Adapting the Hill estimator to distributed inference: dealing with the bias

Abstract

The distributed Hill estimator is a divide-and-conquer algorithm for estimating the extreme value index when data are stored in multiple machines. In applications, estimates based on the distributed Hill estimator can be sensitive to the choice of the number of the exceedance ratios used in each machine. Even when choosing the number at a low level, a high asymptotic bias may arise. We overcome this potential drawback by designing a bias correction procedure for the distributed Hill estimator, which adheres to the setup of distributed inference. The asymptotically unbiased distributed estimator we obtained, on the one hand, is applicable to distributed stored data, on the other hand, inherits all known advantages of bias correction methods in extreme value statistics.

Appendix

1.1 Proofs

1.1.1 Preliminary

Lemma 1

Let $Y, Y_1,\dots , Y_n$ be i.i.d. Pareto (1) random variables with distribution function $1-1/y, \ y\ge 1$. Let $Y^{(1)} \ge \cdots \ge Y^{(n)}$ be the order statistics of $\left\{ Y_1,\dots ,Y_n \right\}$. Let f be a function such that $\text {Var} \left\{ f(Y) \right\} <\infty$. Then for any $k\ge 1$,

$$\begin{aligned} \frac{1}{k}\sum _{i=1}^k f\left( \frac{Y^{(i)}}{Y^{(k+1)}} \right) {\mathop {=}\limits ^{d}}\frac{1}{k}\sum _{i=1}^k f(Y_i^*), \end{aligned}$$

where $Y_1^*, Y_2^*,\ldots , Y_k^*$ are i.i.d. Pareto (1) random variables. Moreover,

$$\begin{aligned} \sqrt{k} \Biggr\{ \frac{1}{k}\sum _{i=1}^k f\left( \frac{Y^{(i)}}{Y^{(k+1)}} \right) -\mathbb {E} f(Y) \Biggl\} \end{aligned}$$

is independent of $Y^{(k+1)}$ and asymptotically normally distributed with mean zero and variance $\text {Var} \left\{ f(Y) \right\}$ as $n \rightarrow \infty$, provided that $k=k(n) \rightarrow \infty$ and $k/n \rightarrow 0$.

Proof of Lemma 1

This Lemma follows directly from Lemma 3.2.3 in de Haan and Ferreira (2006) with the fact that $\log Y$ follows a standard exponential distribution. $\square$

Lemma 2

Let $Y_1,\dots ,Y_n$ be i.i.d. Pareto (1) random variables and $Y^{(1)}\ge \cdots \ge Y^{(n)}$ be the order statistics of $\left\{ Y_1,\dots ,Y_n \right\}$. Then for any $\rho <0$,

$$\begin{aligned} \mathbb {E}\Biggl\{ \left( \frac{k}{n}Y^{(k+1)} \right) ^{\rho } \Biggr\} =g(k,n,\rho ), \end{aligned}$$

where $g(k,n,\rho )$ is defined in (3). Moreover, if k is a fixed integer, then $g(k,n,\rho ) \rightarrow k^{\rho }\Gamma (k-\rho +1)/\Gamma (k+1)$ as $n \rightarrow \infty$. If k is an intermediate sequence, i.e. $k\rightarrow \infty , k/n \rightarrow 0$ as $n \rightarrow \infty$, then,

$$\begin{aligned} g\left(k,n,\rho \right)=1+\frac{1}{2}\left(\rho ^2-\rho \right)k^{-1}-\frac{1}{2}\left(\rho ^2-\rho \right)\left(n-\rho \right)^{-1}+O\left(k^{-2}\right). \end{aligned}$$

Proof of Lemma 2

$$\begin{aligned} \mathbb {E}\Biggl\{ \left( \frac{k}{n}Y^{(k+1)} \right) ^{\rho } \Biggr\}&= \frac{n!}{(n-k-1)!k!} \int _{1}^{\infty } \left( 1-\frac{1}{y} \right) ^{n-k-1}\left( \frac{1}{y} \right) ^{k+2} \left( \frac{k}{n}y \right) ^{\rho }dy \\&= \left( \frac{k}{n} \right) ^{\rho } \frac{n!}{(n-k-1)!k!}\int _{1}^{\infty } \left( 1-\frac{1}{y} \right) ^{n-k-1}\left( \frac{1}{y} \right) ^{k+2-\rho } dy \\&= \left( \frac{k}{n} \right) ^{\rho } \frac{\Gamma (n+1)\Gamma (k-\rho +1)}{\Gamma (n-\rho +1)\Gamma (k+1)} \\&=g(k,n,\rho ). \end{aligned}$$

We first handle the case when k is a fixed integer. By the Stirling’s formula,

$$\begin{aligned} \Gamma (x)=\sqrt{2\pi \left(x-1\right)} \Bigl \{ e^{-1}\left(x-1\right) \Bigr \} ^{x-1}\left\{ 1+\left(x-1\right)^{-1}/12+O\left(1/x^2\right) \right\} \end{aligned}$$

as $x\rightarrow \infty$, we have that, as $n \rightarrow \infty$,

$$\begin{aligned} \begin{aligned} \Gamma (n+1)\sim (2\pi n)^{1/2}\left( \frac{n}{e} \right) ^{n}, \quad \Gamma (n-\rho +1)\sim \left\{ 2\pi (n-\rho ) \right\} ^{1/2}\left( \frac{n-\rho }{\rho } \right) ^{n-\rho }, \end{aligned} \end{aligned}$$

which leads to

$$\begin{aligned} g(k,n,\rho )\rightarrow k^{\rho }\frac{\Gamma (k-\rho +1)}{\Gamma (k+1)}. \end{aligned}$$

Next, we handle the case when k is an intermediate sequence. By the Stirling’s formula, we have that, as $n \rightarrow \infty$,

$$\begin{aligned} \begin{aligned} g(k,n,\rho )&=\left( 1-\frac{\rho }{k} \right) ^{k-\rho +1/2}\left( 1+\frac{\rho }{n-\rho } \right) ^{n-\rho +1/2} \frac{1+n^{-1}/12+O(n^{-2})}{1+(n-\rho )^{-1}/12+O(n^{-2})} \frac{1+(k-\rho )^{-1}/12+O(k^{-2})}{1+k^{-1}/12+O(k^{-2})}\\&=\left( 1-\frac{\rho }{k} \right) ^{k-\rho +1/2}\left( 1+\frac{\rho }{n-\rho } \right) ^{n-\rho +1/2}\left\{ 1+O(n^{-2}) \right\} \left\{ 1+O(k^{-2}) \right\} .\\ \end{aligned} \end{aligned}$$

