and Ferreira, 2006, Lemma 1.2.9 and Corollary 1.2.10 pp.22-23). This yields

$$ \frac{\widehat{ES}_n(\tau_n) - X_{[n\tau_n]:n}}{a((1-\tau_n)^{-1})} - \frac{X_{[n\tau_n]:n} M_n^{(1)}}{a((1-\tau_n)^{-1})} = \frac{1}{\sqrt{n(1-\tau_n)}} \left( \frac{\mu}{(1-\gamma)(1-2\gamma)} + o_P(1) \right). \quad (A.64) $$

To conclude the proof of (i), write

$$ \begin{aligned} & \sqrt{n(1-\tau_n)} \frac{\widehat{ES}_n(\tau_n) - ES(\tau_n)}{a((1-\tau_n)^{-1})} \\ &= \sqrt{n(1-\tau_n)} \left( \frac{\widehat{ES}_n(\tau_n) - X_{[n\tau_n]:n}}{a((1-\tau_n)^{-1})} - \frac{1}{1-\gamma} \right) + \sqrt{n(1-\tau_n)} \frac{X_{[n\tau_n]:n} - q(\tau_n)}{a((1-\tau_n)^{-1})} \\ & - \sqrt{n(1-\tau_n)} \left( \frac{ES(\tau_n) - q(\tau_n)}{a((1-\tau_n)^{-1})} - \frac{1}{1-\gamma} \right) \\ &= \sqrt{n(1-\tau_n)} \left( \frac{X_{[n\tau_n]:n} M_n^{(1)}}{a((1-\tau_n)^{-1})} - \frac{1}{1-\gamma} \right) + \sqrt{n(1-\tau_n)} \frac{X_{[n\tau_n]:n} - q(\tau_n)}{a((1-\tau_n)^{-1})} \\ & - \frac{\lambda}{(1-\gamma)(1-\gamma-\rho)} + \frac{\mu}{(1-\gamma)(1-2\gamma)} + o_P(1) \end{aligned} $$

by (A.64) and Proposition 1(ii).

(ii) Using (i), straightforward calculations show that it suffices to prove that, again when $k = n(1 - \tau_n)$ is a sequence of integers,

$$ \sqrt{k} \left( \frac{X_{n-k:n} M_n^{(1)}}{a(n/k)} - \frac{1}{1-\gamma}, \left\{ \frac{X_{n-k:n}}{a(n/k)} \right\}^2 M_n^{(2)} - \frac{2}{(1-\gamma)(1-2\gamma)}, \frac{X_{n-k:n} - q(1-k/n)}{a(n/k)} \right) \\ \xrightarrow{d} \mathcal{N} \left( \left( \frac{\lambda}{(1-\gamma)(1-\gamma-\rho)} - \frac{\mu}{(1-\gamma)(1-2\gamma)}, \frac{2(3-4\gamma-2\rho)\lambda}{(1-\gamma)(1-2\gamma)(1-\gamma-\rho)(1-2\gamma-\rho)} - \frac{6\mu}{(1-\gamma)(1-2\gamma)(1-3\gamma)}, 0 \right), \begin{pmatrix} \frac{1+\gamma+2\gamma^2}{(1-\gamma)(1-2\gamma)} & \frac{4(1+\gamma^2-3\gamma^3)}{(1-\gamma)^2(1-2\gamma)(1-3\gamma)} & \frac{\gamma}{1-\gamma} \\ \frac{4(1+\gamma^2-3\gamma^3)}{(1-\gamma)^2(1-2\gamma)(1-3\gamma)} & \frac{4(5-\gamma+2\gamma^2-24\gamma^3)}{(1-\gamma)^2(1-2\gamma)(1-3\gamma)(1-4\gamma)} & \frac{4\gamma}{(1-\gamma)(1-2\gamma)} \\ \frac{\gamma}{1-\gamma} & \frac{4\gamma}{(1-\gamma)(1-2\gamma)} & 1 \end{pmatrix} \right) \quad (A.65) $$

Indeed, if this convergence is true, then a simple linearization of $\hat{\gamma}_{n,-}^{\text{Mom}} - \gamma$ yields

$$ \hat{\gamma}_{n,-}^{\text{Mom}} - \gamma = (1-\gamma)^2(1-2\gamma) \left\{ -2 \left( \frac{X_{n-k:n}}{a(n/k)} M_n^{(1)} - \frac{1}{1-\gamma} \right) + \frac{1-2\gamma}{2} \left( \left\{ \frac{X_{n-k:n}}{a(n/k)} \right\}^2 M_n^{(2)} - \frac{2}{(1-\gamma)(1-2\gamma)} \right) \right\} + o_P \left( \frac{1}{\sqrt{k}} \right) $$
and straightforward matrix algebra gives the desired convergence. With the notation of Lemma A.8, the random vector on the left-hand side of (A.65) has the same distribution as

$$ \sqrt{k} \left( \frac{U(Y_{n-k:n})}{a(n/k)} \mathfrak{M}_n^{(1)} - \frac{1}{1-\gamma} \right) \left\{ \frac{U(Y_{n-k:n})}{a(n/k)} \right\}^2 \mathfrak{M}_n^{(2)} - \frac{2}{(1-\gamma)(1-2\gamma)} \frac{U(Y_{n-k:n}) - U(n/k)}{a(n/k)} $$

Now, by Lemma A.5 applied to the function $a$ (which satisfies the required second-order inequality, see (A.21)), Lemma A.8 and (A.58),

$$ \begin{aligned} & \sqrt{k} \left( \frac{U(Y_{n-k:n})}{a(n/k)} \mathfrak{M}_n^{(1)} - \frac{1}{1-\gamma} \right) \\ &= \sqrt{k} \left( \frac{U(Y_{n-k:n})}{a(Y_{n-k:n})} \mathfrak{M}_n^{(1)} - \frac{1}{1-\gamma} \right) + \sqrt{k} \left( \frac{a(Y_{n-k:n})}{a(n/k)} - 1 \right) \frac{U(Y_{n-k:n})}{a(Y_{n-k:n})} \mathfrak{M}_n^{(1)} \\ &= \sqrt{k} \left( \frac{U(Y_{n-k:n})}{a(Y_{n-k:n})} \mathfrak{M}_n^{(1)} - \frac{1}{1-\gamma} \right) + \frac{\gamma}{1-\gamma} \times \sqrt{k} \frac{U(Y_{n-k:n}) - U(n/k)}{a(n/k)} + o_P(1). \end{aligned} $$

Likewise,

$$ \begin{aligned} & \sqrt{k} \left( \left\{ \frac{U(Y_{n-k:n})}{a(n/k)} \right\}^2 \mathfrak{M}_n^{(2)} - \frac{2}{(1-\gamma)(1-2\gamma)} \right) \\ &= \sqrt{k} \left( \left\{ \frac{U(Y_{n-k:n})}{a(Y_{n-k:n})} \right\}^2 \mathfrak{M}_n^{(2)} - \frac{2}{(1-\gamma)(1-2\gamma)} \right) \\ & \quad + \sqrt{k} \left( \frac{a(Y_{n-k:n})}{a(n/k)} - 1 \right) \left( \frac{a(Y_{n-k:n})}{a(n/k)} + 1 \right) \left\{ \frac{U(Y_{n-k:n})}{a(Y_{n-k:n})} \right\}^2 \mathfrak{M}_n^{(2)} \\ &= \sqrt{k} \left( \left\{ \frac{U(Y_{n-k:n})}{a(Y_{n-k:n})} \right\}^2 \mathfrak{M}_n^{(2)} - \frac{2}{(1-\gamma)(1-2\gamma)} \right) \\ & \quad + \frac{4\gamma}{(1-\gamma)(1-2\gamma)} \times \sqrt{k} \frac{U(Y_{n-k:n}) - U(n/k)}{a(n/k)} + o_P(1). \end{aligned} $$

Convergence (A.65) now follows immediately from Lemma A.8 and straightforward, if burdensome, calculations. □

Our final auxiliary result will be useful when establishing the asymptotic properties of moment-based extrapolated Expected Shortfall estimators. This lemma can be seen as a complement to Theorem 3.5.4 p.104 in de Haan and Ferreira (2006), because it provides a unified representation of the bias term of $\hat{\gamma}_n^{\text{Mom}}$.

**Lemma A.10 (Convergence of $\hat{\gamma}_n^{\text{Mom}}$).** Suppose that $X$ satisfies condition $C_2(\gamma, a, \rho, A)$ with $0 < U(\infty) = q(1) \le \infty$. Let $\tau_n \uparrow 1$ be such that $n(1 - \tau_n) \to \infty$, $\sqrt{n(1 - \tau_n)}A((1 - \tau_n)^{-1}) \to \lambda \in \mathbb{R}$ and $\sqrt{n(1 - \tau_n)}(a((1 - \tau_n)^{-1})/q(\tau_n) - \gamma_+) \to \mu \in \mathbb{R}$. Then $\sqrt{n(1 - \tau_n)}(\hat{\gamma}_n^{\text{Mom}} - \gamma)$ is
asymptotically Gaussian with mean $\lambda b_1(\gamma, \rho) + \mu b_2(\gamma)$, where

$$b_1(\gamma, \rho) = \begin{cases} \frac{1 + \gamma + \gamma^2 - \rho\gamma^2}{(1 + \gamma)^2(1 - \rho)^2} & \text{if } \gamma \ge 0, \\ \frac{(1 - \gamma)(1 - 2\gamma)}{(1 - \gamma - \rho)(1 - 2\gamma - \rho)} & \text{if } \gamma < 0, \end{cases}$$

$$\text{and } b_2(\gamma) = \begin{cases} \frac{\gamma}{(1 + \gamma)^2} & \text{if } \gamma \ge 0, \\ -\frac{\gamma(1 + \gamma)}{(1 - \gamma)(1 - 3\gamma)} & \text{if } \gamma < 0, \end{cases}$$

and variance

$$\begin{cases} \gamma^2 + 1 & \text{if } \gamma \ge 0, \\ \frac{(1 - \gamma)^2(1 - 2\gamma)(1 - \gamma + 6\gamma^2)}{(1 - 3\gamma)(1 - 4\gamma)} & \text{if } \gamma < 0. \end{cases}$$

