Extremes of Markov random fields on block graphs: Max-stable limits and structured Hüsler–Reiss distributions

Abstract

We study the joint occurrence of large values of a Markov random field or undirected graphical model associated to a block graph. On such graphs, containing trees as special cases, we aim to generalize recent results for extremes of Markov trees. Every pair of nodes in a block graph is connected by a unique shortest path. These paths are shown to determine the limiting distribution of the properly rescaled random field given that a fixed variable exceeds a high threshold. The latter limit relation implies that the random field is multivariate regularly varying and it determines the max-stable distribution to which component-wise maxima of independent random samples from the field are attracted. When the sub-vectors induced by the blocks have certain limits parametrized by Hüsler–Reiss distributions, the global Markov property of the original field induces a particular structure on the parameter matrix of the limiting max-stable Hüsler–Reiss distribution. The multivariate Pareto version of the latter turns out to be an extremal graphical model according to the original block graph. Thanks to these algebraic relations, the parameters are still identifiable even if some variables are latent.

Appendices

A. Proofs for Section 3

1.1 A.1. Proof of Theorem 3.1

The proof follows the lines of the one of Theorem 1 in Segers (2020). To show (9) it is sufficient to show that for a real bounded Lipschitz function f, for any fixed $u\in V$ it holds that

$$\begin{aligned} \lim _{t \rightarrow \infty } \mathbb {E}[f(X_{V\setminus u}/t)\mid X_u=t] = \mathbb {E}[f(A_{u,V\setminus u})], \end{aligned}$$

(24)

(van der Vaart 1998, Lemma 2.2). Without loss of generality, we assume that $0 \le f(x) \le 1$ and $|f(x)-f(y)|\le L\sum _{j}|x_j-y_j|$ for some constant $L>0$.

We proceed by induction on the number of cliques, m. When there is only one clique ($m = 1$) the convergence happens by Assumption 3.1 with $s = u$: the distribution of $A_{u,V \setminus u}$ is equal to $\nu _{C,s}$ in the assumption, with $C = V$ and $u = s$.

Assume that there are at least two cliques, $m \ge 2$. Let the numbering of the cliques be such that the last clique, $C_m$, is connected to the subgraph induced by $\bigcup _{i=1}^{m-1}C_i$ only through one node, which is the minimal separator between $C_m$ and $\bigcup _{i=1}^{m-1}C_i$. Let $s\in \mathcal {S}$ denote this node and introduce the set

$$C_{1:m-1}=(C_1 \cup \cdots \cup C_{m-1}) {\setminus } u.$$

Note that $\{s\} = (C_1 \cup \cdots \cup C_{m-1}) \cap C_m$. We need to make a distinction between two cases: $s = u$ or $s \ne u$. The case $s = u$ is the easier one, since then $X_{C_m \setminus u}$ and $X_{C_{1:m-1}}$ are conditionally independent given $X_u$ whereas the shortest paths from u to nodes in $C_m \setminus u$ just consist of single edges, avoiding $C_{1:m-1}$ altogether. So we only consider the case $s \ne u$ henceforth. In that case, paths from u to nodes in $C_m \setminus s$ pass through s and possibly other nodes in $C_{1:m-1}$.

The induction hypothesis is that as $t\rightarrow \infty$, we have

$$\begin{aligned} (X_{C_{1:m-1}}/t)\mid X_u=t{\mathop {\longrightarrow }\limits ^{d}}A_{u,C_{1:m-1}}, \end{aligned}$$

(25)

or also that for every continuous and bounded function $h: \mathbbm {R}_+^{C_{1:m-1}} \rightarrow \mathbbm {R}$, we have

$$\begin{aligned} \lim _{t \rightarrow \infty } \mathbb {E}\left[ h(X_{C_{1:m-1}}/t)\mid X_u=t\right] = \mathbb {E}[h(A_{u,C_{1:m-1}})]. \end{aligned}$$

(26)

To prove the convergence in (24) we start with the following inequality: for $\delta > 0$,

$$\begin{aligned} \begin{aligned} &\Bigl | \mathbb {E}[f(X_{V\setminus u}/t)\mid X_u=t] -\mathbb {E}[f(A_{u,V\setminus u})]\Bigr | \\ {}&\le \Bigl |\mathbb {E}\left[ f(X_{V\setminus u}/t) \mathbbm {1}(X_{s}/t\ge \delta )\mid X_u=t\right] - \mathbb {E}\left[ f(A_{u,V\setminus u}) \mathbbm {1}(A_{us}\ge \delta )\right] \Bigr | \end{aligned} \end{aligned}$$

(27)

$$\begin{aligned} + \Bigl |\mathbb {E}\left[ f(X_{V\setminus u}/t) \mathbbm {1}(X_{s}/t<\delta )\mid X_u=t\right] - \mathbb {E}\left[ f(A_{u,V\setminus u}) \mathbbm {1}(A_{us}<\delta )\right] \Bigr |. \end{aligned}$$

(28)

We let $\delta > 0$ be a continuity point of $A_{us}$. Later on, we will take $\delta$ arbitrarily close to zero, which we can do, since the number of atoms of $A_{us}$ is at most countable.

Analysis of (27). We first deal with (27). The first expectation is equal to

$$\int _{[0,\infty )^{V{\setminus } u}} f(x/t) \, \mathbbm {1}(x_s/t\ge \delta ) \, \mathbb {P}(X_{V{\setminus } u}\in \textrm{d}x \mid X_u=t)\,.$$

Because of the global Markov property, $X_{C_m\setminus s}$ is conditionally independent of the variables in the set $C_{1:m-1}$ given $X_s$. As a consequence, the conditional distribution of $X_{C_m\setminus s}$ given $X_{C_{1:m-1}}$ is the same as the one of $X_{C_m \setminus s}$ given $X_s$. Hence we can write the integral as

$$\int _{[0,\infty )^{C_{1:m-1}}} \mathbb {E}\left[ f\left( x/t, X_{C_m{\setminus } s}/t\right) \mid X_{s}=x_{s}\right] \mathbbm {1}(x_s/t \ge \delta ) \mathbb {P}(X_{C_{1:m-1}}\in \textrm{d}x\mid X_u=t).$$

