Testing symmetry for bivariate copulas using Bernstein polynomials

Abstract

In this work, tests of symmetry for bivariate copulas are introduced and studied using empirical Bernstein copula process. Three statistics are proposed and their asymptotic properties are established. Besides, a multiplier bootstrap Bernstein version is investigated for implementation purpose. The simulation study demonstrated the superior performance of the Bernstein tests compared to tests based on empirical copulas. Furthermore, in real data applications, these tests consistently yielded similar conclusions across a diverse range of scenarios.

Appendices

Appendix A: Proof of Theorem 2

Proof

This proof is an adaption of Genest et al. (2012, Proposition 3). For the convergence of $nR_{n, m}$ and $n^{1/2}T_{n, m}$, one can directly apply continuous mapping theorem combined with Theorem 1. For the convergence of $nS_{n, m}$, the functional delta method was used. To this end, let $\mathscr {C}[0, 1]^2$ denote the space of continuous functions on $[0, 1]^2$, $\mathscr {D}[0, 1]^2$ denote the space of functions with continuity from upper right quadrant and limits from other quadrants on $[0, 1]^2$, equipped with sup-norm. Further, denote $\textrm{BV}_1[0, 1]^2$ by the subspace of $\mathscr {D}[0, 1]^2$ where functions with total variation bounded by 1. By continuous mapping theorem,

$$\begin{aligned} \left( \mathbb {S}^2_{n, m}, \mathbb {B}_{n, m}\right) \rightsquigarrow \left( \mathbb {S}_C^2, \mathbb {B}_C\right) \end{aligned}$$

in the space $\ell ^{\infty }[0, 1]^2\times \ell ^{\infty }[0, 1]^2$. Rewrite it as

$$\begin{aligned} \left( \mathbb {S}^2_{n, m}, \mathbb {B}_{n, m}\right) =n^{1/2}\{(A_{n, m}, \widehat{C}_{n, m})-(A, C)\}, \end{aligned}$$

where $A\equiv 0$ and $A_{n, m}:=n^{1/2}(\widehat{C}_{n, m}-\widehat{C}_{n, m}^{\top })^2$, where $\widehat{C}_{n, m}^{\top }(u,v) = \widehat{C}_{n, m}(v,u)$. Then, consider the map $\phi : \ell ^{\infty }[0, 1]^2\times \textrm{BV}_1[0, 1]^2\rightarrow {\mathbb {R}}$ defined by

$$\begin{aligned} \phi (a, b)=\int _{(0, 1]^2}a {\textrm{d}}b, \end{aligned}$$

Clearly,

$$\begin{aligned} nS_{n, m}&=\phi \left( \mathbb {S}^2_{n, m}, \mathbb {B}_{n, m}\right) \\&=n^{1/2}\{\phi (A_{n, m}, \widehat{C}_{n, m})-\phi (A, C)\}. \end{aligned}$$

To conclude the proof, by Carabarin-Aguirre and Ivanoff (2010, Lemma 4.3), $\phi$ is Hadamard differentiable tangentially to $\mathscr {C}[0, 1]^2\times \mathscr {D}[0, 1]^2$ at each $(\alpha , \beta )$ in $\ell ^{\infty }[0, 1]^2\times \textrm{BV}_1[0, 1]^2$ such that $\int |d\alpha |<\infty$ with derivative

$$\begin{aligned} \phi '_{(\alpha , \beta )}(a, b)=\int \alpha {\textrm{d}}b+\int a {\textrm{d}}\beta . \end{aligned}$$

Then by applying the Functional Delta Method (van der Vaart and Wellner 1996, Theorem 3.9.4), $nS_{n, m}\rightsquigarrow \phi '_{(A, C)}\left( \mathbb {S}_C^2, \mathbb {B}_C \right)$, where

$$\begin{aligned} \phi '_{(A, \,C)}\left( \mathbb {S}_C^2, \mathbb {B}_C \right)&=\int _{(0, 1]^2}A {\textrm{d}}\mathbb {B}_C+\int _{(0, 1]^2}\mathbb {S}_C^2 {\textrm{d}}C\\&=\int _{(0, 1]^2}\mathbb {S}_C^2 {\textrm{d}}C. \end{aligned}$$