By the Taylor’s formula and some direct calculation, we obtain that, as $n\rightarrow \infty$,

$$\begin{aligned} \left( 1-\frac{\rho }{k} \right) ^{k-\rho +1/2} = e^{-\rho }\left\{ 1+\frac{1}{2}(\rho ^2-\rho )k^{-1}+O\left(k^{-2}\right) \right\} , \end{aligned}$$

and

$$\begin{aligned} \left( 1+\frac{\rho }{n-\rho } \right) ^{n-\rho +1/2} =e^{\rho }\left\{ 1-\frac{1}{2}\left(\rho ^2-\rho \right)\left(n-\rho \right)^{-1}+O\left(n^{-2}\right) \right\}. \end{aligned}$$

It follows that, as $n \rightarrow \infty$,

$$\begin{aligned} g(k,n,\rho )=1+\frac{1}{2}(\rho ^2-\rho )k^{-1}-\frac{1}{2}(\rho ^2-\rho )(n-\rho )^{-1}+O(k^{-2}). \end{aligned}$$

$\square$

Lemma 3

Let $Y_1,\dots ,Y_n$ be i.i.d. Pareto (1) random variables and $Y^{(1)}\ge \cdots \ge Y^{(n)}$ be the order statistics of $\left\{ Y_1,\dots ,Y_n \right\}$. Define for $\rho <0$,

$$\begin{aligned} Z_k=\frac{1}{k}\sum _{i=1}^k \frac{\left( Y^{(i)}/Y^{(k+1)} \right) ^{\rho }-1}{\rho }. \end{aligned}$$

Then, the following results hold.

• (i) For fixed k, $\mathbb {E}(Z_k^a)<\infty$, for $a=1,2,3,4$. Moreover, $\mathbb {E}\left( Z_k^2 \right) -\left\{ \mathbb {E}\left( Z_k \right) \right\} ^2>0$.

• (ii) For intermediate k, i.e., $k = k(n)\rightarrow \infty , k/n \rightarrow 0$ as $n\rightarrow \infty$, and $a=1,2,3,4$,

  $$\begin{aligned} \mathbb {E}\left( Z_k^a \right) = \frac{1}{(1-\rho )^a}\left\{ 1+\frac{a(a-1)}{2(1-2\rho )}\frac{1}{k}+O\left(k^{-2}\right) \right\} . \end{aligned}$$

Proof of Lemma 3

By Lemma 1, we have that,

$$\begin{aligned} Z_k {\mathop {=}\limits ^{d}} \frac{1}{k}\sum _{i=1}^k \frac{\left( Y_i^* \right) ^{\rho }-1}{\rho }, \end{aligned}$$

where $Y_1^*,\dots ,Y_k^*$ are i.i.d. Pareto (1) random variables. Denote $T_i=\left\{ (Y_i^*)^{\rho }-1 \right\} /\rho$, for $i=1,\dots ,k$ and $Z_k= k^{-1}\sum _{i=1}^k T_i$. Then, $T_i, i=1,\dots ,k$ follows the generalized Pareto distribution with the cumulative distribution function $F(t) = 1-(1+\rho t)^{-1/\rho }$. Thus, we have that for $a=1,2,3,4$,

$$\begin{aligned} \mathbb {E}(T_i^a)=\frac{a!}{(1-a\rho )\cdots (1-\rho )}. \end{aligned}$$

First, we handle the case when k is fixed. The result is obvious since $kZ_k$ is a finite sum of i.i.d. generalized Pareto random variables with shape parameter $\rho <0$.

Next, we handle the case when k is an intermediate sequence. For $a=1$, we have that, $E(Z_k)=E(T_i)=(1-\rho )^{-1}$.

For $a=2$, we have that,

$$\begin{aligned} \begin{aligned} \mathbb {E}\left( Z_k^2 \right)&= \frac{1}{k^2}\left\{ \sum _{i=1}^k E\left( T_i^2 \right) +\sum _{i\ne j}\mathbb {E} \left( T_i \right) \mathbb {E}\left( T_j \right) \right\}\\&=\frac{1}{k^2}\left[k E\left( T_i^2 \right) +k(k-1) \left\{ \mathbb {E} \left( T_i \right) \right\} ^2 \right] \\&= \frac{1}{(1-\rho )^2}+\frac{1}{k}\frac{1}{(1-2\rho )(1-\rho )^2}. \end{aligned} \end{aligned}$$

For $a=3$, we have that

$$\begin{aligned} \mathbb {E}\left( Z_k^3 \right)&= \frac{1}{k^2} \left \{ \sum _{i=1}^k \mathbb {E}\left( T_i^3 \right) +\sum _{i = j \ne l}\mathbb {E} \left( T_i T_j \right) \mathbb {E}\left( T_l \right) + \sum _{i\ne j \ne l}\mathbb {E} \left( T_i \right) \mathbb {E}\left( T_j \right) \mathbb {E}\left( T_l \right) \right\} \\&=\frac{1}{k^3}\left[k \mathbb {E}\left( T_i^3 \right) +3k(k-1) \mathbb {E} \left( T_i^2 \right) \mathbb {E}\left( T_i \right) +k(k-1)(k-2) \left\{ \mathbb {E}(T_i) \right\} ^3 \right]\\&= \frac{1}{(1-\rho )^3}+\frac{1}{k}\frac{3}{(1-2\rho )(1-\rho )^3}+O(k^{-2}). \end{aligned}$$

The term $\mathbb {E} \left( Z_k^4 \right)$ can be handled in a similar way as that for handling $\mathbb {E}\left( Z_k^3 \right)$. $\square$

