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Robust quantile estimation under bivariate extreme value models


In risk quantification of extreme events in multiple dimensions, a correct specification of the dependence structure among variables is difficult due to the limited size of effective data. This paper studies the problem of estimating quantiles for bivariate extreme value distributions, considering that an estimated Pickands dependence function may deviate from the truth within some fixed distance. Our method thus finds optimal upper and lower bounds for the true but unknown dependence function, based on which robust quantile bounds are obtained. A simulation study shows the usefulness of our robust estimates that can supplement traditional error estimation methods.

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The authors thank all the anonymous reviewers and the Editor-in-Chief, Thomas Mikosch, for comments and suggestions that helped improve and clarify this manuscript. The work of K. Kim was supported by the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education (NRF-2019R1A2C1003144).

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Correspondence to Heelang Ryu.

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For any convex function \(A\in \mathcal {A}\), which includes Aref, define \(A^{\prime }(0-)=-\infty \) and \(A^{\prime }(1+)=\infty \).

A.1 Proof of Theorem 1

Fix any s in \(\left [(b-1)/c_{0}, (b-1)/(c_{0}-1)\right ]\) and define w1,s and w2,s as in Eqs. 15 and 16. Then the line f(t) := s(tc0) + b satisfies f(0) ≤ Aref(0) = 1, f(1) ≤ Aref(1) = 1, and f(c0) = b > Aref(c0) by assumption. This implies that Aref(t) = f(t) should have at least two distinct solutions, where one is smaller than c0 and the other is larger than c0. This observation guarantees the existence of w1,s, w2,s in their corresponding intervals. Thus, f(t) ≥ Aref(t) on [w1,s, w2,s].

For the convexity of As, note that f(t) ≤ Aref(t) on t ∈ [0,w1,s] ∪ [w2,s, 1] and, therefore, \(A_{s}(t) = \max \nolimits (A_{{\textsf {ref}}}(t),f(t))\). Being the maximum of two convex functions, As is then convex. For this, we start by observing that

$$ A_{{\textsf{ref}}}^{\prime}(\tau+)\le A_{{\textsf{ref}}}^{\prime}(w_{1,s}+)\le s\le A_{{\textsf{ref}}}^{\prime}(w_{2,s}-)\le A_{{\textsf{ref}}}^{\prime}(\zeta-), $$

for any τ ∈ [0,w1,s] and ζ ∈ [w2,s, 1]. The first inequality comes from the convexity of Aref. The second inequality is valid because Aref(w1,s) = f(w1,s) and Areff on [w1,s, w2,s]. The other two inequalities are derived for similar reasons. Then, we observe for 0 ≤ tw1,s,

$$ \begin{array}{@{}rcl@{}} A_{{\textsf{ref}}}(t) &=& A_{{\textsf{ref}}}(w_{1,s}) - {\int}_{t}^{w_{1},s} A_{{\textsf{ref}}}^{\prime}(\tau+)\mathrm{d} \tau \\ &\geq &f(w_{1,s}) - {\int}_{t}^{w_{1,s}} f^{\prime}(\tau)\mathrm{d} \tau = f(t). \end{array} $$

Likewise, we get Areff on [w2,s, 1].

Now consider any s ∈ [A(c0−),A(c0+)]. Then, we see that

$$ A(c_{0}) - A(0) = {\int}_{0+}^{c_{0}} A^{\prime}(t-)\mathrm{d} t \leq c_{0} A^{\prime}(c_{0}-) \leq c_{0}s. $$

Since A(c0) = b and A(0) = 1 by assumption, s ≥ (b − 1)/c0. On the other hand,

$$ A(1) - A(c_{0}) = {\int}_{c_{0}}^{1-} A^{\prime}(t+)\mathrm{d} t \geq (1-c_{0}) A^{\prime}(c_{0}+) \geq (1-c_{0})s. $$

Since A(1) = 1 by assumption, s ≤ (1 − b)/(1 − c0). This means for such s, As is well defined and convex by previous arguments.