*Proof of Lemma A.10.* Recall that $\hat{\gamma}_n^{\text{Mom}} = M_n^{(1)} + \hat{\gamma}_{n,-}^{\text{Mom}}$. Set $k = \lfloor n(1 - \tau_n) \rfloor$: in particular $k \to \infty$, $k/n \to 0$, $\sqrt{k}A(n/k) \to \lambda$ (because of the regular variation property of A) and $\sqrt{k}(a(n/k)/U(n/k) - \gamma_+) \to \mu$ (using Equations (A.22) and (A.23)). Note that

$$(M_n^{(1)} - \gamma_+, \hat{\gamma}_{n,-}^{\text{Mom}} - \gamma_-) \stackrel{\text{d}}{=} \left( \frac{a(Y_{n-k:n}) \overline{\mathfrak{M}}_n^{(1)}}{U(Y_{n-k:n})} - \gamma_+, 1 - \frac{1}{2} \left( 1 - \frac{(\overline{\mathfrak{M}}_n^{(1)})^2}{\overline{\mathfrak{M}}_n^{(2)}} \right)^{-1} - \gamma_- \right)$$

where, letting the $Y_i$ be independent copies of a unit Pareto random variable Y,

$$\overline{\mathfrak{M}}_n^{(j)} = \left\{ \frac{U(Y_{n-k:n})}{a(Y_{n-k:n})} \right\}^j \times \frac{1}{k} \sum_{i=1}^{k} \log^j \frac{U(Y_{n-i+1:n})}{U(Y_{n-k:n})}.$$

Recalling (A.59), a straightforward linearization then yields

$$(M_n^{(1)} - \gamma_+, \hat{\gamma}_{n,-}^{\text{Mom}} - \gamma_-) \stackrel{\text{d}}{=} \left( \gamma_+ \left( \overline{\mathfrak{M}}_n^{(1)} - \frac{1}{1-\gamma_-} \right) + \frac{1}{\sqrt{k}} \frac{\mu}{1-\gamma_-}, \right. \\
\left. (1-\gamma_-)^2 (1-2\gamma_-) \left\{ -2 \left( \overline{\mathfrak{M}}_n^{(1)} - \frac{1}{1-\gamma_-} \right) + \frac{1-2\gamma_-}{2} \left( \overline{\mathfrak{M}}_n^{(2)} - \frac{2}{(1-\gamma_-)(1-2\gamma_-)} \right) \right\} \right) \\
+ o_{\mathbb{P}} \left( \frac{1}{\sqrt{k}} \right). \qquad (A.66)$$

Conclude that

$$\sqrt{n(1-\tau_n)}(\hat{\gamma}_n^{\text{Mom}} - \gamma) = \frac{\mu}{1-\gamma} + \sqrt{n(1-\tau_n)}(\hat{\gamma}_{n,-}^{\text{Mom}} - \gamma) + o_{\mathbb{P}}(1) \quad (A.67)$$

when $\gamma < 0$, so that the result is an immediate consequence of Lemma A.9 in this case. When $\gamma \ge 0$, Equation (A.66) yields

$$\sqrt{n(1-\tau_n)}(\hat{\gamma}_n^{\text{Mom}} - \gamma) \stackrel{\text{d}}{=} (\gamma - 2) (\overline{\mathfrak{M}}_n^{(1)} - 1) + \frac{1}{2} (\overline{\mathfrak{M}}_n^{(2)} - 2) + \mu + o_{\mathbb{P}}(1)$$

and the result follows from Lemma A.8 and straightforward calculations. $\square$
Proof of Theorem 4. We consider three cases: $\gamma < 0$, then $0 \le \gamma < 1/2$ and finally $1/2 \le \gamma < 1$. As a preliminary step we note that when $\gamma \ge 0$ and $\rho < 0$ one has, according to Theorem 4.2.1 p.131 in de Haan and Ferreira (2006),

$$ \sqrt{n(1 - \tau_n)} \left( \frac{\hat{a}_n^{\text{Mom}}((1 - \tau_n)^{-1})}{a((1 - \tau_n)^{-1})} - 1 \right) = o_P(1). \quad (A.68) $$

Case $\gamma < 0$: Recall that

$$ \hat{a}_n^{\text{Mom}}((1 - \tau_n)^{-1}) = X_{[n\tau_n]:n} M_n^{(1)}(1 - \hat{\gamma}_{n,-}^{\text{Mom}}). $$

It immediately follows from Lemma A.9 that

$$ \sqrt{n(1 - \tau_n)} \left( \frac{\hat{a}_n^{\text{Mom}}((1 - \tau_n)^{-1})}{a((1 - \tau_n)^{-1})} - 1 \right) = (1 - \gamma) \times \sqrt{n(1 - \tau_n)} \left( \frac{X_{[n\tau_n]:n} M_n^{(1)}}{a((1 - \tau_n)^{-1})} - \frac{1}{1 - \gamma} \right) \\ - \frac{1}{1 - \gamma} \times \sqrt{n(1 - \tau_n)} (\hat{\gamma}_{n,-}^{\text{Mom}} - \gamma) + o_P(1). \quad (A.69) $$

Recalling (A.67) and applying Lemma A.9 then yields, after straightforward matrix algebra,

$$ \sqrt{n(1 - \tau_n)} \left( \frac{\widehat{\text{ES}}_n(\tau_n) - \text{ES}(\tau_n)}{a((1 - \tau_n)^{-1})}, \frac{\hat{a}_n^{\text{Mom}}((1 - \tau_n)^{-1})}{a((1 - \tau_n)^{-1})} - 1, \hat{\gamma}_n^{\text{Mom}} - \gamma \right) \xrightarrow{d} (Z_{\text{loc}}, Z_{\text{scale}}, Z_{\text{shape}}) $$

where the vector $(Z_{\text{loc}}, Z_{\text{scale}}, Z_{\text{shape}})$ is trivariate Gaussian with mean vector defined as

$$ \mathbb{E}(Z_{\text{loc}}) = 0, $$

$$ \mathbb{E}(Z_{\text{scale}}) = -\lambda \frac{\rho}{(1 - \gamma - \rho)(1 - 2\gamma - \rho)} + \mu \frac{\gamma}{(1 - 2\gamma)(1 - 3\gamma)}, $$

$$ \text{and } \mathbb{E}(Z_{\text{shape}}) = \lambda \frac{(1 - \gamma)(1 - 2\gamma)}{(1 - \gamma - \rho)(1 - 2\gamma - \rho)} - \mu \frac{\gamma(1 + \gamma)}{(1 - \gamma)(1 - 3\gamma)}, $$

and with covariance matrix

$$ \left( \begin{array}{c} \frac{2}{(1 - \gamma)(1 - 2\gamma)} \\ \frac{1 - 4\gamma - \gamma^2 + 6\gamma^3}{(1 - \gamma)(1 - 2\gamma)(1 - 3\gamma)} \\ \frac{2\gamma}{1 - 3\gamma} \end{array} \begin{array}{c} \frac{1 - 4\gamma - \gamma^2 + 6\gamma^3}{(1 - \gamma)(1 - 2\gamma)(1 - 3\gamma)} \\ \frac{2 - 16\gamma + 51\gamma^2 - 69\gamma^3 + 50\gamma^4 - 24\gamma^5}{(1 - 2\gamma)(1 - 3\gamma)(1 - 4\gamma)} \\ \frac{(1 - \gamma)^2(1 - 4\gamma + 12\gamma^2)}{(1 - 3\gamma)(1 - 4\gamma)} \end{array} \begin{array}{c} \frac{2\gamma}{1 - 3\gamma} \\ \frac{(1 - \gamma)^2(1 - 4\gamma + 12\gamma^2)}{(1 - 3\gamma)(1 - 4\gamma)} \\ \frac{(1 - \gamma)^2(1 - 2\gamma)(1 - \gamma + 6\gamma^2)}{(1 - 3\gamma)(1 - 4\gamma)} \end{array} \right). $$

The result now follows from a direct application of Corollary 1.

Case $0 \le \gamma < 1/2$: In this setting, the result is an immediate consequence of Corollary 2, whose assumptions are fulfilled thanks to Theorem 1, (A.68) and Lemma A.10.

Case $1/2 \le \gamma < 1$: Let, as in the statement of Theorem 3, $d_n = (1 - \tau_n)/(1 - \tau'_n)$. By Equation (A.31),

$$ \int_{1}^{d_n} s^{\gamma-1} \log(s) ds = \frac{1}{\gamma} d_n^{\gamma} \log(d_n) (1 + o(1)) $$
and since $1/d_n = O(1/(n(1 - \tau_n))) \to 0$, one obviously has

$$ \frac{1}{\int_{1}^{d_n} s^{\gamma-1} \log(s) ds} = o(d_n^{-\gamma}) = o((n(1 - \tau_n))^{-\gamma}). $$

Consequently $\sqrt{n(1 - \tau_n)}/\int_1^{d_n} s^{\gamma-1} \log(s) ds \to 0$. Combining Theorems 1 and 3 with (A.68) and Lemma A.10, we get

$$ \sqrt{n(1 - \tau_n)} \frac{\widehat{\mathrm{ES}}_n^*(\tau_n') - \mathrm{ES}(\tau_n')}{a((1 - \tau_n)^{-1}) \int_1^{d_n} s^{\gamma-1} \log(s) ds} = \frac{1}{1 - \gamma} \sqrt{n(1 - \tau_n)} (\hat{\gamma}_n - \gamma) + o_P(1). $$

The conclusion again follows from Lemma A.10.

