A Tree-form Constant Market Share Model for Growth Causes in International Trade Based on Multi-level Classification

Abstract

This paper introduces a tree-form constant market share (CMS) model for analyzing growth causes in international trade based on multi-level classification. The tree-form CMS is a collection of CMS models at different levels, including the entire, branch- and leaf-models, which consists of a large amount of information and has a wide application spectrum. Basic properties of this model are investigated in detail. It is shown that the tree-form CMS model is superior to other CMS models in the literature. It is also shown that well known CMS formulations are special cases of a linear class with two parameters, which control how the interaction term is divided into the demand growth and competitive terms. Application to bilateral trade between China and Germany shows that the growth causes in different periods are clearly different. It is shown that the outputs of the tree-form CMS model can be used for further suitable statistical analysis. Furthermore, our theoretical findings are also confirmed by those data examples.

Appendix: Proofs of the results

1.1 Proof of Theorem 1

• Case 1:

  • a) Since $s_1^0=s_2^0=\ldots =s_n^0$, we have ${s^0}=s_1^0$. This results in

    $$ \begin{array}{*{20}c} {E_1^D=\sum\limits_i {s_i^0\varDelta {Q_i}-{s^0}\varDelta Q=s_1^0\varDelta {Q_1}+s_2^0\varDelta {Q_2}+\ldots +s_n^0\varDelta {Q_n}-{s^0}\varDelta Q} } \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad =s_1^0\varDelta {Q_1}+s_1^0\varDelta {Q_2}+\ldots +s_1^0\varDelta {Q_n}-s_1^0\varDelta Q} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad =s_1^0\left( {\varDelta {Q_1}+\varDelta {Q_2}+\ldots +\varDelta {Q_n}} \right)-s_1^0\varDelta Q} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad =s_1^0\varDelta Q-s_1^0\varDelta Q=0.} \hfill \\ \end{array} $$

  • b) Since $s_1^1=s_2^1=\ldots =s_n^1$, we have ${s^1}=s_1^1$. This results in

    $$ \begin{array}{*{20}c} {E_1^C=\sum\limits_i {\varDelta {s_i}Q_i^0-\varDelta s{Q^0}=\left( {s_1^1-s_1^0} \right)Q_1^0+\left( {s_2^1-s_2^0} \right)Q_2^0+\ldots +\left( {s_n^1-s_n^0} \right)Q_n^0-\left( {{s^1}-{s^0}} \right){Q^0}} } \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad =s_1^1Q_1^0-s_1^0Q_1^0+s_2^1Q_2^0-s_2^0Q_2^0+\ldots +s_n^1Q_n^0-s_n^0Q_n^0-{s^1}{Q^0}+{s^0}{Q^0}} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad =s_1^1Q_1^0+s_1^1Q_2^0+\ldots +s_1^1Q_n^0-s_1^1{Q^0}-q_1^0-q_2^0-\ldots -q_n^0+{q^0}} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad =s_1^1\left( {Q_1^0+Q_2^0+\ldots +Q_n^0} \right)-s_1^1{Q^0}-{q^0}+{q^0}} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad =s_1^1{Q^0}-s_1^1{Q^0}=0.} \hfill \\ \end{array} $$

  • c) Under the conditions of a) and b), we have $E_1^D=0$ and $E_1^C=0$.

    Hence $E_1^I=0$, because $E_1^D + E_1^C+E_1^I=0.$

• Case 2: Since $Q_1^0:Q_2^0:\ldots :Q_n^0=Q_1^1:Q_2^1:\ldots :Q_n^1=1:{C_2}:\ldots :{C_n}$, we have