Lastly, we claim that AsA on [w1,s, w2,s]. Indeed, for all τ ∈ [w1,s, c0], the convexity of A implies A(τ−) ≤ A(c0−) ≤ s. By integrating this over [t,c0], we have

$$ b - A(t) \leq s(c_{0}-t) \Rightarrow A_{s}(t) =f(t) \leq A(t) $$

on [w1,s, c0]. By re-iterating the same arguments, we can prove As(t) ≤ A(t) on [c0, w2,s] as well. Consequently, we have the pointwise inequality Aref(t) ≤ As(t) ≤ A(t) on [w1,s, w2,s], which yields

$$ {\textsf{d}}_{[w_{1,s},w_{2,s}]}(A_{s},A_{{\textsf{ref}}}) \le {\textsf{d}}_{[w_{1,s},w_{2,s}]}(A,A_{{\textsf{ref}}}). $$

Since \({\textsf {d}}_{[0,w_{1,s}]}(A_{s}, A_{{\textsf {ref}}}) = {\textsf {d}}_{[w_{2,s},1]}(A_{s},A_{{\textsf {ref}}})=0\), we conclude that

$$ {\textsf{d}}(A_{s},A_{{\textsf{ref}}})\le {\textsf{d}}(A,A_{{\textsf{ref}}}). $$

A.2 Proof of Theorem 2

We are given Aref, c0, and b. Assume b < Aref(c0). Our target is to find the form of minimal \(A \in {\mathcal A}\) with A(c0) = b.

Minimal A(t)-form for t ∈ [0,c0]

Define \(v_{1} = \sup \mathcal {C}\) with

$$ \mathcal{C}=\left\{t\in [0,c_{0}): \frac{A_{{\textsf{ref}}}(t)-b}{t-c_{0}} \in [A_{{\textsf{ref}}}^{\prime}(t-),A_{{\textsf{ref}}}^{\prime}(t+)]\right\}, $$

and a convex function \({A_{w}^{l}}\) for any fixed 0 ≤ wv1,

$$ {A_{w}^{l}}(t) = \left\{\begin{array}{ll} A_{{\textsf{ref}}}(t) & \text{for} \ 0\le t < w\\ \frac{A_{{\textsf{ref}}}(w)-b}{w-c_{0}}(t-c_{0}) + b \ &\text{for} \ w\le t\le c_{0}. \end{array}\right. $$

The existence of v1 and the convexity of \({A_{w}^{l}}\) are ensured by Lemmas 1 and 2 and thereby \({A_{w}^{l}}\) is a well-defined convex function on [0,c0].

Lemma 1

The set\(\mathcal {C}\)isnonempty and\(\sup \mathcal {C} < c_{0}\).


The proof is given in the electronic supplementary material. □

Lemma 2

For any \(w\in [0,\sup \mathcal {C}]\) , \({A_{w}^{l}}\) is a convex function.


The proof is given in the electronic supplementary material. □

It remains to prove that \({A_{w}^{l}}\) minimizes the distance \({\textsf {d}}_{[0,c_{0}]}\) from Aref if w is suitably chosen. Specifically, we will prove that there exists w1 in \([0, \sup \mathcal {C}]\) such that \(\textsf {d}_{[0,c_{0}]}(A_{w_{1}}^{l},A_{\textsf {ref}})\le \textsf {d}_{[0,c_{0}]}(A,A_{\textsf {ref}})\) via the following three steps.

Step 1: The straightforward case is when A(t) ≤ Aref(t) on [0,c0]. Then, we simply set w1 = v1, which was defined as \(\sup \mathcal {C}\) above. Then, note that \(A(t) \leq A^{l}_{v_{1}}(t) =A_{\textsf {ref}}(t)\) on [0,v1] by definition. On the other hand, \(A(v_{1}) \leq A^{l}_{v_{1}}(v_{1}) = A_{\textsf {ref}}(v_{1})\) and \(A(c_{0})=A_{v_{1}}^{l}(c_{0})= b\). Due to the convexity of A, the line segment connecting \((v_{1}, A^{l}_{v_{1}}(v_{1}))\) and \((c_{0}, A^{l}_{v_{1}}(c_{0}))\) lies above the graph of A. By definition of \(A^{l}_{v_{1}}\), we get \(A(t) \le A_{v_{1}}^{l}(t)\) on [v1, c0].

