\ln A_{uk}\indep \ln A_{ul} \mid \ln A_{us},
\end{equation}

for some node $s\neq u$ which is on the shortest path between $k$ and $l$. The partial correlation $\rho_{kl\mid s}$ is zero if

\[
\var(A'_{us})\cov(A'_{uk}, A'_{ul})
=\cov(A'_{uk}, A'_{us})\cov(A'_{ul}A'_{us}),
\]

where we have set $A'_{uv}=\ln A_{uv}$.
Using (<ref>) we have

\begin{equation} \label{eqn:2pus}
\end{equation}

It can be shown that the left hand-side is equivalent to

\begin{equation*}
\end{equation*}

where we have either $(p_{us}+p_{sl}-p_{ul})=0$ and/or $(p_{us}+p_{sk}-p_{uk})=0$. If $(p_{us}+p_{sl}-p_{ul})\neq 0$ then $\pth{u}{s}\cup\pth{s}{l}\neq \pth{u}{l}$. This means that there is a path from $u$ to $l$ that does not pass through $s$. But then there can not be a path from $u$ to $k$ that does not pass through $s$ because otherwise there will be a cycle between non-adjacent nodes which is impossible in block graphs. To show this, consider a path from $u$ to a node, say $f\neq s$, which is on the shortest path between $s$ and $k$. The nodes $l$ and $f$ are non-adjacent but there will ba a path $\{l, \ldots, s, \ldots, f\}$ and another path $\{l, \ldots, u, \ldots, f\}$, hence a cycle. So we have that all paths from $u$ to $k$ pass through $s$ and hence $\pth{u}{s}\cup\pth{s}{k}\neq \pth{u}{k}$. Then we have $(p_{us}+p_{sk}-p_{uk})=0$ and the equality (<ref>) is satisfied. The analogous reasoning holds when $(p_{us}+p_{sk}-p_{uk})\neq 0$. This completes the proof of (<ref>).

The second step is to prove (<ref>). Let $f_A$ denote $f_A(x_A)$, the density of $(A'_{uv},v\in A\subset V)$. Expression (<ref>) holds if and only if the following equations hold

\begin{equation*}
f_{k,l\mid V\setminus\{u,k,l\}}
=f_{k\mid V\setminus\{u,k,l\}}\,f_{ l\mid V\setminus\{u,k,l\}}
\end{equation*}

which is equivalent to

\begin{equation} \label{eqn:515}
%	\frac{f_{V\setminus u}}{f_{V\setminus\{u,k,l\}}}&&
%	=
%	\frac{f_{V\setminus \{u,l\}}}{f_{V\setminus \{u,k,l\}}}\frac{f_{V\setminus \{u,k\}}}{f_{V\setminus \{u,k,l\}}}\\
%	\iff&
f_{V\setminus u}\,f_{V\setminus\{u,k,l\}}
=f_{V\setminus \{u,l\}}\,f_{V\setminus \{u,k\}}.
\end{equation}

We will prove (<ref>). Let $V\setminus u$ be divided into disjoint subsets $(K,L,\{a\})$ with $k\in K, l\in L$ and let $a$ be the separator

of $K$ and $L$. For fixed partition $K,L$ such a node exists uniquely thanks to the property that every minimal clique separator is a singleton.

The distribution of $(A'_{uv}, v\in {K\cup L})\mid A'_{ua}$ is multivariate normal and because of (<ref>) it follows that for any pair $x\in K, y\in L$, we have $\cov(A'_{ux},A'_{uy}\mid A'_{ua})=0$. Hence the multivariate density of $(A'_{uv}, v\in K\cup L\mid A'_{ua})$ can be factorized in the density of $(A'_{uv}, v\in K)$ and the density of $(A'_{uv}, v\in L)$

\begin{equation*}
f_{\{K\cup L\}\mid a}=f_{K\mid a}\,f_{L\mid a}.
\end{equation*}

Going back to (<ref>) the lhs can be represented using the following results:

\begin{align*}
f_{V\setminus u}&
f_{V\setminus \{u,a\}\mid a}\,f_a
f_{\{K\cup L\}\mid a}\,f_a
f_{K\mid a}\,f_{L\mid a}\,f_a\, ,\\
f_{V\setminus \{u,k,l,a\}\mid a}\,f_a
f_{\{K\setminus k\cup L\setminus l\}\mid a}\,f_a
f_{\{K\setminus k\}\mid a}\,f_{\{L\setminus l\}\mid a}\,f_a\,.
\end{align*}

