Modelling long-range dependence and trends in duration series: an approach based on EFARIMA and ESEMIFAR models

Abstract

Duration series often exhibit long-range dependence and local nonstationarities. Here, exponential FARIMA (EFARIMA) and exponential SEMIFAR (ESEMIFAR) models are introduced. These models capture simultaneously nonstationarities in the mean as well as short- and long-range dependence, while avoiding the complication of unobservable latent processes. The models can be thought of as locally stationary long-memory extensions of exponential ACD models. Statistical properties of the models are derived. In particular the long-memory parameter in the original and the log-transformed process is the same. For Gaussian innovations, exact explicit formulas for all moments and autocovariances are given, and the unconditional distribution is log-normal. Estimation and model selection can be carried out with standard software. The approach is illustrated by an application to average daily transaction durations.

Appendix

1.1 Proofs of results

Proof of Lemma 1

Here only a sketched proof will be given. Under the conditions of Lemma 1, $Z_{t}$ is the strictly stationary solution of $Z_{t}=\ln (X_{t}^{*})$. This even holds without the normal assumption on $\epsilon _{t}$. If a stationary solution of $X_{t}^{*}$ exists, then it must be $X_{t}^{*}=\exp (Z_{t})=\prod \nolimits _{i=0}^{\infty }\eta _{t-i} ^{a_{i}}$ as given in (8). This solution can also be obtained through a recursive expansion of $X_{t}^{*}=\lambda _{t}\eta _{t}$ by means of the explicit definition of $\lambda _{t}$ given in (10) and a technique similar to a Volterra expansion. This is however too complex and unnecessary in the current context, because we are dealing with a log-linear process. Details on this point are omitted to save space.

Now, the question hence reduces to check the stationarity of $X_{t}^{*}$ defined in (8). The answer depends on moment properties of $\epsilon _{t}$. Under the assumption that $\epsilon _{t}$ is an i.i.d. Gaussian process with mean zero and variance $\sigma _{\epsilon }^{2}$, the marginal distribution of $Z_{t}$ is $Z_{t}\sim N(0,\sigma ^{2})$, where $\sigma ^{2}=\sigma _{\epsilon }^{2}\sum \nolimits _{i=0}^{\infty }\alpha _{i}^{2}$ is as defined in Theorem 1. Hence the marginal distribution of $X_{t}^{*}$ is $LN(0,\sigma ^{2})$. Moments of $X_{t}^{*}$ of any order are finite. The fact that $X_{t}^{*}$ is strictly stationary follows directly from its time invariant infinite product of independent normal random variables. Straightforward analysis based on this representation shows that $X_{t}^{*}$ is also weakly stationary, because $E(X_{t}^{*})$ and $cov(X_{t}^{*},X_{t+k}^{*})$ exist and do not depend on $t$. Detailed results on the moments and acf of $X_{t}^{*}$ may be found in Theorem 1.

Proof of Theorem 1

• (i) Let $G$ denote a generic log-normal random variable with the scale parameter $\mu$ and shape parameter $\sigma$. Then we have $E(G^{s})=e^{s\mu +\frac{1}{2}s^{2}\sigma ^{2}}$. Note that $X_{t}^{*}\sim LN(0,\sigma ^{2})$. Inserting the two parameters $\mu =0$ and $\sigma ^{2}$ into this formula we obtain $E[(X_{t}^{*})^{s}]=e^{s^{2}\sigma ^{2}/2}$ as given in Theorem 1 (i).

• (ii) We have

  $$\begin{aligned} var(X_{t}^{*})&=E[(X_{t}^{*})^{2}]-[E(X_{t}^{*})]^{2}\\&=e^{4\sigma ^{2}/2}-[e^{\sigma ^{2}/2}]^{2}\\&=e^{\sigma ^{2}}\left( e^{\sigma ^{2}}-1\right) , \end{aligned}$$

  which can also be obtained as a special case of $\gamma _{X^{*}}(k)$ with $k=0$. Furthermore note that $E(X_{t}^{*})E(X_{t+k}^{*})=[E(X_{t}^{*})]^{2}=e^{\sigma ^{2}}$, because $X_{t}^{*}$ is stationary. The expectation of $X_{t}^{*}X_{t+k}^{*}$ can be calculated as follows.

  $$\begin{aligned} E(X_{t}^{*}X_{t+k}^{*})&=E\left( \prod \limits _{i=0}^{\infty }\eta _{t-i}^{a_{i}}\prod \limits _{i=0}^{\infty }\eta _{t-i+k}^{a_{i}}\right) \\&=E\left( \prod \limits _{i=0}^{k-1}\eta _{t-i}^{a_{i}}\prod \limits _{i=0}^{\infty }\eta _{t-i+k}^{a_{i}+a_{i+k}}\right) \\&=\prod \limits _{i=0}^{k-1}E\left( \eta _{t-i}^{a_{i}}\right) \prod \limits _{i=0}^{\infty }E\left( \eta _{t-i+k}^{a_{i}+a_{i+k}}\right) \\&=\prod \limits _{i=0}^{k-1}e^{a_{i}^{2}\sigma _{\epsilon }^{2}/2}\prod \limits _{i=0}^{\infty }e^{(a_{i}+a_{i+k})^{2}\sigma _{\epsilon }^{2}/2}\\&=\prod \limits _{i=0}^{\infty }e^{a_{i}^{2}\sigma _{\epsilon }^{2}/2}\prod \limits _{i=0}^{\infty }e^{a_{i}^{2}\sigma _{\epsilon }^{2}/2}\prod \limits _{i=0}^{\infty }e^{2a_{i}a_{i+k}\sigma _{\epsilon }^{2}/2}\\&=e^{\sigma ^{2}}e^{\sigma _{\epsilon }^{2}\sum \limits _{i=0}^{\infty }a_{i}a_{i+k}}. \end{aligned}$$

