, by Fatou’s lemma and Skorokhod’s theorem, we use the weak convergence of $\widetilde{\kappa}_{n_k}$ to $\widetilde{\kappa}$ to get

\begin{align*} \int_{[0,\infty)} \frac{\mathrm{e}^{x}-1}{x} \widetilde{\kappa}(\mathrm{d}x) \leq \varliminf_{k\rightarrow\infty} \int_{[0,\infty)} \frac{\mathrm{e}^{x}-1}{x} \widetilde{\kappa}_{n_k}(\mathrm{d}x) \leq \sup_{n} \int_{[0,\infty)} \frac{\mathrm{e}^{x}-1}{x} \widetilde{\kappa}_n(\mathrm{d}x) <\infty.\end{align*}

It then follows that, for any $a>0$ ,

(83) \begin{align} \kappa([a,\infty))<\infty \quad \text{and}\quad \sup_{n}{\kappa}_{n} ([a,\infty))<\infty.\end{align}

Note that if B is compact in the topological subspace $(0,\infty)$ then $B\subset[a,b]$ for some $a>0,b<\infty$ . Obviously, $\kappa_n,\kappa$ are Radon measures on $(0,\infty)$ , we say a measure is Radon if the measure of compact sets is always finite. By $\widetilde{\kappa}_{n_k}\xrightarrow[]{\text{w}}\widetilde{\kappa}$ , we find that $\kappa_{n_k}$ converges vaguely on $(0,\infty)$ to $\kappa$ , i.e. for any continuous and nonnegative function $\phi$ on $(0,\infty)$ with compact support,

\begin{align*} \kappa_{n_k} (\phi) \xrightarrow[k\rightarrow\infty]{} \kappa (\phi),\end{align*}

which extends the definition of vague convergences on a complete and separable metric space (see [Reference Resnick38, p.140]) to $(0,\infty)$ . Since $\mathrm{e}^{-\bar{c}_2 x/2}$ and $\mathrm{e}^{-(1+\underline{c}) x} $ are both monotonically decreasing in $x\in(0,\infty)$ and vanishing at infinity, we can choose two sequences of simple functions with support $\subset(L,b]$ (here different simple functions correspond to different values of b) to uniformly approximate

\begin{align*} \mathrm{e}^{-\bar{c}_2 x/2}\mathbf{1}_{(L,\infty)}(x) \quad \text{and} \quad \mathrm{e}^{-(1+\underline{c}) x} \mathbf{1}_{(L,\infty)}(x),\end{align*}

respectively. By (83) and [Reference Resnick38, Proposition 3.12 (ii)] (this proposition still holds when the space E is replaced by $(0,\infty)$ ) we obtain

\begin{align*} &\lim_{k\rightarrow\infty} \int_{(L,\infty)} \mathrm{e}^{-\bar{c}_2 x/2} \kappa_{n_k}(\mathrm{d} x) = \int_{(L,\infty)} \mathrm{e}^{-\bar{c}_2 x/2} \kappa(\mathrm{d} x),\\ &\lim_{k\rightarrow\infty} \int_{(L,\infty)} \mathrm{e}^{-(1+\underline{c}) x} \kappa_{n_k}(\mathrm{d} x) = \int_{(L,\infty)} \mathrm{e}^{-(1+\underline{c}) x} \kappa(\mathrm{d} x).\end{align*}

Therefore,

\begin{align*}\lim_{k\rightarrow\infty} \int_{(L,\infty)} ( \mathrm{e}^{-\bar{c}_2 x/2}-\mathrm{e}^{-(1+\underline{c}) x} ) \kappa_{n_k}(\mathrm{d} x) = \int_{(L,\infty)} ( \mathrm{e}^{-\bar{c}_2 x/2}-\mathrm{e}^{-(1+\underline{c}) x} ) \kappa(\mathrm{d} x).\end{align*}

By (83), we have

(84) \begin{align} \Big| \int_{(L,\infty)} ( \mathrm{e}^{-\bar{c}_2 x/2}-\mathrm{e}^{-(1+\underline{c}) x} ) \kappa(\mathrm{d} x) \Big| \leq 2 \kappa((L,\infty)) \xrightarrow[L\rightarrow\infty]{}0.\end{align}

Hence, the desired result follows from (82).

