The Signaling Effect of Emission Taxes Under International Duopoly

Abstract

We analyze equilibrium emission taxes under international Cournot duopoly in the presence of asymmetric information between firms. When each firm does not know the production cost of its rival located in the other country, the government of each country can use its emission tax as a signal of the competitiveness of its domestic firm. We show that, owing to this signaling effect, the equilibrium emission tax and expected welfare of each country can be higher than those under symmetric information between firms or even higher than those obtained if the governments were to non-strategically set the tax rate equal to the (expected) marginal damage from pollution (i.e., the Pigouvian tax level). Our results suggest that the presence of asymmetric information between firms can mitigate a “race to the bottom” in strategic environmental policymaking.

Appendices

Proof of Lemma 1

In the non-signaling equilibrium, by taking the first-order condition of (5) with respect to $t_i$ and rearranging it, we obtain

$$\begin{aligned} t_i= & {} \lambda k q_i\big (t_i, c_i; t_j\big (c^L\big ),c^L\big )+(1-\lambda )k q_i\big (t_i, c_i; t_j\big (c^H\big ),c^H\big ) \nonumber \\&-\,\frac{1}{2}\big [\lambda q_i\big (t_i, c_i; t_j\big (c^L\big ),c^L\big )+(1-\lambda )q_i\big (t_i, c_i; t_j\big (c^H\big ),c^H\big )\big ], \end{aligned}$$

(15)

which can be further rearranged as (6). In the right-hand side of (15), the first two terms represent the expected marginal damage from pollution and the last term represents the government’s strategic incentive to lower its emission tax rate (i.e., the pre-commitment effect). In the symmetric equilibrium with $t_A(c^m)=t_B(c^m)=t^N(c^m)$, $m=L,H$, the equilibrium expected tax rate, $E(t^N)$, is given as a solution to $E(t^N)=\lambda R^{N}(c^L;E(t^N))+(1-\lambda )R^{N}(c^H;E(t^N))$; thus, we obtain

$$\begin{aligned} E(t^N)=\frac{(2k-1)(a-\bar{c})}{2k+5}. \end{aligned}$$

(16)

By substituting $c_i=c^m$ and $E(t_j)=E(t^N)$ into (6), we obtain (7).

Next, in the Pigouvian equilibrium, government i chooses the tax rate such that

$$\begin{aligned} t_i=\lambda k q_i\big (t_i, c_i; t_j\big (c^L\big ),c^L\big )+(1-\lambda ) k q_i\big (t_i, c_i; t_j\big (c^H\big ),c^H\big ). \end{aligned}$$

This can be rewritten as the following best-response tax rate for government i:

$$\begin{aligned} R^P(c_i;E(t_j))=\frac{k(a-2c_i+\bar{c}+E(t_j))}{2k+3}. \end{aligned}$$

(17)

In the symmetric equilibrium with $t_A(c^m)=t_B(c^m)=t^P(c^m)$, $m=L,H$, the equilibrium expected tax rate, $E(t^P)$, is given as a solution to $E(t^P)=\lambda R^P(c^L; E(t^P))+(1-\lambda )R^P(c^H; E(t^P))$; thus, we obtain $E(t^P)=k(a-\bar{c})/(k+3)$. By substituting $c_i=c^m$ and $E(t_j)=E(t^P)$ into (17), we obtain (8).

Furthermore, since $R^N(c^m;E(t_j))-R^P(c^m;E(t_j))=-3(a-2c^m+\bar{c}+E(t_j))/[4(k+1)(2k+3)]<0$ for $m=L,H$ and $E(t^N)-E(t^P)=-3(a-\bar{c})/[(k+3)(2k+5)]<0$, we obtain

$$\begin{aligned} t^N(c^m)=R^N(c^m;E(t^N))<R^P(c^m;E(t^N))<R^P(c^m;E(t^P))=t^P(c^m), \quad m=L,H. \end{aligned}$$

Finally, the exact expressions for the expected welfare in the non-signaling and Pigouvian equilibria, $EW^N(c^m)$ and $EW^P(c^m)$ for $m=L,H$ respectively, are given by

$$\begin{aligned} EW^e(c^m)&=\lambda \left[ \left( \frac{a-2c^m+c^L+t^e(c^m)+t^e(c^L)}{3}\right) \left( \frac{a-2c^m+c^L-2t^e(c^m)+t^e(c^L)}{3}\right) \right. \nonumber \\&\qquad \left. -\,\frac{k}{2}\left( \frac{a-2c^m+c^L-2t^e(c^m)+t^e(c^L)}{3}\right) ^2\right] \nonumber \\&\qquad +\,(1-\lambda )\left[ \left( \frac{a-2c^m+c^H+t^e(c^m)+t^e(c^H)}{3}\right) \right. \nonumber \\&\qquad \times \left( \frac{a-2c^m+c^H-2t^e(c^m)+t^e(c^H)}{3}\right) \nonumber \\&\qquad \qquad \quad \left. -\,\frac{k}{2}\left( \frac{a-2c^m+c^H-2t^e(c^m)+t^e(c^H)}{3}\right) ^2\right] \end{aligned}$$

for $e=N,P$ and $m=L,H$, where $t^e(\cdot )$ is given by (7) or (8).