This yields to the desired result. $\square$

Appendix B: Proof of Theorem 3

Proof

This proof is an adaption of Genest et al. (2012, Proposition 4). The strongly uniform consistency of $\widehat{C}_{n,m}$ was provided in Janssen et al. (2012, Theorem 1), then by continuous mapping theorem, it follows immediately that $R_{n,m}$ and $T_{n,m}$ converge to $R_C$ and $T_C$ almost surely, respectively. Further, to prove the convergence of $S_{n,m}$, write

$$\begin{aligned} |S_{n, m}-S_C|\le |\gamma _{n, m}|+|\zeta _{n}|, \end{aligned}$$

where

$$\begin{aligned} \gamma _{n, m}&=\int _{0}^1\int _{0}^1\left\{ \widehat{C}_{n, m}(u, v) -\widehat{C}_{n, m}(v, u)\right\} ^2 {\textrm{d}}\widehat{C}_{n}(u, v)\\&\quad -\int _{0}^1\int _{0}^1\left\{ {C}(u, v)-{C}(v, u)\right\} ^2 {\textrm{d}}\widehat{C}_{n}(u, v), \end{aligned}$$

and

$$\begin{aligned} \zeta _{n}&=\int _{0}^1\int _{0}^1\left\{ {C}(u, v)-{C}(v, u)\right\} ^2 {\textrm{d}}\widehat{C}_{n}(u, v)\\&\quad -\int _{0}^1\int _{0}^1\left\{ {C}(u, v)-{C}(v, u)\right\} ^2 {\textrm{d}}{C}(u, v). \end{aligned}$$

Since

$$\begin{aligned}&\left| \left\{ \widehat{C}_{n ,m}(u, v)-\widehat{C}_{n, m}(v, u)\right\} ^2-\left\{ C(u, v)-C(v, u)\right\} ^2\right| \\&\quad =\left| \left[ \widehat{C}_{n, m}(u, v)+C(u, v)\right] -\left[ \widehat{C}_{n ,m}(v, u)+C(v, u)\right] \right| \\&\quad \quad \times \left| \left[ \widehat{C}_{n, m}(u, v)-C(u, v)\right] -\left[ \widehat{C}_{n, m}(v, u)-C(v, u)\right] \right| \\&\quad \le \left[ \left| \widehat{C}_{n, m}(u, v)+C(u, v)\right| +\left| \widehat{C}_{n ,m}(v, u)+C(v, u)\right| \right] \\&\quad \quad \times \left[ \left| \widehat{C}_{n, m}(u, v)-C(u, v)\right| +\left| \widehat{C}_{n, m}(v, u)-C(v, u)\right| \right] \\&\quad \le 8\sup _{(u,v)\in [0, 1]^2}\left| \widehat{C}_{n, m}(u, v)-C(u, v)\right| , \end{aligned}$$

one has

$$\begin{aligned} |\gamma _{n ,m}|\le 8\sup _{(u,v)\in [0, 1]^2}\left| \widehat{C}_{n, m}(u, v)-C(u, v)\right| \xrightarrow {a.s.} 0. \end{aligned}$$

For $\zeta _{n}$, by Genest et al. (1995, Proposition A.1 (i)), one has

$$\begin{aligned}&\int _{0}^1\int _{0}^1\left\{ {C}(u, v)-{C}(v, u)\right\} ^2 {\textrm{d}}\widehat{C}_{n}(u, v)\\&\quad \rightarrow \int _{0}^1\int _{0}^1\left\{ {C}(u, v)-{C}(v, u)\right\} ^2 {\textrm{d}}{C}(u, v), \end{aligned}$$

then $\zeta _{n} \rightarrow 0$. Therefore, $S_{n, m}$ converges to $S_C$ almost surely.

