On the dependence structure of the trade/no trade sequence of illiquid assets

Abstract

In this paper, we propose to consider the dependence structure of the trade/no trade categorical sequence of individual illiquid stocks returns. The framework considered here is wide as constant and time-varying zero returns probability are allowed. The ability of our approach in highlighting illiquid stock’s features is underlined for a variety of situations. More specifically, we show that long-run effects for the trade/no trade categorical sequence may be spuriously detected in presence of a non-constant zero returns probability. Monte Carlo experiments, and the analysis of stocks taken from the Chilean financial market, illustrate the usefulness of the tools developed in the paper.

Keywords:

• Time-varying illiquidity levels
• Categorical financial time series
• Serial dependence

JEL Classification:

• C13
• C22
• C58

Proofs

Proof of Proposition 2.1.

Firstly, note that we have $\overline{a}-P(a_t=1)=O_p\left(n^{-\frac{1}{2}}\right),$ so that $$n^{\frac{1}{2}}{\widehat{\gamma }}_a(h)=n^{\frac{1}{2}}{\overline{\gamma }}_a(h)+o_p(1),\hspace{0.17em}\text{for}\mathrm{ }\text{all}\hspace{0.17em}h\in \{1,\dots ,m\},$$ where ${\overline{\gamma }}_a(h)={(n-h)}^{-1}{\sum }_{t=1+h}^n(a_t-P(a_t=1))(a_{t-h}-P(a_t=1)).$ Let us define ${\overline{A}}_m=({\overline{\gamma }}_a(1),\dots ,{\overline{\gamma }}_a(m))\prime .$ From the Central Limit Theorem (CLT) for martingale difference sequences (see Theorem A.3 in Francq and Zakoïan (2019)), we have $$n^{\frac{1}{2}}{\overline{A}}_m\stackrel{d}{\to }\mathcal{N}(0,\overline{\Sigma }),$$ where $\overline{\Sigma }$ is a m × m dimensional diagonal matrix, with diagonal component $P{(a_t=1)}^2{(1-P(a_t=1))}^2.$ On the other hand, it is easy to see that ${\widehat{\gamma }}_a(0)\stackrel{a.s.}{\to }P(a_t=1)(1-P(a_t=1)).$ Hence, the result follows from the Slutsky Lemma. □

Proof of (4)(4) $$\sqrt{n}{\tilde{A}}_m\stackrel{d}{\to }\left(\underset{u\in (0,1]}{\mathrm{\text{sup}}}|G(u)|,\dots ,\underset{u\in (0,1]}{\mathrm{\text{sup}}}|G(u)|\right)\prime ,$$(4) . Let ${\xi }_{h,t}=(a_t-P(a_t=1))(a_{t-h}-P(a_{t-h}=1))$ and $${\tilde{\Gamma }}_m(u)=\left({\tilde{\gamma }}_a(1,u),\dots ,{\tilde{\gamma }}_a(m,u)\right)\prime .$$

The sequence $({\xi }_{h,t})$ is a martingale difference, such that $V({\xi }_{h,t})=g^2(t/n){(1-g(t/n))}^2+O(n^{-1}),$ from the Lipschitz condition with a finite number of breaks in Assumption 1. Then from Theorem 2.1 of Hansen (1992), we obtain $${\tilde{\Gamma }}_m(u)\stackrel{d}{\to }\left(G(u),\dots ,G(u)\right)\prime .$$

The desired result follows from the Continuous Mapping Theorem. □

Proof of Proposition 2.3.

Note that $E\{(a_t-P(a_t=1))(a_{t-h}-P(a_{t-h}=1))\}=0.$ Then using the Central Limit Theorem for independent but heterogeneous sequences, see Davidson (1994), Theorem 23.6, we have $$\sqrt{n}\left({\tilde{\gamma }}_a(1,1),\dots ,{\tilde{\gamma }}_a(m,1)\right)\prime \stackrel{d}{\to }N(0,\varpi ),$$ where $\varpi ={\int }_0^1{g}^2(s){(1-g^2(s))}^2\mathit{\text{ds}}$ is obtained using some computations, and since $(a_t)$ is independent. On the other hand, from the Kolmogorov SLLN for independent but non-identically random variables (see, for instance, Sen and Singer (1993), Theorem 2.3.10), we have $${\tilde{\gamma }}_a(h,1)\stackrel{a.s.}{\to }{\int }_0^{1}g(s)(1-g(s))\mathit{\text{ds}}.$$ Hence, the first result of Proposition 2.3 follows from the Slutsky Lemma. Now, for the convergence of $\widehat{\omega },$ using again the Kolmogorov SLLN and some computations, we have $$\begin{array}{c}{n}^{-1}\sum _{t=2}^n(a_t-P(a_t=1){)}^2{(a_{t-1}-P(a_{t-1}=1))}^2\stackrel{a.s.}{\to }{\int }_0^1{g}^2(s){(1-g^2(s))}^2\mathit{\text{ds}},\\ n^{-1}\sum _{t=1}^n{\left(a_t-P(a_t=1)\right)}^2\stackrel{a.s.}{\to }{\int }_0^{1}g(s)(1-g(s))\mathit{\text{ds}}.\end{array}$$ □

Proof of Proposition 2.4.