Step 2. For $h\in\mathcal{L}$ , we have $0< h(x)\leq \bar{c}_1<1$ for $x\in\mathbb{R}$ . By the definitions of $U_{h,\infty}(\nu)$ and $\kappa$ , it is easy to get

\begin{align*}\int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})>L\} } U_{h,\infty}(\nu) \widetilde{\gamma}(\mathrm{d} \nu)&\leq \int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})>L\} } \frac{ \mathrm{e}^{G_{\mathbf{Q}_{\text{min}}}[\bar{c}_1] \nu(\mathbb{R}) } - 1 }{ \nu(\mathbb{R}) } \widetilde{\gamma}(\mathrm{d} \nu)\\&= \int_{ (L,\infty) } ( \mathrm{e}^{(G_{\mathbf{Q}_{\text{min}}}[\bar{c}_1]-1) x} -\mathrm{e}^{-x} ) \kappa(\mathrm{d} x)\\&\xrightarrow[L\rightarrow\infty]{}0,\end{align*}

where the last limit follows similarly to that of (84).

Step 3. For $h\in\mathcal{L}$ , by setting

(85) \begin{align}U_{h,\infty}(\mathbf{0})\,:\!=\,\lim_{\nu\rightarrow\mathbf{0}}U_{h,\infty}(\nu)=G_{\mathbf{Q}_{\text{min}}}[h],\end{align}

where $\nu\rightarrow\mathbf{0}$ means that $\nu$ converges weakly to $\mathbf{0}$ , the functional $U_{h,\infty}$ on $\mathfrak{M}^{\circ}(\mathbb{R})$ can be continuously extended (with respect to the weak topology) to $\mathfrak{M}(\mathbb{R})$ . Based on steps 1 and 2, to get (77) we only need to show that

(86) \begin{align}\int_{ \{\nu\in\mathfrak{M}^{\circ}(\mathbb{R})\colon \nu(\mathbb{R})\leq L\} } U_{h,n_k}(\nu) \widetilde{\gamma}_{n_k}(\mathrm{d} \nu) \xrightarrow[k\rightarrow\infty]{} \int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})\leq L\} } U_{h,\infty} (\nu) \widetilde{\gamma} (\mathrm{d} \nu)\end{align}

holds for every $L\geq1$ with $\widetilde{\gamma}(\{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})=L\})=0$ . Recall that $\widetilde{\gamma}_{n_k}(\{\mathbf{0}\})=0$ for any $k\geq0$ . Then by a simple calculation we have

\begin{align*}\bigg| \int_{ \{\nu\in\mathfrak{M}^{\circ}(\mathbb{R})\colon \nu(\mathbb{R})\leq L\} } U_{h,n_k}(\nu) \widetilde{\gamma}_{n_k}(\mathrm{d} \nu) - \int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})\leq L\} } U_{h,\infty} (\nu) \widetilde{\gamma} (\mathrm{d} \nu) \bigg| \leq J_{1,k} + J_{2,k},\end{align*}

where

\begin{align*}J_{1,k} \,:\!=\, \bigg| \int_{ \{\nu\in\mathfrak{M}^{\circ}(\mathbb{R})\colon \nu(\mathbb{R})\leq L\} } [ U_{h,n_k}(\nu) - U_{h,\infty} (\nu) ] \widetilde{\gamma}_{n_k} (\mathrm{d} \nu) \bigg|\end{align*}

and

\begin{align*}J_{2,k} \,:\!=\, \bigg| \int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})\leq L\} } U_{h,\infty}(\nu) \widetilde{\gamma}_{n_k}(\mathrm{d} \nu) - \int_{ \{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})\leq L\} } U_{h,\infty} (\nu) \widetilde{\gamma} (\mathrm{d} \nu) \bigg| .\end{align*}