Proofs of Lemma 2 and Proposition 1

The proofs consist of the following steps. First, supposing that government j employs tax rule $t_j(c_j)$ with $t_j(c^L) \ne t_j(c^H)$, we derive a best-response function for each type of government i, $R^S(c^m;E(t_j))$, $m=L,H$, that reflects the need to signal its type (Sect. B.1). Next, using the obtained best-response functions, we verify that there always exists a unique equilibrium expected tax rate, $E(t^S)$ (Sect. B.2). Then, to complete the proofs of Lemma 2 and Proposition 1, we derive the equilibrium emission tax rates, $t^S(c^m)$, $m=L,H$, and verify that the strategies and beliefs do provide a separating equilibrium for two separate cases (Sects. B.3 and B.4). Finally, we show that all pooling equilibria are eliminated under the refinement of the Intuitive Criterion (Sect. B.5).

1.1 Derivation of the Best-Response Functions

Suppose that government j plays the separating strategy $t_j(c_j)$ with $t_j(c^L) \ne t_j(c^H)$. In this proof, for notational simplicity, we abbreviate $EW_i(t_i, c_i, \mu _j(t_i) ; t_j(c^L),t_j(c^H))$ as $EW_i(t_i, c_i, \mu _j)$.

If there were no signaling considerations, $EW_i(t_i, c^m, 1)$ and $EW_i(t_i, c^m, 0)$ would be maximized, respectively, at

$$\begin{aligned} \tau ^{m1}(E(t_j))&=\frac{(2k-1)(2a-3c^m-c^L+2\bar{c}+2E(t_j))}{8(k+1)}, \quad m=L,H, \end{aligned}$$

(18)

$$\begin{aligned} \tau ^{m0}(E(t_j))&=\frac{(2k-1)(2a-3c^m-c^H+2\bar{c}+2E(t_j))}{8(k+1)}, \quad m=L,H, \end{aligned}$$

(19)

where $E(t_j)=\lambda t_j(c^L)+(1-\lambda )t_j(c^H)$. From Assumption 1, we obtain $\tau ^{L1}>\tau ^{L0}>\tau ^{H1}>\tau ^{H0}$.

Note first that $EW_i(t_i, c_i, \mu _j)$ is not necessarily monotonic with respect to $\mu _j$, because

$$\begin{aligned} \frac{\partial EW_i(t_i, c_i, \mu _j)}{\partial \mu _j}=\frac{c^H-c^L}{36}&\left\{ (2-k)\left[ 2a-3c_i-(\mu _j c^L+(1-\mu _j)c^H)+2\bar{c}+2E(t_j)\right] \right. \nonumber \\&+2(2k-1)t_i \Bigr \} \end{aligned}$$

(20)

is always non-negative when $k \le 2$ but can be either positive or negative when $k > 2$. We here focus on $EW_i(\tau ^{m0},c^m,\mu _j)$, $m=L,H$, and obtain the following lemma.

Lemma B1

$min_{\mu _j}EW_i(\tau ^{m0},c^m,\mu _j)=min\{EW_i(\tau ^{m0},c^m,1), EW_i(\tau ^{m0},c^m,0)\}$, $m=L,H$.

Proof

When $k\le 2$, (20) implies that $\partial EW_i(\tau ^{m0},c^m,\mu _j)/\partial \mu _j$ is always positive. Then, $EW_i(\tau ^{m0},c^m,\mu _j)$ is increasing in $\mu _j \in [0,1]$ and reaches its minimum at $\mu _j=0$. When $k>2$, $EW_i(\tau ^{m0},c^m,\mu _j)$ is concave with respect to $\mu _j$ because $\partial ^2 EW_i/\partial \mu _j^2=-(k-2)(c^H-c^L)^2/36<0$. Thus, $EW_i(\tau ^{m0},c^m,\mu _j)$ reaches its minimum either at $\mu _j=1$ or at $\mu _j=0$. $\square$

For $m=H$, by comparing $EW_i(\tau ^{H0}, c^H, 1)$ and $EW_i(\tau ^{H0}, c^H, 0)$, we obtain

$$\begin{aligned} EW_i\big (\tau ^{H0}, c^H, 1\big )&\gtreqless EW_i\big (\tau ^{H0}, c^H, 0\big ) \\&\Longleftrightarrow a-2c^H+\bar{c}+E(t_j)\\&\gtreqless \frac{1}{9}\big (c^H-c^L\big )(k+1)(k-2). \end{aligned}$$

Therefore, in the following analysis, we consider two possible cases separately.

Case 1$a-2c^H+\bar{c}+E(t_j) > (c^H-c^L)(k+1)(k-2)/9$.

In this case, $EW_i(\tau ^{H0}, c^H, 1) > EW_i(\tau ^{H0}, c^H, 0)$ and Lemma B1 implies that $EW_i(\tau ^{H0},c^H,\mu _j) \ge EW_i(\tau ^{H0},c^H,0)$ for any $\mu _j \in [0,1]$. We first show that the best response for the high-type government is $R^S(c^H;E(t_j))=\tau ^{H0}$ in the separating equilibrium. Suppose instead that the government chose $\tilde{t} \ne \tau ^{H0}$ and obtained $EW_i(\tilde{t},c^H,0)$ in the equilibrium. Then, if the government deviated to $\tau ^{H0}$, it would be strictly better off whatever belief firm j has after observing such a deviation, because $EW_i(\tilde{t},c^H,0)<EW_i(\tau ^{H0},c^H,0) \le EW_i(\tau ^{H0},c^H,\mu _j)$ for any $\mu _j \in [0,1]$, where the first inequality follows because $\tau ^{H0} = \arg \max _{t_i} EW_i(t_i,c^H,0)$. Hence, $\tilde{t} \ne \tau ^{H0}$ cannot be an equilibrium best response for the high-type government and $R^S(c^H;E(t_j))=\tau ^{H0}$ must hold in the equilibrium.