On the other hand, from the definition of \(\mathcal {C}\), we get

$$ \frac{A_{{\textsf{ref}}}(u) - b}{u-c_{0}} \leq A_{{\textsf{ref}}}^{\prime}(u+) \leq A_{{\textsf{ref}}}^{\prime}(\tau+) $$

for each \(u \in \mathcal {C}\) and for any τ ∈ [u,c0]. By integrating the both sides over [u,t] for each t, we get

$$ \begin{array}{@{}rcl@{}} A_{{\textsf{ref}}}(t) - A_{{\textsf{ref}}}(u) & \geq & \frac{A_{{\textsf{ref}}}(u) - b}{u-c_{0}} (t-u) \\ &=& \frac{A_{{\textsf{ref}}}(u)-b}{u-c_{0}}(t-c_{0}) + b - A_{{\textsf{ref}}}(u). \end{array} $$

By taking a limit \(u \rightarrow v_{1}\), we obtain \(A_{v_{1}}^{l} \leq A_{{\textsf {ref}}}\) on [v1, c0]. Combining these observations, we conclude that \(A(t)\le A_{v_{1}}^{l}(t) \le A_{\textsf {ref}}(t)\) on [0,c0], and therefore,

$$ {\textsf{d}}_{[0,c_{0}]}(A_{v_{1}}^{l},A_{{\textsf{ref}}})\le {\textsf{d}}_{[0,c_{0}]}(A,A_{{\textsf{ref}}}). $$

Step 2: Now, suppose there exists t ∈ (0,c0) such that A(t) > Aref(t). Define

$$ t^{*} = \sup\{t\in[0,c_{0}): A(t)\ge A_{{\textsf{ref}}}(t)\}. $$

The continuity of A and Aref implies that A(t) = Aref(t). Assume tv1. Then it continues to work to have w1 = v1. Indeed, by continuity, we have \(A(v_{1}) \leq A_{\textsf {ref}}(v_{1}) = A_{v_{1}}^{l}(v_{1})\). From \(A(c_{0}) = A_{v_{1}}^{l}(c_{0}) = b\), we can conclude \(A(t) \le A_{v_{1}}^{l}(t)\) on [v1, c0] by the convexity of A. Thus,

$$ {\textsf{d}}_{[v_{1},c_{0}]}(A_{v_{1}}^{l},A_{{\textsf{ref}}})\le {\textsf{d}}_{[v_{1},c_{0}]}(A,A_{{\textsf{ref}}}). $$

Since \({\textsf {d}}_{[0,v_{1}]}(A_{v_{1}}^{l}, A_{{\textsf {ref}}}) = 0 \leq {\textsf {d}}_{[0,v_{1}]} (A, A_{{\textsf {ref}}})\), we conclude that

$$ {\textsf{d}}_{[0,c_{0}]}(A_{v_{1}}^{l},A_{{\textsf{ref}}}) \le {\textsf{d}}_{[0,c_{0}]}(A,A_{{\textsf{ref}}}). $$

Step 3: Assume t > v1. Then, for any 0 ≤ τ < t,

$$ A^{\prime}(\tau-) \le A^{\prime}(t^{*}-) \le \frac{b-A(t^{*})}{c_{0}-t^{*}}=\frac{b-A_{{\textsf{ref}}}(t^{*})}{c_{0}-t^{*}}. $$

The first inequality comes from the monotonicity of A and the second one does from A(c0) = b and the convexity of A. For convenience, define a linear function f(t) on t ∈ [0,c0],

$$ f(t) =\frac{b-A_{{\textsf{ref}}}(t^{*})}{c_{0}-t^{*}}(t-t^{*})+A_{{\textsf{ref}}}(t^{*}). $$

Then, we integrate (1) on both sides with τ over [t,t] to conclude fA on [0,t].

In Step 1, we saw that \(A_{v_{1}}^{l} \leq A_{{\textsf {ref}}}\) on the interval [0,c0]. In particular, \(A_{v_{1}}^{l}(t^{*}) \leq f(t^{*})\). From \(A_{v_{1}}^{l}(c_{0}) = f(c_{0}) = b\), we can conclude that the line segment \(A_{v_{1}}^{l}\) is below the line segment f on [v1, c0], leading to Aref(v1) ≤ f(v1). With f(0) ≤ A(0) = Aref(0), the point \(w_{1} = \sup \left \{t\in [0,v_{1}]: A_{\textsf {ref}}(t) \ge f(t) \right \}\) is well defined.

Now, we claim that \(A_{w_{1}}^{l}\) is closer to Aref than A on each of the intervals [w1, t] and [t, c0]. This concludes the proof. In fact, we show that

  • \(A_{{\textsf {ref}}} \leq A_{w_{1}}^{l} \leq A\) on [w1, t];

  • \(A \leq A_{w_{1}}^{l} \leq A_{{\textsf {ref}}}\) on [t, c0].