After the change of variables $x/t=y$, the integral becomes

$$\begin{aligned} \int _{[0,\infty )^{C_{1:m-1}}} \mathbb {E}\left[ f\left( y, X_{C_m\setminus s}/t\right) \, \Big | \, X_{s}=ty_{s}\right] \mathbbm {1}(y_{s}\ge \delta ) \mathbb {P}\left( X_{C_{1:m-1}}/t \in \textrm{d}y \mid X_u=t\right) . \end{aligned}$$

(29)

Define the functions $g_t$ and g on $[0, \infty )^{C_{1:m-1}}$ by

$$\begin{aligned} g_t(y)&:= \mathbb {E}\left[ f\left( y, X_{C_m\setminus s}/t\right) \, \Big | \, X_{s}=ty_{s}\right] \mathbbm {1}(y_{s}\ge \delta ), \\ g(y)&:= \mathbb {E}\left[ f\left( y, y_sZ_{s,C_m\setminus s}\right) \right] \mathbbm {1}(y_{s}\ge \delta ). \end{aligned}$$

Consider points y(t) and y in $[0, \infty )^{C_{1:m-1}}$ such that $\lim _{t \rightarrow \infty } y(t) = y$ and such that $y_s \ne \delta$. We need to show that

$$\begin{aligned} \lim _{t \rightarrow \infty } g_t(y(t)) = g(y). \end{aligned}$$

(30)

If $y_s < \delta$, this is clear since $y_s(t) < \delta$ for all large t and hence the indicators will be zero. So suppose $y_s > \delta$ and thus also $y_s(t) > \delta$ for all large t, meaning that both indicators are (eventually) equal to one. By Assumption 3.1, we have

$$X_{C_m {\setminus } s} / t \mid X_s = ty_s(t) {\mathop {\longrightarrow }\limits ^{d}}y_s Z_{s,C_m {\setminus } s}, \qquad t \rightarrow \infty .$$

Since f is continuous, also

$$f\left( y(t), X_{C_{m} {\setminus } s} / t \right) \, \Big | \, X_s = ty_s(t) {\mathop {\longrightarrow }\limits ^{d}}f\left( y, y_s Z_{s, C_m {\setminus } s}\right) , \qquad t \rightarrow \infty .$$

As the range of f is contained in [0, 1], the bounded convergence theorem implies that we can take expectations in the previous equation and conclude (30).

By the induction hypothesis (25) and Theorem 18.11 in van der Vaart (1998), the continuous convergence in (30) implies

$$\begin{aligned} g_t\left( \frac{X_{C_{1:m-1}}}{t}\right) \, \Big | \, X_u=t {\mathop {\longrightarrow }\limits ^{d}}g(A_{C_{1:m-1}}), \qquad t \rightarrow \infty ; \end{aligned}$$

note that by the choice of $\delta$, the discontinuity set of g receives zero probability in the limit. As $g_t$ and g are bounded (since f is bounded), we can take expectations and find

$$\begin{aligned} \lim _{t \rightarrow \infty } \mathbb {E}\left[ g_t\left( \frac{X_{C_{1:m-1}}}{t}\right) \, \Big | \, X_u=t \right] = \mathbb {E}[g(A_{C_{1:m-1}})]. \end{aligned}$$

(31)

The expectation on the left-hand side of (31) is the integral in (29) while the right-hand side of (31) is equal to

$$\mathbb {E}[f(A_{u,C_{1:m-1}}, A_{us}Z_{s,C_m{\setminus } s}) \mathbbm {1}(A_{us}\ge \delta )] = \mathbb {E}[f(A_{u,V{\setminus } u}) \, \mathbbm {1}(A_{us}\ge \delta )].$$

Thus we have shown that (27) converges to 0 as $t\rightarrow \infty$, for any continuity point $\delta$ of $A_{us}$.

Analysis of (6.5). As f is a function with range [0, 1] we have

$$0\le \mathbb {E}[f(X_{V{\setminus } u}/t)\mathbbm {1}( X_{s}/t<\delta )\mid X_u=t] \le \mathbb {P}[X_{s}/t<\delta \mid X_u=t]$$

as well as

$$0\le \mathbb {E}[f(A_{u,V{\setminus } u})\mathbbm {1}(A_{us}<\delta )] \le \mathbb {P}[A_{us}<\delta ].$$

By the triangle inequality and the two inequalities above, (28) is bounded from above by

$$\begin{aligned} \mathbb {P}[X_{s}/t<\delta \mid X_u=t] + \mathbb {P}[A_{us}<\delta ] . \end{aligned}$$

(32)

By the induction hypothesis

$$\lim _{t\rightarrow \infty }\mathbb {P}[X_{s}/t<\delta \mid X_u=t] = \mathbb {P}[A_{us}<\delta ],$$

and (32) converges to $2 \mathbb {P}[A_{us}<\delta ]$, which goes to 0 as $\delta \downarrow 0$ in case $\mathbb {P}(A_{us}=0)=0$.

Suppose $\mathbb {P}(A_{us}=0)>0$. In this step we will need Assumption 3.2. By the induction hypothesis, we have $A_{us}=\prod _{(a,b)\in ({u} \rightsquigarrow {s})}Z_{ab}$ and the variables $Z_{ab}$ are independent. Hence

$$\begin{aligned} \mathbb {P}(A_{us}=0)=\mathbb {P}\Big (\min _{(a,b)\in ({u} \rightsquigarrow {s})}Z_{ab}=0\Big ) = 1-\prod _{(a,b)\in ({u} \rightsquigarrow {s})}\mathbb {P}(Z_{ab}>0). \end{aligned}$$

If for any $(a,b)\in ({u} \rightsquigarrow {s})$ we have $\mathbb {P}(Z_{ab}=0)>0$ then $P(Z_{ab}>0)<1$ and hence $\mathbb {P}(A_{us}=0)>0$. Therefore the assumption applies when the marginal distribution $\nu _{C,a}^b(\{0\})$ is positive.

