Proof of Lemma 3.1. From [Reference Segers31, Example 1] we have the limit

\[ \sum_{j\in V} b_{uj} \delta_{\left\{\frac{b_{vj}}{b_{uj}}, v\in V_{\tau}\right\}}. \]

Adapting this representation to a model where we have $b_{uj}=0$ for $j\notin \mathrm{An}(u)$ and $b_{ij}=c_{p(j,i)}b_{jj}$ for $j\in \mathrm{An}(i)$ , we obtain

\[ \sum_{j\in \mathrm{An}(u)} b_{uj} \delta_{\left\{\frac{c_{p(j,v)}b_{jj}}{c_{p(j,u)}b_{jj}}, v\in V_{\tau}\right\}}. \]

Recall that $c_{p(i,i)}=1$ and $c_{p(i,j)}=0$ if $i\notin \mathrm{An}(j)$ .

Next we show that $(M_{uv}, v\in V_{\tau})$ are mutually dependent. When u is the source of $\tau$ , for every $j\in \mathrm{An}(u)$ the atom

\[ \bigg(\frac{c_{p(j,v)}}{c_{p(j,u)}}, v\in V_{\tau}\bigg) =\bigg(\frac{c_{p(j,u)}c_{uv}}{c_{p(j,u)}}, v\in V_{\tau}\bigg) =(1;\; c_{uv}, v\in V_{\tau}\setminus u) \]

has probability $\sum_{j\in \mathrm{An}(u)}b_{uj}=1$ . Hence $(M_{uv}, v\in V_{\tau})$ are at the same time perfectly dependent and independent.

For a node u which is not the source node, the general idea is to take a collection of coordinates for which the joint probability is zero while the product of the marginal probabilities is positive, thus showing that the joint probability does not equal the product of the marginal probabilities for the selected possible value of the random vector.

For brevity, let $V_{\tau}=\{1,2,\ldots,m \}$ , with the nodes labelled according to the order of their out-degrees within $\tau$ : the source node of $\tau$ has out-degree $m-1$ (the largest) and is labelled as 1, the node with out-degree $m-2$ is labelled as 2, etc.

Suppose u is the node 2. Thanks to the no-cycle property within a tournament, we have $\mathrm{An}(2)=\mathrm{An}(1)\cup \{2\}$ . For all $j\in \mathrm{An}(1)$ we have

(23) \begin{equation} \bigg(\frac{c_{p(j,v)}}{c_{p(j,2)}}, v=1, \ldots, m \bigg) =\bigg( \frac{1}{c_{12}};\; 1; \frac{c_{p(j,1)}c_{1v}}{c_{p(j,1)}c_{12}}, v = 3, \ldots, m \bigg), \end{equation}

which is an atom of $(M_{2v},v=1,\ldots, m)$ with mass $\sum_{j\in \mathrm{An}(1)}b_{2j}$ . This means that for the marginal distribution of $M_{21}$ we have $\operatorname{\mathbb{P}}(M_{21}=1/c_{12})\geq\sum_{j\in \mathrm{An}(1)}b_{2j}$ . For $j=2$ we have an atom $(0, 1, c_{23}, \ldots, c_{2m})$ with mass $b_{22}$ . This means that for the marginal probabilities of $(M_{23}, \ldots, M_{2m})$ we have $\operatorname{\mathbb{P}}(M_{2v}=c_{2v})\geq b_{22}$ for all $v=3, \ldots, m$ . Take a vector of coordinates $(1/c_{12}, 1, c_{23}, \ldots, c_{2m})$ . Note that this vector cannot be the same as the one in (23). For any $v=3, \ldots, m$ we cannot have $c_{1v}/c_{12}=c_{2v}$ because of the criticality assumption, according to which $c_{1v} > c_{12}c_{2v}$ for any $v = 3,\ldots,m$ . The joint probability of this vector of coordinates is

\[ \operatorname{\mathbb{P}}(M_{21}=1/c_{12}, M_{22}=1, M_{23}=c_{23},\ldots , M_{2m}=c_{2m}) =0. \]

However, the product of the marginal probabilities is positive:

\[ \operatorname{\mathbb{P}}(M_{21}=1/c_{12}) \operatorname{\mathbb{P}}( M_{22}=1) \prod_{v=3}^m\operatorname{\mathbb{P}}( M_{2v}=c_{2v}) \geq \sum_{j\in \mathrm{An}(1)}b_{2j}\times b_{22}^{m-1} >0. \]

Now let $u\geq 3$ . Take the vector of coordinates in (17) corresponding to $j=1$ , which is equal to $(1/c_{1u},c_{12}/c_{1u}, \ldots, c_{1m}/c_{1u} )$ and has probability at least $b_{u1}$ . Consider also the vector of coordinates for $j=u$ , which is $(0, \ldots, 0, 1;\; c_{uv}, v=u+1, \ldots, m)$ with mass at least $b_{uu}$ . Replace the first coordinate by $1/c_{1u}$ . The vector obtained in this way has joint probability zero. For every $j\in \mathrm{pa}(u)$ we have $b_{vj}/b_{uj}=0$ when v is not a child of j, or equivalently, given the order in the node labelling, when $v<j$ . So for fixed $u\geq3$ , for $j=1$ the vector $(b_{vj}/b_{uj}, v=1, \ldots, m)$ has no zeros. For $j=2$ the vector $(b_{vj}/b_{uj}, v=1, \ldots, m)$ has one zero, namely $(0;b_{vj}/b_{uj}, v=2, \ldots, m)$ ; for $j=3$ the vector $(b_{vj}/b_{uj}, v=1, \ldots, m)$ has two zeros, namely $(0,0;b_{vj}/b_{uj}, v=3, \ldots, m)$ ; and so on until $j=u$ with the corresponding vector $(b_{vj}/b_{uj}, v=1, \ldots, m)=(0,\ldots, 0;b_{vj}/b_{uj}, v=u, \ldots, m)$ . If we replace the first coordinate by a non-zero value in this vector, we get an impossible value for the random vector $(M_{uv}, v=1, \ldots, m)$ or a value with probability zero. Considering the univariate marginal distributions of $(M_{uv}, v=1, \ldots, m)$ , we obtain for the product of the marginal probabilities a positive value:

\[ \operatorname{\mathbb{P}}(M_{u1}=1/c_{1u}) \left[\prod_{v=2}^{u-1}\operatorname{\mathbb{P}}( M_{uv}=0)\right] \operatorname{\mathbb{P}}(M_{uu}=1) \prod_{v=u+1}^m\operatorname{\mathbb{P}}( M_{uv}=c_{uv}) \geq b_{u1}\times b_{uu}^{m-1} >0. \]

This shows that for any $u\in V_{\tau}$ the vector $(M_{u1}, \ldots, M_{um})$ has jointly dependent elements.