$\square$

Appendix C: Proof of Lemma 1

Proof

Under these assumptions, one need to use the framework in Segers et al. (2017). Specifically, let $\mu _{m, (u, v)}$ be the law of random vector $(B_1/m, B_2/m)$, where $B_1, B_2$ follow $\textsf {Binomial}(m, u)$ and $\textsf {Binomial}(m, v)$, respectively. The empirical Bernstein copula can be rewritten as

$$\begin{aligned} \widehat{C}_{n, m}(u, v)=\int _{[0, 1]^2}\widehat{C}_n(x, y)\,\textrm{d}\mu _{m, (u, v)}(x, y), \; (x, y)\in [0 ,1]^2. \end{aligned}$$

Moreover, write $(x, y)(t)=(u, v)+t((x, y)-(u, v))$ with $t\in [0, 1]$. Then, the empirical Bernstein copula process is

$$\begin{aligned} \mathbb {B}_{n, m}(u, v)&=\sqrt{n}\left\{ \widehat{C}_{n, m}(u, v)-C(u, v)\right\} \nonumber \\&=\sqrt{n}\left\{ \widehat{C}_{n, m}(u, v)-\int _{[0, 1]^2}C(x, y)\,\textrm{d}\mu _{m, (u, v)}(x, y)\right. \nonumber \\&\quad \left. +\int _{[0, 1]^2}C(x, y)\,\textrm{d}\mu _{m, (u, v)}(x, y)-C(u, v)\right\} \nonumber \\&=\int _{[0, 1]^2}\sqrt{n}\left\{ \widehat{C}_{n, m}(x, y)-C(x, y)\right\} {\textrm{d}}\mu _{m, (u, v)}(x, y)\nonumber \\&\quad +\sqrt{n}\left\{ \int _{[0, 1]^2}C(x, y)\,\textrm{d}\mu _{m, (u, v)}(x, y)-C(u, v)\right\} \nonumber \\&=T_1+T_2. \end{aligned}$$

(C1)

The two terms are dealt with separately.

• For the term $T_1$, according to Segers et al. (2017, Proposition 3.1), one has

  $$\begin{aligned}&\sup \limits _{(u, v)\in [0, 1]^2}\left| \int _{[0, 1]^2}\sqrt{n}\left\{ \widehat{C}_{n, m}(s, t)-C(s, t)\right\} \,\textrm{d}\mu _{m, (u, v)}(s, t)\right. \\&\qquad \left. -\sqrt{n}\left\{ \widehat{C}_{n, m}(u, v)-C(u, v)\right\} \right| =o_p(1). \end{aligned}$$

  And note that, $\sqrt{n}\left\{ \widehat{C}_{n, m}(u, v)-C(u, v)\right\} \rightsquigarrow \mathbb {B}_C(u, v)$ in $\ell ^{\infty }([0, 1]^2)$ under the Assumption 1 (see Segers (2012)). Therefore, $T_1\rightsquigarrow \mathbb {B}_C(u, v)$ as n goes to infinity.

• For the term $T_2$, Let $m=cn^{\alpha }$ for some $c >0$, by Kojadinovic (2022, Lemma 3.1), under Assumption 1-2, one has

  $$\begin{aligned}&\sup _{(u,v)\in [0, 1]^2}\sqrt{n}\bigg |\int _{[0, 1]^2}C(x, y)\,\textrm{d}\mu _{m, (u, v)}(x, y)\\&\qquad -C(u, v)\bigg |=O\left( n^{(3-4\alpha )/6}\right) \end{aligned}$$

  almost surely. Therefore, if $\alpha > 3/4$, $T_2$ goes to zero as n goes to infinity.

Combining above results completes the proof. $\square$

Appendix D: Proof of Proposition 1

Proof

By Rémillard and Scaillet (2009), one has

$$\begin{aligned}&\mathbb {C}_n(u, v)\\&\quad =\sqrt{n}\left\{ C_n(u, v)-C(u, v)\right\} \\&\quad =n^{1/2}\Bigg \{\frac{1}{n}\sum _{i=1}^n\left\{ {\mathbb {I}}\left( U_i\le u, V_i\le v)-C(u, v\right) \right\} \Bigg \}\\&\quad \rightsquigarrow \mathbb {C}(u ,v), \end{aligned}$$

where $C_n(u,v)=\frac{1}{n}\sum _{i=1}^{n}{\mathbb {I}}\left( U_i\le u, V_i\le v\right)$, and

$$\begin{aligned}&\overline{\mathbb {C}}^{(h)}_n(u, v)=n^{1/2}\Bigg \{\frac{1}{n}\sum _{i=1}^n\left( \xi ^{(h)}_i-\bar{\xi }_n^{(h)}\right) {\mathbb {I}}(\widehat{U}_i\le u, \widehat{V}_i\le v) \Bigg \}\\&\quad \rightsquigarrow \mathbb {C}(u ,v). \end{aligned}$$

To end the proof, one needs to show that the difference between $\left( \mathbb {C}_n, \overline{\mathbb {C}}^{(h)}_n\right)$ and $\left( \widetilde{\mathbb {B}}_{n, m}, \overline{\mathbb {B}}^{(h)}_{n, m}\right)$ are asymptotically negligible.

