An estimator of the stable tail dependence function based on the empirical beta copula

Abstract

The replacement of indicator functions by integrated beta kernels in the definition of the empirical tail dependence function is shown to produce a smoothed version of the latter estimator with the same asymptotic distribution but superior finite-sample performance. The link of the new estimator with the empirical beta copula enables a simple but effective resampling scheme.

Appendix A: Proofs of Propositions 3.5 and 3.6

Proof of Proposition 3.5

Fix ε ∈ (0,δ].Since ν_n,k,xis a probability measure, we can bring the termB_n,k(x)inside the integral. Split the integral according to the two cases|y −x|_∞≤ ε or |y −x|_∞ > ε,where |z|_∞ = max(|z₁| ,…, |z_d|)for $\boldsymbol {z} \in \mathbb {R}^{d}$.For x ∈ [0, 1]^d, the absolute value in (3.3) is bounded by

$$\begin{array}{@{}rcl@{}} && \sup \left\{ \left| B_{n,k}(\boldsymbol{y}) - B_{n,k}(\boldsymbol{x}) \right| \; : \; \boldsymbol{y} \in [0, n/k]^{d}, \, \left| \boldsymbol{y} - \boldsymbol{x} \right|_{\infty} \le \varepsilon \right\} \\ &&+ 2 \sup_{\boldsymbol{y} \in [0, n/k]^{d}} \left| B_{n,k}(\boldsymbol{y} ) \right| \cdot \nu_{n,k,\boldsymbol{x}} \left( \{ \boldsymbol{y} \in [0, n/k]^{d} : \left| \boldsymbol{y} - \boldsymbol{x} \right|_{\infty} > \varepsilon \} \right). \end{array} $$

(A.1)

In the first term in (A.1), we have x ∈ [0, 1]^d,y ∈ [0,n/k]^d, and|y −x|_∞≤ ε ≤ δ, whencey ∈ [0, 1 + δ]^d.The supremum is thus bounded by the maximal increment ofB_n,kon[0, 1 + δ]^dbetween points at adistance at most ε apart,i.e.,

$$\omega(B_{n,k}, \varepsilon ) = \sup \{ \left| B_{n,k}(\boldsymbol{y}_{1}) - B_{n,k}(\boldsymbol{y}_{2}) \right| : \boldsymbol{y}_{1}, \boldsymbol{y}_{2} \in [0, 1+\delta]^{d}, \, \left| \boldsymbol{y}_{1} - \boldsymbol{y}_{2} \right|_{\infty} \le \varepsilon \}. $$

By Condition 3.3, wecan find for every η > 0asufficiently small ε > 0 such that

$$\limsup_{n \to \infty} \mathbb{P} \left[ \omega(B_{n,k}, \varepsilon ) > \eta \right] \le \eta. $$

Thefirst term in (A.1) can thus be made arbitrarily small with arbitrarily large probability, uniformlyin x ∈ [0, 1]^dandfor sufficiently large n.

For the second term in (A.1), note first that

$$ \sup_{\boldsymbol{y} \in [0, n/k]^{d}} \left| B_{n,k}(\boldsymbol{y} ) \right| = \mathrm{O}_{\mathbb{P}}(n / \sqrt{k} ), \qquad n \to \infty. $$

(A.2)

Indeed, since ℓ isa stdf, we have 0 ≤ ℓ(y) ≤ y₁ + ⋯ + y_d ≤ dn/k for y ∈ [0,n/k]^d; for thepilot estimator $\hat {\ell }_{n,k}$,use Condition 3.2.

