A moment-based test for extreme-value dependence

Abstract

This paper proposes a new rank-based test of extreme-value dependence. The procedure is based on the first three moments of the bivariate probability integral transform of the underlying copula. It is seen that the test statistic is asymptotically normal and its finite- and large-sample variance are calculated explicitly. Consistent plug-in estimators for the variance are proposed, and a fast algorithm for their computation is given. Although it is shown via counterexamples that no test based on the probability integral transform can be consistent, the proposed procedure achieves good power against common alternatives, both in finite samples and asymptotically.

Appendices

Appendix A

 

Proof of Proposition 1

Since $\mathrm{E}(I_{ij})=\mu _1$, $\mathrm{E}(I_{ij}I_{kj})=\mu _2$ as well as $\mathrm{E}(I_{ij}I_{kj}I_{lj})=\mu _3$, it is immediate that $\mathrm{E}(S_{3n}+1)=4\mu _1+9\mu _2-16\mu _3$. Hence,

$$\begin{aligned} \mathrm{var}(S_{3n})=\mathrm{var}(S_{3n}+1)&= \mathrm{E}(S_n+1)^2-16\mu _{11}-72\mu _{12}+128\mu _{13} \\&-81\mu _{22}+288\mu _{23}-256\mu _{33}. \end{aligned}$$

Next,

$$\begin{aligned} (S_n+1)^2&= \left\{ \frac{4}{n(n-1)}\sum _{i\ne j}I_{ij}+\frac{9}{n(n-1)(n-2)}\sum _{i\ne j\ne k}I_{ij}I_{kj}-\frac{16}{n(n-1)(n-2)(n-3)}\sum _{i\ne j\ne k\ne \ell }I_{ij}I_{kj}I_{\ell j}\right\} ^2 \end{aligned}$$

can be expanded as follows:

$$\begin{aligned}&\frac{16}{n^2(n-1)^2}\left(\sum _{i\ne j}I_{ij}\right)\left(\sum _{k\ne \ell }I_{k\ell }\right)\\&\quad + \frac{72}{n^2(n-1)^2(n-2)}\left(\sum _{i\ne j\ne k}I_{ij}I_{kj}\right)\left(\sum _{\ell \ne m}I_{\ell m}\right)\\&\quad + \frac{81}{n^2(n-1)^2(n-2)^2}\left(\sum _{i\ne j\ne k}I_{ij}I_{kj}\right)\left(\sum _{p\ne q\ne r}I_{pq}I_{rq}\right)\\&\quad -\frac{128}{n^2(n-1)^2(n-2)(n-3)}\left(\sum _{i\ne j\ne k\ne \ell }I_{ij}I_{kj}I_{\ell j}\right)\left(\sum _{p\ne q}I_{pq}\right)\\&\quad -\frac{288}{n^2(n-1)^2(n-2)^2(n-3)}\left(\sum _{i\ne j\ne k\ne \ell }I_{ij}I_{kj}I_{\ell j}\right)\left(\sum _{p\ne q\ne r}I_{pq}I_{rq}\right)\\&\quad + \frac{256}{n^2(n-1)^2(n-2)^2(n-3)^2}\left(\sum _{i\ne j\ne k\ne \ell }I_{ij}I_{kj}I_{\ell j}\right)\left(\sum _{p\ne q\ne r\ne s}I_{pq}I_{rq}I_{sq}\right). \end{aligned}$$

Expectations of the first three terms are computed in Appendix A of Ben Ghorbal et al. (2009) and lead to the terms involving $A_n$, $B_n$ and $C_n$. Expectations of the remaining terms can be calculated analogously by multiplying out the sums, distinguishing cases according to the number of distinct indices and taking expectations. For example, applying these steps to the fourth term leads to Table 6. Tallying the terms listed therein, one finds

$$\begin{aligned} \mathrm{E}\left\{ \frac{128}{n^2(n-1)^2(n-2)(n-3)}\left(\sum _{i\ne j\ne k\ne \ell }I_{ij}I_{kj}I_{\ell j}\right)\left(\sum _{p\ne q}I_{pq}\right) \right\} = \frac{128 D_n}{n(n-1)}. \end{aligned}$$