(i) The asymptotic behavior of $J_{2,k}$ . Recall that $U_{h,\infty}$ is defined by (75) and (85),

\begin{align*} \widetilde{\kappa}_n(\mathrm{d} x) = \widetilde{\gamma}_n(\{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})\in\mathrm{d} x\}),\quad \widetilde{\kappa}(\mathrm{d} x) = \widetilde{\gamma}(\{\nu\in\mathfrak{M}(\mathbb{R})\colon \nu(\mathbb{R})\in\mathrm{d} x\}).\end{align*}

Hence, we have

\begin{align*} J_{2,k} = \bigg| \int_{ [0,L] } \widetilde{f}(x) \widetilde{\kappa}_{n_k}(\mathrm{d} x) - \int_{ [0,L] } \widetilde{f}(x) \widetilde{\kappa} (\mathrm{d} x) \bigg|,\end{align*}

where

\begin{align*} \widetilde{f}(x) \,:\!=\,\left\{\begin{array}{ll} \frac{\mathrm{e}^{G_{\mathbf{Q}_{\text{min}}}[h] x}-1}{x} & \text{if}\ x >0,\\[2pt] G_{\mathbf{Q}_{\text{min}}}[h] & \text{if}\ x=0. \end{array}\right.\end{align*}

It is easily seen that $\widetilde{f}$ is a bounded continuous function on [0, L]. Thus, $\lim_{k\rightarrow\infty}J_{2,k}=0$ follows by the weak convergence of $\widetilde{\kappa}_{n_k}$ .

(ii) The asymptotic behavior of $J_{1,k}$ . From the definitions of $U_{h,n}(\nu)$ and $U_{h,\infty}(\nu)$ , then for any $\nu\in\mathfrak{M}^{\circ}(\mathbb{R})$ , we have

(87) \begin{align} & U_{h,n_k}(\nu)-U_{h, \infty}(\nu) \nonumber \\&\quad= \frac{(1-q_{n_k})^{q_{n_k}^{-1} \nu(\mathbb{R})} - \mathrm{e}^{-\nu(\mathbb{R})} }{\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}}\nonumber\\&\hskip29pt+ \frac{\exp \{q_{n_k}^{-1} \int_{\mathbb{R}} \log G_{n_k}[h] (x) \nu(\mathrm{d} x)\} - \exp \{(G_{\mathbf{Q}_{\text{min}}}[h]-1) \nu(\mathbb{R})\} } {\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}} .\end{align}

For the second term on the right-hand side of (87), note that

(88) \begin{align} & \int_{ \{\nu\in\mathfrak{M}^{\circ}(\mathbb{R}): \nu(\mathbb{R})\leq L\} } \frac{(1-q_{n_k})^{q_{n_k}^{-1} \nu(\mathbb{R})} - \mathrm{e}^{-\nu(\mathbb{R})} }{\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}} \widetilde{\gamma}_{n_k} (\mathrm{d} \nu) \nonumber \\ &\quad= \int_{ (0,L] } \frac{ (1-q_{n_k})^{q_{n_k}^{-1} x} - \mathrm{e}^{-x} }{x \mathrm{e}^{-x}} \widetilde{\kappa}_{n_k} ( \mathrm{d} x).\end{align}

We denote by $I_{n_k}(x)$ the integrand on the right-hand side of the above equality. Then the function $x \mapsto I_{n_k}(x)$ , continuously extended to [0, L] by setting $I_{n_k}(0)= 1+q_{n_k}^{-1} \log (1-q_{n_k}) $ , converges uniformly on every compact subset of [0, L] to 0. Hence, by Proposition 7 we have

(89) \begin{align}\bigg| \int_{(0,L]} I_{n_k}(x) \widetilde{\kappa}_{n_k} (\mathrm{d} x) \bigg| &= \bigg| \int_{[0,L]} I_{n_k}(x) \widetilde{\kappa}_{n_k} (\mathrm{d} x) \bigg| \nonumber\\&\leq \sup_{x\in[0,L]} \big| I_{n_k}(x) \big| \sup_{k\in\mathbb{N}} \widetilde{\gamma}_{n_k} (\mathfrak{M}(\mathbb{R})) \nonumber\\& \xrightarrow[k\rightarrow\infty]{} 0.\end{align}