Next, we consider the best response for the low-type government in the separating equilibrium, $R^S(c^L;E(t_j))$. We first show the following lemma.

Lemma B2

$min_{\mu _j}EW_i(\tau ^{L0},c^L,\mu _j)=EW_i(\tau ^{L0},c^L,0)$ when $a-2c^H+\bar{c}+E(t_j) > (c^H-c^L)(k+1)(k-2)/9$.

Proof

By Lemma B1, $\min _{\mu _j}EW_i(\tau ^{L0},c^L,\mu _j)\!=\!\min \{EW_i(\tau ^{L0},c^L,1), EW_i(\tau ^{L0},c^L,0)\}$, where $EW_i(\tau ^{L0},c^L,1) > EW_i(\tau ^{L0},c^L,0)$ if and only if $a-2c^H+\bar{c}+E(t_j) > (c^H-c^L)(2k^2-2k-31)/18$. Since $(c^H-c^L)(k+1)(k-2)/9>(c^H-c^L)(2k^2-2k-31)/18$, we obtain $\min _{\mu _j}EW_i(\tau ^{L0},c^L,\mu _j)=EW_i(\tau ^{L0},c^L,0)$ when $a-2c^H+\bar{c}+E(t_j) > (c^H-c^L)(k+1)(k-2)/9$. $\square$

This lemma implies that, by choosing $\tau ^{L0}$, the low-type government can attain at least $EW_i(\tau ^{L0},c^L,0)$ even under the most unfavorable belief by the foreign firm. Thus, in the separating equilibrium, the best-response tax rate $t_i$ for the low-type government must satisfy the following two conditions:

$$\begin{aligned}&EW_i\big (t_i,c^L,1\big ) \ge EW_i\big (\tau ^{L0},c^L,0\big ), \end{aligned}$$

(21)

$$\begin{aligned}&EW_i\big (\tau ^{H0},c^H,0\big ) \ge EW_i\big (t_i,c^H,1\big ). \end{aligned}$$

(22)

Equation (21) states that the low-type government prefers to set $t_i$ and be perceived as a low type rather than allow itself to be perceived as a high type and revert to $\tau ^{L0}$. Equation (22) states that the high-type government prefers to set $\tau ^{H0}$ and be perceived as a high type rather than set $t_i$ and pretend to be a low type.

Equation (21) can be rewritten as $\tau ^{L1}-X^L \le t_i \le \tau ^{L1}+X^L$, where

$$\begin{aligned} X^L(E(t_j))=\frac{3\sqrt{(c^H-c^L)(4a-7c^L-c^H+4\bar{c}+4E(t_j))}}{8(k+1)}>0, \end{aligned}$$

and (22) can be rewritten as $t_i \le \tau ^{H1}-X^H$ or $t_i \ge \tau ^{H1}+X^H$, where

$$\begin{aligned} X^H(E(t_j))=\frac{3\sqrt{(c^H-c^L)(4a-c^L-7c^H+4\bar{c}+4E(t_j))}}{8(k+1)}>0. \end{aligned}$$

Since $\tau ^{L1}>\tau ^{H1}$ and $X^L>X^H$, we have $\tau ^{L1}+X^L>\tau ^{H1}+X^H$. Thus, the separating best response for the low-type government must belong to the interval $[\tau ^{H1}+X^H, \tau ^{L1}+X^L]$.³⁹ In Fig. 4, this interval is shown by the bold line along the horizontal axis.

Fig. 4

$a - 2c^H + \bar{c} + E(t_j) > (c^H-c^L)(k+1)(k-2)/9$. a$a - 2c^H + \bar{c} + E(t_j) \ge (c^H-c^L)(k^2-k)$. b$(c^H-c^L)(k+1)(k-2)/9< a - 2c^H + \bar{c} + E(t_j) < (c^H-c^L)(k^2-k)$

Now, by applying the Intuitive Criterion, we refine this set of equilibrium best responses for the low-type government. First, we show the next lemma.

Lemma B3

$EW_i(t_i,c^H,\mu _j)$ is increasing in $\mu _j \in [0,1]$ when $t_i \ge \tau ^{H1}$.

Proof

From (20), we can infer that $\partial ^2 EW_i(t_i,c^H,\mu _j)/\partial t_i \partial \mu _j=(c^H-c^L)(2k-1)/18>0$ and

$$\begin{aligned} \frac{\partial EW_i\left( \tau ^{H1},c^H,\mu _j\right) }{\partial \mu _j} = \frac{c^H-c^L}{144(k+1)}&\left\{ (2k-1)^2(1-\mu _j)(c^H-c^L)\right. \\&+9\left[ 2a-4c^H+2\bar{c}+\mu _j (c^H\!-\!c^L)+2E(t_j)\right] \Bigr \}\!>\!0. \end{aligned}$$

Therefore, $\partial EW_i(t_i,c^H,\mu _j)/\partial \mu _j>0$ for any $\mu _j \in [0,1]$ and $t_i \ge \tau ^{H1}$. $\square$

Then, for any $t_i> \tau ^{H1}+X^H(> \tau ^{H1})$ and $\mu _j \in [0,1]$, we can infer that

$$\begin{aligned} EW_i(\tau ^{H0},c^H,0)>EW_i(t_i,c^H,1) \ge EW_i(t_i,c^H,\mu _j), \end{aligned}$$

where the first inequality follows because (22) holds with strict inequality for $t_i > \tau ^{H1}+X^H$, and the second inequality follows from Lemma B3. This implies that the high-type government could never be made better off by deviating from $\tau ^{H0}$ to $t_i > \tau ^{H1}+X^H$ irrespective of firm j’s belief after observing such a deviation.