Then by adding and subtracting terms and using the triangle inequality, we have the following upper bound for the term in (28):

$$\begin{aligned}&\left| \mathbb {E}\left[ f\left( \frac{X_{C_{1:m-1}}}{t},\frac{X_{C_m\setminus s}}{t} \right) \mathbbm {1}\left( \frac{X_{s}}{t}<\delta \right) \Big | X_u=t\right] -\mathbb {E}\left[ f\left( \frac{X_{C_{1:m-1}}}{t},0\right) \mathbbm {1}\left( \frac{X_{s}}{t}<\delta \right) \Big | X_u=t\right] \right| \end{aligned}$$

(33)

$$\begin{aligned}&+ \left| \mathbb {E}\left[ f(A_{u,C_{1:m-1}},A_{u,C_m\setminus s} ) \mathbbm {1}(A_{u,s}<\delta )\right] - \mathbb {E}\left[ f(A_{u,C_{1:m-1}},0) \mathbbm {1}(A_{u,s}<\delta )\right] \right| \end{aligned}$$

(34)

$$\begin{aligned}&+ \left| \mathbb {E}\left[ f\left( \frac{X_{C_{1:m-1}}}{t},0\right) \mathbbm {1}\left( \frac{X_{s}}{t}<\delta \right) \Big | X_u=t\right] - \mathbb {E}\left[ f(A_{u,C_{1:m-1}},0) \mathbbm {1}(A_{u,s}<\delta )\right] \right| . \end{aligned}$$

(35)

We treat each of the three terms in turn.

Equation (35) converges to 0 by the induction hypothesis; note again that the set of discontinuities of the integrand receives zero probability in the limit.

Next we look at expression (33). From the assumptions of f, namely that it ranges in [0, 1] and that $|f(x)-f(y)|\le L \Vert x-y\Vert _1$ for some constant $L>0$, where $\Vert z\Vert _1 = \sum _j |z_j|$ for a Euclidean vector z, the term in (33) is bounded by

$$\begin{aligned} \mathbb {E}\left[ \left| f\left( \frac{X_{C_{1:m-1}}}{t},\frac{X_{C_m{\setminus } s}}{t} \right) - f\left( \frac{X_{C_{1:m-1}}}{t},0\right) \right| \mathbbm {1}\left( \frac{X_{s}}{t}<\delta \right) \, \Bigg | \, X_u=t\right] \\ \le \mathbb {E}\left[ \mathbbm {1}\left( X_s/t<\delta \right) \min \left( 1,L \Vert X_{C_{m {\setminus } s}}/t\Vert _1 \right) \, \mid \, X_u=t\right] . \end{aligned}$$

We need to show that the upper bound converges to 0 as $t\rightarrow \infty$. Because the variables in $C_m\setminus s$ are independent of $X_u$ conditionally on $X_s$, the previous integral is equal to

$$\begin{aligned} \int _{[0,\delta ]} \mathbb {E}\left[ \min \left( 1,L\Vert X_{C_m \setminus s}/t\Vert _1\right) \, \mid \, X_{s} / t = x_{s} \right] \mathbb {P}\left( X_{s}/{t}\in \textrm{d}x_{s} \mid X_u=t \right) . \end{aligned}$$

(36)

For $\eta >0$, the inner expectation is equal to

$$\begin{aligned}&\mathbb {E}\left[ \min \left( 1,L\Vert X_{C_m \setminus s}/t\Vert _1\right) \, \mathbbm {1}\left\{ \forall v \in C_m \setminus s : X_v/t \le \eta \right\} \mid X_{s}/t = x_{s} \right] \end{aligned}$$

(37)

$$\begin{aligned}&+ \mathbb {E}\left[ \min \left( 1,L\Vert X_{C_m \setminus s}/t\Vert _1 \right) \, \mathbbm {1}\left\{ \exists v \in C_m\setminus s : X_v / t > \eta \right\} \mid X_{s} / t = x_{s}\right] . \end{aligned}$$

(38)

The integrand in (37) is either zero because of the indicator function or, if the indicator is one, it is bounded by $L \left| C_m \setminus s\right| \eta$. The expression in (38) is clearly smaller than or equal to $\mathbb {P}\left( \exists v\in C_m{\setminus } s: X_v / t > \eta \mid X_s/t = x_s \right)$.

Going back to the integral in (36) we can thus bound it by

$$\begin{aligned}&\int _{[0,\delta ]} \left[ L \left| C_m\setminus s\right| \eta + \mathbb {P}\left( \exists v \in C_m \setminus s : X_v/t > \eta \mid X_s/t = x_s \right) \right] \mathbb {P}\left( X_s/t \in \textrm{d}x_s \mid X_u=t\right) . \end{aligned}$$

(39)

Consider the supremum of the probability in the integrand over the values $x_{s}\in [0,\delta ]$ to bound the integral further. Hence (39) is smaller than or equal to

$$L \left| C_m{\setminus } s \right| \eta + \sup _{x_{s}\in [0,\delta ]} \mathbb {P}\left( \exists v \in C_m {\setminus } s: X_v/t > \eta \mid X_s/t = x_s \right) .$$

Using Assumption 3.2 and the fact that $\eta$ can be chosen arbitrarily small we conclude that (33) converges to 0 as $t\rightarrow \infty$.

Finally we look at the term in (34). As f has range contained in [0, 1] and is Lipschitz continuous, the expression in (34) is smaller than or equal to

$$\begin{aligned}&\mathbb {E}\Big [ \mathbbm {1}(A_{us}<\delta ) \min \Big (1, \, {L\textstyle \sum _{v\in C_m\setminus s}A_{uv}}\Big )\Big ]. \end{aligned}$$

(40)

From $(A_{uv}, v\in C_m\setminus s)=A_{u,C_m\setminus s}=A_{us}Z_{s,C_m\setminus s}$ we can write (40) as

$$\begin{aligned}&\mathbb {E}\Big [ \mathbbm {1}(A_{us}<\delta ) \min \Big (1,\textstyle {LA_{us}\sum _{v\in C_m\setminus s}Z_{sv}}\Big )\Big ]. \end{aligned}$$

The random variable inside the expectation is bounded by 1 for any value of $\delta >0$ and it converges to 0 as $\delta \downarrow 0$. By the bounded convergence theorem, the expectation in (40) converges to 0 as $\delta \downarrow 0$. $\square$