Similarly the terms in the rhs of (<ref>) can be rewritten as

\begin{align*}
f_{V\setminus \{u,l\}}&
f_{V\setminus \{u,l,a\}\mid a}\,f_a
f_{\{K\cup L\setminus l\}\mid a}\,f_a
f_{K\mid a}f_{\{L\setminus l\}\mid a}\,f_a\, ,\\
f_{V\setminus \{u,k\}}&
f_{V\setminus \{u,k,a\}\mid a}\,f_a
f_{\{K\cup L\setminus k\}\mid a}\,f_a
f_{L\mid a}f_{\{K\setminus k\}\mid a}\,f_a\, .
\end{align*}

This completes the proof of (<ref>) and hence of (<ref>).

§.§ Proof of Proposition <ref>

Let $v \in \bar{U}$ have clique degree $\cd(v)$ 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 $\G$. There are two cases: $\cd(v) = 1$ and $\cd(v) = 2$.

Suppose first that $\cd(v) = 1$.
Then $v$ belongs only to a single clique, say $C \in \mathcal{C}$.
For any $i, j \in U$, the shortest path $\pth{i}{j}$ 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 $\cd(v) = 2$.
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 $\pth{a}{b}$ 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 $\pth{a}{b}$ 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 $(\delta_{vi}^2 - \eta) + (\delta_{vj}^2 + \eta) = \delta_{vi}^2 + \delta_{vj}^2$.

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 $(a,b) \in \pth{a}{\bar{\imath}}$.
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 $(a,j)\in \pth{a}{\bar{\jmath}}$ [$(a,y)\in \pth{a}{\bar{y}}$].
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 <cit.>.
It follows that

\begin{align*}
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{align*}

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 $\delta_{ab}^2 = p_{a\bar{\imath}} - p_{b\bar{\imath}}$.

Old proof follows. To be replaced by new proof above?

(Necessity)	Suppose we have a node $v\in \bar{U}$ which makes part of only one clique. This means that $v$ is adjacent to nodes only of the clique to which it belongs. The parameters $\delta^2_{vi}=\delta^2_{iv}$ for $(v,i)\in E$ are in every path sum $p_{vu}=p_{uv}, u\in V$ but $\Lambda_U=\{\lambda_{ij}\}_{i,j\in U}\propto \{p_{ij}\}_{i,j\in U}$, so $\delta^2_{vi}$ for $(v,i)\in E$ does not appear in $\Lambda_U$. For any value of the parameters $(\delta_{vi}^2,(i,v)\in V)$ the distribution determined by the matrix $\Lambda_U$ will be the same.

Now suppose the node $v\in \bar{U}$ is connected to two cliques. Let the nodes from one of the cliques, excluding $v$, be $I$ and the nodes from the other clique, excluding $v$, be $J$. Suppose that all nodes in $I,J$ are observable. We will use induction on the number of nodes in $I$ for fixed number of nodes in set $J$, i.e., $|J|=n$.

When $I=\{i\}$, i.e., $|I|=1$ there are $|J|=n$ paths $(p_{ij}, j\in J)$ available through the matrix coefficients $\lambda_{ij}\propto p_{ij}$ that are given. Hence for fixed $i\in I$ we have the vector of path sums

\begin{equation*}
(p_{ij}=\delta_{iv}^2+\delta_{vj}^2, j\in J),
\end{equation*}

and we can write $p^{n+1}$ as a system of linear equations $D^{n+1} \delta^{n+1}=p^{n+1}$ with $n\times (n+1)$-dimensional matrix
$D^{n+1}=\begin{bmatrix} \mathcal{I}_n & \ind_n \end{bmatrix}$ where $\mathcal{I}_n$ is an identity matrix of size $n$ and $\ind_n$ is a vector of ones of length $n$, and vector $\delta^{n+1}=(\delta_{vj_1}^2, \ldots, \delta_{vj_n}^2, \delta_{iv}^2)$. Therefore  the system of equation $D^{n+1}\delta^{n+1}=p^{n+1}$ has one free parameter and the matrix $D^{n+1}$ has column rank $n$.