Then, we show that under the Intuitive Criterion, the equilibrium best response for the low-type government is $R^S(c^L;E(t_j))=\max \{\tau ^{H1}+X^H, \tau ^{L1}\}$, where

$$\begin{aligned} \tau ^{H1}+X^H \gtreqless \tau ^{L1} \ \ \Longleftrightarrow \ \ a - 2c^H + \bar{c} + E(t_j) \gtreqless (c^H-c^L)(k^2-k). \end{aligned}$$

(23)

We first consider the case when $\tau ^{H1}+X^H \ge \tau ^{L1}$, or equivalently $a - 2c^H + \bar{c} + E(t_j) \ge (c^H-c^L)(k^2-k)$ (see Fig. 4a). Suppose that the low-type government chose $\tilde{t} \in (\tau ^{H1}+X^H,\tau ^{L1}+X^L]$ and obtained $EW_i(\tilde{t},c^L,1)$ in the equilibrium. Then, for any $\tilde{\tilde{t}} \in (\tau ^{H1}+X^H,\tilde{t})$, the high-type government could never be better off by deviating from $\tau ^{H0}$ to $\tilde{\tilde{t}}$, as discussed above. By contrast, since $EW_i(t_i,c^L,1)$ is decreasing for $t_i >\tau ^{H1}+X^H (\ge \tau ^{L1})$, the low-type government could be better off by deviating from $\tilde{t}$ to $\tilde{\tilde{t}}$ if firm j reasonably infers that $\mu _j=1$ on observing such a deviation. Hence, the equilibrium with $\tilde{t} \in (\tau ^{H1}+X^H,\tau ^{L1}+X^L]$ violates the Intuitive Criterion, and, thus, the best response for the low-type government must be $R^S(c^L;E(t_j))=\tau ^{H1}+X^H$ in the equilibrium.

We next consider the case when $\tau ^{H1}+X^H < \tau ^{L1}$, or equivalently ($(c^H-c^L)(k+1)(k-2)/9<$) $a - 2c^H + \bar{c} + E(t_j) < (c^H-c^L)(k^2-k)$ (see Fig. 4b). Suppose that the low-type government chose $\tilde{t} \in [\tau ^{H1}+X^H,\tau ^{L1}+X^L]{\setminus } \{\tau ^{L1}\}$ and obtained $EW_i(\tilde{t},c^L,1)$ in the equilibrium. Note that, since $\tau ^{L1}>\tau ^{H1}+X^H$, the high-type government could never be better off by deviating from $\tau ^{H0}$ to $\tau ^{L1}$, as discussed above. By contrast, since $\tau ^{L1} = \arg \max _{t_i}EW_i(t_i,c^L,1)$, the low-type government could be better off by deviating from $\tilde{t}$ to $\tau ^{L1}$ if firm j reasonably infers that $\mu _j=1$ on observing such a deviation. Hence, the equilibrium with $\tilde{t} \in [\tau ^{H1}+X^H,\tau ^{L1}+X^L]{\setminus } \{\tau ^{L1}\}$ violates the Intuitive Criterion, and, thus, the best response for the low-type government must be $R^S(c^L;E(t_j))=\tau ^{L1}$ in the equilibrium.

Finally, we show that even when $[\tau ^{L1}-X^L, \tau ^{H1}-X^H]$ is not empty, any equilibrium where the low-type government chooses $\tilde{t} \in [\tau ^{L1}-X^L, \tau ^{H1}-X^H]$ violates the Intuitive Criterion. Since $EW_i(t_i,c^L,1)$ is a quadratic function with its maximum at $t_i=\tau ^{L1}$, we have $EW_i(\tau ^{H1}-X^H,c^L,1)=EW_i(2\tau ^{L1}-(\tau ^{H1}-X^H),c^L,1)$. Moreover, since $\tau ^{H1}-X^H<\tau ^{L1}$ and $2\tau ^{L1}-(\tau ^{H1}-X^H)>\tau ^{L1}$, $EW_i(t_i,c^L,1)$ is increasing in $t_i \in [\tau ^{L1}-X^L, \tau ^{H1}-X^H]$ and decreasing in $t_i$ at $t_i=2\tau ^{L1}-(\tau ^{H1}-X^H)$. Then, for any $\tilde{t} \in [\tau ^{L1}-X^L, \tau ^{H1}-X^H]$ and an infinitesimally small number $\epsilon$, we obtain

$$\begin{aligned} EW_i(\tilde{t},c^L,1)&\le EW_i(\tau ^{H1}-X^H,c^L,1) \nonumber \\&=EW_i(2\tau ^{L1}-(\tau ^{H1}-X^H),c^L,1)\nonumber \\&<EW_i(2\tau ^{L1}-(\tau ^{H1}-X^H)-\epsilon ,c^L,1). \end{aligned}$$

(24)

Note that, since $2\tau ^{L1}-(\tau ^{H1}-X^H)-\epsilon >\tau ^{H1}+X^H$, the high-type government never prefers to deviate from $\tau ^{H0}$ to $2\tau ^{L1}-(\tau ^{H1}-X^H)-\epsilon$, as discussed above. By contrast, (24) implies that the low-type government prefers to deviate from $\tilde{t}$ to $2\tau ^{L1}-(\tau ^{H1}-X^H)-\epsilon$ if firm j reasonably infers that $\mu _j=1$ on observing such a deviation. Hence, we can conclude that any equilibrium with $\tilde{t} \in [\tau ^{L1}-X^L, \tau ^{H1}-X^H]$ violates the Intuitive Criterion.