For the first term on the right-hand side of (87), based on (79), by using the inequality $|\mathrm{e}^x-\mathrm{e}^y|\leq |x-y|$ for $x,y\leq0$ it is easily seen that

(90) \begin{align}& \bigg| \exp \bigg\{q_{n_k}^{-1} \int_{\mathbb{R}} \log G_{n_k}[h] (x) \nu(\mathrm{d} x)\bigg\} - \exp \{(G_{\mathbf{Q}_{\text{min}}}[h]-1) \nu(\mathbb{R})\} \bigg| \nonumber\\&\quad\leq \bigg| q_{n_k}^{-1} \int_{\mathbb{R}} \log \bigg[1-q_{n_k}\bigg(1-\frac{G_{n_k}[h](x)-(1-q_{n_k})}{q_{n_k}}\bigg)\bigg] \nu(\mathrm{d} x) - (G_{\mathbf{Q}_{\text{min}}}[h]-1) \nu(\mathbb{R}) \bigg| \nonumber\\&\quad\leq \bigg| q_{n_k}^{-1} \int_{\mathbb{R}} \bigg\{ \log \bigg[ 1-q_{n_k}\bigg(1-\frac{G_{n_k}[h](x)-(1-q_{n_k})}{q_{n_k}}\bigg)\bigg] \nonumber\\&\quad\quad +q_{n_k} \bigg(1-\frac{G_{n_k}[h] (x)-(1-q_{n_k})}{q_{n_k}}\bigg) \bigg\} \nu(\mathrm{d} x) \bigg| \nonumber\\&\quad\quad + \bigg| \int_{\mathbb{R}} \bigg[ \frac{G_{n_k}[h](x)-(1-q_{n_k})}{q_{n_k}} - G_{\mathbf{Q}_{\text{min}}}[h] \bigg] \nu(\mathrm{d} x) \bigg|.\end{align}

Note that $q_{n_k}\rightarrow0$ as $k\rightarrow\infty$ , then by (80) and Taylor’s expansion of $\log (1-x)$ we see that, for sufficiently large $k\geq 1$ ,

\begin{align*}\log \bigg[1-q_{n_k}\bigg(1-\frac{G_{n_k}[h](x)-(1-q_{n_k})}{q_{n_k}}\bigg)\bigg] = -\sum_{i=1}^{\infty} \frac{q_{n_k}^i}{i}\bigg(1-\frac{G_{n_k}[h] (x)-(1-q_{n_k})}{q_{n_k}}\bigg)^i.\end{align*}

It then follows that

\begin{align*}& \bigg| \log \bigg[ 1-q_{n_k}\bigg(1-\frac{G_{n_k}[h](x)-(1-q_{n_k})}{q_{n_k}}\bigg)\bigg] + q_{n_k} \bigg(1-\frac{G_{n_k}[h] (x)-(1-q_{n_k})}{q_{n_k}}\bigg) \bigg| \\&\quad\leq \sum_{i=2}^{\infty} q_{n_k}^i.\end{align*}

Consequently,

(91) \begin{align}& \bigg| q_{n_k}^{-1} \int_{\mathbb{R}} \bigg\{ \log \bigg[ 1-q_{n_k}\bigg(1-\frac{G_{n_k}[h](x)-(1-q_{n_k})}{q_n}\bigg)\bigg] \nonumber\\&\quad\quad + q_{n_k} \bigg(1-\frac{G_{n_k}[h] (x)-(1-q_{n_k})}{q_{n_k}}\bigg) \bigg\} \nu(\mathrm{d} x) \bigg| \nonumber\\&\quad\leq \int_{\mathbb{R}} \sum_{i=2}^{\infty} q_{n_k}^{i-1} \nu(\mathrm{d} x)\nonumber\\&\quad= \frac{q_{n_k}}{1-q_{n_k}} \nu(\mathbb{R}).\end{align}