From the Kolmogorov SLLN for independent but non identically random variables (see, Sen and Singer (1993), Theorem 2.3.10), and using some computations, we have $${\tilde{\gamma }}_a(h,1)\stackrel{a.s.}{\to }{\int }_0^1{g}_h(s)\mathit{\text{ds}}-{\left({\int }_0^{1}g(s)\mathit{\text{ds}}\right)}^2,$$

from the dominated convergence Theorem, and for any $h\in \{1,\dots ,m\}.$ Hence, under ${\tilde{H}}_1$ the result follows. □

Proof of Proposition 2.5.

In this proof similar arguments to that of the proof of Theorem 2 in Xu and Phillips (2008) are considered. As the break number is finite, we assume that the function $g(·)$ is continuous, without a loss of generality. Let us introduce the short notations $${\widehat{p}}_t=\sum _{i=1}^n{w}_{\mathit{\text{ti}}}{a}_i,{\overline{p}}_t=\sum _{i=1}^n{w}_{\mathit{\text{ti}}}{p}_i,\hspace{1em}{p}_t=P(a_t=1),$$

and $z_i=a_i-p_i.$ We can write $${\widehat{p}}_t-{\overline{p}}_t=\frac{\frac{1}{\mathit{\text{nb}}}\sum _{i=1}^n{K}_{\mathit{\text{ti}}}{z}_i}{\frac{1}{\mathit{\text{nb}}}\sum _{j=1}^n{K}_{\mathit{\text{tj}}}}.$$

From ${\tilde{H}}_0,\hspace{0.17em}(z_i)$ is an independent process, such that $E(z_i)=0.$ In addition, from Lemma A(c) in Xu and Phillips (2008), we have $(1/\mathit{\text{nb}}){\sum }_{i=1}^n{K}_{\mathit{\text{ti}}}\to 1.$ In view of the above arguments, deduce that $$\begin{array}{c}E{\left(\frac{1}{\mathit{\text{nb}}}\sum _{i=1}^n{K}_{\mathit{\text{ti}}}{z}_i\right)}^2=\frac{1}{{(\mathit{\text{nb}})}^2}\sum _{i=1}^n{K}_{\mathit{\text{ti}}}^{2}E(z_i^2)\\ \le \left(\frac{1}{\mathit{\text{nb}}}\right)\left(\mathit{\text{su}}{p}_i{K}_{\mathit{\text{ti}}}\right)\left(\frac{1}{\mathit{\text{nb}}}\sum _{i=1}^n{K}_{\mathit{\text{ti}}}\right)\\ =O\left(\frac{1}{\mathit{\text{nb}}}\right),\end{array}$$ Thus, we write (9) $${\widehat{p}}_t-{\overline{p}}_t=O_p\left(\frac{1}{\sqrt{\mathit{\text{nb}}}}\right).$$(9)