Lemma 4

Assume that the distribution function F satisfies the third order condition (4). Then there exist two functions $A_0(t)\sim A(t)$ and $B_0(t)=O\left\{ B(t) \right\}$ as $t\rightarrow \infty$, such that for any $\delta >0$, there exists a $t_0 = t_0(\delta )>0$, for all $t\ge t_0$ and $tx\ge t_0$,

$$\begin{aligned} \left|\frac{\frac{\log U(tx)-\log U(t)-\gamma \log x}{A_0(t)}-\frac{x^{\rho }-1}{\rho }}{B_0(t)}-\frac{x^{\rho +\tilde{\rho }}-1}{\rho +\tilde{\rho }} \right|\le \delta x^{\rho +\tilde{\rho }}\max (x^{\delta },x^{-\delta }). \end{aligned}$$

Proof of Lemma 4

This lemma follows from applying Theorem B.3.10 in de Haan and Ferreira (2006) to the function $f(t):=\log U(t)-\gamma \log t$. $\square$

1.2 Proofs for Section 3

Recall that $U=\left\{ 1/(1-F) \right\} ^{\leftarrow }$. Then $X{\mathop {=}\limits ^{d}}U(Y)$, where Y follows the Pareto (1) distribution. Since we have i.i.d. observations $\left\{ X_1,\dots ,X_N \right\}$, we can write $X_i{\mathop {=}\limits ^{d}}U(Y_i)$, where $\left\{ Y_1,\dots ,Y_N \right\}$ is a random sample of Y. Recall that the N observations are stored in m machines with n observations each. For machine j, let $Y_j^{(1)} \ge \cdots \ge Y_j^{(n)}$ denote the order statistics of the n Pareto (1) distributed variables corresponding to the n observations in this machine. Then $M_j^{(i)}{\mathop {=}\limits ^{d}}U(Y_j^{(i)}), i=1,\dots ,n, j=1,\dots ,m$.

Proof of Proposition 1

We intend to replace t and tx in Lemma 4 by n/k and $Y_j^{(i)}, i=1,\dots ,k+1, j=1,\dots ,m$, respectively. For this purpose, we introduce the set

$$\begin{aligned} \mathcal {F}_{t_0}:=\left\{ Y_j^{(k+1)}\ge t_0, \ for \ all \ 1\le j\le m \right\} . \end{aligned}$$

By Lemma S.2 in the supplementary material of Chen et al. (2021), we have that for any $t_0>1$, if condition (2) holds, then $\lim _{N\rightarrow \infty }\mathbb {P}\left( \mathcal {F}_{t_0} \right) =1$. Then, we can apply the intended replacement to get that, as $N\rightarrow \infty$,

$$\begin{aligned} \begin{aligned} \log U\left(Y_j^{(i)}\right) -\log U(n/k)&=-\gamma \log \left( kY_j^{(i)}/n \right) -A_0(n/k)\left\{ \left( kY_j^{(i)}/n \right) ^{\rho }-1 \right\} /\rho \\&\quad + A_0(n/k)B_0(n/k)\left\{ \left( kY_j^{(i)}/n \right) ^{\rho +\tilde{\rho }}-1 \right\} /\left( \rho +\tilde{\rho } \right) \\&\quad +o_P(1)A_0(n/k)B_0(n/k)\left( kY_j^{(i)}/n \right) ^{\rho +\tilde{\rho } \pm \delta }, \end{aligned} \end{aligned}$$

(11)

where the $o_P(1)$ term is uniform for all $1\le i\le k+1$ and $1\le j\le m$. By applying (11) twice for a general i and $i=k+1$ and the inequality $x^{\rho \pm \delta }/y^{\rho \pm \delta }\le (x/y)^{\rho \pm \delta }$ for any $x,y>0$, we get that as $N \rightarrow \infty$,

$$\begin{aligned} \begin{aligned}\log U\left( Y_{j}^{(i)}\right) -\log U\left( Y_{j}^{(k+1)}\right) =& \ \gamma \left( \log Y_{j}^{(i)}-\log Y_{j}^{(k+1)}\right) \\&+ A_{0}(n / k) \left( k Y_{j}^{(k+1)} / n\right) ^{\rho } \left\{ \left( Y_{j}^{(i)} / Y_{j}^{(k+1)} \right) ^{\rho }-1 \right\} /{\rho }\\& + A_{0}(n / k) B_{0}(n / k)\left( k Y_{j}^{(k+1)} / n\right) ^{\rho +\tilde{\rho }}\left\{ \left( Y_{j}^{(i)} / Y_{j}^{(k+1)} \right) ^{\rho +\tilde{\rho }}-1 \right\} /\left( \rho +\tilde{\rho } \right) \\& + o_P(1)A_{0}(n / k)B_0(n/k) \left( k Y_{j}^{(k+1)}/n\right) ^{\rho +\tilde{\rho } \pm \delta }\left\{ \left( Y_{j}^{(i)}/Y_j^{(k+1)}\right) ^{\rho +\tilde{\rho } \pm \delta }+1 \right\} . \end{aligned} \end{aligned}$$

(12)

By taking the average across i and j, we obtain that

$$\begin{aligned} \begin{aligned}\sqrt{km}\left( R_k^{(1)}-\gamma \right) = & \ \gamma \sqrt{km}\frac{1}{m}\frac{1}{k}\sum _{j=1}^m \sum _{i=1}^k \left\{ \log \left( Y_j^{(i)}/Y_j^{(k+1)} \right) -\gamma \right\} \\& + \sqrt{km}A_0(n/k)\frac{1}{m}\sum _{j=1}^m \left( k Y_{j}^{(k+1)} / n\right) ^{\rho } \rho ^{-1} \frac{1}{k}\sum _{i=1}^k \left\{ \left( Y_{j}^{(i)} / Y_{j}^{(k+1)} \right) ^{\rho }-1 \right\} \\& + \sqrt{km}A_0(n/k)B_0(n/k)\frac{1}{m}\sum _{j=1}^m \left( k Y_{j}^{(k+1)} / n\right) ^{\rho +\tilde{\rho }} (\rho +\tilde{\rho })^{-1}\frac{1}{k}\sum _{i=1}^k \left\{ \left( Y_{j}^{(i)} / Y_{j}^{(k+1)} \right) ^{\rho +\tilde{\rho }}-1 \right\} \\& +o_P(1)\sqrt{km}A_{0}(n / k)B_0(n/k) \frac{1}{m}\sum _{j=1}^m \left( k Y_{j}^{(k+1)} / n\right) ^{\rho +\tilde{\rho }\pm \delta }\frac{1}{k}\sum _{i=1}^k \left\{ \left( Y_{j}^{(i)}/Y_j^{(k+1)}\right) ^{\rho +\tilde{\rho } \pm \delta }+1 \right\} \\ =:& \ I_1+I_2+I_3+I_4. \end{aligned} \end{aligned}$$

