Subsampling for Heavy Tailed, Nonstationary and Weakly Dependent Time Series

Abstract

We present new results on estimation of the periodic mean function of the periodically correlated time series that exhibits heavy tails and long memory under a weak dependence condition. In our model that is a generalization of the work of McElroy and Politis [35, 42] we show that the estimator of the periodic mean function has an asymptotic distribution that depends on the degree of heavy tails and the degree of the long memory. Such an asymptotic distribution clearly poses a problem while trying to build the confidence intervals. Thus the main point of this research is to establish the consistency of one of the resampling methods - the subsampling procedure - in the considered model. We obtain such consistency under relatively mild conditions on time series at hand. The selection of the block length plays an important role in the resmapling methodology. In the article we discuss as well one of the possible ways of selection the length the subsampling window. We illustrate our results with simulated data as well as with real data set corresponding to Nord Spool data. For such data we consider practical issues of constructing the confidence band for the periodic mean function and the choice of the subsampling window.

Appendix

In the appendix the proofs of some results are presented.

Proof of Fact 1

The $\{GG_{s+pT}\}_{s\in {Z}},$ for each $s=1,\ldots ,T$ is a second ordered stationary process.

Assume that the process $\{GG_{s+pT}\}$ satisfies

• $\alpha -$mixing condition

• completely regular condition and

• completely linear regular condition

For Gaussian processes the relationship between above coefficient is as follows [36]: $\rho (k)=r(k),$ and $\alpha (k)\le r(k)\le 2\pi \alpha (k).$ From the theorem below the Gaussian - Gegenbauer process can’t be strong mixing. [19] Since $\{\varepsilon _{t}\}_{t\in {Z}}$ in Eq. (1) is a Gaussian process, the long memory stationary k-factor Gegenbauer process (1) is not completely regular and hence is not strong mixing. From [15] we know that the Gaussian - Gegenbauer process has weak dependence properties.    $\square$

Proof of Theorem 3

The proof of the Theorem 3 is in [22], here only main points will be repeated.

Let: $N(\hat{\eta }_{N}(s)-\eta (s) )=\sum _{p=0}^{N-1}Y_{s+pT},\ \ s = 1,2,...,T$ where $Y_{t}=X_{t}-\eta (t) =\sigma _{t}GG_{t}.$

First we need to show that $N^{-\zeta }\sum _{p=0}^{N-1}Y_{s+pT}$ converge weakly to some random variable.

Let $\mathscr {E}$ be the $\sigma -$field: $\mathscr {E} = \sigma (\varepsilon ) = \sigma (\varepsilon _{t}, t \in {Z}).$ Let $\mathscr {G}$ be the $\sigma -$field: $\mathscr {G} = \sigma (GG) = \sigma (GG_{t}, t \in {Z}).$ From the assumption A in the definition of the model (2) the $\sigma -$fields $\mathscr {E}$ and $\mathscr {G}$ are independent with respect to the probability measure P. The properties of the characteristic function of normal variable sum are used in the proof.

$$Eexp \{i\nu N^{-1/\alpha }\sum _{p=0}^{N-1} Y_{s+pT}\}=E[E[exp\{i\nu N^{-1/\alpha } \sum _{p=0}^{N-1}\sigma _{s+pT}GG_{s+pT}\}|\mathscr {E}]]$$

where $\nu$ is any real number and $s = 1,2,...,T.$

The inner conditional characteristic function, from the properties of Gaussian characteristic function is

$$\begin{aligned}&E[exp\{ i\nu N^{-1/\alpha } \sum _{p=0}^{N-1}\sigma _{s+pT}GG_{s+pT}\}|\mathscr {E}]\\&\quad = exp\{-\frac{({\nu N^{-1/\alpha })^{2}}}{2}\sum _{p,q = 0}^{N-1}\sigma _{s+pT}\sigma _{s+qT}f_{s+pT}f_{s+qT}\gamma _{N}(T(p-q))\},\ s=1,...,T. \end{aligned}$$

This double sum is divided into the diagonal and off-diagonal terms:

$$\begin{aligned} N^{-\frac{2}{\alpha }} \big (\sum _{p=0}^{N-1}\sigma ^{2}_{s+pT}\gamma _{G}(0)+\sum _{p\not = q}\sigma _{s+pT}\sigma _{s+qT}\gamma _{G}((p-q)T)\big ) \end{aligned}$$

(9)

In the case $1/\alpha > (\beta + 1)/2$ the off-diagonal part of (9) is $O_P(N^{1-2/\alpha }N^{\beta })$ which tends to zero as $N \rightarrow \infty .$ The characteristic function of the diagonal part of (9) is the characteristic function of a $S\alpha S$ variable with scale $\sqrt{\gamma _{G}(0)/2}=|f_{s}|\sqrt{\gamma _{N}(0)/2},$ for $s=1,2,...,T$. (see [42]).