This combined with (90) gives

\begin{align*}& \bigg| \exp \bigg\{q_{n_k}^{-1} \int_{\mathbb{R}} \log G_{n_k}[h] (x) \nu(\mathrm{d} x)\bigg\} - \exp \big\{(G_{\mathbf{Q}_{\text{min}}}[h]-1) \nu(\mathbb{R})\big\} \bigg|\\&\quad\leq\bigg( \bar{a}_{n_k}(h) + \frac{q_{n_k}}{1-q_{n_k}} \bigg)\nu(\mathbb{R}),\end{align*}

where

\begin{align*}\bar{a}_{n_k}(h)\,:\!=\,\sup_{x\in\mathbb{R}}\bigg| \frac{G_{n_k}[h](x)-f_{n_k}(0)}{1-f_{n_k}(0)} - G_{\mathbf{Q}_{\text{min}}}[h] \bigg|.\end{align*}

By Proposition 8 we see that $\lim_{k\rightarrow\infty}\bar{a}_{n_k}(h)=0$ for every $h\in\mathcal{V}_0(\mathbb{R})$ . Proposition 7 implies that $\sup_{n\in\mathbb{N}} \widetilde{\gamma}_{n} (\mathfrak{M}(\mathbb{R})) <\infty$ . It then follows that

\begin{align*}& \bigg| \int_{ \{\nu\in\mathfrak{M}^{\circ}(\mathbb{R}): \nu(\mathbb{R})\leq L\} } \frac{ \exp \Big\{q_{n_k}^{-1} \int_{\mathbb{R}} \log G_{n_k}[h] (x) \nu(\mathrm{d} x)\Big\} - \exp \Big\{(G_{\mathbf{Q}_{\text{min}}}[h]-1) \nu(\mathbb{R})\Big\} }{ \nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})} }\widetilde{\gamma}_{n_k} (\mathrm{d} \nu) \bigg|\\&\quad \leq \mathrm{e}^L \bigg( \bar{a}_{n_k}(h) + \frac{q_{n_k}}{1-q_{n_k}} \bigg) \sup_{k\in\mathbb{N}} \widetilde{\gamma}_{n_k} (\mathfrak{M}(\mathbb{R})) \xrightarrow[k\rightarrow\infty]{} 0.\end{align*}

Then (87)–(89) and the above inequality imply that $J_{1,k}\rightarrow0$ as $k\rightarrow\infty$ . Summing up, from the asymptotic behaviors of $J_{2,k}$ (see (i)) and $J_{1,k}$ (see (ii)), (86) holds for any $h\in\mathcal{L}$ .

Given $h\in\mathcal{V}_0(\mathbb{R})$ we take $h_n(x)\,:\!=\,h(x)\mathbf{1}_{\{h(x)<1-{1}/{n}\}}+(1-{1}/{n}) \mathbf{1}_{\{1-{1}/{n}\leq h(x)\leq1\}}$ so that $h_n\in\mathcal{L}$ and $h_n(x)\uparrow h(x)$ as $n\rightarrow\infty$ . Then (77) holds for $h\in\mathcal{V}_0(\mathbb{R})$ by monotone convergence. Therefore, by the definition of $\kappa$ and $\widetilde{\kappa}$ , (77) can be written as

(92) \begin{align}G_{\mathbf{Q}_{\alpha}}[h] = \int_{(0,\infty)} \big( \mathrm{e}^{(G_{\mathbf{Q}_{\text{min}}}[h]-1) x}-\mathrm{e}^{-x} \big) \kappa(\mathrm{d} x)+\widetilde{\kappa}(\{0\}) G_{\mathbf{Q}_{\text{min}}}[h],\quad h\in\mathcal{V}_0(\mathbb{R}).\end{align}

Furthermore, the theory of Laplace transforms gives $\kappa$ and $\widetilde{\kappa}(\{0\})$ , hence, $\widetilde{\kappa}$ is uniquely determined by (92), so that $\widetilde{\kappa}_{n}$ converges weakly to $\widetilde{\kappa}$ . Consequently, $\kappa_{n}$ converges vaguely to $\kappa$ on $(0,\infty)$ .