Firstly, we handle $I_1$. By Lemma 1, we have that,

$$\begin{aligned} I_1 {\mathop {=}\limits ^{d}} \gamma \sqrt{km}\left( \frac{1}{km}\sum _{j=1}^m\sum _{i=1}^k \log Y_{i}^{j,*}-1 \right) , \end{aligned}$$

where $Y_{i}^{j,*}, i=1,\dots ,k,j=1,\dots ,m$ are independent and identically distributed Pareto (1) random variables. The central limit theorem yields that as $N \rightarrow \infty$, $I_1 = \gamma P_N^{(1)} +o_P(1),$ where $P_N^{(1)}\sim N(0,1)$.

For $I_2$, write $\delta _{j,n}=\left( k Y_{j}^{(k+1)} / n\right) ^{\rho }(k\rho )^{-1}\sum _{i=1}^k \left\{ \left( Y_{j}^{(i)} / Y_{j}^{(k+1)} \right) ^{\rho }-1 \right\}$. Then we have that $I_2=\sqrt{km}A_0(n/k)m^{-1}\sum _{j=1}^m \delta _{j,n}$, where $\delta _{j,n}, j=1,\dots ,m$ are i.i.d. random variables.

We are going to show that, as $N \rightarrow \infty$,

$$\begin{aligned} \sqrt{km} \left\{ \frac{1}{m}\sum _{j=1}^m \delta _{j,n} -\mathbb {E}\left( \delta _{j,n} \right) \right\} =O_P(1). \end{aligned}$$

(13)

If k is fixed, (13) follows directly from Lemma 3 (i) and the Lyapunov central limit theorem for triangular array.

Next, we handle the case when k is an intermediate sequence. In this case, in order to apply the Lyapunov central limit theorem with 4-th moment, we need to calculate $\text {Var}\left( \delta _{j,n} \right)$ and $\mathbb {E}[\left\{ \delta _{j,n}-\mathbb {E}\left( \delta _{j,n} \right) \right\} ^4 ]$. Denote $m_{n}^{(a)}:=\mathbb {E}\left\{ \left( \delta _{j,n} \right) ^a \right\} ,\ a=1,2,3,4$. By Lemma 1, we have that,

$$\begin{aligned} m_n^{(a)}= g(k,n,a\rho )\mathbb {E} \left[\left\{ \frac{1}{k}\sum _{i=1}^k\frac{\left( Y_{j}^{(i)} / Y_{j}^{(k+1)} \right) ^{\rho }-1}{\rho } \right\} ^a \right]. \end{aligned}$$

First, we calculate $\text {Var}\left( \delta _{j,n} \right)$. By Lemma 3, we have that,

$$\begin{aligned} \begin{aligned} \text {Var}(\delta _{j,n})&=m_{n}^{(2)}-\left( m_{n}^{(1)} \right) ^2 \\&= g(k,n,2\rho )\left\{ \frac{1}{(1-\rho )^2}+\frac{1}{k}\frac{1}{(1-2\rho )(1-\rho )^2}+O(k^{-2}) \right\} -\left\{ g(k,n,\rho ) \right\} ^2\left\{ \frac{1}{(1-\rho )^2}+O(k^{-2}) \right\} \\&= \frac{1}{k} g(k,n,2\rho )\frac{1}{(1-2\rho )(1-\rho )^2}+\left[g(k,n,2\rho )-\left\{ g(k,n,\rho ) \right\} ^2 \right]\frac{1}{(1-\rho )^2}+O(k^{-2}), \end{aligned} \end{aligned}$$

here in the last step, we used the fact that as $n\rightarrow \infty$, $g(k,n,\rho )\rightarrow 1$ and $g(k,n,2\rho )\rightarrow 1$. By Lemma 2, we have that, as $n \rightarrow \infty$,

$$\begin{aligned} g(k,n,2\rho )-\left\{ g(k,n,\rho ) \right\} ^2&=1+\frac{1}{2}\left( 4\rho ^2-2\rho \right) \frac{1}{k} +o(k^{-1})-\left\{ 1+\frac{1}{2}\left( \rho ^2-\rho \right) \frac{1}{k}+o(k^{-1}) \right\} ^2\\&=\frac{1}{k}\rho ^2+o(k^{-1}). \end{aligned}$$

Hence, as $n\rightarrow \infty$, $\text {Var}\left( \delta _{j,n} \right) =k^{-1}(1-\rho )^{-2}\left( \left( 1-2\rho \right) ^{-1}+\rho ^2 \right) +o(k^{-1})$.

Next, we calculate $\mathbb {E}[\left\{ \delta _{j,n}-\mathbb {E}\left( \delta _{j,n} \right) \right\} ^4 ]$. By Lemma 2 and Lemma 3, we have that, for $a=3,4$, as $N\rightarrow \infty$,

$$\begin{aligned} \begin{aligned} m_{n}^{(a)}&=\left(1-\rho \right)^{-a}\left\{ 1+\frac{1}{2}\frac{1}{k}\frac{a(a-1)}{1-2\rho }+O\left(k^{-2}\right) \right\} \left\{ 1+\frac{1}{2}\left(a^2\rho ^2-a\rho \right)k^{-1}-\frac{1}{2}\left(a^2\rho ^2-a\rho \right)\left(n-a\rho \right)^{-1}+O(k^{-2}) \right\} \\&=(1-\rho )^{-a}\left\{ 1+k^{-1}\frac{1}{2}\frac{a(a-1)}{1-2\rho }+\frac{1}{2}\left(a^2\rho ^2-a\rho \right)k^{-1}-\frac{1}{2}\left(a^2\rho ^2-a\rho \right)\left(n-a\rho \right)^{-1}+O\left(k^{-2}\right) \right\} . \end{aligned} \end{aligned}$$

