Optimal Emission Tax with Endogenous Location Choice of Duopolistic Firms

Abstract

This study explores an optimal (pre-committed or ex-ante) environmental tax policy in a three-stage game in which polluting firms strategically choose the location of their plants after the government has chosen the optimal emission tax rate. We show not only that the optimal emission tax is non-decreasing with the declining cost of relocation (e.g., setup or fixed costs), or else, the progress of globalization but also that the firms may move back their relocated plants to the home country, causing the resulting welfare to decline. As a consequence, the domestic welfare varies in a non-monotonic way. We also show that such a counterintuitive non-monotonic relationship does not arise under time-consistent (ex-post) emission taxes.

Appendix

Lemma 1

When $0\le d\le 2/3$, $W_{\textit{HH}}$ is the largest for $t\in \left[ 0,A/2\right]$.¹²

Proof

It follows from (13) that $g_{3}(t;d)=18(W_{\textit{HH}}-W_{\textit{HF}})=0$ has one negative root, $t_{3}$, and one positive root, $t_{6\text {, }}$for $0\le d\le 2/3$. Hence, $g_{3}(t;d)\ge 0$ for $t\in \left[ t_{3},t_{6} \right]$, while it is negative for $t\notin \left[ t_{3},t_{6}\right]$. Moreover, it is easy to see that $0<W_{\textit{FF}}<W_{\textit{HF}}$ $(0;d)\le W_{\textit{HH}}(0;d)$ and that $W_{\textit{HF}}(A/2;d)<W_{\textit{FF}}<W_{\textit{HH}}(A/2;d)$. $W_{\textit{HH}}(t;d)$ is a concave function of $t$ and, $W_{\textit{FF}}$ is a constant function. Taken together, $W_{\textit{HH}}$ is the largest in $t\in \left[ 0,A/2\right]$. $\square$

Lemma 2

When $2/3<d\le d^{*}$, $W_{\textit{HF}}$ is the largest for $t\in \left[ 0,t_{3} \right]$, while $W_{\textit{HH}}$ is the largest if $t\in \left[ t_{3},t_{5}\right]$ , where $d^{*}$ represents the threshold value at which $W_{\textit{HH}}(t_{3};d^{*})=W_{\textit{HF}}(t_{3};d^{*})=W_{\textit{FF}}$.

Proof

Differentiating $W_{\textit{HH}}$ and $W_{\textit{HF}}$ with respect to $d$ yields

$$\begin{aligned} \frac{\partial W_{\textit{HH}}(t;d)}{\partial d}=-\frac{2}{9}(t-A)^{2}<0\text { and } \frac{\partial W_{\textit{HF}}(t;d)}{\partial d}=-\frac{1}{18}(A-2t)^{2}<0, \end{aligned}$$

(20)

for $t<A/2$. This implies that the functions $W_{\textit{HH}}(t;d)$ and $W_{\textit{HF}}(t;d)$ are decreasing in $d$, whereas the function $W_{\textit{FF}}$ takes a constant value independently of $d$. Hence, when $d=d^{*}$, the functions $W_{\textit{HH}}(t;d)$ and $W_{\textit{HF}}(t;d)$ take a minimum value for each $t$. We compare the welfare functions at $d=d^{*}$. Because when $2/3<d^{*}$, the quadratic equation $g_{3}(t;d^{*})=0$ has two positive real roots, $0<t_{3}<t_{6}$ , and because the graph of $g_{3}(t;d^{*})$ is inverse U-shaped in $t$, $g_{3}(t;d^{*})\equiv 18\left[ W_{\textit{HH}}(t;d^{*})-W_{\textit{HF}}(t;d^{*}) \right] \le 0$ for $t\in \left[ 0,t_{3}\right]$ implies that $W_{\textit{HF}}(t;d^{*})\ge W_{\textit{HH}}(t;d^{*})$ for $t\in \left[ 0,t_{3}\right]$. Next, we compare between $W_{\textit{HF}}(t;d^{*})$ and $W_{\textit{FF}}$. When $t=0$,

$$\begin{aligned} W_{\textit{HF}}(0;d^{*})=\frac{1}{18}\left( 6-d^{*}\right) A^{2}>\frac{2}{9} A^{2}=W_{\textit{FF}}, \end{aligned}$$

where the inequality follows from the fact that $d^{*}<15/7$, while $W_{\textit{HF}}(t_{3};d^{*})=W_{FF\text { }}$ by definition of $d^{*}$. Taken together, $W_{\textit{HF}}(t;d)$ is the largest for $t\in \left[ 0,t_{3}\right]$.