<figure><img src="image_14.png" /></figure>

*Proof of Theorem 5.* We split the proof into two parts: we first consider the case $\gamma < 0$ and then the case $0 \le \gamma < 1$. As a starting point we note that in general, if $\widehat{\mathrm{ES}}_n(\tau_n)$ is defined as

$$ \widehat{\mathrm{ES}}_n(\tau_n) = X_{[n\tau_n]:n} + \hat{a}_n((1 - \tau_n)^{-1}) \frac{1}{1 - \hat{\gamma}_n}, $$

where it is assumed that

$$ \sqrt{n(1 - \tau_n)} \left( \frac{X_{[n\tau_n]:n} - q(\tau_n)}{a((1 - \tau_n)^{-1})}, \frac{\hat{a}_n((1 - \tau_n)^{-1})}{a((1 - \tau_n)^{-1})} - 1, \hat{\gamma}_n - \gamma \right) \xrightarrow{d} (N_{\mathrm{loc}}, N_{\mathrm{scale}}, N_{\mathrm{shape}}), $$

in which the trivariate random vector $(N_{\mathrm{loc}}, N_{\mathrm{scale}}, N_{\mathrm{shape}})$ has a nondegenerate distribution, then a simple combination of Theorem 2 and Corollary 1 entails

$$ \sqrt{n(1 - \tau_n)} \frac{\widehat{\mathrm{ES}}_n^*(\tau_n') - \mathrm{ES}(\tau_n')}{a((1 - \tau_n)^{-1}) \int_1^{(1-\tau_n)/(1-\tau_n')} s^{\gamma-1} \log(s) ds} \xrightarrow{d} \gamma_-^2 N_{\mathrm{loc}} - \gamma_- N_{\mathrm{scale}} + \frac{1}{1-\gamma_+} N_{\mathrm{shape}} - \lambda \frac{\gamma_-}{\gamma_- + \rho}. \quad (A.70) $$

*Case* $\gamma < 0$: Use Lemma A.9 in conjunction with (A.67) and (A.69) to obtain that

$$ \sqrt{n(1 - \tau_n)} \left( \frac{X_{[n\tau_n]:n} - q(\tau_n)}{a((1 - \tau_n)^{-1})}, \frac{\hat{a}_n^{\mathrm{Mom}}((1 - \tau_n)^{-1})}{a((1 - \tau_n)^{-1})} - 1, \hat{\gamma}_n^{\mathrm{Mom}} - \gamma \right) \xrightarrow{d} (N_{\mathrm{loc}}, N_{\mathrm{scale}}, N_{\mathrm{shape}}) $$

where the vector $(N_{\mathrm{loc}}, N_{\mathrm{scale}}, N_{\mathrm{shape}})$ is trivariate Gaussian with mean vector defined as

$$ \mathbb{E}(N_{\mathrm{loc}}) = 0, $$

$$ \mathbb{E}(N_{\mathrm{scale}}) = -\lambda \frac{\rho}{(1 - \gamma - \rho)(1 - 2\gamma - \rho)} + \mu \frac{\gamma}{(1 - 2\gamma)(1 - 3\gamma)}, $$

$$ \text{and } \mathbb{E}(N_{\mathrm{shape}}) = \lambda \frac{(1 - \gamma)(1 - 2\gamma)}{(1 - \gamma - \rho)(1 - 2\gamma - \rho)} - \mu \frac{\gamma(1 + \gamma)}{(1 - \gamma)(1 - 3\gamma)}, $$

and with covariance matrix

$$ \begin{pmatrix} 1 & \gamma & 0 \\ \gamma & \frac{2 - 16\gamma + 51\gamma^2 - 69\gamma^3 + 50\gamma^4 - 24\gamma^5}{(1 - 2\gamma)(1 - 3\gamma)(1 - 4\gamma)} & -\frac{(1 - \gamma)^2(1 - 4\gamma + 12\gamma^2)}{(1 - 3\gamma)(1 - 4\gamma)} \\ 0 & -\frac{(1 - \gamma)^2(1 - 4\gamma + 12\gamma^2)}{(1 - 3\gamma)(1 - 4\gamma)} & \frac{(1 - \gamma)^2(1 - 2\gamma)(1 - \gamma + 6\gamma^2)}{(1 - 3\gamma)(1 - 4\gamma)} \end{pmatrix}. $$
Combine this convergence with (A.70) to obtain the result after straightforward calculations.

Case $0 \le \gamma < 1$: It is a particular consequence of Lemma A.8 that

$$ \sqrt{n(1 - \tau_n)} \frac{X_{\lfloor n\tau_n \rfloor:n} - q(\tau_n)}{a((1 - \tau_n)^{-1})} = O_P(1). \quad (A.71) $$

Combine (A.68), (A.71) and Lemma A.10 to get

$$ \sqrt{n(1 - \tau_n)} \frac{\overline{\mathrm{ES}}_n(\tau_n) - \mathrm{ES}(\tau_n)}{a((1 - \tau_n)^{-1})} = O_P(1) \quad \text{and} \quad \sqrt{n(1 - \tau_n)} \left( \frac{\hat{a}_n^{\mathrm{Mom}}((1 - \tau_n)^{-1})}{a((1 - \tau_n)^{-1})} - 1 \right) = O_P(1). $$

Use this in conjunction with Lemma A.10 and Corollary 2 to obtain the desired result. □

## B Compatibility between Proposition 2 and existing second-order theory

According to the discussion on p.103 in de Haan and Ferreira (2006), if condition $C_2(\gamma, a, \rho, A)$ holds with $0 < U(\infty) = q(1) \le \infty$, $\gamma \ne \rho$ and $\rho < 0$ when $\gamma > 0$, then $\log U$ satisfies a similar second-order condition:

$$ \forall x > 0, \lim_{t \to \infty} \frac{1}{B(t)} \left( \frac{\log U(tx) - \log U(t)}{a(t)/U(t)} - \int_{1}^{x} s^{\gamma-1} ds \right) = \int_{1}^{x} s^{\gamma-1} \left( \int_{1}^{s} u^{\rho'-1} du \right) ds. \quad (B.72) $$

According to Lemma B.3.16 p.398 in de Haan and Ferreira (2006), the quantities $\rho'$ and $B$ are defined as

$$ \rho' = \begin{cases} \rho & \text{if } \gamma < \rho \le 0 \text{ or } 0 < -\rho \le \gamma \text{ or } (0 < \gamma < -\rho \text{ and } l = 0), \\ -|\gamma| & \text{if } \rho < \gamma \le 0 \text{ or } (0 < \gamma < -\rho \text{ and } l \ne 0), \end{cases} $$

$$ \text{and } B(t) = \begin{cases} A(t) & \text{if } \gamma < \rho \le 0, \\ \frac{\rho}{\gamma + \rho} A(t) & \text{if } 0 < -\rho < \gamma \text{ or } (0 < \gamma < -\rho \text{ and } l = 0), \\ \gamma_+ - \frac{a(t)}{U(t)} & \text{if } \rho < \gamma \le 0 \text{ or } \gamma = -\rho \text{ or } (0 < \gamma < -\rho \text{ and } l \ne 0), \end{cases} $$

where for $\gamma + \rho < 0$ the quantity $l$ is defined as $l = \lim_{t \to \infty} U(t) - a(t)/\gamma$ (see Theorem 2.1 in Fraga Alves et al. (2007) for the existence of this limit). Theorem 2.1 in Fraga Alves et al. (2007) also shows that actually, when $\gamma \ne \rho$,

$$ \lim_{t \to \infty} \frac{1}{A(t)} \left( \frac{a(t)}{U(t)} - \gamma_+ \right) = \begin{cases} 0 & \text{if } \gamma < \rho \le 0, \\ \frac{\gamma}{\gamma + \rho} & \text{if } 0 < -\rho < \gamma \text{ or } (0 < \gamma < -\rho \text{ and } l = 0), \\ \pm\infty & \text{if } \rho < \gamma \le 0 \text{ or } \gamma = -\rho \text{ or } (0 < \gamma < -\rho \text{ and } l \ne 0). \end{cases} $$
This means that the second-order condition (B.72) on $\log U$ is an immediate consequence of Proposition 2 when either $A(t)$ or $a(t)/U(t) - \gamma_+$ dominates the other, that is, in any of the four situations $\gamma < \rho \le 0$, $\rho < \gamma \le 0$, $\gamma = -\rho$ and $(0 < \gamma < -\rho$ and $l \ne 0$). When this is not the case, that is, when $0 < -\rho < \gamma$ or $(0 < \gamma < -\rho$ and $l = 0$), then Proposition 2 and the above result from Fraga Alves et al. (2007) yield, as $t \to \infty$,

$$ \frac{\log U(tx) - \log U(t)}{a(t)/U(t)} - \int_{1}^{x} s^{\gamma-1}ds \\ = A(t) \left\{ -\frac{\gamma}{\gamma + \rho} \int_{1}^{x} s^{-1} \left( \int_{1}^{s} u^{-\gamma-1}du \right) ds + x^{-\gamma} \int_{1}^{x} s^{\gamma-1} \left( \int_{1}^{s} u^{\rho-1}du \right) ds + o(1) \right\} $$

pointwise in $x$. A straightforward calculation then shows that

$$ \frac{\log U(tx) - \log U(t)}{a(t)/U(t)} - \int_{1}^{x} s^{\gamma-1}ds = \frac{\rho}{\gamma + \rho} A(t) \int_{1}^{x} s^{-1} \left( \int_{1}^{s} u^{\rho-1}du \right) ds + o(|A(t)|) $$

which, again, is nothing but the second-order condition (B.72). Proposition 2 therefore extends this second-order condition on $\log U$ by avoiding the restrictions $\gamma \ne \rho$ and $\rho < 0$ if $\gamma > 0$.

# C Technical details of the construction of the corrected inference procedures

## C.1 The Expectrem R package

We have implemented our methods in an R package called Expectrem, freely available on GitHub<fnref n="2" />. The objective of this package is to provide methods for the estimation of certain expectation-type quantities, such as expectiles and the Expected Shortfall. We have added the following objects to the existing version of this R package:

* **momentindex**: For a dataset X given in input, this function returns the moment estimator $\hat{\gamma}_n^{\text{Mom}}$ using the $k_n$ (argument k) top log-spacings. The asymptotic Gaussian confidence interval (with confidence level ci.level, default 0.95) derived from Theorem 3.5.4 p.104 in de Haan and Ferreira (2006) is also returned.