Case 2$a-2c^H+\bar{c}+E(t_j) \le (c^H-c^L)(k+1)(k-2)/9$.

In this case, $EW_i(\tau ^{H0}, c^H, 1) \le EW_i(\tau ^{H0}, c^H, 0)$. Let $\bar{t}$ denote the best response of the high-type government in the separating equilibrium, that is, $R^S(c^H;E(t_j))=\bar{t}$. Let us denote the larger solution of $EW_i(\bar{t},c^H,0)=EW_i(t_i,c^H,1)$ with respect to $t_i$ by $\tau ^{H1}+Y$ with $Y>0$. Then, the high-type government could never be better off by deviating from $\bar{t}$ to $t_i > \tau ^{H1}+Y$ irrespective of firm j’s belief after observing such a deviation, because for any $t_i > \tau ^{H1}+Y$ and $\mu _j \in [0,1]$,

$$\begin{aligned} EW_i(\bar{t},c^H,0) = EW_i(\tau ^{H1}+Y,c^H,1) > EW_i(t_i,c^H,1) \ge EW_i(t_i,c^H,\mu _j), \end{aligned}$$

where the second inequality follows because $EW_i(t_i,c^H,1)$ is decreasing for $t_i \ge \tau ^{H1}$ and the last inequality follows from Lemma B3. Moreover, we have the following lemma.

Lemma B4

$\tau ^{L1}>\tau ^{H1}+Y$.

Proof

Note first that $EW_i(\bar{t},c^H,0) \ge EW_i(\tau ^{H0},c^H,1)$ must hold in the equilibrium. This is because Lemma B1 implies that $EW_i(\tau ^{H0},c^H,\mu _j) \ge EW_i(\tau ^{H0},c^H,1)$ for any $\mu _j \in [0,1]$ in the current Case 2; thus, the high-type government can always attain at least $EW_i(\tau ^{H0},c^H,1)$ by choosing $\tau ^{H0}$. Moreover, since $EW_i(t_i,c^H,1)$ is a quadratic function with its maximum at $t_i=\tau ^{H1}$, we have $EW_i(\tau ^{H1}+(\tau ^{H1}-\tau ^{H0}),c^H,1)=EW_i(\tau ^{H0},c^H,1)$ (see Fig. 5). Then,

$$\begin{aligned} EW_i(\tau ^{H1}+(\tau ^{H1}-\tau ^{H0}),c^H,1)= & {} EW_i(\tau ^{H0},c^H,1)\\\le & {} EW_i(\bar{t},c^H,0)=EW_i(\tau ^{H1}+Y,c^H,1). \end{aligned}$$

Thus, given that $EW_i(t_i,c^H,1)$ is decreasing for $t_i \ge \tau ^{H1}$, we have $\tau ^{H1}+Y \le \tau ^{H1}+(\tau ^{H1}-\tau ^{H0})$. By using this condition, we obtain

$$\begin{aligned} \tau ^{H1}+Y \le \tau ^{H1}+(\tau ^{H1}-\tau ^{H0})<\tau ^{L0}+(\tau ^{H1}-\tau ^{H0})=\tau ^{L0}+(\tau ^{L1}-\tau ^{L0})=\tau ^{L1}, \end{aligned}$$

where the third equality follows from $\tau ^{H1}-\tau ^{H0}=\tau ^{L1}-\tau ^{L0}=(2k-1)(c^H-c^L)/(8(k+1))$. $\square$

Fig. 5

$a - 2c^H + \bar{c} + E(t_j) \le (c^H-c^L)(k+1)(k-2)/9$

Thus, the high-type government never prefers to deviate from $\bar{t}$ to $\tau ^{L1}$. Then, we can show that, under the Intuitive Criterion, the best response for the low-type government must be $R^S(c^L;E(t_j))=\tau ^{L1}$ in the equilibrium. Suppose instead that the government chose $\tilde{t} \ne \tau ^{L1}$ and obtained $EW_i(\tilde{t},c^L,1)$. Then, since $\tau ^{L1} = \arg \max _{t_i}EW_i(t_i,c^L,1)$, the low-type government would prefer to deviate from $\tilde{t}$ to $\tau ^{L1}$ if firm j reasonably infers that $\mu _j=1$ on observing such a deviation. Hence, any equilibrium with $\tilde{t} \ne \tau ^{L1}$ violates the Intuitive Criterion.

Finally, we show that, under the Intuitive Criterion, $R^S(c^H;E(t_j))=\bar{t}=\tau ^{H0}$ must hold for the high-type government in the equilibrium. We first show the following lemma.

Lemma B5

$EW_i(\tau ^{H0},c^L,\mu _j) < EW_i(\tau ^{L1},c^L,1)$ for any $\mu _j \in [0,1]$.