It is well-known (for example, see Deheuvels (1979)) that, as $n\rightarrow \infty$,

$$\begin{aligned} \Vert C_n(u, v)-C(u, v)\Vert =O\left( n^{-1/2}(\log \log n)^{1/2}\right) , \end{aligned}$$

almost surely and using the same techniques in Janssen et al. (2012) gives

$$\begin{aligned} \Vert C_{n,m}(u, v)-C(u, v)\Vert =O\left( n^{-1/2}(\log \log n)^{1/2}\right) , \end{aligned}$$

almost surely, it immediately follows that

$$\begin{aligned} \Vert C_{n,m}(u, v)-C_n(u, v)\Vert = O\left( n^{-1/2}(\log \log n)^{1/2}\right) , \end{aligned}$$

almost surely. Therefore, one can conclude that $\widetilde{\mathbb {B}}_{n,m}(u, v)\rightsquigarrow \mathbb {C}(u, v)$.

Further, by Janssen et al. (2012, Lemma 1), as $n\rightarrow \infty$,

$$\begin{aligned} \Vert \widehat{C}_{n}(u ,v)-C(u, v)\Vert&=O\left( n^{-1/2}(\log \log n)^{1/2}\right) , \end{aligned}$$

almost surely. By Lemma 1, under the assumptions,

$$\begin{aligned} \Vert \widehat{C}_{n,m}(u ,v)-C(u, v)\Vert =o_p(1). \end{aligned}$$

Hence

$$\begin{aligned} \Vert \widehat{C}_{n,m}(u ,v)-\widehat{C}_n(u ,v)\Vert = o_p(1), \end{aligned}$$

one has that $\overline{\mathbb {B}}_{n,m}^{(h)}(u, v) \rightsquigarrow \mathbb {C}(u,v)$. $\square$

Appendix E: Proof of Proposition 2

Proof

We only show the uniform consistency for

$$\begin{aligned} \frac{\partial \widehat{C}_{n,m}(u, v)}{\partial u}&=m\sum _{k=0}^{m-1}\sum _{\ell =0}^m\bigg \{\widehat{C}_n\left( \frac{k+1}{m}, \frac{\ell }{m}\right) -\widehat{C}_n\left( \frac{k}{m}, \frac{\ell }{m}\right) \bigg \}\\&\quad \cdot P_{m-1, k}(u)P_{m, \ell }(v). \end{aligned}$$

The result for the other partial derivative can be obtained similarly. For any $u\in [b_n, 1-b_n], v\in [0, 1]$, one has

$$\begin{aligned}&\left| \frac{\partial \widehat{C}_{n,m}(u, v)}{\partial u}-\dot{C}_1(u, v)\right| \\&\quad \le \Bigg |m\sum _{k=0}^{m-1}\sum _{\ell =0}^m\bigg \{\widehat{C}_n\left( \frac{k+1}{m}, \frac{\ell }{m}\right) -\widehat{C}_n\left( \frac{k}{m}, \frac{\ell }{m}\right) \\&\quad \quad -{C}\left( \frac{k+1}{m}, \frac{\ell }{m}\right) +{C}\left( \frac{k}{m}, \frac{\ell }{m}\right) \bigg \}P_{m-1, k}(u)P_{m,\ell }(v)\Bigg |\\&\qquad +\Bigg |m\sum _{k=0}^{m-1}\sum _{\ell =0}^m\bigg \{{C}\left( \frac{k+1}{m}, \frac{\ell }{m}\right) -{C}\left( \frac{k}{m}, \frac{\ell }{m}\right) \bigg \}\\&\qquad \cdot P_{m-1, k}(u)P_{m, \ell }(v)-\dot{C}_1(u, v)\Bigg |\\&\quad =A_1+A_2. \end{aligned}$$