* **extES**: At a given probability level $\tau'_n$ (argument tau), and for a dataset X given in input, compute an extrapolated estimator of the Expected Shortfall at level $\tau'_n$ using the $k_n$ top order statistics (argument k). Setting the argument estim to "Hill" returns the estimator $\hat{\text{ES}}_n^W(\tau'_n)$ if method="direct", and $\hat{\text{ES}}_n^W(\tau'_n)$ if method="indirect". Setting the argument estim to "Moment" leads to $\hat{\text{ES}}_n^*(\tau'_n)$ if method="direct", and to $\hat{\text{ES}}_n^*(\tau'_n)$

<footnote n="2">²https://github.com/AntoineUC/Expectrem</footnote>
if method="indirect". The associated confidence intervals at an asymptotic confidence level 1 − α (argument ci.level, default 0.95), which are ̂ I$_{1}^{*}(α), ̂ I$_{1}^{*}(α), ̂ I$_{4}^{*}(α)$ and ̂ I$_{4}^{*}(α), respectively, are also returned.

• ESexp, ESfrechet, ESgumbel, ESkumaraswamy, ESlnorm, ESlogis, ESnorm, ESpareto, ESrev_frechet, ESt, ESweibull: These functions give the values of the Expected Shortfall (at a vector of levels specified by the argument `probs`) of the exponential, Fréchet, Gumbel, Kumaraswamy, log-normal, logistic, Gaussian, Pareto, reverse-Fréchet, Student and Weibull distributions.

• flood_data: The OpenFEMA flood claim amounts dataset used in Section 5.2 (and Section E of the Appendix).

## C.2 Inference at intermediate levels

We consider here the intermediate setting when $\tau_n = 1 - k_n/n$ with $k_n \to \infty$ and $k_n/n \to 0$ as $n \to \infty$. In this situation, $\mathrm{ES}(\tau_n)$ may be estimated either by the empirical Expected Shortfall $\widehat{\mathrm{ES}}_n(\tau_n)$, or by its quantile-based semiparametric estimator $\widehat{\mathrm{ES}}_n(\tau_n)$. To the best of our knowledge, the literature on $\widehat{\mathrm{ES}}_n(\tau_n)$ has previously focused on the case when $X$ is heavy-tailed and has a finite variance, that is, $\gamma \in (0, 1/2)$. In this case, Remark 1 suggests the following (competitor) confidence interval at level $1 - \alpha$:

$$ \hat{I}_1(\alpha) = \left[ \widehat{\mathrm{ES}}_n(1 - k_n/n) \exp \left( \pm \frac{\hat{\gamma}_n^H}{\sqrt{k_n}} \sqrt{\frac{2(1 - \hat{\gamma}_n^H)}{1 - 2\hat{\gamma}_n^H}} z_{1-\alpha/2} \right) \right], $$

where $z_{1-\alpha/2}$ is the quantile of level $1-\alpha/2$ of the standard normal distribution, and $\hat{\gamma}_n^H = M_n^{(1)}$ is the Hill (1975) estimator. This confidence interval is naturally restricted to the situation $\gamma \in (0, 1/2)$. A more general inference procedure can be obtained as follows: note that up to a negligible bias term,

$$ \frac{\widehat{\mathrm{ES}}_n(\tau_n) - \mathrm{ES}(\tau_n)}{a((1 - \tau_n)^{-1})} \approx \frac{1}{k_n} \sum_{i=1}^{k_n} \frac{X_{n-i+1:n} - X_{n-k_n:n}}{a(n/k_n)} - \frac{1}{1 - \gamma} + \frac{X_{n-k_n:n} - q(1 - k_n/n)}{a(n/k_n)} $$

Therefore

$$ \frac{\widehat{\mathrm{ES}}_n(\tau_n) - \mathrm{ES}(\tau_n)}{a((1 - \tau_n)^{-1})} \approx \frac{a(Y_{n-k_n:n})}{a(n/k_n)} \times \frac{1}{k_n} \sum_{i=1}^{k_n} \left( \frac{U(Y_{n-i+1:n}) - U(Y_{n-k_n:n})}{a(Y_{n-k_n:n})} - \frac{1}{1-\gamma} \right) \\ + \frac{1}{1-\gamma} \left( \frac{a(Y_{n-k_n:n})}{a(n/k_n)} - 1 \right) + \frac{U(Y_{n-k_n:n}) - U(n/k_n)}{a(n/k_n)} $$

Recall that $Y$ is unit Pareto if and only if $\log(Y)$ is unit exponential. Using the regular variation property of the function $a$, condition $C_2(\gamma, a, \rho, A)$ and the Rényi representation
(see (A.15) and (A.16) in the Appendix), we get, with the notation $D_r(x) = \int_1^x s^{r-1}ds$,

$$ 
\begin{aligned}
\frac{\widehat{ES}_n(\tau_n) - ES(\tau_n)}{a((1-\tau_n)^{-1})} &\approx \left(\frac{k_n}{n}Y_{n-k_n:n}\right)^\gamma \times \frac{1}{k_n} \sum_{i=1}^{k_n} \left(D_\gamma\left(Y_{n-i+1:n}/Y_{n-k_n:n}\right) - \frac{1}{1-\gamma}\right) \\
&\quad + \frac{1}{1-\gamma} \left(\left(\frac{k_n}{n}Y_{n-k_n:n}\right)^\gamma - 1\right) + D_\gamma\left(\frac{k_n}{n}Y_{n-k_n:n}\right) \\
&\stackrel{\mathrm{d}}{=} \left(\frac{k_n}{n}\exp\left(\sum_{i=k_n+1}^n \frac{\log(Y_i)}{i}\right)\right)^\gamma \left(\frac{1}{k_n}\sum_{i=1}^{k_n} D_\gamma(Y_i) - \frac{1}{1-\gamma}\right) \\
&\quad + \frac{1}{1-\gamma}D_\gamma\left(\frac{k_n}{n}\exp\left(\sum_{i=k_n+1}^n \frac{\log(Y_i)}{i}\right)\right) =: -G_n^{(1)}(Y_1, \dots, Y_n, \gamma).
\end{aligned}
$$

Besides

$$ \frac{\hat{a}_n((1-\tau_n)^{-1})}{a((1-\tau_n)^{-1})} = \frac{X_{n-k_n:n} - U(n/k_n)}{a(n/k_n)} M_n^{(1)}(1-\hat{\gamma}_{n,-}) + \left\{ \frac{U(n/k_n)}{a(n/k_n)} M_n^{(1)} \right\} (1-\hat{\gamma}_{n,-}) $$

and as such, it follows from (A.58), (A.60), (A.61) and the convergence $M_n^{(1)}(1-\hat{\gamma}_{n,-}) \xrightarrow{\mathbb{P}} \gamma_+$ that

$$ \frac{\hat{a}_n((1-\tau_n)^{-1})}{a((1-\tau_n)^{-1})} \stackrel{\mathrm{d}}{\approx} \gamma_+ D_\gamma \left( \frac{k_n}{n} Y_{n-k_n:n} \right) + \frac{1}{2} \left( \frac{1}{k_n} \sum_{i=1}^{k_n} D_{\gamma_{-}}(Y_{n-i+1:n}/Y_{n-k_n:n}) \right) \left( 1 - \frac{\left( \frac{1}{k_n} \sum_{i=1}^{k_n} D_{\gamma_{-}}(Y_{n-i+1:n}/Y_{n-k_n:n}) \right)^2}{\frac{1}{k_n} \sum_{i=1}^{k_n} (D_{\gamma_{-}}(Y_{n-i+1:n}/Y_{n-k_n:n}))^2} \right)^{-1} $$

Conclude by the Rényi representation again (with the same variables $Y_i$) that

$$ \frac{\hat{a}_n((1-\tau_n)^{-1})}{a((1-\tau_n)^{-1})} \stackrel{\mathrm{d}}{\approx} \gamma_+ D_\gamma \left( \frac{k_n}{n} \exp \left( \sum_{i=k_n+1}^n \frac{\log(Y_i)}{i} \right) \right) + \frac{1}{2} \left( \frac{1}{k_n} \sum_{i=1}^{k_n} D_{\gamma_{-}}(Y_i) \right) \left( 1 - \frac{\left( \frac{1}{k_n} \sum_{i=1}^{k_n} D_{\gamma_{-}}(Y_i) \right)^2}{\frac{1}{k_n} \sum_{i=1}^{k_n} (D_{\gamma_{-}}(Y_i))^2} \right)^{-1} =: G_n^{(2)}(Y_1, \dots, Y_n, \gamma) $$

and therefore

$$ \frac{\widehat{ES}_n(\tau_n) - ES(\tau_n)}{\hat{a}_n((1-\tau_n)^{-1})} \stackrel{\mathrm{d}}{\approx} -\frac{G_n^{(1)}(Y_1, \dots, Y_n, \gamma)}{G_n^{(2)}(Y_1, \dots, Y_n, \gamma)} =: -G_n(Y_1, \dots, Y_n, \gamma). $$

For any $\gamma < 1$, the random quantity $G_n(Y_1, \dots, Y_n, \gamma)$ is straightforward to simulate, meaning that its distribution can be tabulated. This leads to the following confidence interval for $ES(1 - k_n/n)$:

$$ \hat{I}_2(\alpha) = \left[ \widehat{ES}_n(1 - k_n/n) + \hat{a}_n(n/k_n)g_{n,\alpha/2}(\hat{\gamma}_n), \widehat{ES}_n(1 - k_n/n) + \hat{a}_n(n/k_n)g_{n,1-\alpha/2}(\hat{\gamma}_n) \right] $$
where $g_{n,\tau}(\gamma)$ is the $\tau$th quantile of $G_n(Y_1, \dots, Y_n, \gamma)$. Our experience with this interval is that it provides excellent results for $\gamma \le 1/4$, but when $\gamma > 1/4$, its performance is adversely affected by the finite-sample uncertainty in the plug-in step of replacing $\gamma$ by $\hat{\gamma}_n$. To put it differently, for large positive values of $\gamma$, the distributions of $G_n(Y_1, \dots, Y_n, \gamma)$ and $G_n(Y_1, \dots, Y_n, \hat{\gamma}_n)$ may look substantially different in finite samples. Since $\sqrt{k_n}(\hat{\gamma}_n - \gamma) \stackrel{d}{\approx} \mathcal{N}(0, v_2(\gamma))$, we propose to deal with this issue of uncertainty quantification by computing directly the quantiles of $G_n(Y_1, \dots, Y_n, \tilde{\gamma}_n)$ for $\tilde{\gamma}_n = \hat{\gamma}_n + Z\sqrt{v_2(\hat{\gamma}_n)/k_n}$, where $Z \sim \mathcal{N}(0, 1)$ is independent from the data; in addition, we retain only those values of $\tilde{\gamma}_n$ that are smaller than 1, i.e. we resample given $\tilde{\gamma}_n < 1$. This gives rise to an approximation for the distribution of $G_n(Y_1, \dots, Y_n, \hat{\gamma}_n)$ and, accordingly, to a different interval $\hat{I}_3(\alpha)$. A step-by-step description of the construction of both $\hat{I}_2(\alpha)$ and $\hat{I}_3(\alpha)$ can be found in Algorithm 2 (where $\Phi$ denotes the standard normal distribution function).

