Tax haven, pollution haven or both?

Abstract

This paper studies the interplay between two groups of countries, large and small, which compete sequentially on corporate taxes and environmental regulations to attract imperfectly mobile firms. We show that in general, the small countries undercut the large countries in terms of corporate taxes. The small countries choose to be both tax and pollution havens when they are less concerned about the environment than the large countries are and capital integration is low. The large countries never act as both tax havens and pollution havens. Finally, we find that higher firm mobility narrows the tax gap between the large and the small countries but does not affect the optimal environmental policy: tax competition immunizes countries against the detrimental effect of globalization on emission caps.

Appendix

1.1 Number of firms in each country

The number of firms in country m of group i can be described as follows, in terms of the flow of firms between the two groups of countries and the taxes within each group:

$$\begin{aligned} \text {Number of firms in }m_{i}=\left\{ \begin{array}{ll} \left. \begin{array}{ll} s_{i}(1-x_{m_{i}})\; &{} if\;x_{m_{i}}>0\ \\ &{} \\ s_{i}\left( 1-\frac{\sum ^{N_{j}}x_{m_{j}}}{N_{j}}\right) \; &{} if\;x_{m_{j}}>0 \\ &{} \end{array} \right\} \; if \; t_{m_{i}}=t_{m_{i}^{\prime }} &{} \\ &{} \\ \left. \begin{array}{ll} N_{i}s_{i}\; &{} if\;x_{m_{i}}>0 \\ &{} \\ s_{i}\left( 1-\sum ^{N_{j}}x_{m_{j}}\right) \; &{} if\;x_{m_{j}}>0 \\ &{} \end{array} \right\} \; if \; t_{m_{i}}<t_{m_{i}^{\prime }}\;\forall \;m^{\prime }i &{} \\ &{} \\ \left. \begin{array}{ll} 0 \; &{} \quad \quad \quad \quad \quad \quad \quad \forall x_{m_i}, x_{m_j} \end{array} \right. \quad \quad if\;t_{m_{i}}>t_{m_{i}^{\prime }}&\end{array} \right. \end{aligned}$$

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Note that if $x_{m_{i}}>0$ then $x_{m_{j}}=0$ and conversely.

1.2 Tax competition subgame

Maximizing $G_{L}(\alpha _{L},t_{L},\alpha _{S},t_{S})$ and $G_{S}(\alpha _{L},t_{L},\alpha _{S},t_{S})$ w.r.t $t_{L}$ and $t_{S}$ in the tax competition subgame yields:

$$\begin{aligned} \frac{dG_{i}}{dt_{i}}=\frac{dR_{i}}{dt_{i}}-\frac{de_{i}}{dt_{i}}=0, \; i=S,L \end{aligned}$$

with

$$\begin{aligned} \frac{dR_{L}}{dt_{L}}= & {} \frac{N_Ls_{L}-N_is_i x}{N_L}- \frac{s_iN_i}{kN_{L}} t_{L}=\overbrace{s_L}^{\text {tax level effect}}-\overbrace{\frac{s_{i}N_{i}}{ N_L}\left( x+\frac{t_{L}}{k}\right) }^{\text {tax base effect}} \\ \frac{dR_{S}}{dt_{S}}= & {} \frac{N_Ss_{S}+N_is_i x}{N_S}-\left( \frac{ s_{i}N_{i}}{kN_{S}}\right) t_{S} =s_S+\frac{s_{i}N_{i}}{N_S}\left( x- \frac{ t_{S}}{k}\right) \end{aligned}$$

and

$$\begin{aligned} \frac{de_{L}}{dt_{L}}= & {} \frac{q}{2k}\frac{s_iN_i}{ N_{L}} (N_L\alpha _S^2\phi _L-\alpha _L^2(\gamma _L+(N_L-1)\phi _L)) \\ \frac{de_{S}}{dt_{S}}= & {} \frac{q}{2k}\frac{s_iN_i}{ N_{S}} (N_S\alpha _L^2\phi _S-\alpha _S^2(\gamma _S+(N_S-1)\phi _S)) \end{aligned}$$

for $i=L$ when $x>0$, and $i=S$ when $x<0$.

The tax reaction functions are given by:

For $i=L$ when $x>0$ and $i=S$ for $x<0$:

$$\begin{aligned} t_L(t_S)= \frac{1}{4}\left( \frac{2 k N_L s_L}{N_i s_i}+q(\alpha _L^2 N_L \phi _L-\alpha _S^2 N_L \phi _L+\alpha _L^2 \gamma _L -\alpha _L^2 \phi _L+2 \alpha _L -2 \alpha _S) +2 t_S\right) \end{aligned}$$

$$\begin{aligned} t_S(t_L)=\frac{1}{4}\left( \frac{2 k N_S s_S}{ N_i s_i} +q(- \alpha _S^2 N_S \phi _S+\alpha _S^2 N_S \phi _S+\alpha _S^2 \gamma _S -\alpha _S^2 \phi _S-2 \alpha _L +2 \alpha _S) +2 t_L\right) \end{aligned}$$

then the slope of the two tax reaction functions is $\frac{1}{2}$.

