Proof. If $\beta\neq \boldsymbol{0}$ then there exists $j\in [\![1, J]\!]$ such that $\beta_j>0$ . This gives that a.s. on the event $\{ {\mathcal T}_{\theta, \beta}\not \in {\mathbb T}_0\}$ we have $L_{A_j}({\mathcal T}_{\theta, \beta})=+\infty $ and, thus,

\begin{align*} {\mathbb P}({\mathcal T}_{\theta, \beta}\not \in {\mathbb T}_0,\, \sup_{j\in {\mathcal J}^*} L_{A_j}({\mathcal T}_{\theta, \beta})<+\infty )=0. \end{align*}

If $\beta=\boldsymbol{0}$ , we deduce from Remark 3.3 (iv) that if $(\theta, \boldsymbol{0})\in \operatorname{Pa}(p, {\mathcal A})$ then we have either $\theta=\theta_{\min}>0$ , $(\theta, \boldsymbol{0})\in \operatorname{Pa}^*(p, {\mathcal A})$ , $\tilde p_{\theta, \boldsymbol{0}}$ is subcritical; or $\theta>\theta_{\min}$ , $(\theta, \boldsymbol{0})\not \in \operatorname{Pa}^*(p, {\mathcal A})$ and $\tilde p_{\theta, \boldsymbol{0}}$ is supercritical. In the former case, we get ${\mathbb P}({\mathcal T}_{\theta, \boldsymbol{0}}\not \in {\mathbb T}_0)=0$ and in the latter case,

\begin{align*} {\mathbb P}({\mathcal T}_{\theta, \boldsymbol{0}}\not \in {\mathbb T}_0,\, \sup_{j\in {\mathcal J}^*} L_{A_j}({\mathcal T}_{\theta, \boldsymbol{0}})<+\infty )= {\mathbb P}({\mathcal T}_{\theta, \boldsymbol{0}}\not \in {\mathbb T}_0)>0. \end{align*}

This gives the result.

Proof. By construction, we have $\textrm{supp}\,(\tilde p_{\theta, \beta})\subset \textrm{supp}\,(p)$ . Since $p(0)>0$ , there exists $j_0\in [\![0, J]\!]$ such that $0\in A_{j_0}$ .

Let $\theta\in (0, +\infty )$ . The condition $0\in A_{\mathcal J}$ in Lemma 3.3 (i) is equivalent to $\beta_{j_0}>0$ and, thus, to $\tilde p_{\theta, \beta}(0)>0$ . The condition $p\left(A_{\mathcal J} \cap \{0, 1\}^c\right)>0$ is clearly equivalent to $\tilde p_{\theta, \beta}\big(\{0, 1\}^c\big)>0$ . Therefore, conditions in Lemma 3.3 (i) are equivalent to $\tilde p_{\theta, \beta}$ satisfying the nondegeneracy condition (5).

Let $\theta=0$ (and, thus, $0\not \in A_0$ ). The condition $0\in A_{{\mathcal J}^*}$ in Lemma 3.3 (ii) is equivalent to $\beta_{j_0}>0$ and to $\tilde p_{\theta, \beta}(0)>0$ , as $\min A_{j_0}=0$ . Since $\{\min A_j \colon j \in {\mathcal J}^*\}\subset \textrm{supp}\,( \tilde p_{\theta, \beta}) \subset \{\min A_j \colon j \in {\mathcal J}^*\} \cup\{1\}$ , we deduce that $ \max_{j\in{\mathcal J}^*}\min (A_j)>1$ is equivalent to $\tilde p_{\theta, \beta}\big(\{0, 1\}^c\big)>0$ . Thus, conditions in Lemma 3.3 (ii) are equivalent to $\tilde p_{\theta, \beta}$ satisfying (5).

Let $\theta=+\infty $ . The conditions $j_0\in {\mathcal J}^*$ and $A_{j_0}=\{0\}$ in Lemma 3.3 (iii) are equivalent to $\tilde p_{\theta, \beta}(0)>0$ (note that, by (17), $\tilde p_{\theta, \beta}(A_0)>0$ if and only if $1\in A_0$ , and then $0\in A_0$ implies that $\tilde p_{\theta, \beta}(0)=0$ ). Eventually the condition $ \max_{j\in{\mathcal J}^*}\max (A_j)>1$ is also clearly equivalent to $\tilde p_{\theta, \beta}\big(\{0, 1\}^c\big)>0$ . Thus, conditions in Lemma 3.3 (iii) are equivalent to $\tilde p_{\theta, \beta}$ satisfying (5).