  $$ \begin{array}{*{20}c} {\sum\limits_i {s_i^0} \varDelta {Q_i}=s_1^0\varDelta {Q_1}+s_2^0\varDelta {Q_2}+\ldots +s_n^0\varDelta {Q_n}=s_1^0\left( {Q_1^1-Q_1^0} \right)+s_2^0\left( {Q_2^1-Q_2^0} \right)+\ldots +s_n^0\left( {Q_n^1-Q_n^0} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = s_1^0\left( {Q_1^1-Q_1^0} \right)+s_2^0\left( {{C_2}Q_1^1-{C_2}Q_1^0} \right)+\ldots +s_n^0\left( {{C_n}Q_1^1-{C_n}Q_1^0} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = s_1^0\left( {Q_1^1-Q_1^0} \right)+{C_2}s_2^0\left( {Q_1^1-Q_1^0} \right)+\ldots +{C_n}s_n^0\left( {Q_1^1-Q_1^0} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \left( {Q_1^1-Q_1^0} \right)\left( {s_1^0+{C_2}s_2^0+\ldots +{C_n}s_n^0} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \left( {Q_1^1-Q_1^0} \right)\left( {\frac{{q_1^0}}{{Q_1^0}}+{C_2}\frac{{q_2^0}}{{Q_2^0}}+\ldots +{C_n}\frac{{q_n^0}}{{Q_n^0}}} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \left( {Q_1^1-Q_1^0} \right)\left( {\frac{{q_1^0}}{{Q_1^0}}+{C_2}\frac{{q_2^0}}{{{C_2}Q_1^0}}+\ldots +{C_n}\frac{{q_n^0}}{{{C_n}Q_1^0}}} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \left( {Q_1^1-Q_1^0} \right)\left( {\frac{{q_1^0+q_2^0+\ldots +q_n^0}}{{Q_1^0}}} \right)=\left( {Q_1^1-Q_1^0} \right)\frac{{{q^0}}}{{{Q^0}}}\frac{{{Q^0}}}{{Q_1^0}}} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = {s^0}\left( {Q_1^1-Q_1^0} \right)\frac{{Q_1^0+Q_2^0+\ldots +Q_n^0}}{{Q_1^0}}} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = {s^0}\left( {Q_1^1-Q_1^0} \right)\left( {1+{C_2}+\ldots +{C_n}} \right).} \hfill \\ \end{array} $$
  $$ \begin{array}{*{20}c} {\mathrm{And}\quad {s^0}\varDelta Q={s^0}\left( {{Q^1}-{Q^0}} \right)={s^0}\left( {Q_1^1+Q_2^1+\ldots +Q_n^1-Q_1^0-Q_2^0-\ldots -Q_n^0} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = {s^0}\left[ {\left( {Q_1^1-Q_1^0} \right)+\left( {Q_2^1-Q_2^0} \right)+\ldots +\left( {Q_n^1-Q_n^0} \right)} \right]} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = {s^0}\left[ {\left( {Q_1^1-Q_1^0} \right)+{C_2}\left( {Q_1^1-Q_1^0} \right)+\ldots +{C_n}\left( {Q_1^1-Q_1^0} \right)} \right]} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = {s^0}\left( {Q_1^1-Q_1^0} \right)\left( {1+{C_2}+\ldots +{C_n}} \right).} \hfill \\ \end{array} $$