For the first part, note that the continuity of Aref implies Aref(w1) = f(w1) if w1 < v1. Even when w1 = v1, this is valid because Aref(v1) ≤ f(v1). Also by definition, we have \(A_{w_{1}}^{l}(w_{1}) = A_{\textsf {ref}}(w_{1})\) and \(A_{w_{1}}^{l}(c_{0}) = f(c_{0}) = b\). As a result \(A_{w_{1}}^{l} = f\) on [w1, c0]. On the other hand, the convexity together with Aref(w1) = f(w1) and Aref(t) = f(t) result in Areff on [w1, t]. We then conclude the first claim by recalling that fA on [0,t]. In terms of d, we have

$$ {\textsf{d}}_{[w_{1}, t^{*}]}(A_{w_{1}}^{l},A_{{\textsf{ref}}})\le {\textsf{d}}_{[w_{1}, t^{*}]}(A,A_{{\textsf{ref}}}). $$

For the second part, the first inequality is simply due to the facts that A and f have the same function values at t and c0 and that \(A_{w_{1}}^{l} = f\) on [w1, c0]. For the second inequality, as an intermediate step, we claim that

$$ \frac{A_{{\textsf{ref}}}(t^{*})-b}{t^{*}-c_{0}} \leq A_{{\textsf{ref}}}^{\prime}(t^{*}+). $$

Otherwise, the function value φ+(t) < 0 from Lemma 1. This eventually leads to a contradiction to v1 < t. Consequently, \(f^{\prime } \leq A_{{\textsf {ref}}}^{\prime }\) on [t, c0]. By integrating the both sides over [t, t] together with f(t) = Aref(t), we obtain fAref on [t, c0]. In terms of d,

$$ {\textsf{d}}_{[t^{*},c_{0}]}(A_{w_{1}}^{l},A_{{\textsf{ref}}})\le {\textsf{d}}_{[t^{*},c_{0}]}(A,A_{{\textsf{ref}}}). $$

Finally, \(A_{w_{1}}^{l} = A_{{\textsf {ref}}}\) on [0,w1] by definition, and thus \({\textsf {d}}_{[0,c_{0}]}(A_{w_{1}}^{l},A_{{\textsf {ref}}})\le {\textsf {d}}_{[0,c_{0}]}(A,A_{{\textsf {ref}}})\).

Minimal A(t)-form for t ∈ [c0, 1]

As done in the previous subsection, we now define v2 as

$$ v_{2}=\inf\left\{t\in (c_{0},1]: \frac{b-A_{{\textsf{ref}}}(t)}{c_{0}-t} \in [A_{{\textsf{ref}}}^{\prime}(t-),A_{{\textsf{ref}}}^{\prime}(t+)]\right\}. $$

We also set a function \({A_{w}^{r}}\) for any fixed v2w ≤ 1 as

$$ {A_{w}^{r}}(t) = \left\{\begin{array}{ll} \frac{A_{{\textsf{ref}}}(w)-b}{w-c_{0}}(t-c_{0}) + b \ &\text{for} \ c_{0}\le t\le w,\\ A_{{\textsf{ref}}}(t) & \text{for} \ w<t\le 1. \end{array}\right. $$

One can then repeat the proofs of the previous lemmas by considering Aref(1 − t) instead of Aref(t). Hence, without proofs, we state that v2 is well defined and \({A_{w}^{r}}\) is convex on [c0, 1]. Furthermore, above argument will be valid to prove that for a given \(A \in {\mathcal A}\) with A(c0) = b, there exists w2 ∈ [v2, 1] such that

$$ {\textsf{d}}_{[c_{0},1]}({A_{w}^{r}},A_{{\textsf{ref}}})\le {\textsf{d}}_{[c_{0},1]}(A,A_{{\textsf{ref}}}). $$

The only remaining part is that \(A_{w_{1},w_{2}}\) is convex on [0, 1].

Lemma 3

For any pair (w1, w2) ∈ [0,v1] × [v2, 1],\(A_{w_{1},w_{2}}\)isconvex on [0, 1] and\(A_{w_{1},w_{2}}\in \mathcal A\).


The proof is given in the electronic supplementary material. □

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Kim, S., Kim, K. & Ryu, H. Robust quantile estimation under bivariate extreme value models. Extremes 23, 55–83 (2020).

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  • Extreme value theory
  • Extremal dependence misspecification
  • Bivariate quantile
  • Robust risk measure

AMS 2000 Subject Classifications

  • 60G70
  • 62G32
  • 62G35
  • 62H12