The impact of asynchronous trading on Epps effect on Warsaw Stock Exchange

Abstract

The main goal of the analysis is the verification of whether asynchrony in transaction times is a considerable cause of the Epps effect on the Warsaw Stock Exchange among the most liquid assets. A method for compensating for the impact of asynchrony in trading on the Epps effect is presented. The method is easily applicable. Calculations are made using the exact time of transactions and prices of the assets. The estimation is not biased by intervals during which no transactions have taken place. Among all the analyzed stock pairs, asynchrony turns out to be the main cause of the Epps effect. However, the corrected correlation estimator seems to be more volatile than the regular estimator of the correlation. The presented analysis can be reproduced for the same data or replicated for another dataset; all R codes used in the process of writing this article are available upon request. The main novelty/value added of this paper is the application to an emerging market of a new method for compensating for asynchrony in trading.

Appendix

Proof of Theorem 4.1

Let us first calculate the observed normalized process $g_{\Delta t}^i (t)$ in terms of the underlying process $\tilde{g} ^{i}(t)$, where $i=1,2$. The operator $\langle \cdots \rangle$ means the average over the analyzed period $\left[ {0,\hbox {T}}\right]$.

$$\begin{aligned} g_{\Delta t}^i (t)= & {} \frac{{r}_{\Delta t}^i (t)-\langle {r}_{\Delta t}^i\rangle }{\sigma _{\Delta t}^i }=\frac{\mathop \sum \nolimits _{j=1}^{N_{\Delta t}^i (t)}\tilde{r} ^{i}\left( {\gamma ^{i}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) -\langle {r}_{\Delta t}^i \rangle }{\sigma _{\Delta t}^i }\\= & {} \frac{\mathop \sum \nolimits _{j=1}^{N_{\Delta t}^i (t)} \left( {\tilde{\sigma }^{i}\tilde{g} ^{i}\left( {\gamma ^{i}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) +\langle \tilde{r}^{i}\rangle } \right) -\langle {r}_{\Delta t}^i\rangle }{\sigma _{\Delta t}^i}\\= & {} \frac{\tilde{\sigma }^{i}\mathop \sum \nolimits _{j=1}^{N_{\Delta t}^i (t)} \tilde{g}^{i}\left( {\gamma ^{i}\left( \hbox {t} \right) +j\Delta \tilde{t}}\right) +N_{\Delta t}^i (t)\langle \tilde{r}^{i}\rangle -\langle {r}_{\Delta t}^i \rangle }{\sigma _{\Delta t}^i} \end{aligned}$$

Under Assumption A1:

$$\begin{aligned} \langle {r}_{\Delta t}^i (t)\rangle =\langle {N}_{\Delta t}^i\rangle \langle \tilde{r}^{i}\rangle ;\left( {\sigma _{\Delta t}^i } \right) ^{2}=\langle {N}_{\Delta t}^i \rangle \left( {\tilde{\sigma }^{i}} \right) ^{2} \end{aligned}$$

Thus,

$$\begin{aligned} g_{\Delta t}^i (t)=\frac{1}{\sqrt{\langle {N}_{\Delta t}^i \rangle }}\mathop \sum \limits _{j=1}^{N_{\Delta t}^i (t)} \tilde{g}^{i}\left( {\gamma ^{i}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) +\frac{\langle \tilde{r}^{i}\rangle \left( {N_{\Delta t}^i (t)-\langle {N}_{\Delta t}^i \rangle } \right) }{\sigma _{\Delta t}^i } \end{aligned}$$

Let us now calculate the observed correlation $\hbox {corr}\left( {r_{\Delta t}^1 ,r_{\Delta t}^2 } \right)$ in terms of the correlation of underlying processes $\hbox {corr}_{t_j } \left( {\tilde{g}^{1},\tilde{g}^{2}} \right)$.