<caption>**Algorithm 2** Confidence intervals for $\mathrm{ES}(1 - k_n/n)$ - Empirical estimator</caption>

**Require:** $N \ge 1$, $\alpha \in (0, 1)$, $\hat{\mathrm{ES}}_n(1 - k_n/n)$, $\hat{a}_n(n/k_n) = \hat{a}_n^{\mathrm{Mom}}(n/k_n)$, $\hat{\gamma}_n = \hat{\gamma}_n^{\mathrm{Mom}}(k_n)$
**Ensure:** $\hat{\gamma}_n < 1$

1: Simulate $N$ replications $U_1, \dots, U_N$ of a uniform distribution on $[0, 1]$

2: **for** $i \in \{1, \dots, N\}$ **do**

3: Calculate $\tilde{\gamma}_{n,i} = \hat{\gamma}_n + \sqrt{\frac{v_2(\hat{\gamma}_n)}{k_n}} \Phi^{-1} (U_i \Phi ((1 - \hat{\gamma}_n) \sqrt{\frac{k_n}{v_2(\hat{\gamma}_n)}}))$

4: Simulate $n$ replications $Y_1, \dots, Y_n$ of a unit Pareto distribution

5: Compute $G_i = G_n(Y_1, \dots, Y_n, \hat{\gamma}_n)$ and $\tilde{G}_i = G_n(Y_1, \dots, Y_n, \tilde{\gamma}_{n,i})$

6: **end for**

7: Compute $\begin{cases} G_{\mathrm{up}} = G_{\mathrm{up}}(\alpha) = G_{\lfloor N(1-\alpha/2) \rfloor:N} \\ G_{\mathrm{down}} = G_{\mathrm{down}}(\alpha) = G_{\lfloor N\alpha/2 \rfloor:N} \end{cases}$ and $\begin{cases} \tilde{G}_{\mathrm{up}} = \tilde{G}_{\mathrm{up}}(\alpha) = \tilde{G}_{\lfloor N(1-\alpha/2) \rfloor:N} \\ \tilde{G}_{\mathrm{down}} = \tilde{G}_{\mathrm{down}}(\alpha) = \tilde{G}_{\lfloor N\alpha/2 \rfloor:N} \end{cases}$

8: **return** $\begin{cases} \hat{I}_2(\alpha) = [\hat{\mathrm{ES}}_n(1 - k_n/n) + \hat{a}_n(n/k_n)G_{\mathrm{down}}, \hat{\mathrm{ES}}_n(1 - k_n/n) + \hat{a}_n(n/k_n)G_{\mathrm{up}}] \\ \hat{I}_3(\alpha) = [\hat{\mathrm{ES}}_n(1 - k_n/n) + \hat{a}_n(n/k_n)\tilde{G}_{\mathrm{down}}, \hat{\mathrm{ES}}_n(1 - k_n/n) + \hat{a}_n(n/k_n)\tilde{G}_{\mathrm{up}}] \end{cases}$

We turn to the quantile-based estimator $\tilde{\mathrm{ES}}_n(\tau_n)$. We first note that in the heavy-tailed setting $\gamma \in (0, 1)$, owing to the convergence $a(t)/U(t) \to \gamma$ as $t \to \infty$, the following simpler version of the quantile-based estimator has been considered by El Methni and Stupfler (2017):

$$ \tilde{\mathrm{ES}}_n^H (1 - k_n/n) = \frac{X_{n-k_n:n}}{1 - \hat{\gamma}_n^H}, $$

where $\hat{\gamma}_n^H$ is again the Hill estimator, see Section 3.2 therein. Since, by (A.16) with $f = \log$,
$X_{n-k_n:n}$ and $\hat{\gamma}_n^H$ are asymptotically independent, it is straightforward to prove that

$$
\begin{aligned}
\sqrt{k_n} \log \left( \frac{\widetilde{\mathrm{ES}}_n^H (1 - k_n/n)}{\mathrm{ES}(1 - k_n/n)} \right) &= \sqrt{k_n} \left( \frac{\widetilde{\mathrm{ES}}_n^H (1 - k_n/n)}{\mathrm{ES}(1 - k_n/n)} - 1 \right) + o_P(1) \\
&\xrightarrow{d} \mathcal{N} \left( 0, \frac{\gamma^2}{(1 - \gamma)^2} (1 + (1 - \gamma)^2) \right).
\end{aligned}
$$

The following naive (competitor) asymptotic confidence interval at level 1 − α follows:

$$
\tilde{I}_1(\alpha) = \left[ \widetilde{\mathrm{ES}}_n^H (1 - k_n/n) \exp \left( \pm \frac{\hat{\gamma}_n^H}{1 - \hat{\gamma}_n^H} \sqrt{1 + (1 - \hat{\gamma}_n^H)^2 \frac{z_{1-\alpha/2}}{\sqrt{k_n}}} \right) \right].
$$

This interval is only valid for $\gamma \in (0, 1)$. Using the semiparametric estimator $\widetilde{\mathrm{ES}}_n$ (with the moment estimators $\hat{a}_n(n/k_n) = \hat{a}_n^{\mathrm{Mom}}(n/k_n)$ of the scale and $\hat{\gamma}_n = \hat{\gamma}_n^{\mathrm{Mom}}$ of the shape extreme value parameters, respectively) instead of $\widetilde{\mathrm{ES}}_n^H$ and applying Theorem 2 leads to the alternative confidence interval

$$
\tilde{I}_2(\alpha) = \left[ \widetilde{\mathrm{ES}}_n(1 - k_n/n) \pm \frac{\hat{a}_n(n/k_n)}{\sqrt{k_n}} \sqrt{ \frac{(1 + \hat{\gamma}_n)(1 - \hat{\gamma}_n)^3 + v_1(\hat{\gamma}_n)(1 - \hat{\gamma}_n)^2 + 2c(\hat{\gamma}_n)(1 - \hat{\gamma}_n) + v_2(\hat{\gamma}_n)}{(1 - \hat{\gamma}_n)^4} z_{1-\alpha/2} } \right].
$$

This confidence interval is, in theory, valid for any $\gamma < 1$ but its finite-sample performance is often disappointing, particularly when $\gamma$ is close to 1. We correct this interval so as to push its finite-sample coverage close to the nominal level while keeping it Gaussian in nature. Set

$$
\tilde{Z}_n = \sqrt{k_n} \frac{\widetilde{\mathrm{ES}}_n(1 - k_n/n) - \mathrm{ES}(1 - k_n/n)}{\hat{a}_n(n/k_n)} = \sqrt{k_n} \frac{\widetilde{\mathrm{ES}}_n(1 - k_n/n) - \mathrm{ES}(1 - k_n/n)}{a(n/k_n)} \frac{a(n/k_n)}{\hat{a}_n(n/k_n)}.
$$

Theorem 2 suggests the finer approximation

$$
\tilde{Z}_n \stackrel{d}{\approx} \left( N_{\text{loc}} + \frac{1}{1-\gamma} N_{\text{scale}} + \frac{1}{(1-\gamma)^2} N_{\text{shape}} \right) \left( 1 - \frac{N_{\text{scale}}}{\sqrt{k_n}} \right) \stackrel{d}{=} \mathbf{u}^{\top} \mathbf{N} + \frac{\mathbf{N}^{\top} \mathbf{S} \mathbf{N}}{\sqrt{k_n}},
$$

where $\mathbf{N} = (N_{\text{loc}}, N_{\text{scale}}, N_{\text{shape}})^{\top}$ follows a trivariate centered normal distribution with co-
variance matrix $\mathbf{\Sigma}$,

$$
\mathbf{S} = \mathbf{S}(\gamma) = \begin{pmatrix} 0 & -\frac{1}{2} & 0 \\ -\frac{1}{2} & -\frac{1}{1-\gamma} & -\frac{1}{2(1-\gamma)^2} \\ 0 & -\frac{1}{2(1-\gamma)^2} & 0 \end{pmatrix} \quad \text{and} \quad \mathbf{u} = \mathbf{u}(\gamma) = \begin{pmatrix} 1 \\ \frac{1}{1-\gamma} \\ \frac{1}{(1-\gamma)^2} \end{pmatrix}.
$$

Straightforward calculations show that, if $\mathbf{Z} = (Z_1, \dots, Z_p)$ is a $p$-dimensional Gaussian random vector made of independent centered unit Gaussian random variables and $\mathbf{M}$ is a $p \times p$
symmetric matrix, one has $\mathbb{E}(\mathbf{Z}^\top \mathbf{M} \mathbf{Z}) = \text{tr}(\mathbf{M})$ and $\mathbb{E}((\mathbf{Z}^\top \mathbf{M} \mathbf{Z})^2) = 2 \text{tr}(\mathbf{M}^2) + (\text{tr}(\mathbf{M}))^2$. The mean $m(\gamma)$ and variance $s^2(\gamma)$ of the random variable $\mathbf{u}^\top \mathbf{N} + \mathbf{N}^\top \mathbf{S} \mathbf{N} / \sqrt{k_n}$ are then

$$m(\gamma) = \frac{\text{tr}(\mathbf{S}\mathbf{\Sigma})}{\sqrt{k_n}} \quad \text{and} \quad s^2(\gamma) = \mathbf{u}^\top\mathbf{\Sigma}\mathbf{u} + 2\frac{\text{tr}(\mathbf{S}\mathbf{\Sigma}\mathbf{S}\mathbf{\Sigma})}{k_n}.$$