Note that,

$$\begin{aligned} \mathbb {E}\left[\left\{ (\delta _{j,n}-\mathbb {E}\left( \delta _{j,n} \right) \right\} ^4 \right]=m_n^{(4)}-4m_n^{(3)}m_n^{(1)}+6m_n^{(2)}\left( m_n^{(1)} \right) ^2-3\left( m_n^{(1)} \right) ^4. \end{aligned}$$

By some direct calculation, all terms of order $k^{-1}$ and $n^{-1}$ are cancelled out. Thus, as $N \rightarrow \infty$, $\mathbb {E}[\left\{ (\delta _{j,n}-\mathbb {E}\left( \delta _{j,n} \right) \right\} ^4 ] =O(k^{-2})$. Combining $\text {Var}(\delta _{j,n})$ and $\mathbb {E}[\left\{ \delta _{j,n}-\mathbb {E}\left( \delta _{j,n} \right) \right\} ^4 ]$, we conclude that the sequences $\left\{ \delta _{j,n} \right\} _{j=1}^m$ satisfy the Lyapunov’s condition. Then, (13) follows by the central limit theorem. Applying (13), we obtain that, as $N\rightarrow \infty$,

$$\begin{aligned} I_2 = \sqrt{km}A_0(n/k) \left\{ \mathbb {E}\left( \delta _{j,n} \right) +O_P\left(1/\sqrt{km}\right) \right\} =\frac{g(k,n,\rho )}{1-\rho }\sqrt{km}A_0(n/k)+o_P(1). \end{aligned}$$

For $I_3$, by using the weak law of large numbers for triangular array, we have that, as $N \rightarrow \infty$,

$$\begin{aligned} \begin{aligned} I_3&= \frac{\sqrt{km}A_0(n/k)B_0(n/k)}{1-\rho -\tilde{\rho }}\mathbb {E}\left\{ \left( kY_1^{(k+1)}/n \right) ^{\rho +\tilde{\rho }} \right\} \left\{ 1+o_P(1) \right\} \\&=\sqrt{km}A_0(n/k)B_0(n/k)\frac{g(k,n,\rho +\tilde{\rho })}{1-\rho -\tilde{\rho } }+o_P(1), \end{aligned} \end{aligned}$$

where the last equality follows by the condition $\sqrt{km}A(n/k)B(n/k)=O(1)$.

For $I_4$, by similar arguments as for $I_3$, we obtain that, as $N\rightarrow \infty$, $I_4{\mathop {\rightarrow }\limits ^{P}} 0$. Combining $I_1,I_2,I_3$ and $I_4$, we have proved (i).

Next, we handle $R_k^{(2)}$. By (12), we obtain that, as $N\rightarrow \infty ,$

$$\begin{aligned} \begin{aligned}\sqrt{km}\left( R_k^{(2)}-2\gamma ^2 \right) =& \ \gamma ^2 \frac{1}{mk}\sum _{j=1}^m \sum _{i=1}^k \left\{ \log ^2 \left( Y_j^{(i)}/Y_j^{(k+1)} \right) -2 \right\} \\& + 2\gamma \sqrt{km}A_0(n/k)\frac{1}{km}\sum _{j=1}^m \left( kY_j^{(k+1)}/n \right) ^{\rho }\sum _{i=1}^k \log \left( Y_j^{(i)}/Y_j^{(k+1)} \right) \left\{ \left( Y_j^{(i)}/Y_j^{(k+1)} \right) ^{\rho }-1 \right\} /\rho \\& +\sqrt{km} A_0^2(n/k) \frac{1}{km}\sum _{j=1}^m \left( kY_j^{(k+1)}/n \right) ^{2\rho }\sum _{i=1}^k \left\{ \left( Y_j^{(i)}/Y_j^{(k+1)} \right) ^{\rho }-1 \right\} ^2/\rho ^2 \\& +2\gamma \sqrt{km}A_0(n/k) B_0(n/k) \frac{1}{km} \sum _{j=1}^m \left( kY_j^{(k+1)}/n \right) ^{\rho +\tilde{\rho }}\sum _{i=1}^k \log \left( Y_j^{(i)}/Y_j^{(k+1)} \right) \frac{\left( Y_j^{(i)}/Y_j^{(k+1)} \right) ^{\rho +\tilde{\rho }}-1}{\rho +\tilde{\rho }}\\&\quad + o_P(1)\\=:& \ I_5+I_6+I_7+I_8+o_P(1). \end{aligned} \end{aligned}$$

For $I_5$, by Lemma 1, we have that

$$\begin{aligned} I_5{\mathop {=}\limits ^{d}}\gamma ^2\sqrt{km}\left\{ \frac{1}{km}\sum _{j=1}^m\sum _{i=1}^k\left( \log Y_{i}^{j,*} \right) ^2-2 \right\} . \end{aligned}$$

The central limit theorem yields that as $N \rightarrow \infty$, $I_5 = \gamma ^2 P_N^{(2)}+o_P(1)$, where $P_N^{(2)}\sim N(0,20)$. In addition, the covariance of $P_N^{(1)}$ and $P_N^{(2)}$ is equal to the covariance of $\log Y_i^{j,*}$ and $\left( \log Y_i^{j,*} \right) ^2$, where $Y_{i}^{j,*}$ follows the Pareto (1) distribution. Hence, $\text {Cov}(P_N^{(1)},P_N^{(2)})=4.$

For $I_6$, we write $I_6 = 2\sqrt{km}A_0(n/k)m^{-1}\sum _{j=1}^m \eta _{j,n}$, where

$$\begin{aligned} \eta _{j,n}=\left( k Y_{j}^{(k+1)} / n\right) ^{\rho }(k\rho )^{-1}\sum _{i=1}^k \log \left( Y_{j}^{(i)} / Y_{j}^{(k+1)} \right) \left\{ \left( Y_{j}^{(i)} / Y_{j}^{(k+1)} \right) ^{\rho }-1 \right\} \end{aligned}$$

are i.i.d. random variables for $j=1,2,\dots ,m$. We can verify the Lyapunov’s condition for the series $\left\{ \eta _{j,n} \right\} _{j=1}^m$ following similar steps as those for $\left\{ \delta _{j,n} \right\} _{j=1}^m$. Then by applying the central limit theorem and Lemma 2, we obtain that

$$\begin{aligned} I_6=2\gamma \sqrt{km}A_0(n/k) g(k,n,\rho )\frac{1}{\rho }\left\{ \frac{1}{(1-\rho )^2}-1 \right\} +o_P(1). \end{aligned}$$