Table 6 Twenty one terms involved in the calculation of $D_n$

The same strategy leads to the expectations of the remaining two terms. There are 71 and 209 cases to be distinguished in the computation of the expectation of the fifth and the sixth term, respectively. A detailed list of all cases fills several pages and is available from the authors upon request. However, many indicator products are identical up to index relabeling and have the same expectation. For example,

$$\begin{aligned} \mathrm{E} (I_{ij}I_{kj}I_{\ell j}I_{pi}I_{ri}) = \mathrm{E}(I_{ij}I_{kj}I_{\ell j}I_{pk}I_{rk}) = \mathrm{E}(I_{ij}I_{kj}I_{\ell j}I_{p\ell }I_{r\ell }) = \theta _{12}. \end{aligned}$$

Tables 7 and 8 present a list of all parameters involved in the formula for $E_n$ and $F_n$ respectively, along with typical terms having these parameters as expectations. The number of terms leading to the same expectation is also given. Summing up, one finds

$$\begin{aligned}&\mathrm{E} \left\{ \frac{288}{n^2(n-1)^2(n-2)^2(n-3)}\left(\sum _{i\ne j\ne k\ne \ell }I_{ij}I_{kj}I_{\ell j}\right)\left(\sum _{p\ne q\ne r}I_{pq}I_{rq}\right) \right\} \\&\quad \quad = \frac{288 E_n}{n(n-1)(n-2)} \end{aligned}$$

as well as

$$\begin{aligned}&\mathrm{E} \left\{ \frac{256}{n^2(n-1)^2(n-2)^2(n-3)^2}\left(\sum _{i\ne j\ne k\ne \ell }I_{ij}I_{kj}I_{\ell j}\right)\left(\sum _{p\ne q\ne r\ne s}I_{pq}I_{rq}I_{sq}\right)\right\}&\quad \quad = \frac{256 F_n}{n(n-1)(n-2)(n-3)}. \end{aligned}$$

This completes the proof. $\square$

Table 7 Summary of the 73 terms involved in the calculation of $E_n$

Table 8 Summary of the 209 terms involved in the calculation of $F_n$

Proof of Proposition 2

From Proposition 1, one may write $n\cdot \text{ var}(S_{3n})=\alpha _n+\beta _n+\gamma _n$, where

$$\begin{aligned} \alpha _n&= \frac{16\mu _1}{n-1} + \frac{72(2\mu _2+2\theta _1)}{n-1} - \frac{128(3\mu _3+6\theta _4)}{n-1}\\&+ \frac{81\{ 2\mu _2+(n-3)(4\mu _3+8\theta _3+2\theta _8)\} }{(n-1)(n-2)}\\&- \frac{288\{ 6\mu _3+6\theta _3+(n-4)(6\mu _4+12\theta _6+6\theta _{11}+6\theta _{15})\} }{(n-1)(n-2)}\\&+\frac{256\{ 6\mu _3+(n-4)(18\mu _4+36\theta _{11}+6\theta _{19})\}}{(n-1)(n-2)(n-3)} \\&+\frac{256 (n-4)(n-5)(9\mu _5+36\theta _{14}+18\theta _{18}) }{(n-1)(n-2)(n-3)}, \end{aligned}$$

as well as

$$\begin{aligned} \beta _n&= \frac{16(n-2)(\mu _2+2\theta _1+\theta _2)}{n-1} + \frac{72(n-3)(\mu _3+\theta _3+2\theta _4+2\theta _5)}{n-1} \\&+\frac{81(n-3)(n-4)(\mu _4+4\theta _6+4\theta _7)}{(n-1)(n-2)} - \frac{128(n-4)(\mu _4+3\theta _9+3\theta _{10}+\theta _{11})}{n-1}\\&- \frac{288(n-4)(n-5)(\mu _5+3\theta _{12}+6\theta _{13}+2\theta _{14})}{(n-1)(n-2)}\\&+ \frac{256(n-4)(n-5)(n-6)(\mu _6+9\theta _{16}+6\theta _{17})}{(n-1)(n-2)(n-3)} \end{aligned}$$