1.2 A.2. Proof of Proposition 3.1

The quantile function of the unit-Fréchet distribution is $u \mapsto -1/\ln (u)$ for $0< u < 1$. In view of Sklar’s theorem and the identity (3), the copula, K, of G is

$$K(u) = G(-1/\ln u_1, \ldots , -1/\ln u_d) = \exp \bigl ( - \ell (- \ln u_1,\ldots ,-\ln u_d) \bigr ), \qquad u \in (0, 1)^d.$$

It follows that the partial derivative $\dot{K}_1$ of K with respect to its first argument exists, is continuous on $(0, 1)^2$ and is given by

$$\dot{K}_1(u) = \frac{K(u)}{u_1} \, \dot{\ell }_1(-\ln u_1,\ldots ,-\ln u_d),$$

for $u \in (0, 1)^d$. The stdf is homogeneous: for $t > 0$ and $x \in (0, \infty )^d$, we have

$$\ell (tx_1,\ldots ,tx_d) = t \, \ell (x_1,\ldots ,x_d).$$

Taking the partial derivative with respect to $x_1$ on both sides and simplifying yields the identity

$$\dot{\ell }_1(tx_1,\ldots ,tx_d) = \dot{\ell }_1(x_1,\ldots ,x_d).$$

Let $F(x) = \exp (-1/x)$, for $x > 0$, denote the unit-Fréchet cumulative distribution function. Note that $-\ln F(x) = 1/x$ for $x > 0$. For $t > 0$ and $x = (x_2,\ldots ,x_d)\in (0, \infty )^{d-1}$, we find

$$\begin{aligned} \mathbb {P}(\forall j \ge 2, X_j \le t x_j \mid X_1 = t)&= \dot{K}_1 \bigl ( F(t),F(tx_2),\ldots ,F(tx_d) \bigr ) \\&= \frac{K \bigl ( F(t),F(tx_2),\ldots ,F(tx_d) \bigr )}{F(t)} \, \dot{\ell }_1 (1/t, 1/(tx_2), \ldots , 1/(tx_d)) \\&= \frac{K \bigl ( F(t),F(tx_2),\ldots ,F(tx_d) \bigr )}{F(t)} \, \dot{\ell }_1 (1, 1/x_2, \ldots , 1/x_d). \end{aligned}$$

As $t \rightarrow \infty$, the first factor on the right-hand side tends to one, whence

$$\lim _{t \rightarrow \infty } \mathbb {P}(\forall j \ge 2, X_j \le t x_j \mid X_1 = t) = \dot{\ell }_1 (1, 1/x_2, \ldots , 1/x_d).$$

To show that the right-hand side of the previous equation is indeed the cumulative distribution function of a $(d-1)$-variate probability measure on Euclidean space, it is sufficient to show that, for every $j \in \{2, \ldots , d\}$, the family of conditional distributions $(X_j/t \mid X_1 = t)$ as t ranges over $[t_0, \infty )$ for some large $t_0 > 0$ is uniformly tight. Indeed, the family of joint conditional distributions $\left( (X_2,\ldots ,X_d)/t \mid X_1 = t\right)$ for $t \in [t_0, \infty )$ is then uniformly tight as well, and by Prohorov’s theorem (van der Vaart 1998, Theorem 2.4), we can find a sequence $t_n \rightarrow \infty$ such that the joint conditional distributions $(X/t_n \mid X_1 = t_n)$ converge weakly as $n \rightarrow \infty$, the limiting cumulative distribution function then necessarily being equal to the one stated above. It suffices to consider the case $d = j = 2$. By the first part of the proof above,

$$\lim _{t \rightarrow \infty } \mathbb {P}(X_2/t > x_2 \mid X_1 = t) = 1 - \dot{\ell }_1(1, 1/x_2).$$

Since $\ell : [0, \infty )^2 \rightarrow [0, \infty )$ is convex, the functions $y_1 \mapsto \ell (y_1, y_2)$ depend continuously on the parameter $y_2 \ge 0$. Since they are also convex, Attouch’s theorem (Rockafellar and Wets 1998, Theorem 12.35), implies that their derivatives depend continuously in $y_2$ as well, at least in points $y_1$ where $y_1 \mapsto \ell (y_1, y_2)$ is continuously differentiable. But since $\ell (y_1, 0) = y_1$, we find that $\dot{\ell }_1(1, 1/x_2) \rightarrow \dot{\ell }_1(1, 0) = 1$ as $x_2 \rightarrow \infty$. For any $\epsilon > 0$, we can thus find $x_2(\epsilon ) > 0$ such that $1 - \dot{\ell }_1(1, 1/x_2(\epsilon )) < \epsilon / 2$ and then we can find $t(\epsilon ) > 0$ such that $\mathbb {P}(X_2 / t > x_2(\epsilon ) \mid X_1 = t)< \epsilon /2 + 1 - \dot{\ell }_1(1,1/x_2(\epsilon )) < \epsilon$ for all $t > t(\epsilon )$. The uniform tightness follows. $\square$

B. Proofs for Section 4

1.1 B.1. Proof of Proposition 4.1

By Assumption 4.1, the random vector X satisfies Assumption 3.1 and it is Markov with respect to the graph $\mathcal {G}$. Assumption 3.2 is void (i.e., there is nothing to check), since, for each edge $(i, j) \in E$, the limiting distribution of $X_j/t \mid X_i = t$ as $t \rightarrow \infty$ is log-normal by Segers (2020, Example 4) and Example 3.3 here and therefore does not have an atom at zero. We can thus apply Theorem 3.1 to conclude that $(X_v/t, v \in V \setminus u \mid X_u = t)$ converges weakly as $t \rightarrow \infty$. By the continuous mapping theorem, the same then holds true for $(\ln (X_v/t), v \in V \setminus u \mid X_u = t)$. It remains to calculate the limit distribution.