By the weak law of large numbers for triangular array, we have that

$$\begin{aligned} I_7=\sqrt{km}A_0^2(n/k)\frac{g(k,n,2\rho )}{\rho ^2}\left\{ \frac{1}{1-2\rho }-\frac{2}{1-\rho }+1 \right\} +o_P(1), \end{aligned}$$

and

$$\begin{aligned} I_8=2\gamma \sqrt{km}A_0(n/k)B_0(n/k)\frac{g(k,n,\rho +\tilde{\rho })}{\rho +\tilde{\rho }}\left\{ \frac{1}{(1-\rho -\tilde{\rho })^2}-1 \right\} +o_P(1). \end{aligned}$$

Combining the results for $I_5,I_6,I_7$ and $I_8$, we have proved (ii).

Finally, we handle $R_k^{(3)}$. Also, by (12), we have that

$$\begin{aligned}\sqrt{km}\left( R_k^{(3)}-6\gamma ^3 \right) =& \ \gamma ^3 \frac{1}{mk}\sum _{j=1}^m \sum _{i=1}^k \left\{ \log ^3 \left( Y_j^{(i)}/Y_j^{(k+1)} \right) -6 \right\} \\& + 3\gamma ^2\sqrt{km}A_0(n/k)\frac{1}{km}\sum _{j=1}^m \left( kY_j^{(k+1)}/n \right) ^{\rho }\\&\quad\sum _{i=1}^k \left\{ \log \left( Y_j^{(i)}/Y_j^{(k+1)} \right) \right\} ^2\frac{\left( Y_j^{(i)}/Y_j^{(k+1)} \right) ^{\rho }-1}{\rho } \\& + 3\gamma \sqrt{km}A_0^2(n/k)\frac{1}{km}\sum _{j=1}^m \left( kY_j^{(k+1)}/n \right) ^{2\rho }\\&\quad\sum _{i=1}^k \log \left( Y_j^{(i)}/Y_j^{(k+1)} \right) \left\{ \frac{\left( Y_j^{(i)}/Y_j^{(k+1)} \right) ^{\rho }-1}{\rho } \right\} ^2\\& +3\gamma ^2\sqrt{km}A_0(n/k) B_0(n/k) \frac{1}{km} \sum _{j=1}^m \left( kY_j^{(k+1)}/n \right) ^{\rho +\tilde{\rho }}\\&\quad\sum _{i=1}^k \left\{ \log \left( Y_j^{(i)}/Y_j^{(k+1)} \right) \right\} ^2\frac{\left( Y_j^{(i)}/Y_j^{(k+1)} \right) ^{\rho +\tilde{\rho }}-1}{\rho +\tilde{\rho }} +o_P(1)\\=:& \ I_9+I_{10}+I_{11}+I_{12}+o_P(1). \end{aligned}$$

By similar steps as for handling the four items $I_5, I_6,I_7$ and $I_8$, we can show that $I_9 = \gamma ^3 P_N^{(3)}+o_P(1)$, where $P_N^{(3)}\sim N(0,684)$ and $\text {Cov}(P_N^{(1)},P_N^{(3)})=18, \text {Cov}(P_N^{(2)},P_N^{(3)})=98$. And

$$\begin{aligned} \begin{aligned} I_{10}&=6\gamma ^2 \sqrt{km}A_0(n/k)\frac{g(k,n,\rho )}{\rho }\left\{ \frac{1}{(1-\rho )^3}-1 \right\} +o_P(1),\\ I_{11}&=3\gamma \sqrt{km} A^2_0(n/k) \frac{g(k,n,2\rho )}{\rho ^2}\left\{ \frac{1}{(1-2\rho )^2}-\frac{2}{(1-\rho )^2}+1 \right\} +o_P(1),\\ I_{12}&=6\gamma ^2\sqrt{km}A_0(n/k)B_0(n/k)\frac{g(k,n,\rho +\tilde{\rho })}{\rho +\tilde{\rho }}\left\{ \frac{1}{(1-\rho -\tilde{\rho })^3}-1 \right\} +o_P(1), \end{aligned} \end{aligned}$$

which yields (iii). $\square$

Proof of Theorem 1

Applying Proposition 1 with $k=k_{\rho }$, we have that, as $N \rightarrow \infty$,

$$\begin{aligned} R_{k_{\rho }}^{(1)}=&\ \gamma +\frac{\gamma }{\sqrt{k_{\rho }m}} P_N^{(1)} + \frac{g(k_{\rho },n,\rho )}{1-\rho }A_0(n/k_{\rho })+\frac{g(k_{\rho },n,\rho +\tilde{\rho })}{1-\rho -\tilde{\rho }}A_0(n/k_{\rho })B_0(n/k_{\rho })+\frac{1}{\sqrt{k_{\rho }m}}o_P(1),\\ R_{k_{\rho }}^{(2)}=& \ 2\gamma ^2+\frac{\gamma ^2}{\sqrt{k_{\rho }m}} P_N^{(2)} +2 \gamma A_0(n/k_{\rho })\frac{g(k_{\rho },n,\rho )}{\rho }\left\{ \frac{1}{(1-\rho )^2}-1 \right\} \\&+A_0^2(n/k_{\rho })\dfrac{g(k_{\rho },n,2\rho )}{\rho ^2}\left( \frac{1}{1-2\rho }-\frac{2}{1-\rho }+1 \right) \\&+2\gamma A_0(n/k_{\rho })B_0(n/k_{\rho })\frac{g(k_{\rho },n,\rho +\tilde{\rho })}{\rho +\tilde{\rho }}\left\{ \frac{1}{(1-\rho -\tilde{\rho })^2}-1 \right\} +\frac{1}{\sqrt{k_{\rho }m}}o_P(1),\\ R_{k_{\rho }}^{(3)}=& \ 6\gamma ^3+\frac{\gamma ^3}{\sqrt{k_{\rho }m}}P_N^{(3)}+6\gamma A_0(n/k_{\rho })\frac{g(k_{\rho },n,\rho )}{\rho }\left\{ \frac{1}{(1-\rho )^3}-1 \right\} \\& +3A_0^2(n/k_{\rho })\frac{g(k_{\rho },n,2\rho )}{\rho ^2}\left\{ \frac{1}{(1-2\rho )^2}-\frac{2}{(1-\rho )^2}+1 \right\} \\&+6\gamma A_0(n/k_{\rho })B_0(n/k_{\rho })\frac{g(k_{\rho },n,\rho +\tilde{\rho })}{\rho +\tilde{\rho }}\left\{ \frac{1}{(1-\rho -\tilde{\rho })^3}-1 \right\} +\frac{1}{\sqrt{k_{\rho }m}}o_P(1). \end{aligned}$$