Next we show the distribution of a single element $M_{uv}, v\in V_{\tau}\setminus u$ :

1. 1. Consider first the case when u is the source node in $\tau$ . Since $(u, v) \in E$ , we have $\mathrm{An}(u) \subset \mathrm{An}(v)$ and thus $\mathrm{An}(v) \cap \mathrm{An}(u) = \mathrm{An}(u)$ . We have $b_{vj}>0$ , $j\in \mathrm{An}(u)$ ; hence zero is not a possible value of $M_{uv}$ . For $j\in \mathrm{An}(u)$ ,

   \[ \frac{b_{vj}}{b_{uj}}=\frac{c_{p(j,u)}c_{uv}b_{jj}}{c_{p(j,u)}b_{jj}}=c_{uv}, \]
   and since $\sum_{j\in \mathrm{An}(u)}b_{uj}=1$ we obtain the desired result under 1.(a).

   When u is not the source node in $\tau$ , not all shortest paths to v pass through u; hence for $j\in\mathrm{An}(u)$ we have

   \[ \frac{b_{vj}}{b_{uj}}=\frac{c_{p(j,v)}b_{jj}}{c_{p(j,u)}b_{jj}} =\frac{c_{p(j,v)}}{c_{p(j,u)}} >0 \]
   with mass $b_{uj}$ . This yields the result in 1.(b). Note that zero is not a possible value, as we still have $\mathrm{An}(u)\subset \mathrm{An}(v)$ . Also, $c_{p(u,u)}=1$ by convention.

2. 2. Let us now consider $(v,u)\in E$ . In this case $\mathrm{An}(u)\setminus\mathrm{An}(v)$ is not empty, because it contains at least the node u, so zero is a possible value of $M_{uv}$ . We need to distinguish only the zero atoms from the non-zero ones. When v is a source node in $\tau$ , for $j\in \mathrm{An}(v)$ we have

   \[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,v)}b_{jj}}{c_{p(j,u)}b_{jj}} =\frac{c_{p(j,v)}}{c_{p(j,v)}c_{vu}} =\frac{1}{c_{vu}} >0, \]
   which is an atom with probability
   \[ \sum_{j\in \mathrm{An}(v)}b_{uj} =\sum_{j\in \mathrm{An}(v)}c_{p(j,v)}c_{vu}b_{jj} =c_{vu}\sum_{j \in \mathrm{An}(v)} c_{p(j,v)} b_{jj} =c_{vu}\sum_{j \in \mathrm{An}(v)} b_{vj} = c_{vu}. \]
   The probability of the zero atom is $\sum_{j\in \mathrm{An}(u)\setminus \mathrm{An}(v)}b_{uj}=\sum_{j\in \mathrm{An}(u)}b_{uj}-\sum_{j\in \mathrm{An}(v)}b_{uj}=1-c_{vu}$ . This shows 2.(a).

   When v is not a source node of $\tau$ , we have for $j\in \mathrm{An}(v)$

   \[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,v)}b_{jj}}{c_{p(j,u)}b_{jj}} =\frac{c_{p(j,v)}}{c_{p(j,u)}}>0, \]
   an atom with mass $b_{uj}$ and zero atom with probability $\sum_{j\in \mathrm{An}(u)\setminus \mathrm{An}(v)}b_{uj}$ . This shows 2.(b).

Proof. We have already seen that from [Reference Segers31, Example 1] we have the limit

\[ \mathcal{L}\bigg(\frac{X_v}{X_u}, v\in V\mid X_u>t\bigg) \stackrel{d}{\longrightarrow} \sum_{j\in V} b_{uj} \delta_{\left\{\frac{b_{vj}}{b_{uj}}, v\in V\right\}}. \]

Adapting this representation to a model where we have $b_{uj}=0$ for $j\notin \mathrm{An}(u)$ and $b_{ij}=c_{p(j,i)}b_{jj}$ for $j\in \mathrm{An}(i)$ , we obtain

\[ \sum_{j\in \mathrm{An}(u)} b_{uj} \delta_{\left\{\frac{c_{p(j,v)}}{c_{p(j,u)}}, v\in V\right\}}. \]

Recall that $c_{p(i,i)}=1$ and $c_{p(i,j)}=0$ if $i\notin \mathrm{An}(j)$ . For a single $v\in V\setminus u$ we have the marginal distribution

(25) \begin{equation} \mathcal{L}(A_{uv})=\sum_{j\in\mathrm{An}(u)}b_{uj} \delta_{\left\{\frac{b_{vj}}{b_{uj}}\right\}}. \end{equation}

The distribution of $A_{uv}$ depends deterministically on properties of the TTT. When $\mathcal{T}$ has a unique source, according to Lemma 2.3-2 there are three possible shortest trails between two nodes. In addition we have also the property under Lemma 2.3-1. We look at the different distributions of $A_{uv}$ that arise from these two properties of the TTT.

First we deal with 1.(a). Since $\mathrm{An}(u)\subset \mathrm{An}(v)$ , all atoms in (25) are positive and zero is not a possible value of $A_{uv}$ . All paths from $\mathrm{An}(u)$ to v pass through u, because u is a source in $\tau$ and because, by Property (P2) of a TTT, no cycle involving several tournaments is allowed. The case is illustrated by the graph below.

Hence for all $j\in \mathrm{An}(u)$ we have

\[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,v)}b_{jj}}{c_{p(j,u)}b_{jj}} =\frac{c_{p(j,u)}c_{p(u,v)}}{c_{p(j,u)}} =c_{p(u,v)}>0, \]

with mass $\sum_{j\in \mathrm{An}(u)}b_{uj}=1$ .

Next we show 1.(b). Because $\mathrm{An}(u)\subset \mathrm{An}(v)$ , zero is not a possible value of $A_{uv}$ . Not all shortest paths from $\mathrm{An}(u)$ to v pass through u, because u is not a source in $\tau$ . However, all paths from $\mathrm{An}(u)$ to v pass through r, as shown in the picture. Paths from $\mathrm{An}(u)$ to v other than these passing through u or r are impossible because of Property (P2) of a TTT.

For $j\in \mathrm{An}(u)$ we have

\[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,v)}b_{jj}}{c_{p(j,u)}b_{jj}} =\frac{c_{p(j,r)}}{c_{p(j,u)}}c_{p(r,v)}>0, \]

with mass $b_{uj}$ , hence the expression in 1.(b).