and

$$\begin{aligned} \gamma _n&= 16\, \mu _{11}\frac{(n-2)(n-3)-n(n-1)}{n-1} + 72\, \mu _{12}\frac{(n-3)(n-4)-n(n-1)}{n-1} \\&+ 81 \, \mu _{22}\frac{(n-3)(n-4)(n-5)-n(n-1)(n-2)}{(n-1)(n-2)} \\&- 128\, \mu _{13}\frac{(n-4)(n-5)-n(n-1)}{n-1}\\&- 288\, \mu _{23}\frac{(n-4)(n-5)(n-6)-n(n-1)(n-2)}{(n-1)(n-2)} \\&+ 256\, \mu _{33}\frac{(n-4)(n-5)(n-6)(n-7)-n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)} \text{.} \end{aligned}$$

It is easy to see that as $n\rightarrow \infty$, $\alpha _n \rightarrow 0$ while

$$\begin{aligned} \beta _n&\rightarrow 16(\mu _2+2\theta _1+\theta _2) + 72(\mu _3+\theta _3+2\theta _4+2\theta _5) \\&+81(\mu _4+4\theta _6+4\theta _7) -128(\mu _4+3\theta _9+3\theta _{10}+\theta _{11}) \\&-288(\mu _5+3\theta _{12}+6\theta _{13}+2\theta _{14}) +256(\mu _6+9\theta _{16}+6\theta _{17}) \end{aligned}$$

and

$$\begin{aligned} \gamma _n&\rightarrow -16(4\mu _{11})-72(6\mu _{12})-81(9\mu _{22})+128(8\mu _{13})\\&+288(12\mu _{23})-256(16\mu _{33}). \end{aligned}$$

Collecting the terms yields the result. $\square$

Appendix B: An algorithm for computing the plug-in estimators of the finite and large-sample variance of $S_{3n}$

Let $(I_{ij})$ be an $n\times n$ matrix with entries $I_{ij}=\mathbf{1}(X_i\le X_j,Y_i\le Y_j)$. Let also $J$ be the $n\times n$ identity matrix and write $\mathbf{1}$ for an $n\times 1$ vector of 1’s. Given arbitrary $k\times \ell$ matrices $A=(A_{ij})$ and $B=(B_{ij})$, denote the sum of all elements of $A$ by

$$\begin{aligned} \Sigma {(A)}=\sum _{i=1}^k\sum _{j=1}^\ell {A_{ij}} \end{aligned}$$

and the Hadamard (or entrywise) product of $A$ and $B$ by $A\star B$.

To compute the plug-in estimators of the finite- and large-sample variance of $S_{3n}$, respectively, proceed as follows.

• Step 0. Compute $M=(I_{ij})-J$, $A=M\mathbf{1}$, $B=M^\top \mathbf{1}$, $N=M^\top M$, $P=MM^\top -\mathrm{diag}(MM^\top )$, $Q=N-\mathrm{diag}(N)$.

• Step 1. For $k\in \{1, \ldots ,6\}$, set $b_k=\Sigma {(B_k)}$, where $B_k=B\star \cdots \star B$ is the $k$-fold Hadamard product of $B$. Let also $a_2=\Sigma {(A\star A)}$, $c=\Sigma {(A\star B_2)}$, $d=\Sigma {(A\star B_3)}$. Also, for $i\in \{2,3\}$, compute $N_i=N\star \cdots \star N$ as the $i$-fold Hadamard product of $N$.

• Step 2. Set $m_1=b_1$ and compute $t_1=\Sigma {(A\star B)}$, $t_2=a_2-b_1$.

• Step 3. Compute sequentially

  $$\begin{aligned} \begin{array}{ll} m_2&= b_2-m_1,\\ m_3&= b_3-m_1-3m_2,\\ m_4&= b_4-m_1-7m_2-6m_3,\\ m_{11}&= m_1^2-b_2-2t_1-t_2,\quad \end{array}\quad \begin{array}{ll} t_3&= c-t_1,\\ t_4&= B^\top MB-2t_1,\\ t_5&= A^\top MB-b_2-t_1-t_2,\\ t_8&= \Sigma (PP^\top )-m_2. \end{array} \end{aligned}$$