If S is a Bin(n,u) random variable, Bennett’s inequality van der Vaart and Wellner (1996, Proposition A.6.2) states that

$$\mathbb{P}[ \sqrt{n} \left| S/n - u \right| \ge \lambda ] \le 2 \exp \left\{ - nu \, h \left( 1 + \frac{\lambda}{\sqrt{n} u} \right) \right\}, \qquad \lambda > 0, $$

where$h(1+\eta ) = {\int }_{0}^{\eta } \log (1+t) \,\mathrm {d} t$ forη ≥ 0. Note that$h(1+\eta ) > \frac {1}{3} \eta ^{2}$ forη ∈ [0, 1]. Itfollows that

$$\begin{array}{@{}rcl@{}} \nu_{n,k,\boldsymbol{x}} \left( \{ \boldsymbol{y} \in [0, n/k]^{d} : \left| \boldsymbol{y} - \boldsymbol{x} \right|_{\infty} > \varepsilon \} \right) &\le& \sum\limits_{j = 1}^{d} \mathbb{P}\left[ \left| \operatorname{Bin}(n, \frac{k}{n} x_{j}) / k - x_{j} \right| >\varepsilon\right]\\ &=& \sum\limits_{j = 1}^{d} \mathbb{P}\left[ \sqrt{n} \left| \operatorname{Bin}(n, \frac{k}{n} x_{j}) / n - \frac{k}{n} x_{j} \right| > k\varepsilon/\sqrt{n}\right]\\ \end{array} $$

(A.3)

$$\begin{array}{@{}rcl@{}} && \le \sum\limits_{j = 1}^{d} 2 \exp \left\{ - kx_{j} \, h \left( 1 + \frac{k\varepsilon/\sqrt{n}}{\sqrt{n} kx_{j}/n} \right) \right\} \\ &&= \sum\limits_{j = 1}^{d} 2 \exp \left\{ - kx_{j} \, h (1 + \varepsilon/x_{j} ) \right\}. \end{array} $$

(A.4)

As ∂{x h(1 + ε/x)}/∂x < 0for0 < x < 1, we have inf x∈[0,1]{x h(1 + ε/x)} = h(1 + ε). We concludethat

$$\nu_{n,k,\boldsymbol{x}} \left( \!\{ \boldsymbol{y} \in [0, n/k]^{d} : \left| \boldsymbol{y} \,-\, \boldsymbol{x} \right|_{\infty} > \varepsilon \} \right) \le 2d \exp \{ - k \, h(1+\varepsilon) \} \le 2d \exp \!\left( - \frac{1}{3} k \varepsilon^{2} \right)\!. $$

Together, thesupremum over x ∈ [0, 1]^dof the second term in (A.1) is of the order

$$ \mathrm{O}_{\mathbb{P}} \left( \frac{n}{\sqrt{k}} \exp \left( - \frac{1}{3} k \varepsilon^{2} \right) \right), \qquad n \to \infty. $$

(A.5)

It therefore converges to zero in probability since log(n) = o(k)byassumption. □

Remark A.1

In Proposition 3.5, the condition log(n) = o(k)was imposed to control the remainder term in (A.5). That term arose from an application ofBennett’s inequality in (A.4), producing an upper bound to the binomial tail probability theline before. We now show that the same probability also admits a lower bound that would yield the same condition on k. Indeed, starting from the left-hand side of (A.3), we have

$$\begin{array}{@{}rcl@{}} \nu_{n,k,\boldsymbol{x}} \left( \{ \boldsymbol{y} \in [0, n/k]^{d} : \left| \boldsymbol{y} - \boldsymbol{x} \right|_{\infty} > \varepsilon \} \right) &\ge& \mathbb{P}\left[ \left| \operatorname{Bin}\left( n, \frac{k}{n} x_{1}\right) / k - x_{1} \right| > \varepsilon \right] \\ &\ge& \mathbb{P}\left[ \operatorname{Bin}\left( n, \frac{k}{n} x_{1}\right) > k(x_{1}+ \varepsilon) \right] \\ &\ge& \mathbb{P}\left[ \operatorname{Bin}\left( n, \frac{k}{n} x_{1}\right) = m \right] \end{array} $$

where m = m_n = ⌊k(x₁ + ε) + 1⌋and⌊ ⋅ ⌋ is the floor function. Stirling’s formula says that $n! = \sqrt {2\pi n} (n/e)^{n} \{1 + \mathrm {o}(1)\}$ asn →∞ and thus, since k = k_n →∞,also

$$\frac{n!}{m!(n-m)!} = (2\pi m)^{-1/2} (n/m)^{m} (1 - m/n)^{-(n-m)} \{1 + \mathrm{o}(1)\}. $$