Next we show 2.(a). When the directed path is from v to u, the set $\mathrm{An}(u)\setminus \mathrm{An}(v)$ contains at least u; hence we have $b_{vj}=0$ for all $j\in \mathrm{An}(u)\setminus \mathrm{An}(v)$ . This means that zero is a possible value of $A_{uv}$ . All shortest paths from $j\in \mathrm{An}(v)$ to u pass through v, as v is a source in $\tau$ . Otherwise, there would be a cycle involving multiple tournaments, which is not allowed under Property (P2) of a TTT.

For $j\in \mathrm{An}(v)$ , the non-zero atom is given by

\[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,v)}b_{jj}}{c_{p(j,u)}b_{jj}} =\frac{c_{p(j,v)}}{c_{p(j,v)}c_{p(v,u)}}=\frac{1}{c_{p(v,u)}}>0, \qquad j\in\mathrm{An}(v), \]

with mass

\begin{align*} \sum_{j\in \mathrm{An}(v)}b_{uj} =\sum_{j\in \mathrm{An}(v)}c_{p(j,v)}c_{p(v,u)}b_{jj} &=c_{p(v,u)}\sum_{j\in \mathrm{An}(v)}c_{p(j,v)}b_{jj} \\[5pt] &=c_{p(v,u)}\sum_{j\in \mathrm{An}(v)}b_{vj} =c_{p(v,u)}. \end{align*}

For the zero atom we have probability

\[ \sum_{j\in\mathrm{An}(u)\setminus \mathrm{An}(v)}b_{uj}=\sum_{j\in \mathrm{An}(u)}b_{uj}-\sum_{j\in \mathrm{An}(v)}b_{uj}=1-c_{p(v,u)}. \]

This shows 2.(a).

To show 2.(b) we note that when v is not a source node of $\tau$ , not all shortest paths from $j\in \mathrm{An}(v)$ to u pass through v. However, all paths from $j\in \mathrm{An}(v)$ to u pass through r, as can be seen from the figure below.

Hence for $j\in \mathrm{An}(v)$ we have

\[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,v)}b_{jj}}{c_{p(j,u)}b_{jj}} =\frac{c_{p(j,v)}}{c_{p(j,r)}c_{p(r,u)}}>0, \]

which is an atom with mass $b_{uj}=c_{p(j,r)}c_{p(r,u)}b_{jj}=c_{p(r,u)}b_{rj}$ . The zero atom comes from the fact that $b_{vj}=0$ for all $j\in \mathrm{An}(u)\setminus \mathrm{An}(v)$ , and it has probability

\[ \sum_{j\in\mathrm{An}(u)\setminus \mathrm{An}(v)}b_{uj}. \]

This shows the distribution under 2.(b).

By Lemma 2.3-1, the node r, as part of the path p(r, u), is allowed not to be a source node in $\tau_m$ . A similar statement can be made in relation to the path p(r, v). However, when we combine p(r, u) and p(r, v) in one trail t(u, v), the node r should be a source in at least one of $\tau_m$ and $\tau_n$ . Indeed, if r were not a source either in $\tau_m$ or in $\tau_n$ , then there would be a v-structure. However, Lemma 2.2-2 excludes v-structures when $\mathcal{T}$ has a unique source; hence node r should be a source in at least one of the tournaments $\tau_m$ and $\tau_n$ .

To show 3.(a), we note that all paths from $j\in \mathrm{An}(r)$ to u and to v pass through r, as r is a source in both $\tau_n$ and $\tau_m$ . The case is depicted in the following picture.

Also, we have $b_{vj}=0$ for all $j\in \mathrm{An}(u)\setminus \mathrm{An}(r)$ . For $j\in \mathrm{An}(r)$ we have

\[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,r)}c_{p(r,v)}b_{jj}}{c_{p(j,r)}c_{p(r,u)}b_{jj}} =\frac{c_{p(r,v)}}{c_{p(r,u)}}>0, \]

with probability

\begin{align*} \sum_{j\in \mathrm{An}(r)}b_{uj} =\sum_{j\in \mathrm{An}(r)}c_{p(j,r)}c_{p(r,u)}b_{jj} &=c_{p(r,u)}\sum_{j\in \mathrm{An}(r)}c_{p(j,r)}b_{jj} \\[5pt] &=c_{p(r,u)}\sum_{j\in \mathrm{An}(r)}b_{rj} =c_{p(r,u)}. \end{align*}

The probability of the zero atom is

\begin{align*} \sum_{j\in \mathrm{An}(u)\setminus \mathrm{An}(r)}b_{uj}=\sum_{j\in \mathrm{An}(u)}b_{uj}-\sum_{j\in\mathrm{An}(r)}b_{uj}=1-c_{p(r,u)}. \end{align*}

Next we show 3.(b). Because r is not a source in $\tau_n$ , not all paths from $\mathrm{An}(r)$ to v pass through r, but they do all pass through n. Also, all paths from $\mathrm{An}(r)$ to u pass through r, because r is a source in $\tau_m$ .

Hence, for $j\in \mathrm{An}(r)$ ,

\[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,n)}c_{p(n,v)}b_{jj}}{c_{p(j,r)}c_{p(r,u)}b_{jj}} =\frac{c_{p(j,n)}c_{p(n,v)}}{c_{p(j,r)}c_{p(r,u)}}>0, \]

which is an atom with mass $b_{uj}=c_{p(j,r)}c_{p(r,u)}b_{jj}=b_{rj}c_{p(r,u)}$ . The zero atom has probability equal to $\sum_{j\in \mathrm{An}(u)\setminus \mathrm{An}(r)}b_{uj}$ .

Next we show 3.(c). When r is a source in $\tau_n$ , this means that all paths from $\mathrm{An}(r)$ to v pass through r. Because r is not a source in $\tau_m$ , not all paths from $\mathrm{An}(r)$ to u pass through r, but they do all pass through m.

For $j\in \mathrm{An}(r)$ we have

\[ \frac{b_{vj}}{b_{uj}} =\frac{c_{p(j,r)}c_{p(r,v)}b_{jj}}{c_{p(j,m)}c_{p(m,u)}b_{jj}} =\frac{c_{p(j,r)}c_{p(r,v)}}{c_{p(j,m)}c_{p(m,u)}}>0, \]

which is an atom with mass $b_{uj}=c_{p(j,m)}c_{p(m,u)}b_{jj}=b_{mj}c_{p(m,u)}$ . The zero atom comes from $b_{uj}=0$ for all $j\in\mathrm{An}(u)\setminus \mathrm{An}(r)$ . It has probability $\sum_{j\in \mathrm{An}(u)\setminus \mathrm{An}(r)}b_{uj}$ .