Proof. We suppose that the parameter $(\theta, \beta)$ is $(p,{\mathcal A})$ -compatible (which implies that $\tilde p_{\theta, \beta}$ is nontrivial) and that $\theta\in (0, +\infty )$ . We prove that the probability distribution $\tilde p_{\theta, \beta}$ is $(p,{\mathcal A})$ -compatible. Recall that ${\mathcal J}=\{0\}\cup\{j\in [\![1, J]\!] \colon \beta_j>0\}$ .

Note that ${\mathbb P}\big(L_{{\mathcal A}}({\mathcal T}_{\theta, \beta}) = \mathbf n\big)>0$ implies that $n_j=0$ for all $j\not\in {\mathcal J}^*$ , where $\mathbf n=(n_1, \ldots, n_J)$ . Using the definition of $\tilde{p}_{\theta, \beta}$ , we obtain, for ${\mathbf{t}}\in {\mathbb T}_0$ such that ${\mathbb P}({\mathcal T}_p={\mathbf{t}})>0$ and $L_{A_j}({\mathbf{t}})=0$ for $j\not\in {\mathcal J}$ ,

\begin{align*}{\mathbb P}({\mathcal T}_{\theta, \beta} = {\mathbf{t}})&= \prod_{u\in {\mathbf{t}}} \tilde{p}_{\theta,\beta}(k_u({\mathbf{t}}))\\ &= \prod_{j\in{\mathcal J}} \, \prod_{u \in {\mathcal L}_{A_j}({\mathbf{t}})} \beta_j \theta^{k_u({\mathbf{t}})} p(k_u({\mathbf{t}})) \\&= {\mathbb P}\big({\mathcal T}_p= {\mathbf{t}}\big)\, \theta^{\sharp {\mathbf{t}} -1} \prod_{j\in{\mathcal J}}\beta_j^{L_{A_j}({\mathbf{t}})}\\ &= {\mathbb P}\big({\mathcal T}_p= {\mathbf{t}}\big)\, \theta^{-1} \prod_{j\in{\mathcal J}^*} (\beta_j \theta )^{L_{A_j}({\mathbf{t}})},\end{align*}

where we used (6) for the third equality, and $\sum_{j\in{\mathcal J}} L_{A_j}({\mathbf{t}})= \sharp {\mathbf{t}}$ as well as $\beta_0 \theta=1$ for the fourth equality. As ${\mathbb P}({\mathcal T}_p={\mathbf{t}})=0$ implies that ${\mathbb P}({\mathcal T}_{\theta, \beta}={\mathbf{t}})=0$ , we deduce that, for $\mathbf n\in {\mathbb N}^ J$ ,

\begin{align*} {\mathbb P}(L_{{\mathcal A}}({\mathcal T}_{\theta, \beta}) = \mathbf n) &= \sum_{{\mathbf{t}}\in {\mathbb T}_0,\, L_{{\mathcal A}}({\mathbf{t}}) = \mathbf n} {\mathbb P}({\mathcal T}_{\theta, \beta}={\mathbf{t}}) + {\mathbb P}({\mathcal T}_{\theta, \beta}\not \in {\mathbb T}_0, \, L_{{\mathcal A}}({\mathcal T}_{\theta, \beta}) = \mathbf n) \\ &= \sum_{{\mathbf{t}}\in {\mathbb T}_0,\, L_{{\mathcal A}}({\mathbf{t}}) = \mathbf n} {\mathbb P}({\mathcal T}_{\theta, \beta}={\mathbf{t}})\\&= {\mathbb P}\left(L_{{\mathcal A}}({\mathcal T}_p) = \mathbf n\right)\, \theta^{-1} \prod_{j\in {\mathcal J}^*}(\beta_j\theta)^{n_j},\end{align*}

where we used Lemma 3.2 for the second equality.