1.3 Emission caps

Substituting $t_{1}^{*}(\alpha _{1},\alpha _{2})$ and $t_{2}^{*}(\alpha _{1},\alpha _{2})$ into the payoff functions $G_{1}$ and $G_{2}$ gives

$$\begin{aligned} \frac{dG_{i}}{d\alpha _{i}}=\frac{\partial G_{i}}{\partial \alpha _{i}}+ \overbrace{\frac{\partial G_{i}}{\partial t_{i}^{*}(\alpha _{1},\alpha _{2})}}^{=0}\frac{\partial t_{i}^{*}(\alpha _{1},\alpha _{2})}{\partial \alpha _{i}}+\frac{\partial G_{i}}{\partial t_{j}^{*}(\alpha _{1},\alpha _{2})}\frac{\partial t_{j}^{*}(\alpha _{1},\alpha _{2})}{\partial \alpha _{i}}=0 \end{aligned}$$

Let us set $\Gamma _i=(\phi _i+N_j\gamma _j+(N_i-1)\phi _i)$. The previous equation reduces to:

$$\begin{aligned} \frac{d G_{i}}{d \alpha _{i}}= & {} \frac{q(\alpha _i \Gamma _i-1) \left( qN_L s_L \left( 2(\alpha _j-\alpha _i)+\alpha _i^2 \Gamma _i-\alpha _j^2\Gamma _j\right) -2 k(2 N_i s_i+N_j s_j)\right) }{9 k N_i} \\= & {} \frac{q(\alpha _i \Gamma _i-1) \left( qN_L s_L \Gamma _i \alpha _i^2-2q N_L s_L \alpha _i+qN_Ls_L \alpha _j \left( 2-\alpha _j\Gamma _j\right) -2 k(2 N_i s_i+N_j s_j)\right) }{9 k N_i}=0 \end{aligned}$$

with three solutions:

$$\begin{aligned} \alpha _{i1}= & {} \frac{1}{\Gamma _i}=\frac{-b}{2a} \quad with \quad \overline{ \alpha } _i>\alpha _{i1}>{\underline{\alpha }}_i \\ \alpha _{i2}= & {} \frac{-b+\sqrt{\Delta }}{2a}>{\overline{\alpha }}_i \\ \alpha _{i3}= & {} \frac{-b-\sqrt{\Delta }}{2a}>{\overline{\alpha }}_i \end{aligned}$$

with

$$\begin{aligned} a= & {} qN_L s_L \Gamma _i>0 \\ b= & {} -2q N_L s_L<0 \\ c= & {} qN_L s_L\alpha _j(2-\alpha _j \Gamma _j)-2k(N_js_j+2N_is_i) \\ \Delta= & {} b^2-4ac \end{aligned}$$

The second-order condition (i.e maximum) leads to:

$$\begin{aligned} \frac{d^2G_{i}}{d\alpha _{i}^2}=\frac{q}{9kN_i}\left( a^{\prime } \alpha _i^2-b^{\prime }\alpha _i+c^{\prime }\right) \\ \end{aligned}$$

with

$$\begin{aligned} a^{\prime }= & {} 3q N_L s_L \Gamma _i^2=3 a \Gamma _i>0 \\ b^{\prime }= & {} -6 q N_L s_L \Gamma _i= 3 b \Gamma _i<0 \\ c^{\prime }= & {} qN_L s_L\alpha _j(2-\alpha _j \Gamma _j)\Gamma _i-2k(N_js_j+2N_is_i)\Gamma _i+2qN_Ls_L=c \Gamma _i+2qN_L s_L \\ \end{aligned}$$

which has two solutions for $\frac{d^2G_{i}}{d\alpha _{i}^2}=0$

$$\begin{aligned} {\overline{\alpha }}_i= & {} \frac{-b^{\prime }+\sqrt{\Delta ^{\prime }}}{2a^{\prime }} \\ {\underline{\alpha }}_i= & {} \frac{-b^{\prime }-\sqrt{\Delta ^{\prime }}}{2a^{\prime }} \end{aligned}$$

with $\Delta ^{\prime }=b^2-4a^{\prime }c^{\prime }$.

We can verify that $\Delta ^{\prime }>0\;\iff \;c^{\prime }<\frac{b^{\prime }{}^2}{4a^{\prime }}=qN_Ls_L$ which reduces to:

$k>\frac{q N_Ls_L}{2(2N_is_i+N_js_j)}\frac{\alpha _j(2-\alpha _j\Gamma _j) \Gamma _i-1}{\Gamma _i}$.