Proof

Let us define $\tau ^L(\mu _j)\equiv \arg \max _{t_i}EW_i(t_i,c^L,\mu _j)=(2k-1)[2a-3c^L-(\mu _j c^L+(1-\mu _j)c^H)+2\bar{c}+2E(t_j)]/(8k+8)$. Then, $EW_i(\tau ^{H0},c^L,\mu _j) < EW_i(\tau ^L(\mu _j),c^L,\mu _j) \le EW_i(\tau ^{L1},c^L,1)$ holds for any $\mu _j \in [0,1]$, where the first strict inequality follows from $\tau ^L(\mu _j) \ne \tau ^{H0}$ and the second inequality follows from $dEW_i(\tau ^L(\mu _j),c^L,\mu _j)/d\mu _j=\tau ^L(\mu _j)(c^H-c^L)/(4k-2)>0$ and $\tau ^L(1)=\tau ^{L1}$. $\square$

This lemma implies that the low-type government could never be better off by deviating from $\tau ^{L1}$ to $\tau ^{H0}$ whatever belief firm j has after observing such a deviation. Now, suppose that the high-type government chose $\bar{t} \ne \tau ^{H0}$ and obtained $EW_i(\bar{t},c^H,0)$ in the equilibrium. Then, since $\tau ^{H0} = \arg \max _{t_i}EW_i(t_i,c^H,0)$, the government would prefer to deviate to $\tau ^{H0}$ if firm j reasonably infers that $\mu _j=0$ on observing such a deviation. Hence, any equilibrium with $\bar{t} \ne \tau ^{H0}$ violates the Intuitive Criterion.

Summarizing the results of Cases 1 and 2, we can conclude that the best response in the separating equilibrium for each type of government is as follows:⁴⁰

$$\begin{aligned} R^S(c^L;E(t_j))&= {\left\{ \begin{array}{ll} \tau ^{H1}(E(t_j))+X^H(E(t_j)) \ \ &{}\text {if} \ \ \ a - 2c^H + \bar{c} + E(t_j) > (c^H-c^L)(k^2-k), \\ \tau ^{L1}(E(t_j)) \ \ &{}\text {if} \ \ \ a - 2c^H + \bar{c} + E(t_j) \le (c^H-c^L)(k^2-k), \end{array}\right. } \end{aligned}$$

(25)

$$\begin{aligned} R^S(c^H;E(t_j))&=\tau ^{H0}(E(t_j)). \end{aligned}$$

(26)

1.2 Existence and Uniqueness of the Equilibrium Expected Tax Rate

In the symmetric equilibrium with $t_A(c^m)=t_B(c^m)=t^S(c^m)$, $m=L,H$, the equilibrium expected tax rate, $E(t^S)$, is a solution to the following equation with respect to T:

$$\begin{aligned} T=\lambda R^S(c^L;T)+(1-\lambda ) R^S(c^H;T), \end{aligned}$$

(27)

where $R^S(c^L;T)$ and $R^S(c^H;T)$ are respectively given by (25) and (26) with $E(t_j)=T$. Let us define $H^S(T) \equiv \lambda (\tau ^{H1}(T)+X^H(T))+(1-\lambda ) \tau ^{H0}(T)$ and $H^N(T)\equiv \lambda \tau ^{L1}(T)+(1-\lambda ) \tau ^{H0}(T)$. Then, (27) can be rewritten as follows:

$$\begin{aligned} T= {\left\{ \begin{array}{ll} H^S(T) \ \ &{}\text {if} \ \ \ a - 2c^H + \bar{c} + T > (c^H-c^L)(k^2-k), \\ H^N(T) \ \ &{}\text {if} \ \ \ a - 2c^H + \bar{c} + T \le (c^H-c^L)(k^2-k). \end{array}\right. } \end{aligned}$$

(28)

Since (23) implies that $H^S(T) = H^N(T)$ holds when $a - 2c^H + \bar{c} + T = (c^H-c^L)(k^2-k)$, the right-hand side of (28) is continuous. Note also that $H^S(0)>0$ and since

$$\begin{aligned} \frac{dH^S(T)}{dT}=\frac{2k-1}{4k+4}+\frac{3\lambda }{4k+4}\sqrt{\frac{1}{1-4\lambda +\frac{4}{\Delta -1}+\frac{4T}{c^H-c^L}}}>0, \end{aligned}$$

where $\Delta \equiv (a-c^L)/(a-c^H)$, we obtain $dH^S(T)/dT|_{T=0}<1$ and $d^2H^{S}(T)/dT^2<0$.⁴¹ Similarly, $H^N(0)>0$ and $dH^N(T)/dT=(2k-1)/(4k+4)<1$. Hence, these properties imply that (28) always has a unique solution. In the following two sections, we solve (28) and derive the equilibrium expected tax rate.

1.3 Proof of the Statement of Lemma 2

Suppose that the equilibrium expected tax rate, $E(t^S)$, satisfies $a - 2c^H + \bar{c} + E(t^S) \le (c^H-c^L)(k^2-k)$ and, thus, the best response for each type of government is given by $R^S(c^L;E(t^S))=\tau ^{L1}(E(t^S))$ and $R^S(c^H;E(t^S))=\tau ^{H0}(E(t^S))$ from (25) and (26). Then, $E(t^S)$ must be a solution of $T=H^N(T)$ in (28), which yields

$$\begin{aligned} T^N=\frac{(2k-1)(a-\bar{c})}{2k+5}. \end{aligned}$$

(29)

Note that this is a valid and unique solution of (28) if $a - 2c^H + \bar{c} + T^N \le (c^H-c^L)(k^2-k)$, which can be rewritten as (11). As a result, under condition (11), the unique pair of equilibrium emission taxes is given by $t^S(c^L)=\tau ^{L1}(T^N)$ and $t^S(c^H)=\tau ^{H0}(T^N)$.