So for every $\mathbf n\in {\mathbb N}^J$ such that ${\mathbb P}\big(L_{{\mathcal A}}({\mathcal T}_{\theta, \beta}) = \mathbf n\big)>0$ , we have ${\mathbb P}\big(L_{{\mathcal A}}({\mathcal T}_p) = \mathbf n\big)>0$ , and for every ${\mathbf{t}}\in{\mathbb T}_0$ such that $L_{{\mathcal A}}({\mathbf{t}})=\mathbf n$ , we get

\begin{align*}{\mathbb P}({\mathcal T}_{\theta, \beta}={\mathbf{t}}\mid L_{{\mathcal A}}({\mathcal T}_{\theta, \beta})=\mathbf n)={\mathbb P}({\mathcal T}_p={\mathbf{t}}\mid L_{{\mathcal A}}({\mathcal T}_p)=\mathbf n).\end{align*}

Hence, the probability distribution $\tilde p_{\theta, \beta}$ is $(p, {\mathcal A})$ -compatible.

Proof. We first consider the case $\theta=+\infty $ . For simplicity, we write $p'=\tilde p_{\infty , \beta}$ and ${\mathcal T}'={\mathcal T}_{\infty , \beta}$ . We suppose that the parameter $(\infty , \beta)$ is $(p,{\mathcal A})$ -compatible. As p ^′ is nontrivial, it follows that $\beta\in {\mathbb R}^J_+\backslash\{\boldsymbol{0}\}$ and, thus, ${\mathcal J}^*\neq \emptyset$ . In particular, we have $A_0\subset\{1\}$ , thanks to (19) and Lemma 3.3 (iii). We prove that the probability distribution p ^′ is $(p,{\mathcal A})$ -compatible.

Note that ${\mathbb P}(L_{{\mathcal A}}({\mathcal T}') = \mathbf n)>0$ implies that $n_j=0$ for all $j\not\in {\mathcal J}^*$ , where $\mathbf n=(n_1, \ldots, n_J)$ , and, by (6), that

(26) \begin{equation} \sum_{j\in {\mathcal J}^*}n_j(\max A_j-1)=-1.\end{equation}

To simplify notation, recall that we write $L_k$ for $L_{\{k\}}$ for $k\in {\mathbb N}$ . Fix such $\mathbf n\in {\mathbb N}^ J$ and consider the set ${\mathbb T}_{0, n}=\{ {\mathbf{t}}\in {\mathbb T}_0 \colon L_{{\mathcal A}}({\mathbf{t}})=\mathbf{n}\ \text{and}\ {\mathbb P}({\mathcal T}'={\mathbf{t}}) \gt 0\}$ , which is clearly not empty. Using (6), it follows that, for ${\mathbf{t}}\in {\mathbb T}_{0, n}$ , $ \sum_{j\in {\mathcal J}^*} \sum_{k\in A_j} L_{k}({\mathbf{t}}) (k-1)=-1$ . We deduce from (26) that, for $k\not \in A_0$ , we have $L_{k}({\mathbf{t}})=n_j$ if $j\in {\mathcal J}^*$ and $k=\max A_j$ , and $L_{k}({\mathbf{t}})=0$ otherwise.

Let ${\mathbf{t}}\in {\mathbb T}_{0, n}$ . We distinguish two cases according to 1 belonging to $A_0$ or not. First, we consider the case $1 \in A_0$ , that is, $A_0=\{1\}$ , elementary computation gives

\begin{align*} {\mathbb P}({\mathcal T}'={\mathbf{t}} \, |\, L_{{\mathcal A}}({\mathcal T}')=\mathbf n) = c^{-1} \, p(1)^{L_{1}({\mathbf{t}})}, \end{align*}

with, as ${\mathcal J}^*\neq \emptyset$ , c positive and finite given by

\begin{align*} c=\sum_{{\mathbf{t}}' \in {\mathbb T}_0 \, L_{\mathcal A}({\mathbf{t}}')=\mathbf n} p(1)^{L_{1}({\mathbf{t}}')}+ {\mathbb P}({\mathcal T}'\not \in {\mathbb T}_0,\, L_{{\mathcal A}}({\mathcal T}')=\mathbf n) =\sum_{{\mathbf{t}}' \in {\mathbb T}_0 \, L_{\mathcal A}({\mathbf{t}}')=\mathbf n} p(1)^{L_{1}({\mathbf{t}}')}. \end{align*}

