A closer look at the world business cycle synchronization

Abstract

This paper uses a transformation of the period-by-period index proposed by Cerqueira and Martins (2009), to overcome some of its shortcomings, in a non-parametric estimation to analyze how business cycle synchronization for a sample of 111 countries evolved in the period 1960–2007. The period-by-period index is able to distinguish between negative correlations due to episodes in single years, asynchronous behavior in turbulent times and synchronous behavior over stable periods and the non-parametric approach provides a more detailed analysis than the use of a parametric approach. The results show that since the nineties the synchronization at the world level, within and between country groups, experienced a dramatic increase reaching the highest values ever at the sample end.

Appendices

Appendix A - Proof of Proposition 1

Proposition 1: ρ _ij,t is bounded between 3–2T and 1

Proof:

$$ \max \left( {{\rho_{ij,t }}} \right)=1 $$

As the second component in Eq. 1 is always non-negative, when it is zero we get ρ _ij,t  = 1.

$$ \min \left( {{\rho_{ij,t }}} \right)=3-2T $$

To find the extreme values of a specific value of ρ _ij,t (let’s say t = 1), we analyze the function:

$$ {f_1}({d_{i,1 }},{d_{i,2 }},\ldots,{d_{i,T }},{d_{j,1 }},{d_{j,2 }},\ldots,{d_{j,T }})={{\left( {\frac{{{d_{i,1 }}-{{\overline{d}}_i}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{i,t }}-{{\overline{d}}_i}} \right)}}^2}}}}}-\frac{{{d_{j,1 }}-{{\overline{d}}_j}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{j,t }}-{{\overline{d}}_j}} \right)}}^2}}}}}} \right)}^2} $$

From the derivatives to obtain the stationary points we get:

$$ \frac{{d{f_1}}}{{d\left( {{d_{i,1 }}} \right)}}=2\left( {\frac{{{d_{i,1 }}-{{\overline{d}}_i}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{i,t }}-{{\overline{d}}_i}} \right)}}^2}}}}}-\frac{{{d_{j,1 }}-{{\overline{d}}_j}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{j,t }}-{{\overline{d}}_j}} \right)}}^2}}}}}} \right)\frac{{{C_1}(T)}}{{{D_{1i }}(.)}}\mathop{\sum}\limits_{t=2}^T\mathop{\sum}\limits_{s=t+1}^T{{\left( {{d_{i,t }}-{d_{i,s }}} \right)}^2}=0 $$

$$ \frac{{d{f_1}}}{{d\left( {{d_{j,1 }}} \right)}}=2\left( {\frac{{{d_{i,1 }}-{{\overline{d}}_i}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{i,t }}-{{\overline{d}}_i}} \right)}}^2}}}}}-\frac{{{d_{j,1 }}-{{\overline{d}}_j}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{j,t }}-{{\overline{d}}_j}} \right)}}^2}}}}}} \right)\frac{{{C_1}(T)}}{{{D_{1j }}(.)}}\mathop{\sum}\limits_{t=2}^T\mathop{\sum}\limits_{s=t+1}^T{{\left( {{d_{j,t }}-{d_{j,s }}} \right)}^2}=0 $$

(At this point we do not need to know the remaining 2T-2 derivatives)

Where C ₁ is a function that depends on T, so if T fixed C ₁ is a constant and

${D_{1i }}(.)={{\sqrt{{\frac{1}{T}\mathop{\sum}\limits_{t=1}^T{{{\left( {{d_{i,t }}-{{\overline{d}}_i}} \right)}}^2}}}}^3}$ and ${D_{1j }}(.)={{\sqrt{{\frac{1}{T}\mathop{\sum}\limits_{t=1}^T{{{\left( {{d_{j,t }}-{{\overline{d}}_j}} \right)}}^2}}}}^3}$

So or the first factor is zero, which implies that f ₁ = 0 and ρ _ij,t   = 1,(which is the result for the maximum) or, as the last factor is a sum of squares, each component of the sum is zero. Therefore all observations but the first in each series are equal and, as D _1i (.) ≠ 0 and D _1j (.) ≠ 0 or the function would not be defined, implies that the first observation is different from the others.