We will assume now $|I|=k$ and let the nodes be $\{i_1, \ldots, i_k\}$. In this case we have the vectors of paths $p_{i_1}=(\delta_{i_1v}^2+\delta_{vj}^2, j\in J)$, $p_{i_2}=(\delta_{i_2v}^2+\delta_{vj}^2, j\in J)$, etc., which we substack in a $nk$-dimensional vector $p^{n+k}:=(p_{i_1}, p_{i_2}, \ldots, p_{i_k})$. Now consider the matrix $D^{n+k}$ of size $nk\times (n+k)$ as follows

\begin{equation} \label{eqn:Dnplusk}
\mathcal{I}_n&\ind_n & 0_n&\cdots&0_n\\
\mathcal{I}_n&0_n&\ind_n &\cdots&0_n\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
\mathcal{I}_n&0_n&0_n&\cdots&\ind_n\\
\end{bmatrix},
\end{equation}

and the $nk$-dimensional vector of substacked vectors $p^{n+k}:=(p_{i_1}, p_{i_2}, \ldots, p_{i_k})$ and $(n+k)$-dimensional parameter vector $\delta^{n+k}:=(\delta_{vj_1}^2, \ldots, \delta_{vj_n^2}, \delta_{i_1v}^2, \ldots, \delta_{i_kv}^2)$, i.e., all the weights on edges having one end the node $v$.  Then we have the system of linear equations $D^{n+k}\delta^{n+k}=p^{n+k}$. The induction hypothesis is that the column rank of $D^{n+k}$ is at most $n+k-1$ and we will show that the system of linear equations $D^{n+k+1}\delta^{n+k+1}=p^{n+k+1}$ has column rank $n+k$ meaning that we can not identify all parameters in the vector $\delta^{n+k+1}$. The matrix $D^{n+k+1}$ is equal to

\begin{equation*}
D^{n+k} & 0_{nk}\\
\mathcal{I}_{n}  \, 0_{n\times k} &\ind_n
\end{bmatrix}
\end{equation*}

with $D^{n+k}$ as the one in (<ref>) and $0_{n\times k}$ a $n\times k$-dimensional matrix of zeros. W.l.o.g. take the first $n$ rows of $D^{n+k+1}$, i.e.,

\begin{equation*}
\begin{bmatrix}
\mathcal{I}_n&\ind_n & 0_n&\cdots&0_n
\end{bmatrix}
\end{equation*}

and subtract them from the last $n$ rows

\begin{equation*}
\begin{bmatrix}
\mathcal{I}_n&0_n& \cdots& 0_n& \ind_n
\end{bmatrix}
\end{equation*}

which leads to

\begin{equation*}
\begin{bmatrix}
0_{n\times n}& -\ind_n&0_n&\cdots&0_n&\ind_n
\end{bmatrix}.
\end{equation*}

It is clear that the column rank of this matrix is one and because by the induction hypothesis the column rank of $D^{n+k}$ is $n+k-1$, the column rank of the row reduced matrix

\begin{equation*}
\tilde{D}^{n+k+1}=\begin{bmatrix}
D^{n+k} & 0_{nk}\\
0_{n\times n} \, -\ind_n \, 0_n \,\cdots\, 0_n&\ind_n
\end{bmatrix}
\end{equation*}

is $(n+k-1)+1=n+k$. Hence the system is underdefined and we can not identify all the $n+k+1$ parameters from the available path sums $(p_{ij}, i\in I, j\in J)$.

Any other path sum on a shortest path from $a$ to $b$ with $a,b\in \{V\cap U\}\setminus \{v\cup I\cup J\}$ and passing through $v$ will equal

\begin{equation} \label{eqn:pab}
\end{equation}

for some $i\in I$ and $j\in J$. We can rewrite the equation above as

\begin{equation} \label{eqn:pabbar}
\delta_{iv}^2+\delta_{vj}^2=p_{ab}-p_{ai}-p_{jb}=:\bar{p}_{ab}.
\end{equation}

Hence for any shortest path from $a$ to $b$ with $a,b\in \{V\cap U\}\setminus \{v\cup I\cup J\}$ that passes through $v$ we obtain a system of equations $(\delta_{iv}^2+\delta_{vj}^2=\bar{p}_{ab}, i\in I, j\in J)$ which we have seen that it is not uniquely identified.

If $I,J\in \bar{U}$, then consider all shortest paths that pass through $v$ between nodes with observable variables, say between $a,b\in \{V\cap U\}\setminus \{v\cup I\cup J\}$. The available sumpaths on these shortest paths are of the form in (<ref>). Assuming that for fixed $i\in I$ and $j\in J$ we can identify $p_{ai}$ and $p_{jb}$, by (<ref>) we end up with the same identifiability problem on the system of equations $(\delta_{iv}^2+\delta_{vj}^2=\bar{p}_{ab}, i\in I, j\in J)$.