Proof of Proposition 3.1. First we prove that (i) implies (ii). Assume $\mathcal{T}$ has a unique source. We have to prove that for any $u\in V$ , an element from the limiting vector in (15) is given by (16).

In Lemma B.1 we have seen a number of cases for the distribution of $A_{uv}$ , depending on deterministic properties of the trail between u and v. Below we consider each of these cases again.

Case 1. (a). Let u be the source in $\tau$ . From Lemma B.1-1.(a) we have $P(A_{uv}=c_{p(u,v)})=1$ . Consider the variables $(M_e, e\in p(u,v))$ , which are by construction independent of each other, because they belong to different tournaments. Note that in this case every node $v_1,\ldots, v_{n-1}$ is the source node in the tournament containing that node and the next one in the sequence. This follows from Lemma 2.3-1. Then, by Lemma 3.1 1.(a), for every $M_e, e\in p(u,v)$ we have $\operatorname{\mathbb{P}}(M_e=c_e)=1$ , and hence

\[ \operatorname{\mathbb{P}}\left(\prod_{e\in p(u,v)}M_e=c_{p(u,v)}\right) = \prod_{e\in p(u,v)}\operatorname{\mathbb{P}}(M_{e}=c_e)=1, \]

which shows $A_{uv}=\prod_{e\in p(u,v)}M_e$ .

Case 1. (b). If u is not the source in $\tau$ , the distribution of $M_{ur}$ is as in Lemma 3.1-1.(b). As in Case 1.(a), every node $r=v_2,v_3,\ldots, v_{n-1}$ is the source node in the tournament containing that node and the next one in the sequence. The variables $M_e, e\in p(r,v)$ are degenerate at $c_e$ . As in Case 1.(a), the variables $(M_e, e\in p(u,v))$ are by construction independent of each other, because they are indexed by edges which belong to different tournaments. Then we have

(26) \begin{equation} \begin{split} \mathcal{L}\left(\prod_{e\in p(u,v)}M_e\right) =\mathcal{L}\left(M_{ur}\prod_{e\in p(r,v)}M_e\right) &=\left(\sum_{j\in \mathrm{An}(u)} b_{uj}\delta_{\left\{\frac{c_{p(j,r)}}{c_{p(j,u)}}\right\}}\right) \otimes \delta_{\{c_{p(r,v)}\}} \\[5pt] &=\sum_{j\in \mathrm{An}(u)} b_{uj}\delta_{\left\{\frac{c_{p(j,r)}}{c_{p(j,u)}}c_{p(r,v)}\right\}}. \end{split} \end{equation}

The symbol $\otimes$ denotes multiplication between two discrete probability measures, say $\mu$ and $\nu$ , of two independent variables, say $\xi_1$ and $\xi_2$ , respectively. For two possible values $a_1,a_2$ of $\xi_1, \xi_2$ respectively, we have $\mu(\{a_1\})\nu(\{a_1\})$ as a measure of the event $\{\xi_1\xi_2=a_1a_2\}=\{\xi_1=a_1, \xi_2=a_2\}$ . The last expression in (26) is the distribution of $A_{uv}$ in Lemma B.1-1.(b).

Case 2. (a). Let v be the source in $\tau$ . Consider the random variables $M_{v_{i+1}v_{i}}$ , $i=1,\ldots, n-1,$ whose distributions are as in Lemma 3.1-2.(a). Since this is the unique shortest trail from v to u, all edges on it belong to different tournaments, and the vector $(M_{v_{i+1}v_{i}}, i=1,\ldots, n-1)$ contains independent variables by definition. Then

(27) \begin{equation} \operatorname{\mathbb{P}}\Bigg(\prod_{i=1}^{n-1}M_{v_{i+1}v_{i}}=\frac{1}{c_{p(v,u)}}\Bigg)= \prod_{i=1}^{n-1}\operatorname{\mathbb{P}}\bigg(M_{v_{i+1}v_{i}}=\frac{1}{c_{v_{i}v_{i+1}}}\bigg) =\prod_{i=1}^{n-1}c_{v_{i}v_{i+1}}=c_{p(v,u)}. \end{equation}

For the zero atom we have

(28) \begin{align} \operatorname{\mathbb{P}}\Bigg(\prod_{i=1}^{n-1}M_{v_{i+1}v_{i}}=0\Bigg) =1-\prod_{i=1}^{n-1}\operatorname{\mathbb{P}}(M_{v_{i+1}v_{i}}>0)\nonumber &=1-\prod_{i=1}^{n-1}\operatorname{\mathbb{P}}\bigg(M_{v_{i+1}v_{i}} =\frac{1}{c_{v_{i}v_{i+1}}}\bigg) \\[5pt] &=1-c_{p(v,u)}. \end{align}

The expressions in (27) and (28) represent the distribution of $A_{uv}$ in Lemma B.1-2.(a).

Case 2. (b). If v is not the source in $\tau$ , consider a random variable $M_{rv}$ with distribution as in Lemma 3.1-2.(b) and a random variable $A_{ur}$ constructed as in Case 2.(a) above, i.e., as the product $\prod_{i=2}^{n-1}M_{v_{i+1}v_i}$ . By construction $M_{rv}$ is independent from $A_{ur}$ by the same argument as above. We have

\begin{align*} \mathcal{L}(A_{ur}M_{rv}) &=\left(c_{p(r,u)}\delta_{\left\{\frac{1}{c_{p(r,u)}}\right\}} +(1-c_{p(r,u)})\delta_{\{0\}}\right) \\[5pt] &\otimes \left(\sum_{j\in \mathrm{An}(v)} b_{rj} \delta_{\left\{\frac{c_{p(j,v)}}{c_{p(j,r)}}\right\}} +\sum_{j\in \mathrm{An}(r)\setminus \mathrm{An}(v)}b_{rj}\delta_{\{0\}}\right), \end{align*}

which gives non-zero atoms $c_{p(j,v)}/(c_{p(j,r)}c_{p(r,u)})$ , $j\in \mathrm{An}(v)$ , with masses $b_{rj}c_{p(r,u)}$ , $j\in \mathrm{An}(v)$ . To show the probability of the zero atom, consider

\begin{align*} &\operatorname{\mathbb{P}}(A_{ur}M_{rv}=0) =1-\operatorname{\mathbb{P}}(A_{ur}>0) \cdot \operatorname{\mathbb{P}}(M_{rv}>0) =1-c_{p(r,u)}\sum_{j\in \mathrm{An}(v)}b_{rj} \\[5pt] &=\sum_{j\in \mathrm{An}(u)}b_{uj}-\sum_{j\in \mathrm{An}(v)}c_{p(j,r)}b_{jj}c_{p(r,u)} =\sum_{j\in \mathrm{An}(u)}b_{uj}-\sum_{j\in \mathrm{An}(v)}b_{uj} =\sum_{j\in \mathrm{An}(u)\setminus \mathrm{An}(v)}b_{uj}, \end{align*}

which is what we need to confirm $A_{uv}=A_{ur}M_{rv}$ , where $A_{uv}$ is as in Lemma B.1-2.(b).