By example 3.3 we have, as $t \rightarrow \infty$,

$$\begin{aligned} \bigl ( \ln X_v - \ln X_s, \, v \in C \setminus s \mid X_s > t \bigr ) {\mathop {\longrightarrow }\limits ^{d}}\mathcal {N}_{|C\setminus s|} \bigl ( \mu _{C,s}(\Delta _C), \Psi _{C,s}(\Delta _C) \bigr ), \end{aligned}$$

(41)

where the mean vector is

$$\begin{aligned} \bigl (\mu _{C,s}(\Delta )\bigr )_v = -2\delta _{sv}^2, \qquad v \in C \setminus s, \end{aligned}$$

(42)

and the covariance matrix $\Psi _{C,s}(\Delta )$ is as in (19). It follows that if the random vector $Z_{s,C \setminus s}$ has law $\nu _{C,s}$, then the distribution of $(\ln Z_{sv}, v \in C \setminus s)$ is equal to the limit in (41). In particular, $\nu _{C,s}$ is multivariate log-normal.

For fixed $u \in V$, we will identify the limit $A_{u, V \setminus u}$ in Theorem 3.1. Let $Z = (Z_{s,C \setminus s}, C \in \mathcal {C})$ with $Z_{s,C} = (Z_{sv}, v \in C \setminus s)$ be the random vector constructed in Definition 3.1 by concatenating independent log-normal random vectors with distributions $\nu _{C, s}$. In this concatenation, recall that $s \in C$ and that either s is equal to u or s separates u and $C \setminus s$. We can write $Z = (Z_e, e \in E_u)$ where the $E_u$ is the set of edges $e \in E$ that point away from u: for $e = (s, v) \in E_u$, either s is equal to u or s separates u and v. By construction, the distribution of Z is multivariate log-normal too. By (42), we have $\mathbb {E}[\ln Z_e] = -2\delta _e^2$ where $e = (s, v) \in E_u$. The covariance matrix of $(\ln Z_e, e \in E_u)$ has a block structure: for edges $e, f \in E_u$, the variables $\ln Z_e$ and $\ln Z_f$ are uncorrelated (and thus independent) if e and f belong to different cliques, while if they belong to the same clique, i.e., if $e = (s, i)$ and $f = (s, j)$ with $i,j,s \in C$ for some $C \in \mathcal {C}$, then, by (19), we have

$$\begin{aligned} {\text {cov}}\left( \ln Z_e, \ln Z_f \right) = 2(\delta _{si}^2 + \delta _{sj}^2 - \delta _{ij}^2) \end{aligned}$$

(43)

By Theorem 3.1, we can express the limit $A_{u, V \setminus u}$ of $(X_v / t, v \in V \setminus u \mid X_u = t)$ as $t \rightarrow \infty$ in terms of Z: we have

$$\begin{aligned} \ln A_{uv} = \ln \left( \prod _{e\in ({u} \rightsquigarrow {v})}Z_e\right) = \sum _{e\in ({u} \rightsquigarrow {v})}\ln Z_e, \qquad v \in V \setminus u. \end{aligned}$$

(44)

The distribution of $(\ln A_{uv}, v \in V \setminus u)$ is thus multivariate Gaussian, being the one of a linear transformation of the multivariate Gaussian random vector $(\ln Z_e, e \in E_u)$. The expectation of $\ln A_{uv}$ is readily obtained from (44):

$$\mathbb {E}[\ln A_{uv}] = \sum _{e \in ({u} \rightsquigarrow {v})} \mathbb {E}[\ln Z_e] = \sum _{e \in ({u} \rightsquigarrow {v})} (-2\delta _e^2) = - 2p_{uv}, \qquad v \in V {\setminus } u,$$

which coincides with the element v of the vector $\mu _u(\Delta )$ in (22). It remains to show that the covariance matrix of $(\ln A_{uv}, v \in V \setminus u)$ is $\Sigma _u(\Delta )$ in (23).

Calculating $\Sigma _u(\Delta )$. Let $i, j \in V \setminus u$. By (44) and the bilinearity of the covariance operator, we have

$$\begin{aligned} {\text {cov}}\left( \ln A_{ui}, \ln A_{uj} \right) = \sum _{e \in ({u} \rightsquigarrow {i})} \sum _{f \in ({u} \rightsquigarrow {j})} {\text {cov}}\left( \ln Z_e, \ln Z_f \right) . \end{aligned}$$

Each of the paths $({u} \rightsquigarrow {i})$ and $({u} \rightsquigarrow {j})$ has at most a single edge in a given clique $C \in \mathcal {C}$; otherwise, they would not be the shortest paths from u to i and j, respectively. Let the node $a \in V$ be such that $({u} \rightsquigarrow {i}) \cap ({u} \rightsquigarrow {j}) = ({u} \rightsquigarrow {a})$. It could be that $a = u$, in which case the intersection is empty. Now we need to consider three cases.

• 1. If $a = i$, i.e., if i lies on the path from u to j, then the random variables $\ln Z_f$ for $f \in ({i} \rightsquigarrow {j})$ are uncorrelated with the variables $\ln Z_e$ for $e \in ({u} \rightsquigarrow {i})$. By (43), the covariance becomes

  $$\begin{aligned} {\text {cov}}\left( \ln A_{ui}, \ln A_{uj} \right)&= \sum _{e \in ({u} \rightsquigarrow {i})} {\text {var}}\left( \ln Z_e\right) \\&= \sum _{e \in ({u} \rightsquigarrow {i})} 4 \delta _e^2 = 4 p_{ui} = 2 \left( p_{ui} + p_{uj} - p_{ij}\right) , \end{aligned}$$

  since $p_{ui} + p_{ij} = p_{uj}$, the path from u to j passing by i. This case includes the one where $i = j$, since then $({i} \rightsquigarrow {j})$ is empty and thus $p_{ij} = 0$.

• 2. If $a = j$, the argument is the same as in the previous case.