In the case $1/\alpha < (\beta + 1)/2$ the formula (9) becomes

$$\begin{aligned} N^{-(\beta +1)}\big ( \sum _{p=0}^{N-1}\sigma _{s+pT}^{2}\gamma _{G}(0)+\sum _{q\not =p} \sigma _{s+pT}\sigma _{s+qT}\gamma _{G}((p-q)T) \big ). \end{aligned}$$

(10)

The first term is $O_{P} (N^{2/\alpha -(\beta +1)})$ and tends to zero as $N\rightarrow \infty .$

The limiting characteristic function of the second part is characteristic function of a mean zero Gaussian with variance $\tilde{C}(s)=f_{t}(C(s)-\gamma _{G}(0){I}_{\{\beta =0\}}).$

The case $1/\alpha = (\beta +1)/2$ is the combination of the two above cases. From the Slutsky’s Theorem we get weak convergence the sum of two independent random variables.

The next step is to show joint convergence of the first and second moment of the model (2), for each $s=1,..,N$ and $p=0,...,N-1.$

In the proof the joint Fourier/Laplace Transform of the first and second sample moments [21] is considered.

For any real $\theta$ and $\phi > 0,$

$$\begin{aligned}&Eexp\{i\theta N^{-\zeta }\sum _{p=0}^{N-1}Y_{s+pT}-\phi N^{-2\zeta }\sum _{p=0}^{N-1}Y_{s+pT}^{2}\}\\&\quad = E[exp\{-\frac{1}{2}N^{-\zeta }\sum _{p,q}^{N-1} \sigma _{s+pT}\sigma _{s+qT}\gamma _{G}((p-q)T)(\theta +\sqrt{2\phi }W_{s+pT})(\theta +\sqrt{2\phi }W_{s+qT}) \}].\end{aligned}$$

The sequence of random variables $W_{s}$ is i.i.d. standard normal, and is independent of the $Y_{s}$ series. The information about $W_{s}$ is denoted by $\mathscr {W}.$ The double sum in the Fourier/Laplace Transform is broken into diagonal and off-diagonal terms.

The off-diagonal term is

$$\begin{aligned} N^{-2\zeta }\sum _{|h|>0}^{N-1}\sum _{p=0}^{N-|h|}\sigma _{s+pT}\sigma _{s+pT+hT}(\theta +\sqrt{2\phi }W_{s+pT})(\theta +\sqrt{2\phi }W_{s+pT+hT})\gamma _{G}(hT) \end{aligned}$$

(11)

In the case $2/\alpha > \beta + 1,$ by the Markov inequality the off-diagonal term tends to zero in probability as $N \rightarrow \infty .$

In the case $2/\alpha < \beta + 1$ (11) tends to constant (see the proof of Theorem 1 [42]).

In the case $2/\alpha = \beta + 1,$ the off-diagonal part, for fixed $s=1,2,...,T,$ tends to a constant.

The diagonal term is examined separately (by Dominated Convergence Theorem and above fact). Let $V_{s+pT}=\theta +\sqrt{2\phi }W_{s+pT}$

$$\begin{aligned}&E[exp\{-\frac{1}{2}\gamma _{G}(0)N^{-2\zeta }\sum _{p=0}^{N-1}\sigma _{s+pT}^{2}V_{s+pT}^{2}\}]\\&\quad = E[exp\{-(\gamma _{G}(0)/2)^{\alpha /2}N^{-\alpha \zeta }\sum _{p=0}^{N-1}|V_{s+pT}|^{\alpha }\}|. \end{aligned}$$

While $2/\alpha < \beta + 1$ the sum $N^{-\alpha \zeta }\sum _{p=0}^{N-1}|V_{s+pT}|^{\alpha }{\mathop {\longrightarrow }\limits ^{p}} 0.$

While $2/\alpha \ge \beta + 1$ (by the Law of Large Numbers) $N^{-\alpha \zeta }\sum _{p=0}^{N-1}|V_{s+pT}|^{\alpha }{\mathop {\longrightarrow }\limits ^{p}}E|V|^{\alpha }.$ By the Dominated Convergence Theorem, the limit as $N \longrightarrow \infty$ can be taken through the expectation, so that $E[exp\{-(\gamma _{G}(0)/2)^{\alpha /2}N^{-\alpha \zeta }\sum _{p=0}^{N-1}|V_{s+pT}|^{\alpha }\}]\rightarrow exp\{-(\gamma _{G}(0)/2)^{\alpha /2}E|\theta +\sqrt{2\phi }N|^{\alpha }1_{2/\alpha \ge \beta +1}\}.$ Using the Fourier/Laplace Transform and argumentation as in [42] we get the proof of this part of the Theorem 3.