Case 3. (a). Let r be the source in both $\tau_m$ and $\tau_n$ . Consider random variables $A_{rv}$ as in Lemma B.1-1.(a) and $A_{ur}$ as in Lemma B.1-2.(a). Above we have shown in Cases 1.(a) and 2.(a) that $A_{rv}$ and $A_{ur}$ are factorizable into independent multiplicative increments. By construction $A_{rv}$ and $A_{ur}$ are independent from each other, because the multiplicative increments are independent. We have

\begin{align*} \operatorname{\mathbb{P}}\bigg(A_{ur}A_{rv}=\frac{c_{p(r,v)}}{c_{p(r,u)}}\bigg) =\operatorname{\mathbb{P}}\bigg(A_{ur}=\frac{1}{c_{p(r,u)}}\bigg)\operatorname{\mathbb{P}}(A_{rv}=c_{p(r,v)}) =c_{p(r,u)}. \end{align*}

For the probability of the zero atom we have

\[ \operatorname{\mathbb{P}}(A_{ur}A_{rv}=0)=P(A_{ur}=0)=(1-c_{p(r,u)}). \]

The two displays above represent the distribution of $A_{uv}$ in Lemma B.1-3.(a).

Case 3. (b). Let r be the source in $\tau_m$ , but not in $\tau_n$ . Consider three random variables $A_{ur}, M_{rn}, A_{nv}$ with distributions as in Lemma B.1-2.(a), Lemma 3.1-1.(b), and Lemma B.1-1.(a), respectively. For $A_{ur}$ and $A_{nv}$ , we have shown in Cases 2.(a) and 1.(a) above that they are factorizable into independent multiplicative increments. By construction $M_{rn}$ is independent from the increments in $A_{ur}$ and $A_{nv}$ . Then

\begin{align*} &\mathcal{L}(A_{ur}M_{rn}A_{nv}) \\[5pt] &=\left(c_{p(r,u)}\delta_{\left\{\frac{1}{c_{p(r,u)}}\right\}} +(1-c_{p(r,u)})\delta_{\{0\}}\right) \otimes \left(\sum_{j\in \mathrm{An}(r)}b_{rj} \delta_{\left\{\frac{c_{p(j,n)}}{c_{p(j,r)}}\right\}}\right) \otimes \delta_{\{c_{p(n,v)}\}} \\[5pt] &=\sum_{j\in \mathrm{An}(r)}b_{rj}c_{p(r,u)} \delta_{\left\{\frac{c_{p(j,n)}c_{p(n,v)}}{c_{p(j,r)}c_{p(r,u)}}\right\}} +(1-c_{p(r,u)})\delta_{\{0\}}. \end{align*}

Note that

\begin{align*} \sum_{j\in \mathrm{An}(u)\setminus \mathrm{An}(r)}b_{uj} =\sum_{j\in\mathrm{An}(u)}b_{uj}-\sum_{j\in \mathrm{An}(r)}b_{uj} &=1-\sum_{j\in \mathrm{An}(r)} c_{p(j,r)}c_{p(r,u)}b_{jj} \\[5pt] &=1-c_{p(r,u)}\sum_{j\in \mathrm{An}(r)} b_{rj} =1-c_{p(r,u)}. \end{align*}

This shows that the distribution of $A_{ur}M_{rn}A_{nv}$ is that of $A_{uv}$ in Lemma B.1-3.(b).

Case 3. (c). Let r be the source in $\tau_n$ , but not in $\tau_m$ . Consider variables $A_{um}, M_{mr}, A_{rv}$ with distributions as in Lemma B.1-2.(a), Lemma 3.1-2.(b), and Lemma B.1-1.(a), respectively. The variables $A_{um}$ and $A_{rv}$ have been shown to factorize into independent increments in Cases 2.(a) and 1.(a), respectively, of this proof; hence they are also independent from each other. By construction $M_{mr}$ is independent from $A_{um}$ and $A_{rv}$ . Then we have

\begin{align*} \mathcal{L}(A_{um}M_{mr}A_{rv}) &=\left(c_{p(m,u)}\delta_{\left\{\frac{1}{c_{p(m,u)}}\right\}} +(1-c_{p(m,u)})\delta_{\{0\}}\right) \\[5pt] & \otimes \left(\sum_{j\in \mathrm{An}(r)}b_{mj} \delta_{\left\{\frac{c_{p(j,r)}}{c_{p(j,m)}}\right\}} +\sum_{j\in \mathrm{An}(m)\setminus \mathrm{An}(r)}b_{mj}\delta_{\{0\}}\right) \otimes \delta_{\{c_{p(r,v)}\}}. \end{align*}

The non-zero atoms are $c_{p(j,r)}c_{p(r,v)}/(c_{p(j,m)}c_{p(m,u)})$ for $j\in \mathrm{An}(r)$ with masses

\[ c_{p(m,u)}b_{mj}=c_{p(j,m)}c_{p(m,u)}b_{jj}=b_{uj}, \qquad j\in \mathrm{An}(r). \]

The probability of the zero atom is given by

\begin{align*} &\operatorname{\mathbb{P}}(A_{um}M_{mr}A_{rv}=0) =1-\operatorname{\mathbb{P}}(A_{um}>0)\operatorname{\mathbb{P}}(M_{mr}>0) =1-c_{p(m,u)}\sum_{j\in \mathrm{An}(r)}b_{mj} \\[5pt] &=1-\sum_{j\in \mathrm{An}(r)} c_{p(j,m)}c_{p(m,u)}b_{jj} =\sum_{j\in\mathrm{An}(u)}b_{uj}-\sum_{j\in\mathrm{An}(r)}b_{uj} =\sum_{j\in \mathrm{An}(u)\setminus \mathrm{An}(r)}b_{uj}. \end{align*}

Hence the distribution of $A_{um}M_{mr}A_{rv}$ is that of $A_{uv}$ in Lemma B.1-3.(c). This completes the proof that the statement in (i) implies (ii).

The statement in (iii) holds trivially from (ii).