• Step 4. Compute sequentially

  $$\begin{aligned} t_6&= B_2^\top { MB} - 2c- t_3 -t_4,\\ t_7&= ({ MB}-A)^\top { MB} + b_2-b_3-c-t_3-t_5-t_8,\\ m_{12}&= -2(A+B)^\top { MB} + 2a_2-2b_1-(b_1)^2+3b_2+b_1b_2-b_3+4t_1-t_3,\\ m_{22}&= (2A+2B-4B_2-2{ MB})^\top { MB} - 2b_2 - b_1b_2 + (b_2)^2 + 3b_3-b_4\\&+ 7c - m_{12} - 3t_1 - 2t_7. \end{aligned}$$

• Step 5. Compute sequentially

  $$\begin{aligned} m_5&= b_5-10m_4-25m_3-15m_2-m_1,\\ m_6&= b_6-15m_5-65m_4-90m_3-31m_2-m_1,\\ t_9&= B^\top M(B_2-3B)+2t_1-2t_4,\\ t_{10}&= A^\top M(B_2-3B)+2a_2-m_3-2t_4,\\ t_{11}&= d-c-2t_3,\\ t_{12}&= B_2^\top M(B_2-3B)+2c-2t_3-2t_4-4t_6-t_9,\\ t_{15}&= \Sigma {(B^\top N_2)}-B^\top \mathrm{diag}(N_2)-A^\top { MB} + b_2 -2t_8-t_3, \\ t_{13}&= ({ MB}_2-{ MB}-B_2)^\top { MB}+b_3-b_4-3t_3-t_4-2t_5-t_6\\&-2t_7-2t_8-t_{10}-t_{11}-2t_{15},\\ t_{14}&= (B_3-3B_2+2B)^\top { MB}-4t_{11},\\ t_{17}&= (B_3-3B_2+2B)^\top { MB}_2-16t_{11}-9t_{14},\\ t_{19}&= \Sigma {N_3}-\Sigma (N_2)-2t_8,\\ t_{18}&= B^\top N_2B-B_2^\top \mathrm{diag}(N_2)-({ MB})^\top { MB} \\&+b_3-4t_3-4t_8-2t_{11}-4t_{15}-t_{19},\\ t_{16}&= ({ MB}_2-3{ MB}+2A)^\top { MB}_2-2(B_3-{ MB})^\top { MB} \\&-2(B^\top N_2B-B_2^\top \mathrm{diag}(N_2))+4d-4b_3\\&+3b_4-b_5+8t_3+8t_8-t_{10}-3t_{11}-3t_{13} -2t_{14}+2t_{15}-2t_{18},\\ m_{13}&= b_1(2b_1-3b_2+b_3)-2b_2+3b_3-b_4-6t_4-3t_9-3t_{10}-t_{11},\\ m_{23}&= b_2(2b_1-3b_2+b_3)-2b_3+3b_4-b_5-6c-3B_2^\top M(B_2-3B)\\&-3t_{10}-7t_{11}-6t_{13}-2t_{14} -6t_{15}-m_{13},\\ m_{33}&= (9MB-3MB_2-6A-4B_3+3B_2-2B)^\top { MB}_2+15B_3^\top { MB}\\&+b_3(2b_1-3b_2+b_3)+6b_3-11b_4+6b_5-b_6-18d-9t_{13}-9t_{14}\\&-6t_{16}-2t_{17}-6t_{18}-m_{13}-3m_{23}. \end{aligned}$$

• Step 6. Set $n_1=n_2=3$, $n_3=n_4=n_5=n_8=4$, $n_6=n_7=n_9=n_{10}=n_{11}=n_{15}=n_{19}=5$, $n_{12}=n_{13}=n_{14}=n_{18}=6$, and $n_{16}=n_{17}=7$. Then compute

  $$\begin{aligned} \hat{\mu }_i=m_i\cdot (n)_{i+1},\quad \hat{\mu }_{ij}=m_{ij}\cdot (n)_{i+j+2},\quad \hat{\theta }_{i}=t_{i}\cdot (n)_{n_i}, \end{aligned}$$

  where $(n)_i$ represents the falling factorial $n(n-1)(n-2)\cdots (n-i+1)$.