With the last result, and considering that all observation of series d _i,t and d _j,t for t ≥ 2 are α and β, respectively, the problem simplifies to :

$$ maxf_1^{*}(.)={{\left( {\begin{array}{*{20}c} {\frac{{{d_{i,1 }}-\frac{{{d_{i,1 }}}}{T}-\frac{{\alpha \left( {T-1} \right)}}{T}}}{{\sqrt{{\frac{1}{T}\left( {{{{\left( {{d_{i,1 }}-\frac{{{d_{i,1 }}}}{T}-\frac{{\alpha \left( {T-1} \right)}}{T}} \right)}}^2}+{{{\left( {\alpha -\frac{{{d_{i,1 }}}}{T}-\frac{{\alpha \left( {T-1} \right)}}{T}} \right)}}^2}\left( {T-1} \right)} \right)}}}}-} \\ {} \\ {-\frac{{{d_{j,1 }}-\frac{{{d_{j,1 }}}}{T}-\frac{{\beta (T-1)}}{T}}}{{\sqrt{{\frac{1}{T}\left( {{{{\left( {{d_{j,1 }}-\frac{{{d_{j,1 }}}}{T}-\frac{{\beta (T-1)}}{T}} \right)}}^2}+{{{\left( {\beta -\frac{{{d_{j,1 }}}}{T}-\frac{{\beta \left( {T-1} \right)}}{T}} \right)}}^2}(T-1)} \right)}}}}} \\ \end{array}} \right)}^2} $$

Where $\frac{{df_1^{*}}}{{d\alpha }}=0$ and $\frac{{df_1^{*}}}{{d\beta }}=0$.

Which means that we just have to optimize with respect to the first observation of each series (d _i,1 and d _j,1 ).

As the proposed period-by-period index is invariable if we add to all elements of either series a constant (see proof below), we will work onwards with two series, observed over T periods (d _i and d _j }, where all observations but the first are zero. In this case the problem is:

$$ f_1^{** }(.)={{\left( {\frac{{{d_{i,1 }}-\frac{{{d_{i,1 }}}}{T}}}{{\sqrt{{\frac{1}{T}\left( {{{{\left( {{d_{i,1 }}-\frac{{{d_{i,1 }}}}{T}} \right)}}^2}+{{{\left( {\frac{{{d_{i,1 }}}}{T}} \right)}}^2}\left( {T-1} \right)} \right)}}}}-\frac{{{d_{j,1 }}-\frac{{{d_{j,1 }}}}{T}}}{{\sqrt{{\frac{1}{T}\left( {{{{\left( {{d_{j,1 }}-\frac{{{d_{j,1 }}}}{T}} \right)}}^2}+{{{\left( {\frac{{{d_{j,1 }}}}{T}} \right)}}^2}(T-1)} \right)}}}}} \right)}^2} $$

Now notice that:

$$ f_{{t\geq 2}}^{** }(.)={{\left( {\begin{array}{*{20}c} {\frac{{-\frac{{{d_{i,1 }}}}{T}}}{{\sqrt{{\frac{1}{T}\left( {{{{\left( {{d_{i,1 }}-\frac{{{d_{i,1 }}}}{T}} \right)}}^2}+{{{\left( {\frac{{{d_{i,1 }}}}{T}} \right)}}^2}\left( {T-1} \right)} \right)}}}}-} \\ {-\frac{{-\frac{{{d_{j,1 }}}}{T}}}{{\sqrt{{\frac{1}{T}\left( {{{{\left( {{d_{j,1 }}-\frac{{{d_{j,1 }}}}{T}} \right)}}^2}+{{{\left( {\frac{{{d_{j,1 }}}}{T}} \right)}}^2}(T-1)} \right)}}}}} \\ \end{array}} \right)}^2} $$

This simplifies to:

$$ f_{{t\geq 2}}^{** }(.)=\frac{{{T^2}}}{{{{{\left( {T-1} \right)}}^2}}}\frac{{{{{\left( {{d_{i,1 }}\sqrt{{\frac{1}{{{T^2}}}\left( {T-1} \right)d_{j,1}^2}}-{d_{j,1 }}\sqrt{{\frac{1}{{{T^2}}}\left( {T-1} \right)d_{i,1}^2}}} \right)}}^2}}}{{d_{i,1}^2d_{j,1}^2}}= $$

$$ =\frac{{{T^2}}}{{{{{\left( {T-1} \right)}}^2}}}\frac{{{{{\left( {{d_{i,1 }}\left| {{d_{j,1 }}} \right|\sqrt{{\frac{1}{{{T^2}}}\left( {T-1} \right)}}-{d_{j,1 }}\left| {{d_{i,1 }}} \right|\sqrt{{\frac{1}{{{T^2}}}\left( {T-1} \right)}}} \right)}}^2}}}{{d_{i,1}^2d_{j,1}^2}}= $$

$$ = \left\{ {\begin{array}{*{20}{c}} {0\;if\;{{d}_{{i,1}}}{{d}_{{j,1}}} > 0} \\ {\frac{4}{{T - 1\;}}\;if\;{{d}_{{i,1}}}{{d}_{{j,1}}} < 0} \\\end{array}} \right. $$