As a consequence, we have that, as $N \rightarrow \infty$,

$$\begin{aligned} \begin{aligned} \left( R_{k_{\rho }}^{(1)} \right) ^{\tau }&=\gamma ^{\tau }\left\{ 1+ \frac{\tau }{\sqrt{k_{\rho }m}} P_N^{(1)} +\frac{\tau }{\gamma }\frac{g(k_{\rho },n,\rho )}{1-\rho } A_0(n/k_{\rho })+\frac{\tau }{\gamma }\frac{g(k_{\rho },n,\rho +\tilde{\rho })}{1-\rho -\tilde{\rho }}A_0(n/k_{\rho })B_0(n/k_{\rho }) \right\} \\&\quad +\frac{1}{\sqrt{k_{\rho }m}}o_P(1),\\ \left( R_{k_{\rho }}^{(2)}/2 \right) ^{\tau /2}&=\gamma ^{\tau }\Bigg [1+\frac{\tau }{\sqrt{k_{\rho }m}} P_N^{(2)} +\frac{\tau }{2\gamma }A_0(n/k_{\rho })\frac{g(k_{\rho },n,\rho )}{\rho }\left\{ \frac{1}{(1-\rho )^2}-1 \right\} \\&\quad + \frac{\tau }{4\gamma } A_0^2(n/k_{\rho })\frac{g(k_{\rho },n,2\rho )}{\rho ^2}\left( \frac{1}{1-2\rho }-\frac{2}{1-\rho }+1 \right) \\&\quad +\frac{\tau }{2\gamma }A_0(n/k_{\rho })B_0(n/k_{\rho })\frac{g(k_{\rho },n,\rho +\tilde{\rho })}{\rho +\tilde{\rho }}\left\{ \frac{1}{(1-\rho -\tilde{\rho })^2}-1 \right\} \Bigg ]+\frac{1}{\sqrt{k_{\rho }m}}o_P(1),\\ \left( R_{k_{\rho }}^{(3)}/6 \right) ^{\tau /3}&=\gamma ^{\tau }\Bigg [1+\frac{\tau }{\sqrt{k_{\rho }m}}P_N^{(3)}+\frac{\tau }{3\gamma } A_0(n/k_{\rho })\frac{g(k_{\rho },n,\rho )}{\rho }\left\{ \frac{1}{(1-\rho )^3}-1 \right\} \\&\quad +\frac{\tau }{6\gamma } A_0^2(n/k_{\rho })\frac{g(k_{\rho },n,2\rho )}{\rho ^2}\left\{ \frac{1}{(1-2\rho )^2}-\frac{2}{(1-\rho )^2}+1 \right\} \\&\quad +\frac{\tau }{3\gamma } A_0(n/k_{\rho })B_0(n/k_{\rho })\frac{g(k_{\rho },n,\rho +\tilde{\rho })}{\rho +\tilde{\rho }}\left\{ \frac{1}{(1-\rho -\tilde{\rho })^3}-1 \right\} \Bigg ]+\frac{1}{\sqrt{k_{\rho }m}}o_P(1). \end{aligned} \end{aligned}$$

It follows that, as $N\rightarrow \infty$,

$$\begin{aligned} \gamma ^{-\tau }\left\{ \left( R_{k_{\rho }}^{(1)} \right) ^{\tau } -\left( R_{k_{\rho }}^{(2)}/2 \right) ^{\tau /2} \right\}= & \ \frac{\tau }{\sqrt{k_{\rho }m}}\left( P_N^{(1)}-P_N^{(2)} \right) +\frac{\tau }{\gamma }g(k_{\rho },n,\rho )A_0(n/k_{\rho })\frac{-\rho }{2(1-\rho )^2} \\&+A_0^2\left( n/k_{\rho } \right) O(1)+A_0\left( n/k_{\rho } \right) B_0\left( n/k_{\rho } \right) O(1)+\frac{1}{\sqrt{k_{\rho }m}}o_P(1), \end{aligned}$$

and

$$\begin{aligned} \gamma ^{-\tau }\left\{ \left( R_{k_{\rho }}^{(2)}/2 \right) ^{\tau /2} -\left( R_{k_{\rho }}^{(2)}/6 \right) ^{\tau /3} \right\}= & \ \frac{\tau }{\sqrt{k_{\rho }m}}\left( P_N^{(2)}-P_N^{(3)} \right) +\frac{\tau }{\gamma }g(k_{\rho },n,\rho )A_0(n/k_{\rho })\frac{\rho (\rho -3)}{6(1-\rho )^3} \\&+A_0^2\left( n/k_{\rho } \right) O(1)+A_0\left( n/k_{\rho } \right) B_0\left( n/k_{\rho } \right) O(1)+\frac{1}{\sqrt{k_{\rho }m}}o_P(1). \end{aligned}$$

By the condition (5), the dominating terms in the two expressions above are

$$\begin{aligned} \frac{\tau }{\gamma }g(k_{\rho },n,\rho )A_0(n/k_{\rho })\frac{-\rho }{2(1-\rho )^2} \quad \text {and} \quad \frac{\tau }{\gamma }g(k_{\rho },n,\rho )A_0(n/k_{\rho })\frac{\rho (\rho -3)}{6(1-\rho )^3}, \end{aligned}$$

respectively. Therefore, as $N \rightarrow \infty$,

$$\begin{aligned} \begin{aligned} T_{k_{\rho },\tau }&=3\frac{\rho -1}{\rho -3} \left\{ 1+\frac{\gamma }{\sqrt{k_{\rho }m}}\frac{2(1-\rho )^2}{-\rho A_0(n/k_{\rho })}\left( P_N^{(1)}-P_N^{(2)} \right) -\frac{\gamma }{\sqrt{k_{\rho }m}A_0(n/k_{\rho })} \frac{6(1-\rho )^3}{\rho ^2-3\rho }\left( P_N^{(2)}-P_N^{(3)} \right) \right\} \\&\quad +O_P\left\{ A_0\left( n/k_{\rho } \right) \right\} +O_P\left\{ B_0\left( n/k_{\rho } \right) \right\} +\frac{1}{\sqrt{k_{\rho }m}A_0(n/k_{\rho })}o_P(1). \end{aligned} \end{aligned}$$