Approximating $\tilde{Z}_n$ by a Gaussian random variable with mean $m(\hat{\gamma}_n)$ and variance $s^2(\hat{\gamma}_n)$ suggests the confidence interval

$$\tilde{\mathrm{I}}_3(\alpha) = \left[ \tilde{\mathrm{ES}}_n(1 - k_n/n) - \frac{\hat{a}_n(n/k_n)}{\sqrt{k_n}} m(\hat{\gamma}_n) \pm \frac{\hat{a}_n(n/k_n)}{\sqrt{k_n}} s(\hat{\gamma}_n) z_{1-\alpha/2} \right].$$

Similarly to $\tilde{\mathrm{I}}_2(\alpha)$, the finite-sample performance of this interval suffers from not taking into account the statistical uncertainty of the estimator $\hat{\gamma}_n$ plugged into $m$ and $s^2$. To take this uncertainty into account, we analytically derive the correction term that should be added due to this plug-in step. Let $\mathbf{\Sigma} = \mathbf{\Lambda}\mathbf{\Lambda}^\top$ be the Cholesky decomposition of $\mathbf{\Sigma}$, where

$$\mathbf{\Lambda} = \mathbf{\Lambda}(\gamma) = \begin{pmatrix} 1 & 0 & 0 \\ \gamma & \sqrt{v_1(\gamma) - \gamma^2} & 0 \\ 0 & \frac{c(\gamma)}{\sqrt{v_1(\gamma) - \gamma^2}} & \sqrt{v_2(\gamma) - \frac{c^2(\gamma)}{v_1(\gamma) - \gamma^2}} \end{pmatrix}$$

and note that $\mathbf{N} = \mathbf{\Lambda}\mathbf{Z}$ where $\mathbf{Z}$ is a 3-dimensional Gaussian random vector made of independent centered unit Gaussian random variables. Recalling that $\sqrt{k_n}(\hat{\gamma}_n - \gamma) \approx N_{\text{shape}}$, a Taylor expansion yields

$$\tilde{Z}_n \stackrel{\mathrm{d}}{=} \mathbf{u}(\gamma)^\top \mathbf{\Lambda}(\gamma) \mathbf{Z} + \frac{\mathbf{Z}^\top \mathbf{\Lambda}(\gamma)^\top \mathbf{S}(\gamma) \mathbf{\Lambda}(\gamma) \mathbf{Z}}{\sqrt{k_n}} + o_{\mathbb{P}}\left(\frac{1}{\sqrt{k_n}}\right) \\ \stackrel{\mathrm{d}}{=} \mathbf{u}(\hat{\gamma}_n)^\top \mathbf{\Lambda}(\hat{\gamma}_n) \mathbf{Z} + \frac{\mathbf{Z}^\top \mathbf{\Lambda}(\hat{\gamma}_n)^\top \mathbf{S}(\hat{\gamma}_n) \mathbf{\Lambda}(\hat{\gamma}_n) \mathbf{Z}}{\sqrt{k_n}} - \theta(\hat{\gamma}_n)^\top \frac{N_{\text{shape}}}{\sqrt{k_n}} \mathbf{Z} + o_{\mathbb{P}}\left(\frac{1}{\sqrt{k_n}}\right)$$

$$\text{where } \theta(\gamma) = \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}\gamma}(\gamma)^\top \mathbf{\Lambda}(\gamma) + \mathbf{u}(\gamma)^\top \frac{\mathrm{d}\mathbf{\Lambda}}{\mathrm{d}\gamma}(\gamma).$$

Since $\mathbf{N} = \mathbf{\Lambda}\mathbf{Z}$, one has $N_{\text{shape}} = \Lambda_{23}(\gamma)Z_2 + \Lambda_{33}(\gamma)Z_3 \stackrel{\mathrm{d}}{=} \Lambda_{23}(\hat{\gamma}_n)Z_2 + \Lambda_{33}(\hat{\gamma}_n)Z_3 + o_{\mathbb{P}}(1)$. Then

$$\tilde{Z}_n \stackrel{\mathrm{d}}{=} \mathbf{w}(\hat{\gamma}_n)^\top \mathbf{Z} + \frac{\mathbf{Z}^\top \mathbf{W}(\hat{\gamma}_n) \mathbf{Z}}{\sqrt{k_n}} + o_{\mathbb{P}}\left(\frac{1}{\sqrt{k_n}}\right),$$

where $\mathbf{w}(\gamma) = \mathbf{\Lambda}^\top(\gamma)\mathbf{u}(\gamma)$ and

$$\mathbf{W}(\gamma) = \mathbf{\Lambda}(\gamma)^{\top} \mathbf{S}(\gamma) \mathbf{\Lambda}(\gamma) - \frac{1}{2} \begin{pmatrix} 0 & \theta_1(\gamma)\Lambda_{23}(\gamma) & \theta_1(\gamma)\Lambda_{33}(\gamma) \\ \theta_1(\gamma)\Lambda_{23}(\gamma) & 2\theta_2(\gamma)\Lambda_{23}(\gamma) & \theta_2(\gamma)\Lambda_{33}(\gamma) + \theta_3(\gamma)\Lambda_{23}(\gamma) \\ \theta_1(\gamma)\Lambda_{33}(\gamma) & \theta_2(\gamma)\Lambda_{33}(\gamma) + \theta_3(\gamma)\Lambda_{23}(\gamma) & 2\theta_3(\gamma)\Lambda_{33}(\gamma) \end{pmatrix}.$$
Explicit expressions can of course be given for $\mathbf{w}(\gamma)$ and $\mathbf{W}(\gamma)$; we omit the details for the sake of brevity. As in the construction of the confidence interval $\tilde{I}_3(\alpha)$, we then approximate the distribution of the random variable $\mathbf{w}(\gamma_0)^\top\mathbf{Z}+\mathbf{Z}^\top\mathbf{W}(\gamma_0)\mathbf{Z}/\sqrt{k_n}$ (for any fixed $\gamma_0 < 1$) by a Gaussian distribution with mean $\text{tr}(\mathbf{W}(\gamma_0))/\sqrt{k_n}$ and variance $\|\mathbf{w}(\gamma_0)\|_2^2+2\text{tr}(\mathbf{W}^2(\gamma_0))/k_n$. This suggests our final confidence interval for $\text{ES}(1-k_n/n)$:

$$ \tilde{I}_4(\alpha) = \left[ \tilde{\text{ES}}_n(1-k_n/n) + \frac{\hat{a}_n(n/k_n)}{\sqrt{k_n}} \left( -\frac{\text{tr}(\mathbf{W}(\hat{\gamma}_n))}{\sqrt{k_n}} \pm \sqrt{\|\mathbf{w}(\hat{\gamma}_n)\|_2^2 + 2\frac{\text{tr}(\mathbf{W}^2(\hat{\gamma}_n))}{k_n}} \right) z_{1-\alpha/2} \right]. $$

We shall see that this interval has coverage probabilities close to the nominal level across the full range $\gamma < 1$ in a reasonably large class of models.

## D Simulation study at intermediate levels

We check the quality of our estimators and related inference procedures at an intermediate level on simulated datasets. We consider the same distributions as in the main paper, that is:

* The Kumaraswamy distribution with $1-F(t) = (1-t^\alpha)^\beta$ for $t \in [0, 1]$, and the Reverse-Burr distribution with $1-F(t) = (1+(1-t)^{-\beta/\alpha})^{-1/\beta}$ for $t < 1$ (here $\alpha > 0, \beta > 0$), with respective extreme value indices $\gamma = -1/\beta < 0$ and $\gamma = -\alpha < 0$.

* The Gumbel distribution with $1-F(t) = 1 - \exp(-\exp(-t))$, and the Exponential distribution $1-F(t) = \exp(-t)$ for $t > 0$, both having extreme value index $\gamma = 0$.

* The Pareto distribution with $1-F(t) = t^{-\alpha}$ for $t > 1$, and the Fréchet distribution with $1-F(t) = 1 - \exp(-t^{-\alpha})$ for $t > 0$ (here $\alpha > 0$), both having extreme value index $\gamma = 1/\alpha > 0$.

In each setting, we simulate $N = 10,000$ replications of an i.i.d. sample of size $n = 1,000$ from the chosen distribution. We fix $k_n = 200$ and we estimate and infer the quantity $\text{ES}(1-k_n/n) = \text{ES}(0.8)$ using the estimators $\tilde{\text{ES}}_n(1-k_n/n)$ and $\tilde{\text{ES}}_n(1-k_n/n)$, and the confidence intervals $\tilde{I}_1(\alpha)$, $\tilde{I}_2(\alpha)$, $\tilde{I}_3(\alpha)$, $\tilde{I}_4(\alpha)$ at confidence level $1-\alpha = 0.95$. The true values of $\text{ES}(0.8)$, evaluated using Table E.1, and the empirical coverage probabilities of the competing intervals are provided in Table E.2.

At the intermediate level, it appears that the interval $\tilde{I}_3(\alpha)$ overall performs best among the confidence intervals based on $\tilde{\text{ES}}_n$; in particular, its coverage probability is very close to the nominal level even for the infinite-variance Fréchet distribution with $\gamma = 1/2$. The conclusion is somewhat more complex regarding the intervals based on $\tilde{\text{ES}}_n$: the interval $\tilde{I}_2(\alpha)$, which is purely based on the asymptotic normality of $\tilde{\text{ES}}_n$, seems to perform best for finite-variance distributions, while $\tilde{I}_4(\alpha)$ should be preferred to $\tilde{I}_2(\alpha)$ when the underlying
distribution has an infinite variance. Overall, the confidence intervals $\hat{I}_1(\alpha)$ and $\tilde{I}_1(\alpha)$ behave
very poorly when the assumption of a heavy right tail is violated, and the intervals $\hat{I}_1(\alpha)$ and
$\hat{I}_2(\alpha)$ also have poor coverage when the variance of the underlying distribution is infinite.