The following property of the measure $\kappa$ follows from this convergence: for every compact interval $B\subset(0,\infty)$ whose endpoints are not atoms of $\kappa$ ,

\begin{align*}\kappa(B)=&\ \lim _{n \rightarrow \infty} \gamma_{n+1}(\{\nu \in \mathfrak{M}^{\circ}(\mathbb{R}): \nu(\mathbb{R}) \in B \} )\\=&\ \lim _{n \rightarrow \infty} m^{-\alpha(n+1)} \mathbf{Q}_{\alpha} ( \{\nu\in\mathfrak{N}^{\circ}(\mathbb{R})\colon \nu(\mathbb{R}) \in q_{n+1}^{-1}B\} )\\=&\ m^{-\alpha} \lim _{n \rightarrow \infty} m^{-\alpha n} \mathbf{Q}_{\alpha} \bigg( \bigg\{\nu\in\mathfrak{N}^{\circ}(\mathbb{R})\colon \nu(\mathbb{R}) \in q_{n}^{-1} \bigg(\frac{q_n}{q_{n+1}}\bigg) B\bigg\} \bigg)\\=&\ m^{-\alpha} \lim _{n \rightarrow \infty} m^{-\alpha n} \gamma_{n}\bigg(\bigg\{\nu \in \mathfrak{M}^{\circ}(\mathbb{R})\colon \nu(\mathbb{R}) \in \bigg(\frac{q_n}{q_{n+1}}\bigg)B \bigg\} \bigg)\\=&\ m^{-\alpha} \kappa(m^{-1} B)\quad\text{(by $q_n/q_{n+1}\rightarrow 1/m$)}.\end{align*}

This implies that the measure $\Delta$ defined by $\Delta(\mathrm{d} x)\,:\!=\,x^\alpha \kappa(\mathrm{d} x)$ satisfies $\Delta(m B) =\Delta(B)$ for every $B\in\mathcal{B}((0,\infty))$ .

It remains to investigate which terms in (92) vanish for particular values of $\alpha$ . A first constraint comes from the fact that $\mathbf{Q}_{\alpha}$ is a probability on $\mathfrak{N}^{\circ}(\mathbb{R})$ , then $G_{\mathbf{Q}_{\alpha}}[1]=1$ .

A second constraint comes from the fact that $G_{\mathbf{Q}_{\alpha}}[h]\leq 1<\infty$ for $h\in\mathcal{V}_0(\mathbb{R})$ , then the integral in (92) needs to be finite as well, that is,

\begin{align*}\int_{(0,\infty)}(\mathrm{e}^{(G_{\mathbf{Q}_{\text{min}}}[h]-1) x}-\mathrm{e}^{-x}) \kappa(\mathrm{d} x)<\infty.\end{align*}

Checking the proof of [Reference Maillard35, Corollary 16], one can show that this corollary still holds when the generating function H is replaced by the PGFL $G_{\mathbf{Q}_{\text{min}}}$ . Then the above inequality holds if and only if $\alpha<1$ or $\kappa=0$ .

A third constraint comes from the fact that $G_{\mathbf{Q}_{\alpha}}$ satisfies (9). We first recall the following equation for $G_{\mathbf{Q}_{\text{min}}}$ :

(93) \begin{align}G_{\mathbf{Q}_{\text{min}}}[G[h]({\cdot})]-G_{\mathbf{Q}_{\text{min}}} [f(0)] & =m G_{\mathbf{Q}_{\text{min}}}[h], \quad h\in\mathcal{V}_0(\mathbb{R}),\end{align}

(94) \begin{align}G_{\mathbf{Q}_{\text{min}}}[f(0)] & =1-m,\end{align}

(95) \begin{align}G_{\mathbf{Q}_{\text{min}}}[G[h]({\cdot})]-1 & =m(G_{\mathbf{Q}_{\text{min}}}[h]-1), \quad h\in\mathcal{V}_0(\mathbb{R}) .\end{align}