Similarly, we have

\begin{align*} {\mathbb P}({\mathcal T}_p={\mathbf{t}} \mid L_{{\mathcal A}}({\mathcal T}_p)=\mathbf n) = c^{-1} \, p(1)^{L_{1}({\mathbf{t}})}. \end{align*}

This readily implies that p ^′ is $(p, {\mathcal A})$ -compatible.

Secondly, we consider the case $1\not\in A_0$ , that is, $A_0=\emptyset$ . It follows that the set ${\mathbb T}_{0, n}$ is finite. Similarly to the first case, we show that, with $c=\textrm{Card}\;({\mathbb T}_{0, n})\geq 1$ ,

\begin{align*} {\mathbb P}({\mathcal T}'={\mathbf{t}} \mid L_{{\mathcal A}}({\mathcal T}')=\mathbf n) = {\mathbb P}({\mathcal T}_p={\mathbf{t}} \mid L_{{\mathcal A}}({\mathcal T}_p)=\mathbf n)= c^{-1} .\end{align*}

This readily also implies that p ^′ is $(p, {\mathcal A})$ -compatible.

Eventually, the $\theta=0$ case can be handled similarly using $ \sum_{j\in {\mathcal J}^*}n_j(\min A_j-1)=-1$ instead of (26).

Proof. For $k\in {\mathbb N}$ , let ${\mathbf{t}}_k$ denote the tree with the root having k children, all of them being leaves (that is, $k_\emptyset({\mathbf{t}}_k)=k$ and $\sharp {\mathbf{t}}_k=k+1$ ).

We assume that $0\in A_0$ and p ^′ is $(p, {\mathcal A})$ -compatible. For simplicity, we write ${\mathcal T}'$ for ${\mathcal T}_{p'}$ . We have $p(0)>0$ and $p'(0)>0$ since p and p ^′ satisfy (5). Let $j\in [\![0, J]\!]$ be such that $p'(A_j)>0$ . There exists $k_j\in A_j$ such that $p'(k_j)>0$ . We have

\begin{align*} {\mathbb P}(L_{\mathcal A}({\mathcal T}')=\mathbf e^j)\ge {\mathbb P}({\mathcal T}'={\mathbf{t}}_{k_j})=p'(k_j)p'(0)^{k_j}>0.\end{align*}

This also implies that $ {\mathbb P}(L_{\mathcal A}({\mathcal T}_{p})=\mathbf e^j)>0$ as $\textrm{supp}\,(p')\subset \textrm{supp}\,(p)$ . In particular, thanks to (14), we have, for any $k\in A_j$ ,

\begin{align*}{\mathbb P}({\mathcal T}'={\mathbf{t}}_k\mid L_{\mathcal A}({\mathcal T}')=\mathbf e^j) ={\mathbb P}({\mathcal T}_{p}={\mathbf{t}} _ k\mid L_{\mathcal A}({\mathcal T}_{p})=\mathbf e^j).\end{align*}

This gives

\begin{align*} \frac{p'(k)p'(0)^k}{{\mathbb P}(L_{\mathcal A}({\mathcal T}')=\mathbf e^j)} =\frac{p(k)p(0)^k}{{\mathbb P}(L_{\mathcal A}({\mathcal T}_{p})=\mathbf e^j)},\end{align*}