E OpenFEMA data analysis: Further results

Figure 6 below contains our results linked to the analysis of the data in zones A and B in
2021 (samples of sizes n = 557 and 101, respectively), and the analysis of the data in Zones
A01, A02, A03, A04 and A05 (each corresponding to a so-called Special Flood zone with Base
Flood Elevation on the insurance rate map) in 2012 (the data points in each Zone have been
pooled to form a sample of size n = 812).
<figure><img src="image_15.png" /><figcaption>Figure 6: OpenFEMA Flood Insurance data – Left panels: Histograms of the claims. Middle panels: Extreme value index estimates $\hat{\gamma}_n^{\text{Mom}}$ (solid black) with 95% asymptotic Gaussian confidence intervals (black ribbon), and Hill estimates (solid red) as functions of $k$. Right panels: Expected Shortfall estimates $\hat{\text{ES}}_n^*(0.995)$ and $\tilde{\text{ES}}_n^*(0.995)$ (solid blue and green, respectively) with 95% asymptotic confidence intervals $\hat{I}_4^*(\alpha)$ (blue ribbon) and $\tilde{I}_4^*(\alpha)$ (green ribbon) for $\alpha = 0.05$, and Weissman-Hill Expected Shortfall estimates $\hat{\text{ES}}_n^W(0.995)$ (solid red) and $\tilde{\text{ES}}_n^W(0.995)$ (dotted red) as functions of $k$.</figcaption></figure>
<table><caption>Table E.1: A list of standard, unit-scale continuous distributions and the associated values of the Expected Shortfall. The parameters $\alpha$ and $\beta$ are positive, and for the log-normal distribution, we take $\mu \in \mathbb{R}$ and $\sigma > 0$. We write $\Phi$ for the standard Gaussian distribution function and $\varphi = \Phi'$ for the associated density function, $\Phi_\alpha$ for the Student distribution function with $\alpha$ degrees of freedom, $x \mapsto {}_2F_1(a, b, c; x)$ for the ordinary hypergeometric function found as one of the two fundamental solutions of the differential equation $x(1-x)y''(x) + (c - (a+b+1)x)y'(x) - aby(x) = 0$, $\gamma_x(\alpha) = \int_0^x t^{\alpha-1}e^{-t} dt$ for the lower incomplete Gamma function, $\Gamma_x(\alpha) = \int_x^\infty t^{\alpha-1}e^{-t} dt$ for the upper incomplete Gamma function, $c$ for the Euler-Mascheroni constant, $E_1(x) = \int_x^\infty \{\exp(-t)/t\} dt$ for the exponential integral and $B_x(\alpha, \beta) = \int_0^x t^{\alpha-1}(1-t)^{\beta-1} dt$ for the lower incomplete Beta function.</caption><thead><tr><th>Distribution</th><th>Density function f(t)</th><th>ES(τ)</th></tr></thead><tbody><tr><td>Pareto (α > 1)</td><td>αt<sup>-α-1</sup>, t > 1</td><td><sup>α</sup>&frasl;<sub>α-1</sub>(1 - τ)<sup>-1/α</sup></td></tr><tr><td>Burr (0 < α < 1)</td><td><sup>1</sup>&frasl;<sub>α</sub>t<sup>β/α-1</sup>(1 + t<sup>β/α</sup>)<sup>-1/β-1</sup>, t > 0</td><td><sup>((1-τ)<sup>-β-1</sup>)<sup>α/β</sup></sup>&frasl;<sub>1-α</sub>2F<sub>1</sub>(1, 1 + <sup>1-α</sup>&frasl;<sub>β</sub>, <sup>1</sup>&frasl;<sub>1-τ-β</sub>; <sup>-α</sup>&frasl;<sub>β</sub>)</td></tr><tr><td>Fréchet (α > 1)</td><td>αt<sup>-1-α</sup> exp(-t<sup>-α</sup>), t > 0</td><td><sup>γ<sub>log(1/τ)</sub>(1-1/α)</sup>&frasl;<sub>1-τ</sub></td></tr><tr><td>Student (α > 1)</td><td>Φ'<sub>α</sub>(t) = <sup>Γ(<sup>α+1</sup>&frasl;<sub>2</sub>)</sup>&frasl;<sub>√απΓ(<sup>α</sup>&frasl;<sub>2</sub>)</sub>(1 + <sup>t<sup>2</sup></sup>&frasl;<sub>α</sub>)<sup>-(α+1)/2</sup></td><td><sup>Γ(<sup>α+1</sup>&frasl;<sub>2</sub>)</sup>&frasl;<sub>√απΓ(<sup>α</sup>&frasl;<sub>2</sub>)</sub> <sup>α</sup>&frasl;<sub>α-1</sub>(1 + (<sup>Φ<sub>α</sub><sup>-1</sup>(τ)</sup>&frasl;<sub>α</sub>)<sup>2</sup>)<sup>(1-α)/2</sup></td></tr><tr><td>Exponential</td><td>exp(-t), t > 0</td><td>1 - log(1 - τ)</td></tr><tr><td>Weibull</td><td>βt<sup>β-1</sup> exp(-t<sup>β</sup>), t > 0</td><td><sup>Γ<sub>log(1/(1-τ))</sub>(1+1/β)</sup>&frasl;<sub>1-τ</sub></td></tr><tr><td>Log-normal</td><td><sup>1</sup>&frasl;<sub>tσ√2π</sub> exp(-<sup>((log(t)-μ)<sup>2</sup>)</sup>&frasl;<sub>2σ<sup>2</sup></sub>), t > 0</td><td>exp(<sup>μ + σ<sup>2</sup></sup>&frasl;<sub>2</sub>) <sup>Φ(<sup>σ-Φ<sup>-1</sup>(τ)</sup>&frasl;<sub>1-τ</sub>)</sup>&frasl;<sub>1-τ</sub></td></tr><tr><td>Gumbel</td><td>exp(-t) exp(-exp(-t))</td><td><sup>c+τ log(-log(τ)) + E<sub>1</sub>(-log(τ))</sup>&frasl;<sub>1-τ</sub></td></tr><tr><td>Laplace</td><td><sup>1/2</sup> exp(-|t|)</td><td><sup>min{τ,1-τ}(1-log(2 min{τ,1-τ}))</sup>&frasl;<sub>1-τ</sub></td></tr><tr><td>Logistic</td><td><sup>exp(-t)</sup>&frasl;<sub>(1+exp(-t))<sup>2</sup></sub></td><td>(<sup>τ</sup>&frasl;<sub>1-τ</sub> log(τ) + log(1-τ))</td></tr><tr><td>Normal</td><td>φ(t) = Φ'(t) = <sup>1</sup>&frasl;<sub>√2π</sub> exp(-<sup>t<sup>2</sup></sup>&frasl;<sub>2</sub>)</td><td><sup>φ(<sup>Φ<sup>-1</sup>(τ)</sup>&frasl;<sub>1-τ</sub>)</sup>&frasl;<sub>1-τ</sub></td></tr><tr><td>Kumaraswamy</td><td>αβt<sup>α-1</sup>(1 - t<sup>α</sup>)<sup>β-1</sup>, t ∈ [0, 1]</td><td>β <sup>B(<sup>1</sup>&frasl;<sub>α</sub>+1,β)-B<sup>1-(1-τ)</sup>&frasl;<sub>1-τ</sub>1/β(<sup>1</sup>&frasl;<sub>α</sub>+1,β)</sup>&frasl;<sub>1-τ</sub></td></tr><tr><td>Reverse-Fréchet</td><td>α(1 - t)<sup>α-1</sup> exp(-(1 - t)<sup>α</sup>), t < 1</td><td>1 - <sup>γ- log(τ) <sup>(1</sup>&frasl;<sub>α</sub>) - ατ(-log(τ))<sup>1/α</sup></sup>&frasl;<sub>α(1-τ)</sub></td></tr><tr><td>Triangular</td><td>2t, t ∈ [0, 1] and (2 - t), t ∈ [1, 2]</td><td><table><tbody><tr><td><sup>1-(2τ)<sup>3/2</sup>/3</sup>&frasl;<sub>1-τ</sub></td><td>for τ ≤ 1/2,</td></tr><tr><td>2(<sup>√2(1-τ)</sup>&frasl;<sub>3</sub>)</td><td>for τ ≥ 1/2</td></tr></tbody></table></td></tr></tbody></table>