Note from (6) and (19) that $\tau ^{L1}(E(t^S))=R^{N}(c^L;E(t^S))$ and $\tau ^{H0}(E(t^S))=R^{N}(c^H;E(t^S))$, and from (16) and (29) that $T^N=E(t^N)$. Therefore, we obtain

$$\begin{aligned} t^S(c^m)=\tau ^{mm}(T^N)=R^{N}(T^N,c^m)=R^{N}(E(t^N),c^m)=t^N(c^m), \quad m=L,H. \end{aligned}$$

Finally, suppose that government j plays the strategy $t^S(c^m)$, $m=L,H$, as shown above, and that each firm maintains the beliefs $\mu ^S(t)$ such that $\mu ^S(t)=1$ for $t \ge t^S(c^L)$ and $\mu ^S(t)=0$ for $t<t^S(c^L)$. Then, it is easy to verify that government i$(\ne j)$ would also be willing to play the strategy $t^S(\cdot )$, and the beliefs $\mu ^S(\cdot )$ are consistent with the governments’ strategy. Thus, we can conclude that the strategy and beliefs do provide a separating equilibrium.

1.4 Proof of the Statement of Proposition 1

Suppose that the equilibrium expected tax rate, $E(t^S)$, satisfies $a - 2c^H + \bar{c} + E(t^S) > (c^H-c^L)(k^2-k)$ and, thus, the best response for each type of government is given by $R^S(c^L;E(t^S))=\tau ^{H1}(E(t^S))+X^H(E(t^S))$ and $R^S(c^H;E(t^S))=\tau ^{H0}(E(t^S))$ from (25) and (26). Then, $E(t^S)$ must be a solution of $T=H^S(T)$ in (28), which yields

$$\begin{aligned} T^S=\frac{X+Y+3\lambda \sqrt{(c^H-c^L)\left[ 2X+Y+(2k+5)^2(4a-c^L-7c^H+4\bar{c})\right] }}{2(2k+5)^2}, \end{aligned}$$

(30)

where $X=(2k-1)(2k+5)(2a-3c^H+\bar{c})$ and $Y=9\lambda ^2(c^H-c^L)$. Note that this is a valid and unique solution of (28) if $a - 2c^H + \bar{c} + T^S > (c^H-c^L)(k^2-k)$, which can be rewritten after some cumbersome algebra as (12). As a result, under condition (12), the unique pair of equilibrium emission taxes is given by $t^S(c^L)=\tau ^{H1}(T^S)+X^H(T^S)$ and $t^S(c^H)=\tau ^{H0}(T^S)$.

Then, (13) follows because

$$\begin{aligned} R^{S}(c^L;E(t^S))&=\tau ^{H1}(E(t^S))+X^H(E(t^S)) \\&=\tau ^{L1}(E(t^S))+[\tau ^{H1}(E(t^S))+X^H(E(t^S))-\tau ^{L1}(E(t^S))] \\&=R^{N}(c^L;E(t^S))+[\tau ^{H1}(E(t^S))+X^H(E(t^S))-\tau ^{L1}(E(t^S))], \end{aligned}$$

where the bracketed term can be rearranged to yield the second term of the right-hand side of (13). Similarly, (14) follows from $R^S(c^H;E(t^S))=\tau ^{H0}(E(t^S))=R^{N}(c^H;E(t^S))$. Moreover, as in the proof of Lemma 1, it is easy to verify that the strategy and beliefs, $t^S(\cdot )$ and $\mu ^S(\cdot )$, do provide a separating equilibrium.

Since condition (12) is equivalent to $a - 2c^H + \bar{c} + T^S > (c^H-c^L)(k^2-k)$ as mentioned earlier, (23) implies that $\tau ^{H1}(E(t^S))+X^H(E(t^S))-\tau ^{L1}(E(t^S))>0$ with $E(t^S)=T^S$. Therefore, the second term of the right-hand side of (13) is also strictly positive; that is, $R^S(c^L;E(t^S))>R^N(c^L;E(t^S))$ holds. Then, we obtain

$$\begin{aligned} t^S(c^L)&=R^S(c^L;E(t^S))>R^N(c^L;E(t^S))>R^N(c^L;E(t^N))=t^N(c^L), \end{aligned}$$

(31)

$$\begin{aligned} t^S(c^H)&=R^S(c^H;E(t^S))=R^N(c^H;E(t^S))>R^N(c^H;E(t^N))=t^N(c^H), \end{aligned}$$

(32)

where, in the third inequality in (31) and (32), we also used the fact that $R^N(c^m;E(t_j))$ is strictly increasing in $E(t_j)$ for $m=L,H$.

Finally, note that the exact expression for the expected welfare in this signaling equilibrium, $EW^S(c^m)$, $m=L,H$, is given by (18) with $t^e(\cdot )$ replaced by $t^S(\cdot )$, as derived above.