But then

$$\begin{array}{@{}rcl@{}} \mathbb{P}\left[ \operatorname{Bin}\left( n, \frac{k}{n} x_{1}\right) = m \right] &=& \frac{n!}{m!(n-m)!} \left( \frac{k}{n}x_{1}\right)^{m} \left( 1 - \frac{k}{n}x_{1}\right)^{n-m} \\ &=& \frac{1}{\sqrt{2\pi m}} \left( \frac{kx_{1}}{m} \right)^{m} \left( \frac{1 - kx_{1} / n}{1 - m/n} \right)^{n-m} \{ 1 + \mathrm{o}(1) \} \\ &\ge& \frac{1}{\sqrt{2\pi m}} \left( \frac{kx_{1}}{m} \right)^{m} \{ 1 + \mathrm{o}(1) \}, \end{array} $$

since m ≥ kx₁. For n sufficiently large, k = k_nis such that m_n ≤ k(x₁ + 2ε)and thus (kx₁/m)^m ≥ ρ^kwith $\rho = \{x_{1}/(x_{1}+ 2\varepsilon )\}^{x_{1}+ 2\varepsilon }$. Wefind

$$\mathbb{P}\left[ \operatorname{Bin}\left( n, \frac{k}{n} x_{1}\right) = m \right] \ge \frac{1}{\sqrt{2\pi(x_{1} + 2\varepsilon)}} k^{-1/2} \rho^{k}. $$

Recallthat in the proof of Proposition 3.5, we needed to control the second term in (A.1). Inview of (A.2) and the above lower bound, we need a sequence k such that, for everyε > 0, we have(n/k)ρ^k → 0. Here0 < ρ = ρ(x₁, ε) <  1 approaches 1 as ε↓ 0, for every fixed x₁ > 0. Butthen k = k_nmustbe such that log(n) − log(k) + k log(ρ) →−∞ as n →∞, and since log(ρ) < 0can be arbitrarily close to 0 as ε↓0,we still need that log(n) = o(k).

Proof of Proposition 3.6

Fix x ∈ [0,M]^d.For j ∈{1,…,d}such that x_j = 0, the binomial distribution Bin(n, (k/n)x_j)is concentrated on 0. As a consequence, the integral overy ∈ [0,n/k]^dwith respect to ν_n,k,xcan be restricted to the set of those y ∈ [0,n/k]^dsuch that y_j = 0for all j ∈{1,…,d}for which x_j = 0.Call this set $\mathbb {D}(n,k,\boldsymbol {x})$.

For $\boldsymbol {y} \in \mathbb {D}(n,k,\boldsymbol {x})$, the function$[0, 1] \to \mathbb {R} : t \mapsto f(\boldsymbol {x} + t(\boldsymbol {y}-\boldsymbol {x}))$ is continuouson [0, 1]and is continuously differentiable on (0, 1); indeed, ifx_j = 0, then the jth componentof x + t(y −x)vanishes and thusdoes not depend on t ∈ [0, 1],while if x_j > 0,then that component is (strictly) positive for allt ∈ [0, 1), so that, by assumption,$\dot {f}_{j}(\boldsymbol {x} + t(\boldsymbol {y}-\boldsymbol {x}))$ exists and iscontinuous in t ∈ [0, 1). WritingJ(x) = {j = 1,…,d : x_j > 0}, we find, by the fundamental theorem of calculus,

$$f(\boldsymbol{y} ) - f(\boldsymbol{x} ) = \sum\limits_{j \in J(\boldsymbol{x})} (y_{j} - x_{j}) {{\int}_{0}^{1}} \dot{f}_{j} \left( \boldsymbol{x} + t(\boldsymbol{y} - \boldsymbol{x}) \right) \,\mathrm{d} t, \qquad \boldsymbol{y} \in \mathbb{D}(n,k,\boldsymbol{x}). $$