Indeed, (93) comes from Corollary 7; (94) follows from (93) by setting $h\equiv1$ ; (95) comes from (93) and (94) by reordering terms. Then by (92), for any $h\in\mathcal{V}_0(\mathbb{R})$ ,

where the last equality follows by $\kappa(B)=m^{-\alpha} \kappa(m^{-1}B)$ . Comparing with (9) implies that $\widetilde{\kappa}(\{0\})=0$ unless $\alpha=1$ . Summing up, we have the following constraints for (92):

• • when $\alpha>1$ , there exist no QSDs of Z with eigenvalue $m^{\alpha}$ ,

• • when $\alpha=1, \kappa=0$ and $\widetilde{\kappa}(\{0\})=1$ ,

• • when $\alpha<1, \widetilde{\kappa}(\{0\})=0$ and $\int_{(0,\infty)} (1-\mathrm{e}^{-x}) \kappa(\mathrm{d} x)=1$ .

This proves the necessity part of the Theorem 2.

6.2.2. Proof of the sufficiency part of Theorem 2

We only need to consider the case $\alpha\in(0,1)$ . Suppose that $G_{\mathbf{Q}_{\alpha}}$ is a functional given by (10) with a measure $\Delta$ on $(0,\infty)$ satisfying 3(i) and 3(ii) of Theorem 2. By the above calculations, it is easy to see that $G_{\mathbf{Q}_{\alpha}}$ satisfies (9). It remains to show that $G_{\mathbf{Q}_{\alpha}}$ is the PGFL of a probability measure on $\mathfrak{N}^{\circ}(\mathbb{R})$ . According to Maillard [Reference Maillard35], we can construct a finite point process ( $\neq\mathbf{0}$ ) whose PGFL is $G_{\mathbf{Q}_{\alpha}}$ as follows. For $x\in(0,\infty)$ , suppose that $\{\mathcal{N};\mathbb{P}\}$ is a Poisson random variable with a random parameter drawn according to the measure $x^{-\alpha} \Delta(\mathrm{d}x)$ and conditioned on being positive, that is,

\begin{align*}\mathbb{P}(\mathcal{N}=0)=0,\quad \mathbb{P}(\mathcal{N}=k)=\int_{(0,\infty)} \mathrm{e}^{-x} \frac{x^k}{k!} \frac{1}{x^{\alpha}} \Delta (\mathrm{d}x),\quad k\geq 1.\end{align*}

Therefore, the generating function of the positive random variable $\mathcal{N}$ is

(96) \begin{align}\mathbb{E}\big[ s^{\mathcal{N}} \big] = \int_{(0,\infty)} \big( \mathrm{e}^{(s-1)x}-\mathrm{e}^{-x} \big)\frac{1}{x^{\alpha}} \Delta (\mathrm{d}x),\quad s\in[0,1].\end{align}

Suppose that $\{X_i;\mathbb{P}\}_{i\geq1}$ is a sequence independent and identically distributed point processes with the same PGFL $G_{\mathbf{Q}_{\text{min}}}$ . We assume that $\mathcal{N}$ is independent of $\{X_i;i\geq1\}$ . Let

\begin{align*}\bar{X}\,:\!=\,\sum_{i=1}^{\mathcal{N}} X_i.\end{align*}

Then it is easy to check that $\bar{X}$ is a finite point process ( $\neq \mathbf{0}$ ) and its PGFL is equal to

\begin{align*}\mathbb{E} \big[ \mathrm{e}^{\bar{X}(\!\log h)} \big]&= \mathbb{E} \bigg[ \prod_{i=1}^{\mathcal{N}} \mathbb{E}[ \mathrm{e}^{X_i(\!\log h)} ] \bigg]\\&= \mathbb{E} \big[ G_{\mathbf{Q}_{\text{min}}}[h]^{\mathcal{N}} \big]\\&= \int_{(0,\infty)} \big( \mathrm{e}^{(G_{\mathbf{Q}_{\text{min}}}[h]-1)x}-\mathrm{e}^{-x} \big)\frac{1}{x^{\alpha}} \Delta (\mathrm{d}x),\quad h\in\mathcal{V}_0(\mathbb{R}),\end{align*}

where the first equality comes from the independence and the last one follows by (96). This completes the proof of the sufficiency part.