Consider a case where $t\in \left[ t_{3},t_{5}\right]$. Because $g_{3}(t;d)\ge 0$ for $t\in \left[ t_{3},t_{6}\right]$ and because $t_{5}<t_{6}$, $g_{3}(t;d)\ge 0$ for $t\in \left[ t_{3},t_{5}\right]$. This implies that $W_{\textit{HH}}\ge W_{\textit{HF}}$ for $t\in \left[ t_{3},t_{5}\right]$. In addition, by definition of $d^{*}$, $W_{\textit{FF}}=W_{\textit{HF}}(t_{3}^{*};d^{*})<W_{\textit{HF}}(t;d^{*})$. Combining these facts, $W_{\textit{HH}}(t;d)$ is the largest for $t\in \left[ t_{3},t_{5}\right]$. $\square$

Lemma 3

When $d^{*}<d<5/4$, $W_{\textit{HF}}$ is the largest for $t\in \left[ 0,t_{4} \right]$, $W_{\textit{FF}}$ is the largest for $t\in \left[ t_{4},t_{2}\right]$ and $W_{\textit{HH}}$ is the largest for $t\in \left[ t_{2},t_{5}\right]$.

Proof

For $t\in \left[ 0,t_{4}\right]$ $g_{2}(t;d)\le 0$ implies $W_{\textit{FF}}\le W_{\textit{HF}}$. Because $W_{\textit{HH}}(t;d)$ and $W_{\textit{HF}}(t;d)$ are decreasing functions of $d$, the intersection between the curves $W_{\textit{HH}}(t;d)$ and $W_{\textit{HF}}(t;d)$ is situated below the horizontal line $W_{\textit{FF}}$ for $d^{*}\le d$. As shown in Fig. 5, we have $t_{4}<t_{3}<t_{2}$. An inspection of Fig. 5 delivers the desired result. $\square$

Lemma 4

When $5/4\le d\le 2$, $W_{\textit{HF}}$ is the largest for $t\in \left[ 0,t_{4} \right]$, whereas $W_{\textit{FF}}$ is the largest for $t\in \left[ t_{4},A^{2}/2 \right]$. When $2<d\le 15/7$, $W_{\textit{FF}}$ is the largest for $t\in \left[ 0,t_{1}\right]$ or $t\in \left[ t_{4},A^{2}/2\right]$, while $W_{\textit{HF}}$ is the largest for $t\in \left[ t_{1},t_{4}\right]$.

Proof

It follows from (11) that $W_{\textit{HH}}$ is never the largest, because $W_{\textit{HH}}$ cannot intersect $W_{\textit{FF}}$ (recall that the quadratic equation $g_{1}(t;d)=0$ has complex roots). Hence, we compare $W_{\textit{HF}}$ with $W_{\textit{FF}}$. When $5/4\le d\le 2$, $W_{\textit{HF}}(t;d)$ is the largest for $t\in \left[ 0,t_{4} \right]$ because $g_{2}(t;d)\equiv 18\left[ W_{\textit{FF}}-W_{\textit{HF}}\right] \le 0$ for $t\in \left[ 0,t_{4}\right]$, while $W_{\textit{FF}}$ is the largest for $t\in \left[ t_{4},A/2\right]$ because $g_{2}(t;d)\ge 0$ for $t\ge t_{4}$. When $2<$ $d\le 15/7$, on the other hand, because $g_{2}(t;d)\le 0$ for $t\in [0,t_{1}]$, $W_{\textit{FF}}$ is the largest for $t\in [0,t_{1}]$. The rest of the proof is the same as others. $\square$