Robust quantile estimation under bivariate extreme value models

Abstract

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.

Appendix

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

1.1 A.1 Proof of Theorem 1

Fix any s in $\left [(b-1)/c_{0}, (b-1)/(c_{0}-1)\right ]$ and define w_1,s and w_2,s as in Eqs. 15 and 16. Then the line f(t) := s(t − c₀) + b satisfies f(0) ≤ A_ref(0) = 1, f(1) ≤ A_ref(1) = 1, and f(c₀) = b > A_ref(c₀) by assumption. This implies that A_ref(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 w_1,s, w_2,s in their corresponding intervals. Thus, f(t) ≥ A_ref(t) on [w_1,s, w_2,s].

For the convexity of A_s, note that f(t) ≤ A_ref(t) on t ∈ [0,w_1,s] ∪ [w_2,s, 1] and, therefore, $A_{s}(t) = \max \nolimits (A_{{\textsf {ref}}}(t),f(t))$. Being the maximum of two convex functions, A_s 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,w_1,s] and ζ ∈ [w_2,s, 1]. The first inequality comes from the convexity of A_ref. The second inequality is valid because A_ref(w_1,s) = f(w_1,s) and A_ref ≤ f on [w_1,s, w_2,s]. The other two inequalities are derived for similar reasons. Then, we observe for 0 ≤ t ≤ w_1,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 A_ref ≥ f on [w_2,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, A_s is well defined and convex by previous arguments.

Lastly, we claim that A_s ≤ A on [w_1,s, w_2,s]. Indeed, for all τ ∈ [w_1,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 [w_1,s, c₀]. By re-iterating the same arguments, we can prove A_s(t) ≤ A(t) on [c₀, w_2,s] as well. Consequently, we have the pointwise inequality A_ref(t) ≤ A_s(t) ≤ A(t) on [w_1,s, w_2,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

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

1.2 A.2 Proof of Theorem 2

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

1.2.1 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 ≤ w ≤ v₁,

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

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$\sup \mathcal {C} < c_{0}$.

Proof

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.

Proof

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 A_ref 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) ≤ A_ref(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) > A_ref(t). Define

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

The continuity of A and A_ref implies that A(t^∗) = A_ref(t^∗). Assume t^∗≤ v₁. 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^∗,

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

(1)

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 f ≤ A 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 A_ref(v₁) ≤ f(v₁). With f(0) ≤ A(0) = A_ref(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 A_ref 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 A_ref implies A_ref(w₁) = f(w₁) if w₁ < v₁. Even when w₁ = v₁, this is valid because A_ref(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 A_ref(w₁) = f(w₁) and A_ref(t^∗) = f(t^∗) result in A_ref ≤ f on [w₁, t^∗]. We then conclude the first claim by recalling that f ≤ A 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^∗) = A_ref(t^∗), we obtain f ≤ A_ref 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 ${\textsf {d}}_{[0,c_{0}]}(A_{w_{1}}^{l},A_{{\textsf {ref}}})\le {\textsf {d}}_{[0,c_{0}]}(A,A_{{\textsf {ref}}})$.

1.2.2 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 v₂ ≤ w ≤ 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 A_ref(1 − t) instead of A_ref(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$A_{w_{1},w_{2}}\in \mathcal A$.

Proof

The proof is given in the electronic supplementary material. □