(Sufficiency)	Let all $v\in \bar{U}$ be of clique degree at least three. Let $a\in \bar{U}$ and $\delta^2_{ab}$ be the parameter attached to the edge $(a,b)\in E$. By assumption there are at least three cliques that are connected to $a$, let's call them $I, J, Y$. W.l.o.g. let $b\in I$. If $b\in U$ set $\bar{i}:=b$ and if $b\in \bar{U}$ walk away from $b$ and take the first node with observable variable, $\bar{i}$, such that $(a,b)\in \pth{a}{\bar{i}}$. Similarly for the cliques $J$ and $Y$. Take a node $j\in J$($y\in Y$) and set $\bar{j}:=j$ ($\bar{y}:=y$) if $j\in U$ ($y\in U$), and if $j\in \bar{U}$ ($y\in \bar{U}$) take the first node with observable variable $\bar{j}$ ($\bar{y}$) such that $(a,j)\in \pth{a}{\bar{j}}$ ($(a,y)\in \pth{a}{\bar{y}}$). Because the variables on the nodes $\bar{i}, \bar{j}$ and $\bar{y}$ are observed the sum paths, which are elements of the matrix $\lambda_U$ are given: $p_{\bar{i}\bar{j}}, p_{\bar{i}\bar{y}}, p_{\bar{y}\bar{j}}$. By <cit.> we have

\begin{align*}
\end{align*}

We can identify the three path sums on the rhs and this gives us $	p_{a\bar{i}}=0.5p_{\bar{i}\bar{y}}+0.5p_{\bar{i}\bar{j}}-0.5p_{\bar{y}\bar{j}}
$. Because $b=\bar{i}$ if $b\in U$ we showed how to obtain the parameter associated to $(a,b)$, namely $\delta^2_{ab}=p_{a\bar{i}}$. If $b\in \bar{U}$ we need to repeat the same procedure as above but considering node $b$. We keep the node $\bar{i}$ but the nodes $\bar{j}, \bar{y}$ may be different then as the previous step. After having identified the path $p_{b\bar{i}}$ we can compute $\delta_{ab}^2=p_{a\bar{i}}-p_{b\bar{i}}$.

The bivariate HR distribution with Gumbel margins $F(x)=e^{-e^{-x}}, x\in \reals$ is given in Hüsler and Reiss, 1989 by

\begin{equation}
\end{equation}

To obtain the form of the HR copula, i.e. where the margins are uniform we do

\begin{align}
H^{*}(u,v)&=H^G\big(-\ln(-\ln u), -\ln(-\ln v)\big)
\\&=
\exp{\big\{\ln v\Phi(\lambda+ \ln (\ln v/\ln u)/2\lambda)+
\ln u\Phi(\lambda+ \ln (\ln u/ \ln v)/2\lambda)\big\}}
\end{align}

A HR copula with Fréchet margins is

\begin{align}
H^{F}(x,y)&=H^*\big(e^{-1/x}, e^{-1/y}\big)
\\&=
\exp{-\big\{(1/y)\Phi(\lambda+ \ln (x/y)/2\lambda)+
(1/x)\Phi(\lambda+ \ln (y/x)/2\lambda)\big\}}
\end{align}

To pass from HR with Gumbel margins to HR with Frechet margins we do

\[
H^F(x,y)=H^G(\ln x, \ln y).
\]

§.§ Correspondence between our Proposition <ref> and Example 8.4 in Heffernan and Tawn, 2004

The correspondence will be shown in bivariate case, $X=(X_1,X_2)$. Let $X$ have max-stable copula with unit Fréchet margins. Let $G_X$ denote the max-stable distribution of $X$ and by Heffernan and Tawn, 2004 let $V_X^{(2)}$ denote the exponent measure of $G_X$ and $\ell_X$ the stdf of $X$. We have for $(x_1,x_2)\in [0,\infty]^2\setminus \{(0,0)\}$

\[
\P(X_1\leq x_1, X_2\leq x_2)
\]

Let $Y=(Y_1,Y_2)=(\ln X_1, \ln X_2))$. Then $Y$ has unit Gumbel marginal distributions. For $Y$ with Gumbel margins we have for $(y_1,y_2)\in [-\infty, \infty]^2$

\begin{equation} \label{eqn:Vtoell}
\begin{split}
\P(Y_1\leq y_1, Y_2\leq y_2)
&= \P(\ln X_1\leq y_1, \ln X_2\leq y_2)
= \P(X_1\leq e^{y_1}, X_2\leq e^{y_2})
\\&= G_X(e^{y_1}, e^{y_2})
\end{split}
\end{equation}