  Hence $E_1^D=\sum\limits_i {s_i^0\varDelta {Q_i}} -{s^0}\varDelta Q=0.$

  $$ \begin{array}{*{20}c} {\mathrm{Furthermore},\quad \sum\limits_i {\varDelta {s_i}Q_i^0} =\varDelta {s_1}Q_1^0+\varDelta {s_2}Q_2^0+\ldots +\varDelta {s_n}Q_n^0} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = \varDelta {s_1}Q_1^0+\varDelta {s_2}{C_2}Q_1^0+\ldots +\varDelta {s_n}{C_n}Q_1^0} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left( {\varDelta {s_1}+{C_2}\varDelta {s_2}+\ldots +{C_n}\varDelta {s_n}} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left[ {\left( {\frac{{q_1^1}}{{Q_1^1}}-\frac{{q_1^0}}{{Q_1^0}}} \right)+{C_2}\left( {\frac{{q_2^1}}{{Q_2^1}}-\frac{{q_2^0}}{{Q_2^0}}} \right)+\ldots +{C_n}\left( {\frac{{q_n^1}}{{Q_n^1}}-\frac{{q_n^0}}{{Q_n^0}}} \right)} \right]} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left[ {\left( {\frac{{q_1^1}}{{Q_1^1}}-\frac{{q_1^0}}{{Q_1^0}}} \right)+{C_2}\left( {\frac{{q_2^1}}{{{C_2}Q_1^1}}-\frac{{q_2^0}}{{{C_2}Q_1^0}}} \right)+\ldots +{C_n}\left( {\frac{{q_n^1}}{{{C_n}Q_1^1}}-\frac{{q_n^0}}{{{C_n}Q_1^0}}} \right)} \right]} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left[ {\left( {\frac{{q_1^1}}{{Q_1^1}}+\frac{{q_2^1}}{{Q_1^1}}+\ldots +\frac{{q_n^1}}{{Q_1^1}}} \right)-\left( {\frac{{q_1^0}}{{Q_1^0}}+\frac{{q_2^0}}{{Q_1^0}}+\ldots +\frac{{q_n^0}}{{Q_1^0}}} \right)} \right]} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left( {\frac{{{q^1}}}{{Q_1^1}}-\frac{{{q^0}}}{{Q_1^0}}} \right)=Q_1^0\left( {\frac{{{q^1}}}{{{Q^1}}}\frac{{{Q^1}}}{{Q_1^1}}-\frac{{{q^0}}}{{{Q^0}}}\frac{{{Q^0}}}{{Q_1^0}}} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left( {\frac{{{q^1}}}{{{Q^1}}}\frac{{Q_1^1+Q_2^1+\ldots +Q_n^1}}{{Q_1^1}}-\frac{{{q^0}}}{{{Q^0}}}\frac{{Q_1^0+Q_2^0+\ldots +Q_n^0}}{{Q_1^0}}} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left( {\frac{{{q^1}}}{{{Q^1}}}\frac{{Q_1^1+{C_2}Q_1^1+\ldots +{C_n}Q_1^1}}{{Q_1^1}}-\frac{{{q^0}}}{{{Q^0}}}\frac{{Q_1^0+{C_2}Q_1^0+\ldots +{C_n}Q_1^0}}{{Q_1^0}}} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left[ {\frac{{{q^1}}}{{{Q^1}}}\frac{{Q_1^1\left( {1+{C_2}+\ldots +{C_n}} \right)}}{{Q_1^1}}-\frac{{{q^0}}}{{{Q^0}}}\frac{{Q_1^0\left( {1+{C_2}+\ldots +{C_n}} \right)}}{{Q_1^0}}} \right]} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad = Q_1^0\left( {1+{C_2}+\ldots +{C_n}} \right)\left( {\frac{{{q^1}}}{{{Q^1}}}-\frac{{{q^0}}}{{{Q^0}}}} \right),} \hfill \\ \end{array} $$
  $$ \begin{array}{*{20}c} {\mathrm{and}\quad \varDelta s{Q^0}=\left( {\frac{{{q^1}}}{{{Q^1}}}-\frac{{{q^0}}}{{{Q^0}}}} \right)\left( {Q_1^0+Q_2^0+\ldots +Q_n^0} \right)=\left( {\frac{{{q^1}}}{{{Q^1}}}-\frac{{{q^0}}}{{{Q^0}}}} \right)\left( {Q_1^0+{C_2}Q_1^0+\ldots +{C_n}Q_1^0} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =Q_1^0\left( {1+{C_2}+\ldots +{C_n}} \right)\left( {\frac{{{q^1}}}{{{Q^1}}}-\frac{{{q^0}}}{{{Q^0}}}} \right).} \hfill \\ \end{array} $$

  Hence $E_1^C=\sum\limits_i {\varDelta {s_i}} Q_i^0-\varDelta s{Q^0}=0.$

  Finally we have $E_1^I=0$, because $E_1^D+E_1^C+E_1^I=0.$

1.2 Proof of Corollary 1

• a) When n = 2, we have

  $$ \begin{array}{*{20}c} {E_1^D=s_1^0\varDelta Q{}_1+s_2^0\varDelta Q{}_2-{s^0}\varDelta Q=\frac{{q_1^0}}{{Q_1^0}}\left( {Q_1^1-Q_1^0} \right)+\frac{{q_2^0}}{{Q_2^0}}\left( {Q_2^1-Q_2^0} \right)-\frac{{q_1^0+q_2^0}}{{Q_1^0+Q_2^0}}\left( {Q_1^1+Q_2^1-Q_1^0-Q_2^0} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac{{Q_1^1}}{{Q_1^0}}q_1^0-\frac{{Q_1^1+Q_2^1}}{{Q_1^0+Q_2^0}}q_1^0-\frac{{Q_1^1+Q_2^1}}{{Q_1^0+Q_2^0}}q_2^0+\frac{{Q_2^1}}{{Q_2^0}}q_2^0} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac{{Q_1^1Q_2^0-Q_1^0Q_2^1}}{{Q_1^0}}q_1^0-\frac{{Q_1^1Q_2^0-Q_1^0Q_2^1}}{{Q_2^0}}q_2^0} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\left( {Q_1^0Q_2^1-Q_1^1Q_2^0} \right)\left( {s_2^0-s_1^0} \right).} \hfill \\ \end{array} $$

  Hence $E_1^D$ is positive (zero or negative), iff $\left( {Q_1^0Q_2^1-Q_1^1Q_2^0} \right)\left( {s_1^0-s_2^0} \right)<0\left( {=0\,\mathrm{or} > 0} \right).$