• 3. Suppose a is different from both i and j. Let $e_a$ and $f_a$ be the first edges of the paths $({a} \rightsquigarrow {i})$ and $({a} \rightsquigarrow {j})$, respectively. These two edges may or may not belong to the same clique. All other edges on $({a} \rightsquigarrow {i})$ and $({a} \rightsquigarrow {j})$, however, must belong to different cliques. It follows that

  $$\begin{aligned} {\text {cov}}\left( \ln A_{ui}, \ln A_{uj} \right)&= \sum _{e \in ({u} \rightsquigarrow {a})} {\text {var}}\left( \ln Z_e \right) + {\text {cov}}\left( \ln Z_{e_a}, \ln Z_{f_a} \right) \\&= 4 p_{ua} + {\text {cov}}\left( \ln Z_{e_a}, \ln Z_{f_a} \right) . \end{aligned}$$

  Now we need to distinguish between two further sub-cases.

  • (3.a) Suppose $e_a$ and $f_a$ do not belong to the same clique. Then the covariance between $\ln Z_{e_a}$ and $\ln Z_{f_a}$ is zero, so that

    $$\begin{aligned} {\text {cov}}\left( \ln A_{ui}, \ln A_{uj} \right) = 4 p_{ua}&= 2 \left( \left( p_{ui} - p_{ai}\right) + \left( p_{uj} - p_{aj}\right) \right) \\&= 2 \left( p_{ui} + p_{uj} - \left( p_{ai} + p_{aj} \right) \right) \\&= 2 \left( p_{ui} + p_{uj} - p_{ij} \right) , \end{aligned}$$

    since the shortest path between i and j passes through a.

  • (3.b) Suppose $e_a$ and $f_a$ belong to the same clique; see Fig. 5. Writing $e_a = (a, k)$ and $f_a = (a, l)$, we find, in view of (43),

    $$\begin{aligned} {\text {cov}}\left( \ln A_{ui}, \ln A_{uj} \right)&= 4 p_{ua} + 2 \left( \delta _{ak}^2 + \delta _{al}^2 - \delta _{kl}^2\right) \\&= 2 \left( \left( p_{ua} + \delta _{ak}^2\right) + \left( p_{ua} + \delta _{al}^2\right) - \delta _{kl}^2 \right) \\&= 2 \left( p_{uk} + p_{ul} - \delta _{kl}^2 \right) \\&= 2 \left( \left( p_{ui} - p_{ki}\right) + \left( p_{uj} - p_{lj}\right) - \delta _{kl}^2 \right) \\&= 2 \left( p_{ui} + p_{uj} - \left( p_{ki} + p_{lj} + \delta _{kl}^2 \right) \right) \\&= 2 \left( p_{ui} + p_{uj} - p_{ij} \right) , \end{aligned}$$

    since the shortest path between i and j passes by k and l.

Fig. 5

Calculation of ${\text {cov}}\left( \ln A_{ui}, \ln A_{uj} \right)$ in the proof of Proposition 4.1. The paths from u to i and from u to j have the path from u to a in common. On these two paths, the edges right after node a are $e_a = (a, k)$ and $f_a = (a, l)$, respectively. The picture considers the case (3.b) where the three nodes a, k, l belong to the same clique, say C

We conclude that the covariance matrix of $(\ln A_{uv}, v \in V \setminus u)$ is indeed $\Sigma _u(\Delta )$ in (23).

Positive definiteness of $\Sigma _u(\Delta )$.

Being a covariance matrix, $\Sigma _u(\Delta )$ is positive semi-definite. We need to show it is invertible. The linear transformation in (44) can be inverted to give

$$\ln Z_e = \ln A_{ub} - \ln A_{ua}$$

for an edge $e = (a, b)$ in $E_u$; note that either $a = u$, in which case $A_{ua} = 1$ and thus $\ln A_{ua} = 0$, or a lies on the path from u to b. Also, each edge $e \in E_u$ is uniquely identified by its endpoint v in $V \setminus u$; let e(v) be the unique edge in $E_u$ with endpoint v. It follows that, as column vectors, the random vectors $(\ln Z_{e(v)}, v \in V \setminus u)$ and $(\ln A_{uv}, v \in V \setminus u)$ are related by

$$\left( \ln A_{uv}, v \in V {\setminus } u \right) = M_u \left( \ln Z_{e(v)}, v \in V {\setminus } u \right) ,$$

where $M_u$ is a $(|V|-1) \times (|V|-1)$ matrix indexed by $(v, w) \in (V \setminus u)^2$ whose inverse is

$$(M_u^{-1})_{vw} = {\left\{ \begin{array}{ll} 1 &{} \text {if } w = v, \\ -1 &{} \text {if } (w, v) \in E_u, \\ 0 &{} \text {otherwise.} \end{array}\right. }$$

The covariance matrix of $(\ln A_{uv}, v \in V \setminus u)$ is thus

$$\Sigma _u(\Delta ) = M_u \, \Sigma _u^Z(\Delta ) \, M_u^\top$$

where $\Sigma _u^Z(\Delta )$ is the (block-diagonal) covariance matrix of $(\ln Z_{e(v)}, v \in V \setminus u)$. The blocks in $\Sigma _u^Z$ are given by (43) and are positive definite and thus invertible by the assumption that each parameter matrix $\Delta _C$ is conditionally negative definite. As a consequence, $\Sigma _u^Z(\Delta )$ is invertible too. Writing $\Theta _u^Z(\Delta ) = \bigl ( \Sigma _u^Z(\Delta ) \bigr )^{-1}$, we find that $\Sigma _u(\Delta )$ is invertible as well with inverse

$$\begin{aligned} \Theta _u(\Delta ) = (M_u^{-1})^\top \, \Theta _u^Z(\Delta ) \, M_u^{-1}. \end{aligned}$$

(7.5)

$P(\Delta )$ is conditionally negative definite.