Because we express the result in terms of the partial derivative of the stdf  with respect to its first component, $\dot{\ell}_{X,1}$, and in Heffernan and Tawn, 2004 in terms of the partial derivative of the exponent measure with respect to the first component, $V_{:1}^{(2)}$ ($i=1$), we need to make the relation between $\dot{\ell}_{X,1}$ and $V_{:1}^{(2)}$ for the expressions where the margins are Gumbel. Form the last equation in (<ref>) we have

\[
V_X^{(2)}(e^{x_1}, e^{x_2})=\ell_X(e^{-x_1},e^{-x_2})
\]

Taking derivative with respect to $x_1$ from both sides we obtain

\[
\]

and then

\begin{equation} \label{eqn:dotelltodotV}
\dot{\ell}_{X,1}(e^{-x_1},e^{-x_2})=-e^{2x_1}V_{X,:1}^{(2)}(e^{x_1},e^{x_2}).
\end{equation}

We denote by $C_Y$ the bivariate copula of $Y$. We have

\begin{equation}  \label{eqn:PYCY}
\P(Y_2-y_2\leq t\mid Y_1=t)
=\dot{C}_{Y,1}(F_1(t), F_2(y_2+t)),
\end{equation}

where $\dot{C}_{Y,1}$ is the derivative with respect to the first argument of $C_Y$, and $F_1, F_2$ are standard Gumbel distributions. For the copula of $Y$ we have

\begin{equation} \label{eqn:CYtoGX}
\begin{split}
C_Y(u_1,u_2)&=G_Y(F_1^{-1}(u_1), F_2^{-1}(u_2))
=G_Y\big(-\ln(-\ln u_1), -\ln(-\ln u_2)\big)
\\&= G_X(e^{-\ln(-\ln u_1)}, e^{-\ln(-\ln u_2)})
=G_X(-1/\ln u_1, -1/ \ln u_2)
\\&=\exp\{-\ell_X(-\ln u_1, -\ln u_2)\}
\end{split}
\end{equation}

The partial derivative of $C_Y$ with respect to its first argument considering the expression in (<ref>) is given by

\begin{equation} \label{eqn:CYuGX}
\begin{split}
\dot{C}_{Y,1}(u_1,u_2)=\frac{1}{u_1}G_X(-1/\ln u_1, -1/ \ln u_2)
\dot{\ell}_{X,1}(-\ln u_1, -\ln u_2)
\end{split}
\end{equation}

Consider (<ref>) and (<ref>): replacing $u_1$ par $F_1(t)=e^{-e^{-t}}$ and $u_2$ par $F_2(y_2+t)=e^{-e^{-(y_2+t)}}$ in the expression for $\dot{C}_{Y,1}$ we get

\[
\dot{C}_{Y,1}\big(F_1(t), F_2(y_2+t)\big)
=\frac{1}{F_1(t)}G_X(e^t, e^{y_2+t})\dot{\ell}_{X,1}(e^{-t}, e^{-(y_2+t)}).
\]

The homogeneity of $\ell$ of order one means that $\ell(ax,ay)=a\ell(x,y), a>0$. Taking partial derivative with respect to $x$ from both sides we obtain
\dot{\ell}_1(ax,ay)a=a\dot{\ell}_1(x,y)
and hence $\dot{\ell}_1(ax,ay)=\dot{\ell}_1(x,y)$. From here the following equality follows for $\dot{\ell}_{X,1}$ in the previous display

\[
\dot{\ell}_{X,1}(e^{-t}, e^{-t}e^{-y_2})=\dot{\ell}_{X,1}(1,e^{-y_2}).
\]

The limit of $\frac{1}{F_1(t)}G_X(e^t, e^{y_2+t})$ is 1 as $t\rightarrow\infty$ and hence we have

\[
\lim_{t\rightarrow\infty}\dot{C}_{Y,1}\big(F_1(t), F_2(y_2+t)\big)=\dot{\ell}_{X,1}(1,e^{-y_2}).
\]

This is the expression we would have obtained in Proposition <ref> had we used standard Gumbel margins, instead of Fréchet margins.
From (<ref>) we have

\[
\dot{\ell}_{X,1}(1,e^{-y_2})
=-1^{-1}1^{-1}V_{X,:1}^{(2)}(1^{-1}, (e^{-y_2})^{-1})
\]

The last expression is the one used in Heffernan and Tawn, 2004.