We can immediately deduce from () that $\frac{-b^{\prime }}{ 2a^{\prime }}=\frac{-b}{2a}$ while $\Delta ^{\prime }<9\Gamma _i \Delta$. Then $\frac{\sqrt{\Delta ^{\prime }}}{2a^{\prime }}<\frac{\sqrt{\Delta }}{2a}$ and $\alpha _{i2}>{\overline{\alpha }}_i$ and $\alpha _{i3}<{\underline{\alpha }}_i$. Since $a^{\prime }>0$, we know that $\frac{d^2G_{i}}{d\alpha _{i}^2}<0$ for any $\alpha _i \in [ {\underline{\alpha }}_i,{\overline{\alpha }}_i]$, which is not the case of $\alpha _{i2}$ and $\alpha _{i3}$.

Then the concavity condition excludes solutions 2 and 3, which are minima, and solution $\alpha _{i1}$ is the only maximum: $\alpha _i=\frac{1}{ \Gamma _i}$ for $k>\dfrac{qN_Ls_L(\Gamma _i-\Gamma _j)}{ 2\Gamma _i \Gamma _j(2N_is_i+N_js_j) }$ which is the concavity condition at the equilibrium.

1.4 Equilibrium configurations

Table 1 Sets of equilibria depending on the level of capital integration and the ranking of pollution concern gaps for $k>k_{Max}$ and $k>k_{Min}$ with $\Delta _{i}=\gamma _{i}-\phi _{i}$

Let us recall the values of $k_{Max}$ and $k_{Min}$, with: $k_{Max}\equiv \frac{qN_{S}s_{S}(\Delta _{S}-\Delta _{L})}{2\Gamma _{S}\Gamma _{L}(2N_{L}s_{L}+4N_{S}s_{S})}$

$k_{Min}\equiv \frac{qN_{L}s_{L}(\Delta _{L}-\Delta _{S})}{2\Gamma _{S}\Gamma _{L}(2N_{S}s_{S}+4N_{L}s_{L})}$.

The values of the threshold mobility costs are:

For $i=L$ when $x>0$ and $i=S$ when $x<0$:

${\tilde{k}}\equiv \frac{5qN_is_i(\Delta _L-\Delta _S)}{2 \Gamma _L \Gamma _S \left( N_Ls_L-N_Ss_S\right) }$, and $t_{1}^{*}-t_{2}^{*}>0 \iff k> {\tilde{k}}$.

${\bar{k}}\equiv \frac{qs_{i}N_i (\Delta _S-\Delta _L)}{2 \Gamma _L \Gamma _S \left( N_Ls_L-N_Ss_S\right) }$, and $x^{*}>0\iff k>{\bar{k}}$ when $i=L$ and $x^{*}<0\iff k<{\bar{k}}$ when $i=S$

${\hat{k}}=\frac{qN_is_{i}(\Delta _L-\Delta _{S})}{ \Gamma _L \Gamma _S(N_Ls_L-N_Ss_S)}$ and $N_L\times n_{L}^{*}>N_S \times n_{S}^{*}\iff k>{\hat{k}}$.

Recall that $n_L$ is the number of firms in each large country, i.e. $n_L=s_L(1-x)$ when $x>0$ and $n_L=s_L-xs_S\frac{N_S}{N_L}$ when $x<0$,

and $n_S$ is the number of firms in each small country, i.e. $n_S=s_S+s_L x \frac{N_L}{N_S}$ when $x>0$ and $n_S=s_S(1+ x)$ when $x<0$.

Finally, to compare the total payoffs, we determine $N_{L}^{*}G_{L}^{*}-N_{S}^{*}G_{S}^{*}$:

$$\begin{aligned}&N_{L}^{*}G_{L}^{*}-N_{S}^{*}G_{S}^{*}=\frac{ (N_{L}s_{L}+N_{S}s_{S})}{6\Gamma _{L}^{2}\Gamma _{S}^{2}N_{i}s_{i}}\times \nonumber \\&\left( 2k\Gamma _{L}^{2}\Gamma _{S}^{2}(N_{L}s_{L}-N_{S}s_{S})-3qN_{L}s_{L}(\Gamma _{L}^{2}N_{L}\phi _{L}-\Gamma _{S}^{2}N_{S}\phi _{S})\right) -2qN_{L}s_{L}\Gamma _{S}\Gamma _{L}(\Gamma _{L}-\Gamma _{S}) \end{aligned}$$

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It holds that:

${\check{k}}\equiv \frac{qN_is_{i}}{N_Ls_L-N_Ss_S}\left( \frac{ \Delta _L-\Delta _S}{\Gamma _L\Gamma _S}-\frac{3(N_L\phi _L\Gamma _L^2-N_S\phi _S \Gamma _S^2)}{ 2\Gamma _L^2\Gamma _S^2}\right)$ and $N_LG_{L}^{*}-N_SG_{S}^{*}>0\Longleftrightarrow k>{\check{k}}$.