  This leads to

  $$\begin{aligned} \gamma _{X^{*}}(k)&=e^{\sigma ^{2}}e^{\sigma _{\epsilon }^{2} \sum \limits _{i=0}^{\infty }a_{i}a_{i+k}}-e^{\sigma ^{2}}\nonumber \\&=e^{\sigma ^{2}}\left( e^{\sigma _{\epsilon }^{2}\sum \limits _{i=0}^{\infty }a_{i}a_{i+k}}-1\right) . \end{aligned}$$

  (18)

• (iii) We obtain the formulas of $\rho _{X}(k)$ by inserting results in $ii$) into $\rho _{X}(k)=\gamma _{X^{*}}(k)/var(X_{t}^{*})$.

• (iv) Note that $\sigma _{\epsilon }^{2}\sum \limits _{i=0}^{\infty }a_{i} a_{i+k}=\rho _{Z}(k)\sigma ^{2}$. By means of Taylor expansion of the exponential function we obtain

  $$\begin{aligned} \rho _{X}(k)=\left\{ \sum \limits _{i=1}^{\infty }[\rho _{Z}(k)\sigma ^{2} ]^{i}/i!\right\} \left\{ \sum \limits _{i=1}^{\infty }[\sigma ^{2} ]^{i}/i!\right\} ^{-1}. \end{aligned}$$

  (19)

  Now we will show that the first sum on the right hand side of (19) is dominated by its first term $\rho _{Z}(1)\sigma ^{2}$. Note that $0<\rho _{Z}(k)<1$, if $k$ is large enough. We have

  $$\begin{aligned} \sum \limits _{i=2}^{\infty }[\rho _{Z}(k)\sigma ^{2}]^{i}/i!&<[\rho _{Z}(k)]^{2}\sum \limits _{i=2}^{\infty }[\sigma ^{2}]^{i}/i!\\&=[\rho _{Z}(k)]^{2}(e^{\sigma ^{2}}-1-\sigma ^{2})\\&=O\{[\rho _{Z}(k)]^{2}\}. \end{aligned}$$

  Hence, we have

  $$\begin{aligned} \rho _{X^{*}}(k)=\rho _{Z}[c_{X}^{\mathrm {e}}+o(1)], \end{aligned}$$

  (20)

  where

  $$\begin{aligned} c_{X}^{\mathrm {e}}=\sigma ^{2}\left\{ \sum \limits _{i=1}^{\infty }[\sigma ^{2}]^{i}/i!\right\} ^{-1}. \end{aligned}$$

  (21)

  It is clear that $0<c_{X}^{\mathrm {e}}<1$. Theorem 1 is proved. $\square$

Proof of Lemma 2

Note that $\zeta _{t}=\sum \limits _{i=1}^{\infty }a_{i}\epsilon _{t-i}$ so that $var(\zeta _{t})=\sigma _{\epsilon }^{2} \sum \limits _{i=1}^{\infty }a_{i}^{2}$ and $cov(\zeta _{,}\zeta _{t+k} )=\sigma _{\epsilon }^{2}\sum \limits _{i=1}^{\infty }a_{i}a_{i+k}$. This leads to

$$\begin{aligned} \rho _{\zeta }(k)=\left[ \sum \limits _{i=1}^{\infty }a_{i}a_{i+k}\right] \left[ \sum \limits _{i=1}^{\infty }a_{i}^{2}\right] ^{-1}. \end{aligned}$$

(22)

Furthermore,

$$\begin{aligned} \rho _{\zeta }(k)&=\left[ \sum \limits _{i=0}^{\infty }a_{i}a_{i+k} -a_{k}\right] \left[ \sum \limits _{i=0}^{\infty }a_{i}^{2}-1\right] ^{-1}\\&=\frac{\sum \limits _{i=0}^{\infty }a_{i}a_{i+k}}{\sum \limits _{i=0}^{\infty }a_{i}^{2}}\frac{\sum \limits _{i=0}^{\infty }a_{i}^{2}}{\sum \limits _{i=1} ^{\infty }a_{i}^{2}}-\frac{a_{k}}{\sum \limits _{i=1}^{\infty }a_{i}^{2}}\\&=\rho _{Z}(k)c_{\rho }^{\lambda }[1+o(1)], \end{aligned}$$

because $a_{k}=o[\rho _{Z}(k)]$ for $d>0$, where $c_{\rho }^{\lambda } =\frac{\sum _{i=0}^{\infty }a_{i}^{2}}{\sum _{i=1}^{\infty }a_{i}^{2} }=\frac{\sigma ^{2}}{\sigma _{\lambda }^{2}}>1$ is as defined in Lemma 2. This completes the proof of Lemma 2. $\square$

A sketched proof of Corollary 2 As in the proof of Theorem 1, straightforward calculation leads to the formulas of $\rho _{\lambda }(k)$ in (i). Furthermore, following the proof of Theorem 1 (iv), we can obtaine $\tilde{c}_{\rho }^{\mathrm {e}}=\sigma _{\lambda }^{2}\left\{ \sum \limits _{i=1}^{\infty }[\sigma _{\lambda }^{2}]^{i}/i!\right\} ^{-1}$, which is also a constant between zero and one. Detailed calculation are omitted to save space.