Estimation - Note 6"
In gremes: Estimation of Tail Dependence in Graphical Models

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This estimator differs from the others because the conditioning event does not depend on a particular node $u$ but it depends on the event that the geometric mean exceeds a high threshold.

For application of this estimator, see Vignette "Code - Note 6".

In an unpublished note of Johan Segers (@segers2019mean) it is shown that if $X= (X_1, \ldots, X_d)$ with unit Pareto margins and in the domain of attraction of a Huesler-Reiss distribution with parameter matrix $\Lambda=(\lambda^2){ij}$, then it holds
\begin{equation} \mathcal{L}\Big((Y_v-\bar{Y}){v=1}^d|\bar{Y}>y\Big) \rightarrow \mathcal{N}_d(\bar{\mu}, \bar{\Sigma}), \end{equation} with $Y=(Y_1,\ldots, Y_d)=(\ln X_1,\ldots, \ln X_d )$ and $$\bar{\Sigma} =-M_d\Lambda M_d,\qquad \bar{\mu}=-(1/d)\Lambda 1_d + (1/d)1_d^T \Lambda 1_d 1_d$$ where $M_d=I_d-(1/d)1_d1_d^T$, $I_d$ is an identity matrix of size $d$ and $1_d$ is a a vector of ones of length $d$.

Consider a tree $T=(V,E)$ and edge weights $\theta=(\theta_e, e\in E)$. Under the assumption that $X=(X_v, v\in V)$ is in the domain of attraction of a Huesler-Reiss copula with unit Frechet margins and structured parameter matrix $\Lambda(\theta)$ \begin{equation} \big(\Lambda(\theta)\big){ij} = \lambda^2{ij}(\theta) = \frac{1}{4}\sum_{e \in p(i,j)} \theta_e^2\, , \qquad i,j\in V, \ i \ne j, e\in E. \end{equation} we can employ the method of moments or the composite likelihood method to estimate $\theta=(\theta_e, e\in E)$ from $\bar{\Sigma}(\theta)$.

The method of moments estimator

The method of moments estimator is given by

[ \hat{\theta}^{\mathrm{MMave}}{k,n}=\arg\min{\theta\in (0,\infty)^{|E|}} \|\hat{\Sigma}-\bar{\Sigma}(\theta)\|^2_F ]

$n$ is the number of all observations in the sample
$k$ is the number of the upper order statistics used in the estimation
$\| \cdot \|_F$ is the Frobenius norm
$U\subseteq V$ is the set of observable variables
$\hat{\Sigma}$ is the non-parametric covariance matrix
$\bar{\Sigma}(\theta)$ is the parametric covariance matrix
The parametric matrix $\bar{\Sigma}(\theta)$ is given by \begin{equation} \big(\bar{\Sigma}(\theta)\big){ij} = -M\big(\Lambda(\theta)\big){i,j\in U}M \end{equation} with \begin{equation} \big(\Lambda(\theta)\big){ij} = \lambda^2{ij}(\theta) = \frac{1}{4}\sum_{e \in p(i,j)} \theta_e^2\, , i \ne j, e\in E. \end{equation} (See also the parameterization used for trees in Vignette "Introduction".)
If the sample of the original variables is $\xi_{v,i}, v\in U, i=1,\ldots, n$ consider the transformation using the empirical cumulative distribution function $\hat{F}{v,n}(x)=\big[\sum{i=1}^n\mathbb{1}(\xi_{v,i}\leq x)\big]/(n+1)$. \begin{equation} \hat{X}{v,i} = \frac{1}{1-\hat{F}{v,n}(\xi_{v,i})}, \qquad v \in U, \quad i = 1, \ldots, n. \end{equation} Then consider their logarithm [ \hat{Y}{v,i}=\ln \hat{X}{v,i} ]
For given $k\in {1,\ldots n}$ consider the set of indices [ I = \Big{i = 1,\ldots,n: \overline{\hat{Y}}{i}=(1/|U|)\sum{v\in U} \hat{Y}_{v,i}> n/k \Big} ]
For every $v\in U$ and $i\in I$ compose the differences \begin{equation} \Delta_{v,i} = \hat{Y}{v,i}-\overline{\hat{Y}}{i}. \end{equation}
The vector of means of these differences is given by \begin{equation} \hat{\mu} = \frac{1}{|I|}\sum_{i\in I}(\Delta_{v,i}, v\in U ). \end{equation}
The non-parametric covariance matrix $\hat{\Sigma}$ is given by

\begin{equation} \hat{\Sigma} = \frac{1}{|I|}\sum_{i\in I}(\Delta_{v,i}-\hat{\mu}, v\in U) (\Delta_{v,i}-\hat{\mu}, v\in U)^\top\, . %\end{split} \end{equation}

A non-parametric estimator of this type $\hat{\mu}$ and $\hat{\Sigma}$ has been suggested in @engelke.

The composite likelihood estimator

The composite likelihood estimator is given by [ \hat{\theta}^{\mathrm{MLEave}}{k,n}=\arg\max{\theta\in(0,\infty)^{|E|}} L\Big(\bar{\mu}(\theta), \bar{\Sigma}(\theta); {\Delta_{v,i}, i\in I, v\in U}\Big). ] The likelihood function $L$ above is the one of $|U|$-variate Gaussian probability density function with mean $\bar{\mu}$ and covariance matrix $\bar{\Sigma}$.