On the other hand, we have $$\begin{array}{c}\frac{1}{\mathit{\text{nb}}}\sum _{i=1}^n{K}_{\left[\mathit{\text{nr}}\right]i}{p}_i=\frac{1}{\mathit{\text{nb}}}\sum _{i=1}^{n}K\left(\frac{\left[\mathit{\text{nr}}\right]-i}{\mathit{\text{nb}}}\right)g\left(\frac{i}{n}\right)\\ =\frac{1}{b}[{\int }_{1/n}^{2/n}K\left(\frac{\left[\mathit{\text{nr}}\right]-\left[\mathit{\text{ns}}\right]}{\mathit{\text{nb}}}\right)g\left(\frac{\left[\mathit{\text{ns}}\right]}{n}\right)\mathit{\text{ds}}+\dots \\ +{\int }_1^{(n+1)/n}K\left(\frac{\left[\mathit{\text{nr}}\right]-\left[\mathit{\text{ns}}\right]}{\mathit{\text{nb}}}\right)g\left(\frac{\left[\mathit{\text{ns}}\right]}{n}\right)\mathit{\text{ds}}]\\ =\frac{1}{b}\left({\int }_{1/n}^{(n+1)/n}K\left(\frac{\left[\mathit{\text{nr}}\right]-\mathit{\text{ns}}}{\mathit{\text{nb}}}\right)g\left(s\right)\mathit{\text{ds}}\right)+O\left(\frac{1}{\mathit{\text{nb}}}\right)\\ \stackrel{z=(s¯r)/b}{=}{\int }_{(\frac{1}{n}-r)/b}^{(1+\frac{1}{n}-r)/b}K\left(\frac{\left[\mathit{\text{nr}}\right]-n(r+\mathit{\text{bz}})}{\mathit{\text{nb}}}\right)g\left(r+\mathit{\text{bz}}\right)\mathit{\text{dz}}+O\left(\frac{1}{\mathit{\text{nb}}}\right)\\ ={\int }_{-\infty }^{\infty }K\left(\frac{\left[\mathit{\text{nr}}\right]-\mathit{\text{nr}}}{\mathit{\text{nb}}}-z\right)g\left(r+\mathit{\text{bz}}\right)\mathit{\text{dz}}+O\left(\frac{1}{\mathit{\text{nb}}}\right),\end{array}$$ where the last equality is obtained for small enough b, and since a compact support is assumed for $K(·)$ in Assumption 2(a). Using the Lipschitz condition in Assumption 1, deduce that (10) $$\frac{1}{\mathit{\text{nb}}}\sum _{i=1}^n{K}_{\left[\mathit{\text{nr}}\right]i}{p}_i=g(r)+O(b)+O\left(\frac{1}{\mathit{\text{nb}}}\right).$$(10) Now, writing $$\begin{array}{c}{n}^{-\frac{1}{2}}\sum _{t=1+h}^n(a_t-{\widehat{p}}_t)(a_{t-h}-{\widehat{p}}_{t-h})-n^{-\frac{1}{2}}\sum _{t=1+h}^n(a_t-p_t)(a_{t-h}-p_{t-h})\\ =n^{-\frac{1}{2}}\sum _{t=1+h}^n(a_t-p_t)(p_{t-h}-{\overline{p}}_{t-h})+n^{-\frac{1}{2}}\sum _{t=1+h}^n(a_t-p_t)({\overline{p}}_{t-h}-{\widehat{p}}_{t-h})\\ +n^{-\frac{1}{2}}\sum _{t=1+h}^n(p_t-{\overline{p}}_t)(a_{t-h}-p_{t-h})+n^{-\frac{1}{2}}\sum _{t=1+h}^n(p_t-{\overline{p}}_t)(p_{t-h}-{\overline{p}}_{t-h})\\ +n^{-\frac{1}{2}}\sum _{t=1+h}^n(p_t-{\overline{p}}_t)({\overline{p}}_{t-h}-{\widehat{p}}_{t-h})+n^{-\frac{1}{2}}\sum _{t=1+h}^n({\overline{p}}_t-{\widehat{p}}_t)(a_{t-h}-p_{t-h})\\ +n^{-\frac{1}{2}}\sum _{t=1+h}^n({\overline{p}}_t-{\widehat{p}}_t)(p_{t-h}-{\overline{p}}_{t-h})+n^{-\frac{1}{2}}\sum _{t=1+h}^n({\overline{p}}_t-{\widehat{p}}_t)({\overline{p}}_{t-h}-{\widehat{p}}_{t-h}),\end{array}$$ and using (9)(9) $${\widehat{p}}_t-{\overline{p}}_t=O_p\left(\frac{1}{\sqrt{\mathit{\text{nb}}}}\right).$$(9) and (10)(10) $$\frac{1}{\mathit{\text{nb}}}\sum _{i=1}^n{K}_{\left[\mathit{\text{nr}}\right]i}{p}_i=g(r)+O(b)+O\left(\frac{1}{\mathit{\text{nb}}}\right).$$(10) , the desired result follows. □

Notes

1 For u < 0, the function is set constant, that is $g(u)=\underset{u\downarrow 0}{\mathrm{\text{lim}}}g(u).$ Throughout the paper, the piecewise Lipschitz condition means: there exists a positive integer p and some mutually disjoint intervals $I_1,\dots ,I_p$ with $I_1\cup \cdots \cup I_p=(0,1]$ such that $g(u)={\sum }_{l=1}^p{g}_l(u){\mathbf{1}}_{\{u\in I_l\}},$ $u\in (0,1],$ where $g(·)$ is a Lipschitz smooth function on $I_1,\dots ,I_p,$ respectively.

2 The author is grateful to Andres Celedon for research assistance.

Additional information

Funding

This author acknowledges the ANID funding, Fondecyt 1201898.