<table><caption>Table E.2: Inference about ES(1 − k$_{n}$/n ) = ES(0.8) with sample size n = 1 , 000 − For α = 0 . 05 and each interval among ˆ I$_{1}$ ( α ) , ˆ I$_{2}$ ( α ) , ˆ I$_{3}$ ( α ) , ˆ I$_{4}$ ( α ) , ˆ I$_{5}$ ( α ) , ˆ I$_{6}$ ( α ) and ˆ I$_{7}$ ( α ) having nominal coverage probability 1 − α = 0 . 95 , and in each tested case, we report between square brackets the median values (over the N = 10 , 000 replications) of its lower and upper bounds from which ES(1 − k$_{n}$/n ) has been subtracted, and after subsequent normalization by ES(1 − k$_{n}$/n ) ( i.e. if I ( α ) = [ m ( α ) , M ( α )] is any such interval, we report the empirical median of m ( α ) /ES(1 − k$_{n}$/n ) − 1 as the lower bound and the empirical median of M ( α ) /ES(1 − k$_{n}$/n ) − 1 as the upper bound). The Median Relative Widths (MRW) of each confidence interval thus normalized and its empirical coverage probability (between round brackets) are also reported.</caption><thead><tr><th>Distribution</th><th>γ</th><th>ES(τ<sub>n</sub>)</th><th>Î<sub>1</sub></th><th>Î<sub>2</sub></th><th>Î<sub>3</sub></th><th>Î<sub>4</sub></th><th>Î<sub>5</sub></th><th>Î<sub>6</sub></th><th>Î<sub>7</sub></th><th>Î<sub>8</sub></th><th>Î<sub>9</sub></th><th>Î<sub>10</sub></th><th>Î<sub>11</sub></th><th>Î<sub>12</sub></th><th>Î<sub>13</sub></th><th>Î<sub>14</sub></th></tr></thead><tbody><tr><td>Kumaraswamy<br>α = β = 2</td><td>-0.5</td><td>0.835</td><td>[-0.024,0.024]<br>MRW: 0.048<br>(0.992)</td><td>[-0.015,0.020]<br>MRW: 0.035<br>(0.942)</td><td>[-0.015,0.020]<br>MRW: 0.035<br>(0.946)</td><td>[-0.020,0.028]<br>MRW: 0.048<br>(0.988)</td><td>[-0.020,0.016]<br>MRW: 0.036<br>(0.958)</td><td>[-0.021,0.016]<br>MRW: 0.037<br>(0.958)</td><td>[-0.021,0.016]<br>MRW: 0.037<br>(0.958)</td><td>[-0.021,0.016]<br>MRW: 0.037<br>(0.958)</td><td>[-0.021,0.016]<br>MRW: 0.037<br>(0.958)</td><td>[-0.021,0.016]<br>MRW: 0.037<br>(0.958)</td><td>[-0.021,0.016]<br>MRW: 0.037<br>(0.958)</td><td>[-0.021,0.016]<br>MRW: 0.037<br>(0.958)</td><td>[-0.021,0.016]<br>MRW: 0.037<br>(0.958)</td></tr><tr><td>Reversed Burr<br>α = 1/4, β = 3</td><td>-0.25</td><td>0.465</td><td>[-0.082,0.088]<br>MRW: 0.169<br>(0.998)</td><td>[-0.043,0.055]<br>MRW: 0.098<br>(0.940)</td><td>[-0.044,0.056]<br>MRW: 0.099<br>(0.944)</td><td>[-0.039,0.123]<br>MRW: 0.161<br>(0.941)</td><td>[-0.058,0.044]<br>MRW: 0.102<br>(0.952)</td><td>[-0.059,0.044]<br>MRW: 0.102<br>(0.953)</td><td>[-0.059,0.044]<br>MRW: 0.102<br>(0.953)</td><td>[-0.059,0.044]<br>MRW: 0.102<br>(0.953)</td><td>[-0.059,0.044]<br>MRW: 0.102<br>(0.953)</td><td>[-0.059,0.044]<br>MRW: 0.102<br>(0.953)</td><td>[-0.059,0.044]<br>MRW: 0.102<br>(0.953)</td><td>[-0.059,0.044]<br>MRW: 0.102<br>(0.953)</td></tr><tr><td>Kumaraswamy<br>α = 1, β = 10</td><td>-0.1</td><td>0.226</td><td>[-0.115,0.127]<br>MRW: 0.238<br>(0.999)</td><td>[-0.055,0.069]<br>MRW: 0.123<br>(0.944)</td><td>[-0.057,0.070]<br>MRW: 0.126<br>(0.950)</td><td>[-0.043,0.167]<br>MRW: 0.210<br>(0.916)</td><td>[-0.071,0.059]<br>MRW: 0.129<br>(0.955)</td><td>[-0.071,0.059]<br>MRW: 0.130<br>(0.957)</td><td>[-0.070,0.060]<br>MRW: 0.130<br>(0.957)</td><td>[-0.070,0.060]<br>MRW: 0.130<br>(0.957)</td><td>[-0.070,0.060]<br>MRW: 0.130<br>(0.957)</td><td>[-0.070,0.060]<br>MRW: 0.130<br>(0.957)</td><td>[-0.070,0.060]<br>MRW: 0.130<br>(0.957)</td><td>[-0.070,0.060]<br>MRW: 0.130<br>(0.957)</td></tr><tr><td>Gumbel</td><td>0</td><td>2.556</td><td>[-0.209,0.251]<br>MRW: 0.451<br>(0.834)</td><td>[-0.070,0.085]<br>MRW: 0.155<br>(0.946)</td><td>[-0.073,0.087]<br>MRW: 0.159<br>(0.953)</td><td>[-0.038,0.271]<br>MRW: 0.308<br>(0.830)</td><td>[-0.088,0.073]<br>MRW: 0.161<br>(0.951)</td><td>[-0.089,0.074]<br>MRW: 0.162<br>(0.953)</td><td>[-0.087,0.077]<br>MRW: 0.163<br>(0.955)</td><td>[-0.087,0.077]<br>MRW: 0.163<br>(0.955)</td><td>[-0.087,0.077]<br>MRW: 0.163<br>(0.955)</td><td>[-0.087,0.077]<br>MRW: 0.163<br>(0.955)</td><td>[-0.087,0.077]<br>MRW: 0.163<br>(0.955)</td><td>[-0.087,0.077]<br>MRW: 0.163<br>(0.955)</td></tr><tr><td>Exponential<br>λ = 1</td><td>0</td><td>2.609</td><td>[-0.157,0.179]<br>MRW: 0.329<br>(0.996)</td><td>[-0.068,0.081]<br>MRW: 0.147<br>(0.955)</td><td>[-0.071,0.082]<br>MRW: 0.152<br>(0.960)</td><td>[-0.047,0.209]<br>MRW: 0.256<br>(0.892)</td><td>[-0.081,0.073]<br>MRW: 0.153<br>(0.955)</td><td>[-0.082,0.073]<br>MRW: 0.154<br>(0.956)</td><td>[-0.080,0.076]<br>MRW: 0.155<br>(0.960)</td><td>[-0.080,0.076]<br>MRW: 0.155<br>(0.960)</td><td>[-0.080,0.076]<br>MRW: 0.155<br>(0.960)</td><td>[-0.080,0.076]<br>MRW: 0.155<br>(0.960)</td><td>[-0.080,0.076]<br>MRW: 0.155<br>(0.960)</td><td>[-0.080,0.076]<br>MRW: 0.155<br>(0.960)</td></tr><tr><td>Pareto<br>α = 10</td><td>0.1</td><td>1.305</td><td>[-0.021,0.021]<br>MRW: 0.042<br>(0.951)</td><td>[-0.018,0.022]<br>MRW: 0.040<br>(0.945)</td><td>[-0.019,0.022]<br>MRW: 0.041<br>(0.953)</td><td>[-0.021,0.021]<br>MRW: 0.041<br>(0.957)</td><td>[-0.021,0.020]<br>MRW: 0.041<br>(0.955)</td><td>[-0.021,0.020]<br>MRW: 0.041<br>(0.956)</td><td>[-0.021,0.021]<br>MRW: 0.042<br>(0.957)</td><td>[-0.021,0.021]<br>MRW: 0.042<br>(0.957)</td><td>[-0.021,0.021]<br>MRW: 0.042<br>(0.957)</td><td>[-0.021,0.021]<br>MRW: 0.042<br>(0.957)</td><td>[-0.021,0.021]<br>MRW: 0.042<br>(0.957)</td><td>[-0.021,0.021]<br>MRW: 0.042<br>(0.957)</td></tr><tr><td>Pareto<br>α = 4</td><td>0.25</td><td>1.994</td><td>[-0.060,0.060]<br>MRW: 0.120<br>(0.949)</td><td>[-0.058,0.059]<br>MRW: 0.115<br>(0.949)</td><td>[-0.063,0.062]<br>MRW: 0.123<br>(0.957)</td><td>[-0.057,0.058]<br>MRW: 0.115<br>(0.955)</td><td>[-0.060,0.056]<br>MRW: 0.115<br>(0.950)</td><td>[-0.060,0.057]<br>MRW: 0.117<br>(0.953)</td><td>[-0.058,0.062]<br>MRW: 0.119<br>(0.957)</td><td>[-0.058,0.062]<br>MRW: 0.119<br>(0.957)</td><td>[-0.058,0.062]<br>MRW: 0.119<br>(0.957)</td><td>[-0.058,0.062]<br>MRW: 0.119<br>(0.957)</td><td>[-0.058,0.062]<br>MRW: 0.119<br>(0.957)</td><td>[-0.058,0.062]<br>MRW: 0.119<br>(0.957)</td></tr><tr><td>Fréchet<br>α = 2</td><td>0.5</td><td>4.395</td><td>[-0.378,0.325]<br>MRW: 0.696<br>(0.217), NA: 77.6%</td><td>[-0.271,0.149]<br>MRW: 0.410<br>(0.916)</td><td>[-0.306,0.185]<br>MRW: 0.484<br>(0.938)</td><td>[-0.142,0.208]<br>MRW: 0.351<br>(0.967)</td><td>[-0.177,0.166]<br>MRW: 0.338<br>(0.938)</td><td>[-0.179,0.170]<br>MRW: 0.344<br>(0.941)</td><td>[-0.169,0.207]<br>MRW: 0.370<br>(0.958)</td><td>[-0.169,0.207]<br>MRW: 0.370<br>(0.958)</td><td>[-0.169,0.207]<br>MRW: 0.370<br>(0.958)</td><td>[-0.169,0.207]<br>MRW: 0.370<br>(0.958)</td><td>[-0.169,0.207]<br>MRW: 0.370<br>(0.958)</td><td>[-0.169,0.207]<br>MRW: 0.370<br>(0.958)</td></tr><tr><td>Pareto<br>α = 5/3</td><td>0.6</td><td>6.566</td><td>[-0.525,0.243]<br>MRW: 0.747<br>(0.009), NA: 99.0%</td><td>[-0.471,0.182]<br>MRW: 0.654<br>(0.871)</td><td>[-0.530,0.283]<br>MRW: 0.817<br>(0.918)</td><td>[-0.203,0.245]<br>MRW: 0.448<br>(0.950)</td><td>[-0.266,0.227]<br>MRW: 0.481<br>(0.917)</td><td>[-0.270,0.234]<br>MRW: 0.490<br>(0.921)</td><td>[-0.260,0.316]<br>MRW: 0.558<br>(0.950)</td><td>[-0.260,0.316]<br>MRW: 0.558<br>(0.950)</td><td>[-0.260,0.316]<br>MRW: 0.558<br>(0.950)</td><td>[-0.260,0.316]<br>MRW: 0.558<br>(0.950)</td><td>[-0.260,0.316]<br>MRW: 0.558<br>(0.950)</td><td>[-0.260,0.316]<br>MRW: 0.558<br>(0.950)</td></tr></tbody></table>