Next we prove that (iii) implies (i) by contraposition: we assume that $\mathcal{T}$ has at least two sources and show that it is not possible to obtain the factorization in (16). If $\mathcal{T}$ has at least two sources, then by Lemma 2.2-2 there is at least one v-structure, say on nodes 1,2,3 and involving edges $(1,3), (2,3)\in E$ . Consider the nodes 1,2. For every $u\in V$ we have two possibilities:

1. (a) The v-structure belongs to only one of the trails t(u, 1) and t(u, 2): without loss of generality, $(1,3), (2,3)\in t(u,2)$ and $(1,3), (2,3)\notin t(u,1)$ .

2. (b) Each trail t(u, 1) and t(u, 2) contains one edge of the v-structure: without loss of generality, $(1,3)\in t(u,1)$ and $(2,3)\in t(u,2)$ .

If $u \in \{1, 2\}$ , then we are in the case (a), while if $u = 3$ we are in the case (b). If $u \not\in \{1,2,3\}$ , then node 3 must belong to at least one of the two trails t(u, 1) or t(u, 2), because otherwise the skeleton graph would have a cycle connecting nodes u, 1, 2, 3 and passing through more than one block. The latter is impossible according to Property (P2). The two possibilities are illustrated in Figure 5.

Figure 5. The two possible configurations of the trails t(u, 1) and t(u, 2) when nodes 1, 2, 3 form a v-structure.

Case 3. (c-i). Consider first the case when, without loss of generality, the v-structure belongs to t(u, 2) but not to t(u, 1); see Figure 5(a). Let the trail from 1 to u be on nodes $\{v_1=1, v_2, \ldots, v_n=u\}$ . We can have any direction on the edges of t(1, u). Recall the distribution of $A_{u2}$ :

\[ \mathcal{L}(A_{u2})=\sum_{j\in\mathrm{An}(u)}b_{uj}\delta_{\{b_{2j}/b_{uj}\}}. \]

We have $b_{2j}=0$ for all $j\notin \mathrm{An}(2)$ . We claim that $\mathrm{An}(u)\cap \mathrm{An}(2)=\varnothing$ . According to Property (P2) of a TTT, $\mathcal{T}$ does not contain an undirected cycle involving several tournaments. This means that it is impossible to find a node from which there are paths going to u and to 2. Also, it is impossible to find a path passing through 3 and going to 2, because otherwise there would be either an undirected cycle involving several tournaments, or a cycle within a tournament. Both are impossible for a TTT. This leads to the conclusion that $A_{u2}$ is degenerate at zero. Now we look at the variables $(M_{v_{i+1}v_i}, i=n-1, \ldots, 1;\; M_{13}, M_{32})$ , which by construction we take to be independent, as they belong to different tournaments. Each of them is one of the variables in Lemma 3.1, and none of these is degenerate at zero. Hence their product cannot be degenerate at zero either.

Case 3. (c-ii). Next we consider the second case, when without loss of generality $(1,3)\in t(u,1)$ and $(2,3)\in t(u,2)$ ; see Figure 5(b). Let the trail from node 3 to u be on nodes $\{v_1=3, v_2, \ldots, v_n=u\}$ . First we consider the case when we have at least one $i=1, \ldots, n-1$ for which $(v_{i+1},v_i)\in E$ , i.e., we have at least one edge with direction from u to 3. Because t(u, 3) is a shortest trail, the edges incident to the nodes on the trail belong to different tournaments. The distribution of $(A_{u1}, A_{u2})$ is given by

\[ \mathcal{L}(A_{u1}, A_{u2})= \sum_{j\in\mathrm{An}(u)}b_{uj} \delta_{\left\{\frac{b_{1j}}{b_{uj}},\frac{b_{2j}}{b_{uj}}\right\}}, \]

where $b_{1j}=0$ and $b_{2j}=0$ if $j\notin \mathrm{An}(1)$ and $j\notin \mathrm{An}(2)$ , respectively. When for some $i=1, \ldots, n-1$ we have $(v_{i+1}, v_i)\in E$ , it follows necessarily that $\mathrm{An}(1) \cap \mathrm{An}(u)=\varnothing$ and $\mathrm{An}(2) \cap \mathrm{An}(u)=\varnothing$ . There cannot be a path from $\mathrm{An}(1)$ or $\mathrm{An}(2)$ to any of the nodes $\{v_2,\ldots, v_n=u\}$ , because otherwise there would be a cycle involving several tournaments, in contradiction to the definition of a TTT. Because of the edge $(v_{i+1}, v_i)\in E$ , all nodes in $\mathrm{An}(1)\cup\mathrm{An}(2)$ are not ancestors of u. Also, because of the edges $(1,3), (2,3)\in E$ , all nodes in $\mathrm{An}(u)$ cannot be ancestors of node 1 or 2. Thus, when for some $i=1, \ldots, n-1$ there is a directed edge $(v_{i+1}, v_i)\in E$ , we have $\mathcal{L}(A_{u1},A_{u2})=\delta_{\{0,0\}}$ . We have found a node $v \in V$ such that $A_{uv} = 0$ almost surely, but then the factorization (15)–(16) cannot hold, because these are never degenerate at zero.

Now let the trail from node 3 to u actually be a path. Also, let nodes 1 and 2 be sources with respect to the tournaments shared with node 3, say $1,3\in V_{\tau_1}$ and $2,3\in V_{\tau_2}$ . It is always possible to choose 1 and 2 in such a way that they are sources of $\tau_1$ and $\tau_2$ . This is because node 3 obviously is not a source in $\tau_1$ and $\tau_2$ , so the sources of these must point to 3. We can decompose $\mathrm{An}(u)$ into three disjoint sets, namely $\mathrm{An}(1)$ , $\mathrm{An}(2)$ , and the rest, $\mathrm{An}(u)\setminus \{\mathrm{An}(1)\cup \mathrm{An}(2)\}$ . For the distribution of $(A_{u1}, A_{u2})$ we have

\begin{align*} &\mathcal{L}(A_{u1}, A_{u2}) =\sum_{j\in \mathrm{An}(u)}b_{uj} \delta_{\left\{\frac{b_{1j}}{b_{uj}}, \frac{b_{2j}}{b_{uj}}\right\}} \\[5pt] &=\sum_{j\in \mathrm{An}(1)}b_{uj} \delta_{\left\{\frac{b_{1j}}{b_{uj}}, \frac{b_{2j}}{b_{uj}}\right\}} +\sum_{j\in \mathrm{An}(2)}b_{uj} \delta_{\left\{\frac{b_{1j}}{b_{uj}}, \frac{b_{2j}}{b_{uj}}\right\}} +\sum_{j\in\mathrm{An}(u)\setminus \{\mathrm{An}(1)\cup \mathrm{An}(2)\}}b_{uj} \delta_{\left\{\frac{b_{1j}}{b_{uj}}, \frac{b_{2j}}{b_{uj}}\right\}}. \end{align*}