If d _i,1 d _j,1 > 0 the linear correlation index is equal to 1. This means that:

$$ {\rho_{{ij,t\geq 2}}}=1-\frac{1}{2}f_{{t\geq 2}}^{** }(.)=1 $$

Which is the maximum value for a given ρ _ij,t as argued before.

But, if d _i,1 d _j,1 < 0 the linear correlation index is equal to −1 and

$$ {\rho_{{ij,t\geq 2}}}=1-\frac{1}{2}f_{{t\geq 2}}^{** }(.)=\frac{T-3 }{T-1 } $$

Then as ${{\rho }_{{ij}}} = \frac{{\sum\nolimits_{{t = 1}}^{T} {{{\rho }_{{ij,t}}}} }}{T} = \frac{{{{\rho }_{{ij,1}}} + \sum\nolimits_{{t = 2}}^{T} {{{\rho }_{{ij,t}}}} }}{T} = \frac{{{{\rho }_{{ij,1}}} + \frac{{\left( {T - 1} \right)T - 3}}{{T - 1}}}}{T}$, which leads to:

$$ \frac{{{\rho_{ij,1 }}+T-3}}{T}=-1\mathop{\Leftrightarrow}\limits{\rho_{ij,1 }}=3-2T $$

So over the stationary points ρ _ij,1 takes either the value 1 or 3−2T.

But a stationary point is just a necessary condition to reach a minimum/maximum in a point (its not sufficient). But this also means that if there is global minimum/maximum as the function is continuously differentiable over a open set the maximum/minimum cannot be obtained outside the set of stationary points.

So assuming that the global minimum/maximum exists, the maximum/minimum value the function takes over the set of stationary points is the maximum/minimum of the function. As it only takes two values, the biggest is the global maximum and the smallest is the global minimum. So in fact:

$\max \left( {{\rho_{ij,t }}} \right)=1$ and $\min \left( {{\rho_{ij,t }}} \right)=3-2T$ .

Proposition 3: The purposed period-by-period index is invariable if we add to all elements of either series a constant, being them equal or different across the series

Proof: Lets define the constants by δ and ε, then we get:

$$ \begin{array}{*{20}c} {f_1}\left( {{d_{i,1 }}+\varepsilon,\ldots,{d_{i,T }}+\varepsilon, {d_{j,1 }}+\delta,\ldots,{d_{j,T }}+\delta } \right)= \hfill \\ ={{\left( {\frac{{{d_{i,1 }}+\varepsilon -\overline{{{d_i}+\varepsilon }}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{i,t }}+\varepsilon -\overline{{{d_i}+\varepsilon }}} \right)}}^2}}}}}-\frac{{{d_{j,1 }}+\delta -\overline{{{d_j}+\delta }}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{j,t }}+\delta -\overline{{{d_j}+\delta }}} \right)}}^2}}}}}} \right)}^2}= \hfill \\ ={{\left( {\frac{{{d_{i,1 }}+\varepsilon - {{\overline{d}}_i} - \varepsilon }}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{i,t }}+\varepsilon - {{\overline{d}}_i} - \varepsilon } \right)}}^2}}}}} - \frac{{{d_{j,1 }}+\delta - {{\overline{d}}_j} - \delta }}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{j,t }}+\delta - {{\overline{d}}_j} - \delta } \right)}}^2}}}}}} \right)}^2}= \hfill \\ ={{\left( {\frac{{{d_{i,1 }} - {{\overline{d}}_i}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{i,t }} - {{\overline{d}}_i}} \right)}}^2}}}}} - \frac{{{d_{j,1 }} - {{\overline{d}}_j}}}{{\sqrt{{\frac{1}{T}\mathop{\sum}\nolimits_{t=1}^T{{{\left( {{d_{j,t }} - {{\overline{d}}_j}} \right)}}^2}}}}}} \right)}^2}={f_1}({d_{i,1 }},{d_{i,2 }},\ldots,{d_{i,T }},{d_{j,1 }},{d_{j,2 }},\ldots,{d_{j,T }}) \hfill \\\end{array} $$

Appendix B

Table 1 Country list and classification