It follows that as $N\rightarrow \infty$,

$$\begin{aligned} \sqrt{k_{\rho }m}A_0(n/k_{\rho })\left( T_{k_{\rho },\tau }-3\frac{\rho -1}{\rho -3} \right) =& -\gamma \frac{2(1-\rho )^2}{\rho }\left( P_N^{(1)}-P_N^{(2)} \right) \\&-\gamma \frac{6(1-\rho )^3}{\rho ^2-3\rho }\left( P_N^{(2)}-P_N^{(3)} \right) +O_P(1). \end{aligned}$$

Theorem 1 is thus proved by applying the Cramér’s delta method. $\square$

Proof of Theorem 2

By Proposition 1, as $N \rightarrow \infty$, $R_{k_n}^{(1)}$ has the following asymptotic expansion:

$$\begin{aligned} \sqrt{k_nm}\left( R_{k_n}^{(1)}-\gamma \right) -\gamma P_N^{(1)}-\frac{g(k_n,n,\rho )}{1-\rho }\sqrt{k_nm}A_0(n/k_n)=o_P(1), \end{aligned}$$

which leads to

$$\begin{aligned} \sqrt{k_nm}\left\{ \left( R_{k_n}^{(1)} \right) ^2-\gamma ^2 \right\} -2\gamma ^2 P_N^{(1)}-2\gamma \frac{g(k_n,n,\rho )}{1-\rho }\sqrt{k_nm}A_0(n/k_n)=o_P(1). \end{aligned}$$

Together with the asymptotic expansion of $R_{k_n}^{(2)}$, we have that, as $N\rightarrow \infty$,

$$\begin{aligned} \sqrt{k_nm}\left\{ R_{k_n}^{(2)}-2\left( R_{k_n}^{(1)} \right) ^2 \right\} -\gamma ^2\left( P_N^{(2)}-4P_N^{(1)} \right) -\sqrt{k_nm}A_0(n/k_n) g(k_n,n,\rho )\frac{2\gamma \rho }{(1-\rho )^2}=o_P(1). \end{aligned}$$

Thus, as $N\rightarrow \infty$,

$$\begin{aligned}\sqrt{k_nm}\left( \tilde{\gamma }_{k_n,k_{\rho },\tau }-\gamma \right) =& \ \sqrt{k_nm}\left( R_{k_n}^{(1)}-\gamma \right) - \frac{1}{2R_{k_n}^{(1)}\hat{\rho }_{k_{\rho },\tau } (1-\hat{\rho }_{k_{\rho },\tau })^{-1}}\sqrt{k_n m}\left\{ R_{k_n}^{(2)}-2\left( R_{k_n}^{(1)} \right) ^2 \right\} \\ =& \ \gamma P_N^{(1)} +\sqrt{k_nm}A_0(n/k_n)\frac{g(k_n,n,\rho )}{1-\rho }+o_P(1)\\& -\frac{1}{2R_{k_n}^{(1)}\hat{\rho }_{k_{\rho },\tau } (1-\hat{\rho }_{k_{\rho },\tau })^{-1}}\left\{ \gamma ^2\left( P_N^{(2)}-4P_N^{(1)} \right) \right.\\&\left.+\sqrt{k_nm}A_0(n/k_n) g(k_n,n,\rho )\frac{2\gamma \rho }{(1-\rho )^2}+o_P(1) \right\} \\=& \ \gamma P_N^{(1)}-\frac{\gamma ^2(1-\hat{\rho }_{k_{\rho },\tau })}{R_{k_n}^{(1)}\hat{\rho }_{k_{\rho },\tau }}\left( P_N^{(2)}/2-2P_N^{(1)} \right) \\& +\sqrt{k_n m}A_0(n/k_n)\frac{\rho }{(1-\rho )^2}g(k_n,n,\rho ) \left( \frac{1-\rho }{\rho }-\frac{1-\hat{\rho }_{k_{\rho },\tau }}{\hat{\rho }_{k_{\rho },\tau }} \right) +o_P(1). \end{aligned}$$

The relation $k_n/k_{\rho }\rightarrow 0$ implies that $A(n/k_n)/A(n/k_{\rho })\rightarrow 0$ as $N \rightarrow \infty$. Thus, by Theorem 1, we have that, as $N \rightarrow \infty$,

$$\begin{aligned} \sqrt{k_n m}A_0(n/k_n)\frac{\rho }{(1-\rho )^2}g(k_n,n,\rho )\left( \frac{1-\rho }{\rho }-\frac{1-\hat{\rho }_{k_{\rho },\tau }}{\hat{\rho }_{k_{\rho },\tau }} \right) =o_P(1). \end{aligned}$$

Together with the consistency of $\hat{\rho }_{k_{\rho },\tau }$ and $R_{k_n}^{(1)}$, we have that, as $N \rightarrow \infty$,

$$\begin{aligned} \sqrt{k_nm}\left( \tilde{\gamma }_{k_n,k_{\rho },\tau }-\gamma \right) = \frac{\gamma }{\rho }\left\{ P_N^{(2)}(\rho -1)/2+P_N^{(1)}(2-\rho ) \right\} +o_P(1). \end{aligned}$$

Combining with Proposition 1, we obtain that, as $N\rightarrow \infty$,

$$\begin{aligned} \sqrt{k_nm}\left( \tilde{\gamma }_{k_n,k_{\rho },\tau }-\gamma \right) {\mathop {\rightarrow }\limits ^{d}} N\left[0,\gamma ^2 \left\{ 1+\left( \rho ^{-1}-1 \right) ^2 \right\} \right]. \end{aligned}$$

$\square$