For the atoms in the first summation we have

\[ \frac{b_{1j}}{b_{uj}}=\frac{c_{p(j,1)}b_{jj}}{c_{p(j,u)}b_{jj}} =\frac{c_{p(j,1)}}{c_{p(j,1)}c_{13}c_{p(3,u)}} =\frac{1}{c_{13}c_{p(3,u)}} \]

and $b_{2j}/b_{uj}=0$ , as $b_{2j}=0$ for all $j\in \mathrm{An}(1)$ . Hence we have an atom that does not depend on $j\in \mathrm{An}(1)$ , i.e., $\big(1/(c_{13}c_{p(3,u)}),0\big)$ , and its mass is

\[ \sum_{j\in \mathrm{An}(1)}b_{uj}=\sum_{j\in \mathrm{An}(1)}c_{p(j,1)}c_{13}c_{p(3,u)}b_{jj}=c_{13}c_{p(3,u)}=c_{p(1,u)}. \]

In a similar way, from the second summation in the last display we have an atom $\big(0,1/(c_{23}c_{p(3,u)})\big)$ with mass $c_{23}c_{p(3,u)}=c_{p(2,u)}$ . In the third summation term the atom is (0,0), as $b_{1j}=b_{2j}=0$ for all $j\in\mathrm{An}(u)\setminus \{\mathrm{An}(1)\cup \mathrm{An}(2)\}$ , and its mass is $1-c_{13}c_{p(3,u)}-c_{23}c_{p(3,u)}=1-c_{p(3,u)}(c_{13}+c_{23})$ . Consider now the multiplicative increments $(M_{31};\; M_{32}, M_{v_{i+1}v_i} i=1, \ldots,n-1)$ , which are mutually independent since they belong to different tournaments. Because node 1 is a source node in the tournament $\tau_1$ , the distribution of $M_{31}$ is $c_{13}\delta_{\{1/c_{13}\}}+(1-c_{13})\delta_{\{0\}}$ by Lemma 3.1-2.(a); and similarly for $M_{32}$ . We have

(29) \begin{equation} \begin{split} & \operatorname{\mathbb{P}}\left(M_{31}\prod_{i=1}^{n-1}M_{v_{i+1}v_i}=0, M_{32}\prod_{i=1}^{n-1}M_{v_{i+1}v_i}=0\right) =1-\prod_{i=1}^{n-1}\operatorname{\mathbb{P}}(M_{v_{i+1}v_i}>0) \\[5pt] & +\operatorname{\mathbb{P}}(M_{31}=0)\operatorname{\mathbb{P}}(M_{32}=0) -\left(1-\prod_{i=1}^{n-1}\operatorname{\mathbb{P}}(M_{v_{i+1}v_i}>0)\right) \operatorname{\mathbb{P}}(M_{31}=0)\operatorname{\mathbb{P}}(M_{32}=0). \end{split} \end{equation}

After some rearranging of the expression above we obtain

(30) \begin{equation} 1-\prod_{i=1}^{n-1}\operatorname{\mathbb{P}}(M_{v_{i+1}v_i}>0)(c_{13}+c_{23}-c_{13}c_{23}). \end{equation}

There are two further subcases: either every node $v_1,\ldots,v_{n-1}$ is the source node with respect to the tournament involving the next node in the sequence, or not. In the first subcase, namely when every node in $\{v_1=3, v_2, \ldots, v_{n-1}\}$ is the source node with respect to the tournament involving the next node in the sequence, we have $\operatorname{\mathbb{P}}(M_{v_{i+1}v_i}>0)=c_{v_{i}v_{i+1}}$ for $i=1, \ldots, n-1$ . This means that the probability in (30), and consequently in (29), equals $1-c_{p(3,u)}(c_{13}+c_{23}-c_{13}c_{23})$ , which is different from $\operatorname{\mathbb{P}}(A_{u1}=0, A_{u2}=0)=1-c_{p(3,u)}(c_{13}+c_{23})$ . In the second subcase, i.e., if at least one node from $\{v_1=3, v_2, \ldots, v_{n-1}\}$ is not the source with respect to the tournament involving the next node in the sequence, we have that the possible values for $M_{31}\prod_{i=1}^{n-1}M_{v_{i+1}v_i}$ are not only $\{0,1/c_{p(1,u)}\}$ , which are the only possible values of $A_{u1}$ as we showed in the previous paragraph. Let $i \in \{1,\ldots,n-1\}$ be such that node $v_i$ is not the source node in the tournament shared with $v_{i+1}$ , say $\tau_i$ . This is depicted in the following graph:

Recall the distribution of $M_{v_{i+1}v_i}$ from Lemma 3.1-2.(b):

\[ \mathcal{L}(M_{v_{i+1}v_i}) =\sum_{j\in \mathrm{An}(v_{i})}b_{v_{i+1}j}\delta_{\{b_{v_ij}/b_{v_{i+1}j}\}} +\sum_{j\in \mathrm{An}(v_{i+1})\setminus \mathrm{An}(v_i)}\delta_{\{0\}}. \]

Take for instance a node, say s, which is a parent of $v_i$ and accordingly is in $\mathrm{An}(v_i)$ . Then

\[ \frac{b_{v_is}}{b_{v_{i+1}s}} =\frac{c_{sv_i}}{c_{sv_{i+1}}} \]

is a possible value of $M_{v_{i+1}v_i}$ with positive probability, namely at least $b_{v_{i+1}s}$ . Another possible positive value is for $j=v_i\in \mathrm{An}(v_i)$ , namely

\[ \frac{b_{v_iv_i}}{b_{v_{i+1}v_i}} =\frac{1}{c_{v_iv_{i+1}}} \]

with probability at least $b_{v_{i+1}v_i}$ . The criticality assumption on edge weights guarantees that $\frac{c_{sv_i}}{c_{sv_{i+1}}}\neq 1/c_{v_iv_{v+1}}$ . This means that the product $M_{31}\prod_{i=1}^{n-1}M_{v_{i+1}v_i}$ has at least two different positive values: one involving $\frac{c_{sv_i}}{c_{sv_{i+1}}}$ and another $1/c_{v_iv_{i+1}}$ . However, $A_{u1}$ has only one possible positive value.

Proof of Proposition 3.2. Sufficiency. Assume $\mathcal{T}$ has a unique source. We need to show that, for any disjoint and nonempty sets A, B, S, we have $X_A\perp\!\!\!\perp X_B\mid X_S$ whenever S is a separator of A and B in the skeleton T of $\mathcal{T}$ . We would like to use [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Theorem 5.15], by which we need to show that $A\perp_*B\mid S$ in $\mathcal{D}_S^*$ —that is, that there are no $*$ -connecting paths between any pair of nodes in A and B in the conditional reachability DAG $\mathcal{D}_S^*$ . We will explain these notions further.