Further, let $P_{m, k}'(u)= \frac{1}{u(1-u)}P_{m, k}(u)\left( k -mu\right)$ be the derivative of $P_{m, k}(u)$, then

$$\begin{aligned} A_1&\le \sum _{k=0}^{m}\sum _{\ell =0}^m\bigg |\widehat{C}_n\left( \frac{k}{m}, \frac{\ell }{m}\right) -{C}\left( \frac{k}{m}, \frac{\ell }{m}\right) \bigg |\\&\quad \cdot \left| P_{m, k}'(u)\right| P_{m, \ell }(v)\\&\le \sup _{(u,v)\in [0, 1]^2}\bigg |\widehat{C}_n\left( u, v\right) -{C}\left( u, v\right) \bigg |\cdot \sum _{k=0}^m\left| P_{m, k}'(u)\right| \\&=O\left( m^{1/2}n^{-1/2}(\log \log n)^{1/2}\right) , \end{aligned}$$

almost surely as $n\rightarrow \infty$ and where

$$\begin{aligned} \sum _{k=0}^m\left| P_{m, k}'(u)\right| =O(m^{1/2}), \end{aligned}$$

by Janssen et al. (2014, Lemma 1).

For dealing with $A_2$, let $\nu _{m, (u, v)}$ be the law of random vector $(S_1/(m-1), S_2/m)$, where $S_1, S_2$ follow $\textsf {Binomial}(m-1, u)$ and $\textsf {Binomial}(m, v)$, respectively. Therefore,

$$\begin{aligned}&\sum _{k=0}^{m-1}\sum _{\ell =0}^mC\left( \frac{k+1}{m}, \frac{\ell }{m}\right) P_{m-1, k}(u)P_{m, \ell }(v)\\&\quad =\int _{[0, 1]^2} C\left( \left( x+\frac{1}{m-1}\right) \frac{m-1}{m}, y\right) \\&\qquad {\textrm{d}}\nu _{m, (u, v)}(x, y), \end{aligned}$$

and

$$\begin{aligned}&\sum _{k=0}^{m-1}\sum _{\ell =0}^mC\left( \frac{k}{m}, \frac{\ell }{m}\right) P_{m-1, k}(u)P_{m, \ell }(v)\\&\quad =\int _{[0, 1]^2} C\left( x\frac{m-1}{m}, y\right) {\textrm{d}}\nu _{m, (u, v)}(x, y). \end{aligned}$$

Using the representation in Segers et al. (2017, Proof of Proposition 3.4), for $0< t <1$, one has

$$\begin{aligned} A_2&=\Bigg |m\sum _{k=0}^{m-1}\sum _{\ell =0}^m\bigg \{{C}\left( \frac{k+1}{m}, \frac{\ell }{m}\right) -{C}\left( \frac{k}{m}, \frac{\ell }{m}\right) \bigg \}\nonumber \\&\quad \cdot P_{m-1, k}(u)P_{m, \ell }(v)-\dot{C}_1(u, v)\Bigg |\nonumber \\&=\left| \int _0^1\bigg \{\int _{[0, 1]^2} \left[ \dot{C}_1\left( \frac{m-1}{m}x{+}\frac{1+t}{m}, y\right) {-}\dot{C}_1(u, v)\right] \right. \nonumber \\&\quad \left. \textrm{d}\nu _{m, (u, v)}(x, y) \bigg \} \textrm{d}t\right| . \end{aligned}$$

(E2)

Let $x'(x, t):= \frac{m-1}{m}x+\frac{1+t}{m}$ and $\varepsilon _n= b_n/2$, then one has,

$$\begin{aligned} (\text {E2})&\le \left| \int _0^1\bigg \{\int _{[0, 1]^2} \left[ \dot{C}_1\left( x', y\right) -\dot{C}_1(u, v)\right] \right. \\&\quad \cdot {\mathbb {I}}\left( \max (|x'-u|, |y-v|)\le \varepsilon _n\right) \\&\quad \left. {\textrm{d}}\nu _{m, (u, v)}(x, y) \bigg \} {\textrm{d}}t\right| \\&\quad +\left| \int _0^1\bigg \{\int _{[0, 1]^2} \left[ \dot{C}_1\left( x', y\right) -\dot{C}_1(u, v)\right] \right. \\&\quad \left. \cdot {\mathbb {I}}\left( \max (|x'-u|, |y-v|)> \varepsilon _n\right) {\textrm{d}}\nu _{m, (u, v)}(x, y) \bigg \} \textrm{d}t\right| \\&= A_{21}+A_{22}. \end{aligned}$$

Further, the two terms are dealt with separately using the strategy in Kojadinovic (2022, Proof of Lemma 3.1).