Clearly, $P(\Delta )$ is symmetric and has zero diagonal. For any non-zero vector $a \in \mathbbm {R}^V$, we have, since $\Sigma _u(\Delta )$ is positive definite,

$$\begin{aligned} 0&< a^\top \Sigma _u(\Delta ) a \\&= 2 \sum _{i \in V} \sum _{j \in V} a_i \left( p_{ui} + p_{uj} - p_{ij} \right) a_j \\&= 2 \sum _{i \in V} a_i p_{ui} \sum _{j \in V} a_j + 2 \sum _{i \in V} a_i \sum _{j \in V} p_{uj} - 2 \sum _{i \in V} \sum _{j \in V} a_i p_{ij} u_j. \end{aligned}$$

If $\sum _{i \in V} a_i = 0$, the first two terms on the right-hand side vanish. The last term is $-2 a^\top P(\Delta ) a$. We conclude that $P(\Delta )$ is conditionally negative definite, as required. $\square$

1.2 B.2. Proof of Proposition 4.2

Let $H_{P(\Delta )}$ be the |V|-variate max-stable Hüsler–Reiss distribution in (20) with parameter matrix $P(\Delta )$ in (21). From (15) we have

$$-\ln H_{P(\Delta )}(x_v, v\in V) = \ell (1/x_v, v\in V) = \mathbb {E}\left[ \max _{v \in V} x_v^{-1} A_{uv}\right] .$$

By the maximum–minimums identity, we have

$$\mathbb {E}\left[ \max _{v \in V} x_v^{-1} A_{uv}\right] =\sum _{i=1}^{|V|} (-1)^{i-1} \sum _{W \subseteq V: |W|=i} \mathbb {E}\left[ \min _{v \in V} x_v^{-1}A_{uv}\right]$$

If W is a singleton, $W = \{u\}$, then the expectation is simply $x_u^{-1}$, whereas if W has more than one element, we write the expectation as the integral of the tail probability:

$$\begin{aligned} \mathbb {E}\left[ \min _{v \in W} x_v^{-1} A_{uv} \right]&= \int _0^\infty \mathbb {P}\left[ \forall v \in W : x_v^{-1} A_{uv}> y \right] \textrm{d}y \\&= \int _0^{x_u^{-1}} \mathbb {P}\left[ \forall v \in W \setminus u : A_{uv}> x_v y \right] \textrm{d}y \\&= \int _{\ln x_u}^{\infty } \mathbb {P}\left[ \forall v \in W \setminus u : \ln A_{uv} > \ln (x_v)-z \right] \, e^{-z} \, \textrm{d}z. \end{aligned}$$

If $W = \{u\}$, we interpret the probability inside the integral as equal to one, so that the integral formula is valid for any non-empty $W \subseteq V$. Combining things, we find that

$$\begin{aligned} -\ln H_{P(\Delta )}(x_v, v\in V)\\ = \sum _{i=1}^{|V|} (-1)^{i-1} \sum _{W \subseteq V: |W| = i} \int _{\ln x_u}^{\infty } \mathbb {P}\left[ \forall v \in W {\setminus } u: \ln A_{uv} > \ln (x_v) - z \right] e^{-z} \, \textrm{d}z. \end{aligned}$$

Recall that the distribution of $(\ln A_{uv}, v \in V \setminus u)$ is multivariate normal with mean vector $\mu _u(\Delta )$ and covariance matrix $\Sigma _u(\Delta )$ in (22) and (23), respectively, hence the expression of $H_{P(\Delta )}$ in (20). $\square$

1.3 Proof of Proposition 4.3

By Proposition 4.2, the Markov field X in Assumption 4.1 satisfies (4) with F its joint cumulative distribution function and $G = H_{P(\Delta )}$. It follows that (6) holds too, yielding the weak convergence of high-threshold excesses to a Pareto random vector Y with distribution given in (5). It remains to show that Y is an extremal graphical model with respect to the given graph $\mathcal {G}$ in the sense of Definition 2 in Engelke and Hitz (2020).

Proposition 4.1 implies

$$\left( \ln X_v - \ln X_u, v \in V {\setminus } u \mid X_u > t \right) {\mathop {\longrightarrow }\limits ^{d}}\mathcal {N}_{|V {\setminus } u|} \bigl ( \mu _u(\Delta ), \Sigma _u(\Delta ) \bigr )$$

as $t \rightarrow \infty$. The latter is the distribution of $(\ln Y_v - \ln Y_u, v \in V \setminus u \mid Y_u > 1)$ for the multivariate Pareto random vector Y in (6) associated to the max-stable Hüsler–Reiss distribution with parameter matrix $P(\Delta )$.

To show that Y is an extremal graphical model with respect to the given block graph $\mathcal {G}$, we apply the criterion in Proposition 3 in Engelke and Hitz (2020). Let $\Theta _u(\Delta ) = (\Sigma _u(\Delta ))^{-1}$ be the precision matrix of the covariance matrix $\Sigma _u(\Delta )$ in (23); see (45). For $i, j \in V$ such that i and j are not connected, i.e., (i, j) is not an edge, we need to show that there is $u \in V \setminus \{i, j\}$ such that

$$\bigl (\Theta _u(\Delta )\bigr )_{ij} = 0.$$

Indeed, according to the cited proposition, the latter identity implies that

$$Y_i \perp _{\textrm{e}} Y_j \mid Y_{\setminus \{i, j\}},$$

the relation $\perp _{\textrm{e}}$ being defined in Definition 1 in Engelke and Hitz (2020), and thus that Y is an extremal graphical model with respect to $\mathcal {G}$.

For two distinct and non-connected nodes $i, j \in V$, let $u \in V \setminus \{i, j\}$. We will show that $(\Theta _u(\Delta ))_{ij} = 0$. We have

$$\begin{aligned} \bigl (\Theta _u(\Delta )\bigr )_{ij}&= \sum _{a \in V \setminus u} \sum _{b \in V \setminus u} \bigl ((M_u^{-1})^\top \bigr )_{ia} \bigl (\Theta _u^Z(\Delta )\bigr )_{ab} \bigl (M_u^{-1}\bigr )_{bj} \\&= \sum _{a \in V \setminus u} \sum _{b \in V \setminus u} \bigl (M_u^{-1}\bigr )_{ai} \bigl (\Theta _u^Z(\Delta )\bigr )_{ab} \bigl (M_u^{-1}\bigr )_{bj}. \end{aligned}$$

Now, $\bigl (M_u^{-1}\bigr )_{ai}$ and $\bigl (M_u^{-1}\bigr )_{bj}$ are non-zero only if $a = i$ or $(i, a) \in E_u$ together with $b = j$ or $(j, b) \in E_u$. In neither case can a and b belong to the same clique, because otherwise we would have found a cycle connecting the nodes u, i, a, b, j. But if a and b belong to different cliques, then so do the edges e(a) and e(b) in $E_u$ with endpoints a and b, and thus $\bigl (\Theta _u^Z(\Delta )\bigr )_{ab} = 0$, since $\Sigma _u^Z(\Delta )$ and thus $\Theta _u^Z(\Delta )$ are block-diagonal. $\square$

1.4 B.4. Proof of Proposition 4.4

Necessity. Let $v \in \bar{U}$ have clique degree $\textrm{cd}(v)$ at most two. We show that the full path sum matrix $P(\Delta )$ is not uniquely determined by the restricted path sum matrix $P(\Delta )_U$ and the graph $\mathcal {G}$. There are two cases: $\textrm{cd}(v) = 1$ and $\textrm{cd}(v) = 2$.