Weobtain

$$\begin{array}{@{}rcl@{}} {\Delta}_{n,k}(\boldsymbol{x}) &\!:=& {\int}_{\mathbb{D}(n,k,\boldsymbol{x})} \sqrt{k} \{ f(\boldsymbol{y}) - f(\boldsymbol{x}) \} \,\mathrm{d} \nu_{n,k,\boldsymbol{x}}(\boldsymbol{y} )\\ &=& \sum\limits_{j \in J(\boldsymbol{x})} {\int}_{\boldsymbol{y} \in \mathbb{D}(n,k,\boldsymbol{x})} \sqrt{k} (y_{j} - x_{j}) {{\int}_{0}^{1}} \dot{f}_{j} \left( \boldsymbol{x} + t(\boldsymbol{y} - \boldsymbol{x}) \right) \,\mathrm{d} t \,\mathrm{d} \nu_{n,k,\boldsymbol{x}}(\boldsymbol{y}) \\ &=& \sum\limits_{j \in J(\boldsymbol{x})} {\int}_{\boldsymbol{y} \in \mathbb{D}(n,k,\boldsymbol{x})} \sqrt{k} (y_{j} \,-\, x_{j}) {{\int}_{0}^{1}} \left\{ \dot{f}_{j} \!\left( \boldsymbol{x} \,+\, t(\boldsymbol{y} \,-\, \boldsymbol{x}) \right) \,-\, \dot{f}_{j}(\boldsymbol{x}) \right\} \,\mathrm{\!d} t \,\mathrm{d} \nu_{n,k,\boldsymbol{x}}(\boldsymbol{y}), \end{array} $$

where the last step is justified via $\int y_{j} \,\mathrm {d} \nu _{n,k,\boldsymbol {x}}(\boldsymbol {y}) = \operatorname {\mathbb {E}}[ \operatorname {Bin}(n, (k/n)x_{j})/k ] = x_{j}$.Taking absolute values, we find, forx ∈ [0,M]^d,

$$\left| {\Delta}_{n,k}(\boldsymbol{x}) \right| \le \sum\limits_{j \in J(\boldsymbol{x})} I_{n,k}(\boldsymbol{x}, j) $$

where

$$I_{n,k}(\boldsymbol{x}, j) = {\int}_{\boldsymbol{y} \in \mathbb{D}(n,k,\boldsymbol{x})} \sqrt{k} \left| y_{j} - x_{j} \right| {{\int}_{0}^{1}} \left| \dot{f}_{j} \left( \boldsymbol{x} + t(\boldsymbol{y} \,-\, \boldsymbol{x}) \right) \,-\, \dot{f}_{j}(\boldsymbol{x} ) \right| \,\mathrm{d} t \,\mathrm{d} \nu_{n,k,\boldsymbol{x}}(\boldsymbol{y}). $$

We will find anupper bound for I_n,k(x,j).

Let K > 0be suchthat $\left | \dot {f}_{i} \right | \le K$ forall i ∈{1,…,d}.Choose δ ∈ (0,M]and ε ∈ (0,δ/2]. InI_n,k(x,j), split the integral over$\boldsymbol {y} \in \mathbb {D}(n,k,\boldsymbol {x})$ into two pieces,depending on whether |y −x| ≤ ε or |y −x| > ε, where$\left | \boldsymbol {z} \right | = ({z_{1}^{2}} + {\cdots } + {z_{d}^{2}})^{1/2}$ denotesthe Euclidean norm of $\boldsymbol {z} \in \mathbb {R}^{d}$.

In I_n,k(x,j), the integralover $\boldsymbol {y} \in \mathbb {D}(n,k,\boldsymbol {x})$ for which |y −x| > ε is bounded by

$$\begin{array}{@{}rcl@{}} 2K \sqrt{k} {\int}_{\mathbb{D}(n,k,\boldsymbol{x})} \left| y_{j} - x_{j} \right| \operatorname{\mathbb{1}}\{ \left| \boldsymbol{y} - \boldsymbol{x} \right| > \varepsilon \} \,\mathrm{d} \nu_{n,k,\boldsymbol{x}}(\boldsymbol{y}) &\le& \frac{2K\sqrt{k}}{\varepsilon} {\int}_{\mathbb{D}(n,k,\boldsymbol{x})} \left| \boldsymbol{y} - \boldsymbol{x} \right|^{2} \,\mathrm{d} \nu_{n,k,\boldsymbol{x}}(\boldsymbol{y}) \\ &=& \frac{2K\sqrt{k}}{\varepsilon} \sum\limits_{i = 1}^{d} \frac{1}{k^{2}} \cdot n \cdot \frac{k}{n} x_{i} \cdot \left( 1 - \frac{k}{n} x_{i}\right) \\ &\le& \frac{2KMd}{\varepsilon \sqrt{k}}. \end{array} $$