that is,

(27) \begin{equation} p'(k)=\beta_j \theta^k p(k)\end{equation}

with $\theta=p(0)/p'(0)>0$ and $\beta_j={\mathbb P}(L_{\mathcal A}({\mathcal T}')=\mathbf e^j)/{\mathbb P}(L_{\mathcal A}({\mathcal T}_{p})=\mathbf e^j)>0$ . Note that $\theta$ and $\beta_j$ do not depend on $k\in A_j$ . For $j=0$ , we have $p'(A_0)>0$ as $p'(0)>0$ . For $k=0\in A_0$ , we deduce from (27) that $\beta_0=p'(0)/p(0)=1/\theta$ . For $j\in [\![0, J]\!] $ such that $p'(A_j)=0$ , (27) also holds with $\beta_j=0$ . Note that when $\beta=\boldsymbol{0}$ , we have ${\mathbb P}(L_{\mathcal A}({\mathcal T}')=\boldsymbol{0})=1$ and ${\mathbb P}(L_{\mathcal A}({\mathcal T}_p)=\boldsymbol{0})< 1,$ which entails the fact that $\beta_0={\mathbb P}(L_{\mathcal A}({\mathcal T}')=\boldsymbol{0})/{\mathbb P}(L_{\mathcal A}({\mathcal T}_{p})=\boldsymbol{0})>1$ , or equivalently, $\theta\in (0, 1)$ . This proves that $p'=\tilde p_{\theta, \beta}$ , the latter being defined in (20) as $\theta\in (0, +\infty )$ .

To conclude, note that condition (16) holds as p ^′ is a probability distribution and $\beta_0=1/\theta$ .

Proof of Lemma 4.4. We assume that $0\not \in A_0$ , that is, without loss of generality, $0\in A_1$ , and that p ^′ is $(p, {\mathcal A})$ -compatible. Since p and p ^′ satisfy (5), we get $p(0)>0$ and $p'(0)>0$ . So, we can define $\beta_1=p'(0)/p(0)>0$ . By considering trees with vertices having zero or one child, we get

(30) \begin{equation} 1\in A_0 \, \Longrightarrow \, p'(1)=p(1).\end{equation}

Recall the set ${\mathcal J}^*=\{j\in [\![1, J]\!] \colon p'(A_j)>0\}$ . We set $\beta_j=0$ for $j\in [\![1, J]\!]\setminus {\mathcal J}^*$ in accordance with the definition of the $\tilde{p}_{\theta, \beta}$ . Note that ${\mathcal J}^*$ is nonempty as $1\in {\mathcal J}^*$ . For $j\in {\mathcal J}^*$ , let $k_j$ be an element of $A_j$ such that $p'(k_j)>0$ . For $j\in {\mathcal J}^*\setminus {\mathcal J}^{**}$ , we have $A_j=\{k_j\}$ and, whatever the value of $\theta$ , which will be defined later on, we set $\beta_j= \theta^{-k_j} p'(k_j)/ p(k_j)$ if $\theta\in (0, +\infty )$ and $\beta_j= p'(k_j)/ p(k_j)$ otherwise; this is also in accordance with the definition of the $\tilde p_{\theta, \beta}$ . We now have to consider the value of $p'(\ell)$ for $\ell\in A_j$ and $j\in \{0\} \cup{\mathcal J}^{**}$ .

We first consider the case ${\mathcal J}^{**}=\emptyset$ . In particular, it follows that $A_1=\{0\}$ . Then we have either that $p'(A_0)=p'(A_0\cap\{1\})$ or there exists $k_0 \in A_0$ such that $k_0\geq 2$ and $p'(k_0)>0$ . In the former case, using (30) and $\beta_0=p(1)\, 1_{\{1\in A_0\}}$ , the probability distribution can be written as a $\tilde p_{\theta, \beta}$ with $\theta=0$ . In the latter case, using Lemma 4.5 (i) and (30), we deduce that (20) holds for all $\ell\in A_0$ with a common $\theta\in (0, +\infty )$ (given by the constant value of the map in Lemma 4.5 (i)) and $\beta_0=1/\theta$ ; thus, the probability distribution p ^′ can be written as a $\tilde p_{\theta, \beta}$ with $\theta\in (0, +\infty )$ .

We now assume that ${\mathcal J}^{**}\neq \emptyset$ . From Lemma 4.5 (iii), only the following three cases are possible.

1. (C1) For all $j\in {\mathcal J}^{**}$ , we have $k_j=\min A_j$ , $p'(k_j)>0$ and $p'(A_j\setminus \{k_j\})=0$ .

2. (C2) For all $j\in {\mathcal J}^{**}$ , we have $k_j=\max A_j$ , $p'(k_j)>0$ and $p'(A_j\setminus \{k_j\})=0$ .