Let $A, B,S\subset V$ be nonempty disjoint node sets such that S is a separator of A and B in the skeleton T. Consider Figure 6. According to [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Definition 5.4], a $*$ -connecting path between $a \in A$ and $b \in B$ is one of the five configurations shown in the figure. Our goal is to show that for $a\in A$ and $b\in B$ it is impossible to find a $*$ -connecting path in a certain graph $\mathcal{D}_S^*$ , which is neither $\mathcal{T}$ nor T, but is constructed under particular rules given in [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Definition 5.1].

Figure 6. According to [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Definition 5.4], a $*$ -connected path between a and b relative to S is one of the five configurations above. In the last three graphs we have $s\in S$ .

According to this definition, the conditional reachability graph $\mathcal{D}_S^*$ is on the same vertex set, V. Between two nodes i and j in V, there is an edge (i, j) in $\mathcal{D}_S^*$ if and only if there is a directed path from i to j in $\mathcal{T}$ such that no node on that path belongs to S, except possibly for i and j themselves.

Consider Figure 6. We need to show that in the conditional reachability graph $\mathcal{D}_S^*$ , there is no $*$ -connecting path between a node $a \in A$ and a node $b \in B$ .

To obtain the first configuration in $\mathcal{D}_S^*$ , there must be, in the skeleton T, nodes $a\in A$ and $b\in B$ such that no node on the path from a to b passes through S. But this is impossible, because we assumed that S is a separator of A and B in T. A similar argument holds for the second configuration.

For the other three configurations in Figure 6, consider Figure 7.

Figure 7. The left and right trails in the original graph $\mathcal{T}$ are the only possible trails that give rise to the middle path in the graph $\mathcal{D}_S^*$ .

In Figure 7, the left-hand and right-hand trails in the original graph $\mathcal{T}$ are the only possible ones that give rise to the middle path in Figure 6 with respect to the graph $\mathcal{D}_S^*$ : both the left-hand and right-hand graphs in Figure 7 show existing trails between a and b in $\mathcal{T}$ , trails composed of a directed path from a to s, and a directed path from b to s. The only node on these trails which belongs to S is s. Hence in $\mathcal{D}_S^*$ we put a directed edge from a to s and from b to s. This gives the third $*$ -connecting path in Figure 6. But this configuration cannot occur, for the following reason. On the left-hand trail in Figure 7, the separator node s has parents $u_r$ and $v_q$ in different tournaments. This leads to a v-structure between the nodes $u_r,s,v_q$ , in contradiction to Lemma 2.2-2 and the hypothesis that $\mathcal{T}$ has a unique source.

On the right-hand trail in Figure 7, the node s shares a tournament with its parents $u_r$ and $v_q$ , but only s belongs to S; on the trail $\{a=u_1, u_2, \ldots, u_r, v_q, \ldots, v_2,v_1=b\}$ , none of the nodes are in S. In T, this means that there is a path between A and B that does not pass through S. This is in contradiction to the assumption that S separates A and B in T.

To show that the fourth type of $*$ -connecting path in Figure 6 cannot occur, we can use the reasoning used for the third one by setting $a=a'$ in Figure 7. Then either s has parents from two different tournaments or there is a non-directed path from a ^′ to b which does not pass through S. The first case is excluded by Lemma 2.2-2 and the assumption that $\mathcal{T}$ has a unique source, and the second one by the assumption that S is a separator of A and B in T. The impossibility of the fifth $*$ -connected configuration follows analogously.

Necessity. We will show that if $\mathcal{T}$ has multiple source nodes, there is a triple of disjoint, non-empty sets $A,B,S\subset V$ such that S is a separator of A and B in T, but $X_A$ and $X_B$ are conditionally dependent given $X_S$ . In the case when $\mathcal{T}$ has at least two sources, we have at least one v-structure in $\mathcal{T}$ by Lemma 2.2-2. Take a triple of nodes in a v-structure, say u, v, w, with u and w being parents of v. Then node v separates nodes u and w in T, i.e., $S = \{v\}$ is a separator of $A = \{u\}$ and $B = \{w\}$ in T. All references below are from [Reference Améndola, Klüppelberg, Lauritzen and Tran2].

To show $X_u\not\!\perp\!\!\!\perp X_w\mid X_v$ , we will use [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Theorem 6.18] (context-free completeness). We need to show that there is an effective $*$ -connecting path in the critical DAG $\mathcal{D}^*_S(\theta)$ between nodes u and w, as in Definitions 5.2 and 6.5 of [Reference Améndola, Klüppelberg, Lauritzen and Tran2].

The subgraph on nodes u, v, w of $\mathcal{D}^*_S(\theta)$ is a v-structure, , according to the definition of $\mathcal{D}^*_S(\theta)$ . According to [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Definition 6.4], the $|S| \times |S|$ substitution matrix of $(u,v)\in E$ relative to $S=\{v\}$ is zero, because S is a singleton and by definition all diagonal entries of the substitution matrix are zero, i.e., $\Xi^{vu}_S=0$ . Similarly, $\Xi^{vw}_S=0$ . Because the edges (u, v), (w, v) form a $*$ -connecting path between u, w in $\mathcal{D}^*_S(\theta)$ , the substitution matrix of this path relative to S, say $\Xi_S$ , is zero too:

\[ \Xi_S=\max(\Xi^{vu}_S, \Xi^{vw}_S)=0. \]

To find out whether the edges (u, v), (w, v) form an effective $*$ -connecting path between u, w, we need to compute the tropical eigenvalue of $\max(\Gamma_{SS},\Xi_S)$ , where $\Gamma$ is as in [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Equation (2.3)] and $\Gamma_{SS}$ is the vv-element of $\Gamma$ , i.e., $\{\Gamma\}_{vv}$ . Because $\{\Gamma\}_{ij}>0$ if and only if there is a directed path from j to i, we have $\Gamma_{SS}=\{\Gamma\}_{vv}=0$ and so

\[ \max(\Gamma_{SS}, \Xi_S)=0. \]

The tropical eigenvalue [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Equation (2.7)] of the above matrix is trivially equal to zero and thus smaller than one. By [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Definition 6.5], there is indeed an effective $*$ -connecting path between u, w in $\mathcal{D}^*_{S}(\theta)$ . In view of [Reference Améndola, Klüppelberg, Lauritzen and Tran2, Theorem 6.18], we conclude that $X_u \not\!\perp\!\!\!\perp X_w \mid X_v$ .