Using argumentation as in [42] and in [35] we obtain for $s=1,...,T$

$$\begin{aligned}&N^{-\beta \rho }(\widehat{LM}(\rho ,s))^{\rho }=o_{P}(1) \\&\quad +\Big |N^{-\beta \rho }\sum _{|h|>0}^{[N^{\rho }]}\frac{1}{N-|h|}\sum _{p=0}^{N-|h|}Y_{s+pT}Y_{s+pT+hT}\Big |{\mathop {\longrightarrow }\limits ^{P}}\mu ^{2}C(s).\end{aligned}$$

At last from the Slutsky Theorem we get the convergence 5.

The second convergence in (5) follows from the continuous mapping theorem (denominators are different from zero).    $\square$

Proof of Theorem 4

Let us consider a sequence of statistics $P_{N}(s),$ for fixed $s=1,2,...,T$ and $N=1,2,...$.

$L_{N}(s)(x)=P(Z_{N}(s)\le x)$ is cumulative distribution function of $P_{N}(s).$

There exist

$$r_{N}(s)=sup_{x\in {R}}|L_{N}(s)(x)-L(s)(x)|\longrightarrow 0,\ \ N\rightarrow \infty $$

For overlapping samples the number of subsamples:

$Y_{b_s,p}(s)=(X_{N,s+pT},X_{s+(p+1)T}...,X_{N,s+(p+b_s-1)T}),$ $p=0,1,...,N-b_s$ and the number of subsampling statistics:

$P_{N,b_s,p}(s) = \sqrt{b_s}(\hat{\eta }_{N,b_s,p}(s)-\hat{\eta }_{N}(s))/\hat{\sigma }_{N,b_s,p}(s)$ is $N-b_s+1.$

Above statistics are used to approximate the distributions $L_{N}(s)(x)$ by empirical distribution functions: $L_{N,b_s,p}(s)(x)=\frac{1}{N-b+1}\sum _{p=0}^{N-b_s}{I}_{\{P_{N,b_s,p}(s)\le x\}}.$

For $A_{N}(s)$ let define subsampled statistics:

$$U_{N,b_s,p}(s)(x)=\frac{1}{N-b+1}\sum _{p=0}^{N-b}\varphi {(\frac{A_{N}(s)- x}{\varepsilon _n})}. $$

The sequence $\varepsilon _n$ is decreasing to zero and $\varphi$ is the non-increasing continuous function such that $\varphi = 1$ or 0 according to $x \le 0$ or $x \ge 1$ and which is affine between 0 and 1.

It is enough to investigate the variance of $U_{N,b_s,p}(s),\ \ s=1,...,T$ (Theorem 11.3.1 [46]).

$$Var U_{N,b_s,p}(s)(x)=(N-b_s+1)^{-2}(\sum _{|h|<N-b_s+1}(N-b_s+1-|h|)\gamma (h))$$

here $\gamma (h)=Cov(\varphi (A_{N}(s)),\varphi (A_{N}(s))).$

From the assumption that we have $\lambda -$weak dependency

$$Cov(\varphi (A_{N,p}(s)),\varphi (A_{N,p+h}(s)))\le \sqrt{b_s}\lambda _{h-b+1}/\varepsilon _N$$

For $\varepsilon _N = (N^{2(1-\beta )b})^{1/4}$ above covariance converge to zero. From Cesaro mean argument $Var U_{N,b_s,p}(s)(x)$ also goes to zero.

$$V_{N,b_s,p}(s)(x)=\frac{1}{N-b+1}\sum _{p=0}^{N-b}\varphi {(\frac{A_{N}(s)P_{N,b_s,p}(s)- x}{\varepsilon _n})}. $$

Under condition A1 through A4 and the Theorem 3.1, [50] we have:

$$lim_{N\rightarrow \infty }|E[V_{N,b_s,p}(s)(x)-E[V_{N,b_s,p}(s)(x)]]^{2}|=0.$$

It implies that $Var(V_{N,b_s,p}(s)(x))$ tends to zero, it proves point 1. of the Theorem 4.

To prove the point 2. of the Theorem 4 we also use the Theorem 2 in [14].

$$lim_{N\rightarrow \infty }sup_{x\in {R}}|U_{N,b_s,p}(s)(x)-L(s)(x)|=0,$$

in probability. The proof of point 3. if the 1. holds and under assumption of the model (2) is the same as the proof of 3. in the Theorem 11.3.1 [46].   $\square$

Proof of the Lemma 2

The proof of the Lemma 2 strictly follows from Lemma 2, [5] and Theorem 2, [14].    $\square$