3. (C3) For all $j\in {\mathcal J}^{**}$ , we have $p'(k)>0$ for all $k\in A_j$ .

We investigate each case separately. In (C1) we deduce from Lemma 4.5 (ii) that $p'(\ell)=0$ for all $\ell\in A_0$ with $\ell\geq 2$ . Thus, using (30), it follows that the probability distribution p ^′ can be written as a $\tilde p_{\theta, \beta}$ with $\theta=0$ .

In (C2) since $p'(0)>0$ , we deduce that $A_1=\{0\}$ and that if j belongs to ${\mathcal J}^{**}$ (and, thus, to ${\mathcal J}^*$ ) then $\sup A_j$ is finite. Then we use Lemma 4.5 (ii) to deduce that there is no element $\ell\geq 2$ in $A_0$ , that is, $A_0\subset \{1\}$ . Thus, using (30), it follows that the probability distribution p ^′ can be written as a $\tilde p_{\theta, \beta}$ with $\theta=+\infty $ .

In (C3), to fix ideas, let $i=\min {\mathcal J}^{**}$ and consider that $k_i=\min A_i$ and $\ell=\min A_i \setminus \{k_i\}$ . This uniquely determines $\beta_i $ and the $\theta\in (0, +\infty )$ solution of, for $k'\in \{k_i, \ell\}$ ,

(31) \begin{equation} p'(k')=\beta_i \theta^{k'} p(k'). \end{equation}

Then for $k''\in A_i$ larger than $\ell$ , we use Lemma 4.5 (iii) and, in particular, (29), with $j=i$ , $k_j=k''$ and $k=\ell$ to deduce that p ^′(k ^′′) can also be written as in (31). For $j\in {\mathcal J}^{**}$ with $j\neq i$ , set $k=\min A_j$ and define $\beta_j$ by

(32) \begin{equation} p'(k)= \beta_j \theta^{k} p(k). \end{equation}

Then we use Lemma 4.5 (iii) and, in particular, (29), with $k_j>k$ (and $k_j\in A_j$ ), to deduce that (32) holds for k replaced by $k_j$ . Then for $\ell\geq 2$ in $A_0$ , we use Lemma 4.5 (ii) to obtain $p'(\ell)= \theta^{\ell -1} p(\ell)$ (which is also consistent with (30)). We deduce that the probability distribution p ^′ can be written as a $\tilde p_{\theta, \beta}$ with $\theta\in (0,+\infty)$ . This concludes the proof.

Proof of Lemma 4.5. Without loss of generality, we assume that $0\in A_1$ . In the following description of trees we do not give the precise number of leaves, as it is determined through (7). The argument is based on considering two well-chosen trees ${\mathbf{t}}$ and ${\mathbf{t}}'$ such that $L_{\mathcal A}({\mathbf{t}})=L_{\mathcal A}({\mathbf{t}}')$ . Recall that we write $L_k$ for $L_{\{k\}}$ and $k\in {\mathbb N}$ .

We prove part (i) of the lemma. Assume that there exist $k,k'\in A_0$ such that $\min(k, k')\geq 2$ . Consider a tree ${\mathbf{t}}$ having only leaves and $\alpha=k'-1>0$ vertices of out-degree k (that is, $L_{k}({\mathbf{t}})=\alpha$ ) and a tree ${\mathbf{t}}'$ having only leaves and $\beta=k-1>0$ vertices of out-degree k ^′ (that is, $L_{k'}({\mathbf{t}}')=\beta$ ). Set $a=1+(k'-1)(k-1)> 1$ . Thanks to (7), we get $L_{0}({\mathbf{t}})= L_{0}({\mathbf{t}}')=a$ , and thus,

\begin{align*} L_{\mathcal A}({\mathbf{t}})=L_{\mathcal A}({\mathbf{t}}')= \mathbf n \quad\text{with } \mathbf n=a\, \mathbf e^1.\end{align*}