1.5 Elimination of Pooling Equilibria

In a pooling equilibrium, both types of governments set the same tax rate in each country, that is, $t_A(c^L)=t_A(c^H)=t_A^{pool}$ and $t_B(c^L)=t_B(c^H)=t_B^{pool}$, and each firm infers that its rival’s marginal cost is $c^L$ ($c^H$) with probability $\lambda$ ($1-\lambda$). Note that such a pooling equilibrium fails the Intuitive Criterion if, given government j’s pooling strategy $t_j(c^L)=t_j(c^H)=t_j^{pool}$, there exists a tax rate $t^*$ for government i such that

$$\begin{aligned} EW_i(t^*,c^H,1 ; t_j^{pool})&< EW_i(t_i^{pool},c^H,\lambda ; t_j^{pool}), \end{aligned}$$

(33)

$$\begin{aligned} EW_i(t^*,c^L,1 ; t_j^{pool})&> EW_i(t_i^{pool},c^L,\lambda ; t_j^{pool}). \end{aligned}$$

(34)

Equation (33) states that the high-type government could not be better off by deviating to $t^*$ even if firm j had the most favorable inference, $\mu _j=1$, on observing such a deviation.⁴² By contrast, (34) states that the low-type government prefers to deviate to $t^*$ if firm j reasonably infers that $\mu _j=1$ since the high-type government never prefers such a deviation when (33) is satisfied. In the following proof, we show that, for any $t_i^{pool}$ and $t_j^{pool}$, there exists $t^*$ such that (33) and (34) are both satisfied.

(33) can be rewritten as $t^* < \tau ^{H1}-Z^H$ or $t^* > \tau ^{H1}+Z^H$, where

$$\begin{aligned} Z^H=\sqrt{(\tau ^{H1}-t_i^{pool})^2+\frac{(\bar{c}-c^L)[(2-k)(4a-c^L-6c^H+3\bar{c}+4t_j^{pool})+4(2k-1)t_i^{pool}]}{16(k+1)}}>0, \end{aligned}$$

and (34) can be rewritten as $\tau ^{L1}-Z^L< t^* < \tau ^{L1}+Z^L$, where

$$\begin{aligned} Z^L=\sqrt{(\tau ^{L1}-t_i^{pool})^2+\frac{(\bar{c}-c^L)[(2-k)(4a-7c^L+3\bar{c}+4t_j^{pool})+4(2k-1)t_i^{pool}]}{16(k+1)}}>0. \end{aligned}$$

Below, we prove that $\tau ^{L1}+Z^L>\tau ^{H1}+Z^H$ holds for any $t_i^{pool}$ and $t_j^{pool}$.

Note that $\tau ^{L1}>\tau ^{H1}$, and we can show that $Z^L \gtreqless Z^H$ if and only if $t_i^{pool} \lesseqgtr \bar{\tau }$, where

$$\begin{aligned} \bar{\tau }=\frac{(2k-1)^2(4a-3c^L-3c^H+2\bar{c}+4t_j^{pool})+18(\bar{c}-c^L)}{16(k+1)(2k-1)}. \end{aligned}$$

Hence, when $t_i^{pool} \le \bar{\tau }$ and $Z^L \ge Z^H$, we can easily infer that $\tau ^{L1}+Z^L>\tau ^{H1}+Z^H$ holds.

When $t_i^{pool} > \bar{\tau }$ and $Z^L < Z^H$, on the other hand, $\tau ^{L1}+Z^L>\tau ^{H1}+Z^H$ holds if and only if $(\tau ^{L1}-\tau ^{H1})^2-(Z^H-Z^L)^2$ is positive. The latter expression can be rearranged as

$$\begin{aligned} 2Z^HZ^L-[(Z^H)^2+(Z^L)^2-(\tau ^{L1}-\tau ^{H1})^2], \end{aligned}$$

(35)

where the sign of the bracketed term is ambiguous. If the bracketed term is non-positive, then (35) is positive and $\tau ^{L1}+Z^L>\tau ^{H1}+Z^H$ holds. If the term is positive, the sign of (35) is equivalent to that of $(2Z^HZ^L)^2-[(Z^H)^2+(Z^L)^2-(\tau ^{L1}-\tau ^{H1})^2]^2$, which can be rewritten as

$$\begin{aligned} \frac{81(c^H-c^L)^2(\bar{c}-c^L)[4(2k-1)t_i^{pool}-(2-k)(\bar{c}-c^L)]}{256(k+1)^3}. \end{aligned}$$

This is positive if and only if $t_i^{pool}>(2-k)(\bar{c}-c^L)/[4(2k-1)]$. Note that since

$$\begin{aligned} \bar{\tau }-\frac{(2-k)(\bar{c}-c^L)}{4(2k-1)}=\frac{(2k-1)^2(4a-4c^L-3c^H+3\bar{c}+4t_j^{pool})+9(\bar{c}-c^L)}{16(k+1)(2k-1)}>0 \end{aligned}$$

and $t_i^{pool}>\bar{\tau }$ in the present case, we obtain $t_i^{pool}>\bar{\tau }>(2-k)(\bar{c}-c^L)/[4(2k-1)]$. Therefore, (35) is positive and $\tau ^{L1}+Z^L>\tau ^{H1}+Z^H$ holds.

$\tau ^{L1}+Z^L>\tau ^{H1}+Z^H$ for any $t_i^{pool}$ and $t_j^{pool}$ implies that there always exists $t^* \in (\tau ^{H1}+Z^H,\tau ^{L1}+Z^L)$, for which (33) and (34) are both satisfied. We can thus conclude that no pooling equilibrium exists under the refinement of the Intuitive Criterion.