• For $A_{21}$, under Assumption 1-2, by Segers (2012, Lemma 4.3), for a constant $L>0$, one has

  $$\begin{aligned}&\left| \dot{C}_1\left( x', y\right) -\dot{C}_1(u, v)\right| {\mathbb {I}}\left( \max (|x'-u|, |y-v|)\le \varepsilon _n\right) \\&\quad \le Lb_n^{-1}\left[ |x'-u|+|y-v|\right] . \end{aligned}$$

  Further,

  $$\begin{aligned}&A_{21}\nonumber \\&\quad \le Lb_n^{-1}\int _0^1\bigg \{\int _{[0, 1]^2} \left[ |x'-u|+|y-v|\right] \nonumber \\&\quad \quad \textrm{d}\nu _{m, (u, v)}(x, y) \bigg \} {\textrm{d}}t\nonumber \\&\quad \le Lb_n^{-1} \int _{[0, 1]^2} \int _{0}^{1} \left[ |x-u|+|y-v|+\left| \frac{x}{m}-\frac{1+t}{m}\right| \right] \nonumber \\&\quad \quad {\textrm{d}}t {\textrm{d}}\nu _{m, (u, v)}(x, y)\nonumber \\&\quad \le Lb_n^{-1} \int _{[0, 1]^2} \int _{0}^{1} \left[ |x-u|+|y-v|\right. \nonumber \\&\quad \quad \left. +\frac{x}{m}+\frac{1+t}{m}\right] {\textrm{d}}t {\textrm{d}}\nu _{m, (u, v)}(x, y)\nonumber \\&\quad =Lb_n^{-1} \int _{[0, 1]^2} \left[ |x-u|+|y-v|+\frac{x}{m}\right. \nonumber \\&\quad \quad \left. +\frac{3}{2m}\right] {\textrm{d}}\nu _{m, (u, v)}(x, y). \end{aligned}$$

  (E3)

  By Cauchy-Schwarz inequality,

  $$\begin{aligned} (\text {E3})&\le Lb_n^{-1}\left[ O\left( m^{-1/2}\right) +O(m^{-1})\right] \\&=O\left( b_n^{-1}m^{-1/2}\right) . \end{aligned}$$

• For $A_{22}$, since $0\le \dot{C}_1\le 1$ and using the result of $A_{21}$,

  $$\begin{aligned} A_{22}&\le \left| \int _0^1\bigg \{\int _{[0, 1]^2} \frac{2}{\varepsilon _n}\max (|x'-u|, |y-v|)\right. \\&\quad \left. {\textrm{d}}\nu _{m, (u, v)}(x, y) \bigg \} \textrm{d}t\right| \\&\le \left| \int _0^1\bigg \{\int _{[0, 1]^2}\frac{2}{\varepsilon _n}(|x'-u|+|y-v|)\right. \\&\quad \left. {\textrm{d}}\nu _{m, (u, v)}(x, y) \bigg \} \textrm{d}t\right| \\&\le \varepsilon _n^{-1}\left[ O\left( m^{-1/2}\right) +O(m^{-1})\right] \\&=O\left( b_n^{-1}m^{-1/2}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned}&\sup _{\begin{array}{c} v\in [0, 1],\\ u\in [b_n, 1-b_n] \end{array}}\left| \frac{\partial \widehat{C}_{n,m}(u, v)}{\partial u}-\dot{C}_1(u, v)\right| \\&\quad =O\left( m^{1/2}n^{-1/2}(\log \log n)^{1/2}\right) +O\left( b_n^{-1}m^{-1/2}\right) , \end{aligned}$$

almost surely as $n\rightarrow \infty$, which completes the proof. $\square$