Assume that $p'(k)> 0$ . We have ${\mathbb P}(L_{\mathcal A}({\mathcal T}_{p'})=\mathbf n)\ge{\mathbb P}({\mathcal T}_{p'}={\mathbf{t}})=p'(k)^\alpha p'(0)^a>0$ , and thus, ${\mathbb P}(L_{\mathcal A}({\mathcal T}_{p})=\mathbf n)>0$ . We set

\begin{align*}c=\frac{{\mathbb P}\big(L_{{\mathcal A}}({\mathcal T}_{p'})=\mathbf n\big)}{{\mathbb P}\big(L_{{\mathcal A}}({\mathcal T}_{p})=\mathbf n\big)}>0.\end{align*}

Then, we apply (14) to the trees ${\mathbf{t}}$ and ${\mathbf{t}}'$ to obtain

\begin{align*} p'(k)^\alpha\, p'(0)^{a} =c p(k)^\alpha\, p(0)^{a} \quad\text{and}\quad p'(k')^\beta\, p'(0)^{a} =c p(k')^\beta\, p(0)^{a}.\end{align*}

This readily implies that p ^′(k ^′) is positive (as $k'\in A_0$ implies that $p(k')>0$ ) and that $(p'(k)/p(k))^{1/(k -1)}=(p'(k')/p(k'))^{1/(k' -1)}$ . This gives Lemma 4.5 (i).

We now prove part (ii) of the lemma. Assume that there exist $k'>k\geq 0$ , which are elements of $A_j$ with $j\in{\mathcal J}^{**}$ and $\ell\geq 2$ , which is an element of $A_0$ . Note that j can possibly take the value 1 when $1\in {\mathcal J}^{**}$ (that is, $\textrm{Card}\;(A_1)\geq 2$ ). Consider a tree ${\mathbf{t}}$ having only leaves and $\alpha=\ell-1>0$ vertices of out-degree k ^′ (that is, $L_{k'}({\mathbf{t}})=\alpha$ ) and a tree ${\mathbf{t}}'$ having only leaves, $\beta=k'-k>0$ vertices of out-degree $\ell$ and, if $k>0$ , $\alpha$ vertices of out-degree k (that is, $L_{\ell}({\mathbf{t}}')=\beta$ and, if $k>0$ , $L_{k}({\mathbf{t}}')= \alpha$ ).

We first assume that $k\geq 1$ (which is automatically satisfied if $j\geq 2$ ). Thanks to (7), we get with $a=1+\alpha(k' -1)=1+\alpha(k-1)+ \beta(\ell -1)$ that $ L_{0}({\mathbf{t}})= L_{0}({\mathbf{t}}')=a$ , and thus,

(33) \begin{equation} L_{\mathcal A}({\mathbf{t}})=L_{\mathcal A}({\mathbf{t}}')= \mathbf n\quad\text{with } \mathbf n=a\, \mathbf e^1 + \alpha \mathbf e^j.\end{equation}

Assume that $p'(k')>0$ . We have ${\mathbb P}(L_{\mathcal A}({\mathcal T}_{p'})=\mathbf n)\ge{\mathbb P}({\mathcal T}_{p'}={\mathbf{t}})=p'(k')^\alpha p'(0)^a>0$ , and thus, ${\mathbb P}(L_{\mathcal A}({\mathcal T}_{p})=\mathbf n)>0$ . We set

\begin{align*}c=\frac{{\mathbb P}(L_{{\mathcal A}}({\mathcal T}_{p'})=\mathbf n)}{{\mathbb P}(L_{{\mathcal A}}({\mathcal T}_{p})=\mathbf n)}>0.\end{align*}

Then, we apply (14) to the trees ${\mathbf{t}}$ and ${\mathbf{t}}'$ to get

\begin{align*} p'(k')^\alpha\, p'(0)^{a} =c p(k')^\alpha\, p(0)^{a} \quad\text{and}\quad p'(k)^\alpha\, p'(\ell)^\beta\, p'(0)^{a} =c p(k)^\alpha\, p(\ell)^\beta\, p(0)^{a}.\end{align*}

This readily implies that p ^′(k) and $p'(\ell)$ are positive (as $k\in A_j$ and $\ell\in A_0$ imply that p(k) and $p(\ell)$ are positive). Similarly, assuming that p ^′(k) and $p'(\ell)$ are positive implies that p ^′(k ^′) is positive. Note also that (28) is obvious.