6.3. Proofs of Propositions 6–8

Proof of Proposition 6. In view of (87), it suffices to prove that

(97) \begin{align}\lim_{n\rightarrow\infty} \frac{\exp \{q_n^{-1} \int_{\mathbb{R}} \log G_n[h] (x) \nu(\mathrm{d} x)\} - \exp \{(G_{\mathbf{Q}_{\text{min}}}[h]-1) \nu(\mathbb{R})\} } {\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}} =0\end{align}

and

(98) \begin{align}\lim_{n\rightarrow\infty} \frac{(1-q_n)^{q_n^{-1} \nu(\mathbb{R})} - \mathrm{e}^{-\nu(\mathbb{R})} }{\nu(\mathbb{R}) \mathrm{e}^{-\nu(\mathbb{R})}}=0\end{align}

hold for $h\in\mathcal{V}_0(\mathbb{R})$ and $\nu\in\mathfrak{M}^{\circ}(\mathbb{R})$ . It is not hard to see that $(1-q_n)^{q_n^{-1} \nu(\mathbb{R})}\rightarrow\mathrm{e}^{-\nu(\mathbb{R})}$ as $n\rightarrow\infty$ . Then (98) holds. Since $h\in\mathcal{V}_0(\mathbb{R})$ implies that $h(x)\in(0,1]$ for $x\in\mathbb{R}$ , it is easy to check that (80) still holds for $h\in\mathcal{V}_0(\mathbb{R})$ . Hence, $|q_n^{-1}(G_n[h]-(1-q_n))-G_{\mathbf{Q}_{\text{min}}}[h]|\leq2$ for $n\in\mathbb{N}$ . Note that $\nu(\mathbb{R})<\infty$ , then in view of (66), we can conclude by dominated convergence that, under conditions (C1)–(C3),

\begin{align*}\bigg| \int_{\mathbb{R}} \Big[ \frac{G_{n}[h](x)-(1-q_{n})}{q_{n}} - G_{\mathbf{Q}_{\text{min}}}[h] \Big] \nu(\mathrm{d} x) \bigg| \xrightarrow[n\rightarrow\infty]{} 0.\end{align*}

Therefore, (97) follows from (90) and (91).

Proof of Proposition 7. Suppose that $\widetilde{\gamma}_n$ is a measure on $\mathfrak{M}(\mathbb{R})$ given by (74). From the inequality $1-x\leq \mathrm{e}^{-x}$ for $x\in\mathbb{R}$ , then taking $h\equiv1$ in (73) yields

\begin{align*}U_{1,n}(\nu)=\frac{1-(1-q_n)^{q_n^{-1}\nu(\mathbb{R})}}{\nu(\mathbb{R})\mathrm{e}^{-\nu(\mathbb{R})}} \geq \frac{ \mathrm{e}^{\nu(\mathbb{R})}-1 }{\nu(\mathbb{R}) },\quad \nu\in\mathfrak{M}^{\circ}(\mathbb{R}).\end{align*}

On the other hand, by (72) we get

\begin{align*}\lim_{n\rightarrow\infty} \int_{\mathfrak{M}^{\circ}(\mathbb{R})} U_{1,n}(\nu) \widetilde{\gamma}_n(\mathrm{d} \nu) = G_{\mathbf{Q}_{\alpha}} [1] <\infty.\end{align*}

Recall that $\nu \mapsto { (\mathrm{e}^{\nu(\mathbb{R})}-1 ) }/{\nu(\mathbb{R}) }$ can be continuously extended (with respect to the weak topology) to $\mathfrak{M}(\mathbb{R})$ by setting ${( \mathrm{e}^{\nu(\mathbb{R})}-1 ) }/{\nu(\mathbb{R}) } =1$ for $\nu=\mathbf{0}$ . Note that $\widetilde{\gamma}_{n}(\{\mathbf{0}\})=0$ , so it is easy to check that