Note that:

${\tilde{k}}>0\Leftrightarrow \Delta _L>\Delta _S$.

${\bar{k}}>0\Leftrightarrow \Delta _L>\Delta _S$

${\hat{k}}>0\Leftrightarrow \Delta _L<\Delta _S$

The sign of ${\check{k}}$ is much more complicated to determine and depends on the ranking of $\Gamma _L$ $\Gamma _S$, $\Delta _L$, $\Delta _S$ and on the ranking of $\phi _S$ and $\phi _S$

Let us rank the threshold values ${\hat{k}},{\check{k}}$ and ${\tilde{k}}$, when positive:

$$\begin{aligned} {\tilde{k}}-{\check{k}}\equiv & {} \frac{3qN_is_{i}}{2} \frac{\Gamma _L\Gamma _S( \Gamma _L-\Gamma _S)+(N_L \Gamma _L^2 \phi _L-N_S \Gamma _S^2 \phi _S)}{ \Gamma _L^2\Gamma _S^2(N_Ls_L-N_Ss_S) }> 0 \\\Leftrightarrow & {} N_S \Gamma _S^2 \phi _S+\Gamma _S^2\Gamma _L<N_L \Gamma _L^2 \phi _L+\Gamma _S\Gamma _L^2 \\\Leftrightarrow & {} N_S \Gamma _S^2 \phi _S-N_L \Gamma _L^2 \phi _L<\Gamma _L\Gamma _S (\Delta _L-\Delta _S) \\&\\ {\hat{k}}-{\check{k}}\equiv & {} \frac{3}{2}s_{i}N_i q\frac{N_S \Gamma _S^2 \phi _S-N_L \Gamma _L^2 \phi _L}{\Gamma _L^2\Gamma _S^2(N_Ls_L-N_Ss_S) } \gtreqqless 0\Leftrightarrow N_S \Gamma _S^2 \phi _S>N_L \Gamma _L^2 \phi _L \\&\\ {\tilde{k}}-{\hat{k}}= & {} 3qs_{i}N_i\frac{(\Gamma _{L}-\Gamma _{S})}{2 \Gamma _L \Gamma _S\left( N_L s_{L}-N_S s_S\right) }\gtreqqless 0\Leftrightarrow \Delta _{S}\lesseqqgtr \Delta _{L} \\&\\ {\bar{k}}-{\check{k}}\equiv & {} -\frac{3}{2}s_{i}N_i q\frac{N_S \Gamma _S^2 \phi _S-N_L \Gamma _L^2 \phi _L}{\Gamma _L^2\Gamma _S^2(N_Ls_L-N_Ss_S) } \gtreqqless 0\Leftrightarrow N_S \Gamma _S^2 \phi _S<N_L \Gamma _L^2 \phi _L \\ \end{aligned}$$

So whenever

(i) $\Delta _S>\Delta _L$, it follows ${\bar{k}}>0$, ${\hat{k}}<0$ and ${\tilde{k}} <0$

(ii) $\Delta _L>\Delta _S$, it holds that ${\tilde{k}}>{\hat{k}}>0$ and ${\bar{k}} <0.$

Moreover, we know that $k_{Max }<0$ for $\Delta _L>\Delta _S$ and $k_{Min}<0$ for $\Delta _S>\Delta _L$. Let us check the ranking of $k_{Max }$ and ${\bar{k}}$ when $\Delta _S>\Delta _L$ and the ranking of $k_{Min}$ and ${\hat{k}}$ when $\Delta _L>\Delta _S$:

• Let us compare ${\bar{k}}$ and $k_{Max}$:

  $$\begin{aligned} {\bar{k}}-k_{Max }=\frac{q(N_Ss_S+N_Ls_L)(N_Ss_S+2N_Ls_L)}{ 4(2N_Ss_S+N_Ls_L)(N_Ls_L-N_Ss_S)}\frac{ (\Delta _S-\Delta _L)}{\Gamma _L \Gamma _S} \end{aligned}$$

  Then ${\bar{k}}>k_{Max}$ when $\Delta _S>\Delta _L$.

• Now let us compare ${\hat{k}}$ and $k_{Min}$ when $\Delta _L>\Delta _S$, which implies $x<0$:

  $$\begin{aligned} {\hat{k}}-k_{Min}=\frac{q(3N_Ss_S+N_Ls_L)3N_Ls_L}{ 4(2N_Ss_S+N_Ls_L)(N_Ls_L-N_Ss_S)}\frac{ (\Delta _L-\Delta _S)}{\Gamma _L \Gamma _S} \end{aligned}$$

  Then ${\hat{k}}>k_{Min}$ when $\Delta _L>\Delta _S$.