We now consider the case $k=0$ and, thus, $j=1$ (as $0\in A_1$ ). We have $L_{A_1}({\mathbf{t}})=L_{0}({\mathbf{t}}) + L_{k'}({\mathbf{t}})=1+ \alpha(k'-1)+ \alpha= 1+(\ell-1) k'$ and $ L_{A_1}({\mathbf{t}}')=L_{0}({\mathbf{t}}')=1+\beta (\ell -1)= 1+ (\ell-1) k'$ , which implies that (33) still holds. We then conclude similarly as in the case $k>0$ . This gives Lemma 4.5 (ii).

Finally, we prove part (iii) of the lemma. Assume that there exist $i, j\in {\mathcal J}^{**}$ , $k_j>k\geq 0$ that are elements of $A_j$ , $\ell > k_i\geq 0$ that are elements of $A_i$ , with $p'(k_j)>0$ and $p'(k_i)>0$ . Note that i and j can be possibly equal and can possibly take the value 1 when $1\in {\mathcal J}^{**}$ . Consider a tree ${\mathbf{t}}$ having only leaves, $\alpha=\ell-k_i>0$ vertices of out-degree $k_j$ and, if $k_i>0$ , $\beta=k_j-k>0$ vertices of out-degree $k_i$ (that is, $L_{k_j}({\mathbf{t}})=\alpha$ and, if $k_i>0$ , $L_{k_i}({\mathbf{t}})=\beta$ ) and a tree ${\mathbf{t}}'$ having only leaves, $\beta$ vertices of out-degree $\ell$ , and, if $k>0$ , $\alpha$ vertices of out-degree k (that is, $L_{\ell}({\mathbf{t}}')=\beta$ and, if $k>0$ , $L_{k}({\mathbf{t}}')=\alpha$ ).

We assume that $k\geq 1$ and $k_i\geq 1$ , and leave the cases $k=0$ (and, thus, $j=1$ ) and/or $k_i=0$ (and, thus, $i=1$ ) to the reader as the proof can be handled very similarly; see also the end of the proof of Lemma 4.5 (ii). Thanks to (7), we get with $a=1+\alpha(k_j -1)+ \beta (k_i-1)=1+\alpha(k-1)+ \beta(\ell -1)$ that $ L_{0}({\mathbf{t}})=L_{0}({\mathbf{t}}')=a$ , and thus,

\begin{align*} L_{\mathcal A}({\mathbf{t}})=L_{\mathcal A}({\mathbf{t}}')= \mathbf n \quad\text{with } \mathbf n=a\, \mathbf e^1 + \alpha \mathbf e^j+ \beta \mathbf e^i.\end{align*}

We have ${\mathbb P}(L_{\mathcal A}({\mathcal T}_{p'})=\mathbf n)\ge{\mathbb P}({\mathcal T}_{p'}={\mathbf{t}})=p'(k_j)^\alpha \, p'(k_i)^\beta\, p'(0)^a>0$ ; thus, ${\mathbb P}(L_{\mathcal A}({\mathcal T}_{p})=\mathbf n)>0$ . We set

\begin{align*}c=\frac{{\mathbb P}(L_{{\mathcal A}}({\mathcal T}_{p'})=\mathbf n)}{{\mathbb P}(L_{{\mathcal A}}({\mathcal T}_{p})=\mathbf n)}>0.\end{align*}

Then, we apply (14) to the trees ${\mathbf{t}}$ and ${\mathbf{t}}'$ to get

\begin{align*} p'(k_j)^\alpha\, p'(k_i)^\beta\, p'(0)^{a} =c p(k_j)^\alpha\, p(k_i)^\beta\, p(0)^{a} \end{align*}

and

\begin{align*} p'(k)^\alpha\, p'(\ell)^\beta\, p'(0)^{a} =c p(k)^\alpha\, p(\ell)^\beta\, p(0)^{a}.\end{align*}

This readily implies that p ^′(k) and $p'(\ell)$ are positive (as $k\in A_j$ and $\ell\in A_i$ imply that p(k) and $p(\ell)$ are positive). Equation (29) is then obvious. This gives Lemma 4.5 (iii).