• b) Furthermore, we have

  $$ \begin{array}{*{20}c} {E_1^C=\varDelta {s_1}Q_1^0+\varDelta {s_2}Q_2^0-\varDelta s{Q^0}=\left( {\frac{{q_1^1}}{{Q_1^1}}-\frac{{q_1^0}}{{Q_1^0}}} \right)Q_1^0+\left( {\frac{{q_2^1}}{{Q_2^1}}-\frac{{q_2^0}}{{Q_2^0}}} \right)Q_2^0-\left( {\frac{{q_1^1+q_2^1}}{{Q_1^1+Q_2^1}}-\frac{{q_1^0+q_2^0}}{{Q_1^0+Q_2^0}}} \right)\left( {Q_1^0+Q_2^0} \right)} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac{{Q_1^0}}{{Q_1^1}}q{}_1^1-\frac{{Q_1^0+Q_2^0}}{{Q_1^1+Q_2^1}}q_1^1-\frac{{Q_1^0+Q_2^0}}{{Q_1^1+Q_2^1}}q_2^1+\frac{{Q_2^0}}{{Q_2^1}}q{}_2^1} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{{Q_1^0Q_2^1-Q_1^1Q_2^0}}{{Q_1^1}}q_1^1-\frac{{Q_1^0Q_2^1-Q_1^1Q_2^0}}{{Q_2^1}}q{}_2^1} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \left( {Q_1^0Q_2^1-Q_1^1Q_2^0} \right)\left( {s_1^1-s_2^1} \right).} \hfill \\ \end{array} $$

  Hence $E_1^C$ is positive (zero or negative), iff $\left( {Q_1^0Q_2^1-Q_1^1Q_2^0} \right)\left( {s_1^1-s_2^1} \right)>0\left( {=0\,or < 0} \right)$.

• c) For $E_1^I$, we have

  $$ \begin{array}{*{20}c} {E_1^I=\varDelta {s_1}\varDelta {Q_1}+\varDelta {s_2}\varDelta {Q_2}-\varDelta s\varDelta Q} \\ {=\left( {\frac{{q_1^1}}{{Q_1^1}}-\frac{{q_1^0}}{{Q_1^0}}} \right)\left( {Q_1^1-Q_1^0} \right)+\left( {\frac{{q_2^1}}{{Q_2^1}}-\frac{{q_2^0}}{{Q_2^0}}} \right)\left( {Q_2^1-Q_2^0} \right)-\left( {\frac{{q_1^1+q_2^1}}{{Q_1^1+Q_2^1}}-\frac{{q_1^0+q_2^0}}{{Q_1^0+Q_2^0}}} \right)\left( {Q_1^1+Q_2^1-Q_1^0-Q_2^0} \right)} \\ {=\frac{{Q_1^0Q_2^1-Q_1^1Q_2^0}}{{Q_1^0\left( {Q_1^0+Q_2^0} \right)}}q_1^0-\frac{{Q_1^0Q_2^1-Q_1^1Q_2^0}}{{Q_2^0\left( {Q_1^0+Q_2^0} \right)}}q_2^0-\frac{{Q_1^0Q_2^1-Q_1^1Q_2^0}}{{Q_1^1\left( {Q_1^1+Q_2^1} \right)}}q_1^1+\frac{{Q_1^0Q_2^1-Q_1^1Q_2^0}}{{Q_2^1\left( {Q_1^1+Q_2^1} \right)}}q_2^1} \\ {=\frac{{Q_1^0Q_2^1-Q_1^1Q_2^0}}{{{Q^0}}}\left( {s_1^0-s_2^0} \right)-\frac{{Q_1^0Q_2^1-Q_1^1Q_2^0}}{{{Q^1}}}\left( {s_1^1-s_2^1} \right).} \\ \end{array} $$

  $E_1^I$ is positive (zero or negative), iff $\left( {Q_1^0Q_2^1-Q_1^1Q_2^0} \right)\left[ {\left( {s_1^0-s_2^0} \right){Q^1}-\left( {s_1^1-s_2^1} \right){Q^0}} \right]>0\left( {=0\,or < 0} \right)$.

• d) It is clear the terms $E_1^D$, $E_1^C$ and $E_1^I$ all vanish, if $Q_1^0Q_2^1=Q_1^1Q_2^0$ or $s_1^0=s_2^0$ and $s_1^1=s_2^1$. On the other hand, if $Q_1^0Q_2^1\ne Q_1^1Q_2^0$ and $s_1^0\ne s_2^0$ or $s_1^1\ne s_2^1$ at least one of $E_1^D$ and $E_1^C$ is non-zero. This finishes the proof.