Suppose first that $\textrm{cd}(v) = 1$. Then v belongs only to a single clique, say $C \in \mathcal {C}$. For any $i, j \in U$, the shortest path $({i} \rightsquigarrow {j})$ does not pass through v. Hence the edge weights $\delta _{vw}^2$ for $w \in C \setminus v$ do not show up in any path sum $p_{ij}$ appearing in $P(\Delta )_U$. It follows that these edge weights can be chosen freely (subject to specifying a valid Hüsler–Reiss parameter matrix on C) without affecting $P(\Delta )_U$.

Suppose next that $\textrm{cd}(v) = 2$. Without loss of generality, assume $U = V \setminus v$; this only enlarges the number of visible path sums with respect to the initial problem. We will show that the path sum sub-matrix $(p_{ab}(\Delta ))_{a, b \in V \setminus v}$ does not determine the complete path sum matrix $P(\Delta )$.

By assumption, v is included in two different cliques. Let the set of nodes from one of them, excluding v, be I and let the set of nodes from the other one, excluding v, be J. The sets I and J are non-empty and disjoint. We will show that the edge parameters $\delta _{vi}^2$ for $i \in I$ and $\delta _{vj}^2$ for $j \in J$ are not uniquely determined by the path sums $p_{ab}$ for $a, b \in V \setminus \{v\}$.

• On the one hand, if the path $({a} \rightsquigarrow {b})$ does not pass by v, then the path sum $p_{ab}$ does not contain any of the edge parameters $\delta _{vi}^2$ or $\delta _{vj}^2$ as a summand.

• On the other hand, if the path $({a} \rightsquigarrow {b})$ passes through v, then for some $i \in I$ and $j \in J$ determined by a and b the path sum $p_{ab}$ contains the sum $\delta _{vi}^2 + \delta _{vj}^2$ as a summand. However, sums of the latter form do not change if we decrease each $\delta _{vi}^2$ ($i \in I$) by some small quantity, say $\eta > 0$, and simultaneously increase each $\delta _{vj}^2$ ($j \in J$) by the same quantity, yielding $(\delta _{vi}^2 - \eta ) + (\delta _{vj}^2 + \eta ) = \delta _{vi}^2 + \delta _{vj}^2$.

Sufficiency. Let every node in $\bar{U}$ have clique degree at least three. Let $a \in \bar{U}$ and let $\delta ^2_{ab}$ be the parameter attached to the edge $(a,b)\in E$, with $b \in V \setminus a$. We will show that we can solve $\delta ^2_{ab}$ from the observable path sums $p_{ij}$ for $i, j \in U$.

By assumption there are at least three cliques that are connected to a, say I, J, and Y. Without loss of generality, assume $b\in I$. If $b\in U$ set $\bar{\imath }:=b$, while if $b\in \bar{U}$ walk away from b up to the first node $\bar{\imath }$ in U and this along the unique shortest path between b and $\bar{\imath }$; note that $(a,b) \in ({a} \rightsquigarrow {\bar{\imath }})$. Apply a similar procedure to the cliques J and Y: choose a node $j\in J \setminus a$ (respectively $y\in Y \setminus a$) and if $j\in U$ ($y\in U$) set $\bar{\jmath }:=j$ ($\bar{y}:=y$), while if $j\in \bar{U}$ (respectively $y\in \bar{U}$) take the first node $\bar{\jmath }$ ($\bar{y}$) such that $(a,j)\in ({a} \rightsquigarrow {\bar{\jmath }})$ [$(a,y)\in ({a} \rightsquigarrow {\bar{y}})$]. Because the nodes $\bar{\imath }$, $\bar{\jmath }$ and $\bar{y}$ belong to U, the path sums $p_{\bar{\imath }\bar{\jmath }}$, $p_{\bar{\imath }\bar{y}}$, and $p_{\bar{y}\bar{\jmath }}$ are given. By construction, node a lies on the unique shortest paths between the nodes $\bar{\imath }$, $\bar{\jmath }$ and $\bar{y}$; see also Behtoei et al. (2010, Theorem 1(b)). It follows that

$$\begin{aligned} p_{\bar{\imath }\bar{\jmath }}&= p_{a\bar{\imath }}+p_{a\bar{\jmath }}, \\ p_{\bar{\imath }\bar{y}}&= p_{a\bar{\imath }}+p_{a\bar{y}}, \\ p_{\bar{y}\bar{\jmath }}&= p_{a\bar{\jmath }}+p_{a\bar{y}}. \end{aligned}$$

These are three equations in three unknowns, which can be solved to give, among others, $p_{a\bar{\imath }} = \frac{1}{2} (p_{\bar{\imath }\bar{y}} + p_{\bar{\imath }\bar{\jmath }} - p_{\bar{y}\bar{\jmath }})$. Now we distinguish between two cases, $b \in U$ and $b \in \bar{U}$.

• If $b \in U$ then $\bar{\imath } = b$ and we have written $\delta ^2_{ab} = p_{a\bar{\imath }}$ in terms of the given path sums.

• If $b\in \bar{U}$ we repeat the same procedure as above but starting from node b. We keep the node $\bar{\imath }$, but the nodes $\bar{\jmath }$ and $\bar{y}$ may be different from those when starting from a. After having written $p_{b\bar{\imath }}$ in terms of observable path sums, we can compute $\delta _{ab}^2 = p_{a\bar{\imath }} - p_{b\bar{\imath }}$. $\square$