To analyze the integral in I_n,k(x,j)over those $\boldsymbol {y} \in \mathbb {D}(n,k,\boldsymbol {x})$ forwhich |y −x| ≤ ε, we need to distinguishbetween two cases: x_j < δ and x_j ≥ δ. Incase x_j < δ,the integral is simply bounded by

$$\begin{array}{@{}rcl@{}} 2K {\int}_{\mathbb{D}(n,k,\boldsymbol{x})} \sqrt{k} \left| y_{j} - x_{j} \right| \,\mathrm{d} \nu_{n,k,\boldsymbol{x}}(\boldsymbol{y}) &\le& 2K \sqrt{k} \sqrt{ \frac{1}{k^{2}} \cdot n \cdot \frac{k}{n}x_{j} \cdot \left( 1 - \frac{k}{n}x_{j}\right)} \\ &\le&K \sqrt{x_{j}} < 2K \sqrt{\delta}. \end{array} $$

In case x_j ≥ δ, the inequality |y −x| ≤ ε ≤ δ/2 and the fact that x ∈ [0,M]^d and y ∈ [0, ∞)^d imply that y belongs to theset

$$\mathbb{B}_{j}(M, \delta) = \{ \boldsymbol{z} \in [0, M+\delta/2]^{2} : z_{j} > \delta/2 \}. $$

Let

$$\omega_{j}(M, \delta, \varepsilon) = \sup \{ \left| \dot{f}_{j}(\boldsymbol{z}_{1} ) - \dot{f}_{j}(\boldsymbol{z}_{2} ) \right| : \boldsymbol{z}_{1}, \boldsymbol{z}_{2} \in \mathbb{B}_{j}(M, \delta), \, \left| \boldsymbol{z}_{1} - \boldsymbol{z}_{2} \right| \le \varepsilon \}. $$

The integralin I_n,k(x,j)over$\boldsymbol {y} \in \mathbb {D}(n,k,\boldsymbol {x})$ suchthat |y −x| ≤ ε is bounded by

$$\omega_{j}(M, \delta, \varepsilon) {\int}_{\mathbb{D}(n,k,\boldsymbol{x})} \sqrt{k} \left| y_{j} \,-\, x_{j} \right| \,\mathrm{d} \nu_{n,k,\boldsymbol{x}}(\boldsymbol{y}) \!\le\! \omega_{j}(M, \delta, \varepsilon) \sqrt{x_{j}} \le \omega_{j}(M, \delta, \varepsilon) \sqrt{M} $$

using the Cauchy–Schwarz inequality and the first two moments of the binomial distribution.

Assembling all the pieces, we obtain

$$\sup_{\boldsymbol{x} \in [0, M]^{d}} \left| {\Delta}_{n,k}(\boldsymbol{x}) \right| \\ \le \frac{2d^{2}KM}{\varepsilon \sqrt{k}} + 2dK \sqrt{\delta} + \sqrt{M} \sum\limits_{j = 1}^{d} \omega_{j}(M, \delta, \varepsilon). $$

As a consequence, for every δ ∈ (0,M] and every ε ∈ (0,δ/2], wehave

$$\limsup_{n \to \infty} \sup_{\boldsymbol{x} \in [0, M]^{d}} \left| {\Delta}_{n,k}(\boldsymbol{x}) \right| \le 2dK \sqrt{\delta} + \sqrt{M} \sum\limits_{j = 1}^{d} \omega_{j}(M, \delta, \varepsilon). $$

The function$\dot {f}_{j}$ is continuous and thus uniformly continuous on the compact set $\mathbb {B}_{j}(M, \delta )$. As consequence, inf ε> 0ω_j(M,δ,ε) = 0.The limit superior in the previous display is thus bounded by$2dK \sqrt {\delta }$, for allδ ∈ (0,M], and must thereforebe equal to zero. □