$$\begin{aligned}&\hbox {corr}\left( {r_{\Delta t}^1 ,r_{\Delta t}^2 } \right) =\frac{1}{T}\mathop \sum \limits _{j=1}^T g_{\Delta t}^1 (t)g_{\Delta t}^2 (t)\\&\quad =\frac{1}{T}\mathop \sum \limits _{j=1}^T \left( \left( {\frac{1}{\sqrt{\langle {N}_{\Delta t}^1\rangle }}\mathop \sum \limits _{j=1}^{N_{\Delta t}^1 (t)} \tilde{g}^{1}\left( {\gamma ^{1}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) +\frac{\langle \tilde{r}^{1}\rangle \left( {N_{\Delta t}^1 (t)-\langle {N}_{\Delta t}^1 \rangle } \right) }{\sigma _{\Delta t}^1}} \right) \right. \\&\qquad \left. \times \left( {\frac{1}{\sqrt{\langle {N}_{\Delta t}^2\rangle }}\mathop \sum \limits _{j=1}^{N_{\Delta t}^2 (t)} \tilde{g}^{2}\left( {\gamma ^{2}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) +\frac{\langle \tilde{r}^{2}\rangle \left( {N_{\Delta t}^2 (t)-\langle {N}_{\Delta t}^2\rangle } \right) }{\sigma _{\Delta t}^2 }} \right) \right) \end{aligned}$$

The average of the second component in each factor equals zero, thus under Assumption A1 we have.

$$\begin{aligned} \hbox {corr}\left( {r_{\Delta t}^1 ,r_{\Delta t}^2 } \right)= & {} \frac{1}{T}\mathop \sum \limits _{j=1}^T \left( \frac{1}{\sqrt{\langle {N}_{\Delta t}^1 \rangle }}\mathop \sum \limits _{j=1}^{N_{\Delta t}^1 (t)} \tilde{g} ^{1}\left( {\gamma ^{1}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) \right. \\&\left. \times \frac{1}{\sqrt{\langle {N}_{\Delta t}^2 \rangle }}\mathop \sum \limits _{j=1}^{N_{\Delta t}^2 (t)} \tilde{g} ^{2}\left( {\gamma ^{2}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) \right) \end{aligned}$$

The means of the process $\tilde{g}^{1}$ and $\tilde{g}^{2}$ equal zero, thus under Assumption A2 we have.

$$\begin{aligned}&\hbox {corr}\left( {r_{\Delta t}^1 ,r_{\Delta t}^2 } \right) \\&\quad =\frac{1}{T}\mathop \sum \limits _{j=1}^T \left( {\frac{1}{\sqrt{\langle N_{\Delta t}^1\rangle \langle N_{\Delta t}^2\rangle }}\mathop \sum \limits _{j=1}^{\overline{N}\left( {t_j } \right) } \tilde{g}^{1}\left( {\gamma ^{1}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) \tilde{g}^{2}\left( {\gamma ^{2}\left( \hbox {t} \right) +j\Delta \tilde{t}} \right) } \right) \\&\quad =\frac{1}{T}\mathop \sum \limits _{j=1}^T \left( \frac{ \overline{N} \left( {t_j } \right) }{\sqrt{\langle N_{\Delta t}^1\rangle \langle N_{\Delta t}^2 \rangle }}\hbox {corr}_{t_j } \left( {\tilde{g}^{1},\tilde{g}^{2}} \right) \right) \end{aligned}$$

Thus,

$$\begin{aligned} \hbox {corr}_c \left( {r_{\Delta t}^1 ,r_{\Delta t}^2 } \right) =\frac{1}{T}\mathop \sum \limits _{j=1}^T g_{\Delta t}^1 (t)g_{\Delta t}^2 (t)\frac{\sqrt{\langle N_{\Delta t}^1\rangle \langle N_{\Delta t}^2\rangle }}{\overline{N}\left( {t_j } \right) }; \end{aligned}$$

which had to be proven. $\square$