International Fisheries Agreements with a Shifting Stock

Abstract

When a fish stock shifts from one nation to another nation, e.g., due to climate change, the nation that loses the resource has incentives to deplete it, while the other nation, receiving the resource, has incentives to conserve it. We propose an analytical model to study under which circumstances self-enforcing agreements can align incentives. Our setup allows to distinguish between a fast and a slow shift and between a smooth or a sudden shift in ownership. We show that the shorter the expected duration of the transition, the higher the total equilibrium exploitation rate. Similarly, a sudden shift implies—by and large—more aggressive non-cooperative exploitation than a gradual shift. However, a self-enforcing agreement without side-payments is more likely for a sudden than for a smooth shift. Further, the scope for cooperation increases with the expected duration of the transition, and it decreases with the renewability of the resource and the discount rate. Most importantly, we show that concentrating on in-kind transfers can be very detrimental for shifting renewable resources: In some cases, there is no efficient bargaining solution without side-payments, even when there are only two players.

Appendix

1.1 Parameter Overview

See Table 2.

Table 2 Overview of the used parameters

1.2 Proofs

1.2.1 Proof of Proposition 1

The existence of the first phase follows from the assumption that $s_0 = 1$ so that player A is the sole owner in the first stage of the game and player B cannot extract anything. The existence of the last phase follows from the assumption that $s_T = 0$ and player A cannot extract anything. The second phase may not exist (e.g., when $T = 1$).

Consider the first phase (Eq. 10). Player B’s extraction rate is constrained by the share $1-s_{\tau }$ accessible to him. This share is increasing as $s_{\tau }$ decreases. The extraction rate of player A is given by $a_{\tau } = s_{\tau } \gamma ^A_{\tau }$. This may in principle increase or decrease. Clearly, $s_{\tau }$ is declining with $\tau$, but $\gamma _{\tau }^{A}$ increases with $\tau$:

$$\begin{aligned} \gamma ^A_{\tau +1} - \gamma _{\tau }^{A}> 0&\Leftrightarrow \frac{1}{1+\alpha \beta \varphi _{\tau +1}^A} - \frac{1}{1+\alpha \beta \varphi _{\tau }^A} > 0 \quad \Leftrightarrow \quad \varphi _{\tau }^A > \varphi _{\tau +1}^A \\&\Leftrightarrow q k^{A}_{\tau }+(1-q)k^{A}_{\tau +1} > q k^{A}_{\tau +1}+(1-q)k^{A}_{\tau +2} \end{aligned}$$

The last line is true because $k^{A}_{\tau } > k^{A}_{\tau +1}$, as we have discussed in relation to Eq. (7) on page 9. In spite of this indeterminacy, it is possible to show that the total extraction rate in the first phase, which we denote $d^I_{\tau }$ for clarity, increases with $\tau$:

$$\begin{aligned} d^{I}_{\tau +1} > d^{I}_{\tau }&\Leftrightarrow s_{\tau +1} \gamma ^A_{\tau +1} + 1 - s_{\tau +1} > s_{\tau } \gamma ^A_{\tau } + 1 - s_{\tau } \\&\Leftrightarrow s_{\tau +1} (\gamma ^A_{\tau +1} - 1) > s_{\tau } (\gamma ^A_{\tau } - 1) \quad \Leftrightarrow \quad \frac{s_{\tau +1}}{s_{\tau } } < \frac{\gamma ^A_{\tau }-1}{\gamma ^A_{\tau +1}-1} \end{aligned}$$

The change in sign in the last line occurs because $\gamma _{\tau }^{A}< 1$ for all $\tau$. The left-hand side of the last inequality is smaller than one as $s_{\tau +1} < s_{\tau }$ by construction. The right-hand side, however, is larger than one: $\frac{\gamma ^A_{\tau }-1}{\gamma ^A_{\tau +1}-1} > 1 \Leftrightarrow \gamma ^A_{\tau }-1 < \gamma ^A_{\tau +1}-1 \Leftrightarrow \gamma ^A_{\tau +1} > \gamma _{\tau }^{A}$. The first phase ends when $1-s_{\tau } \ge \frac{\gamma ^B (1-\gamma _{\tau }^A)}{1-\gamma _{\tau }^A\gamma ^B} \Leftrightarrow s_{\tau } \le \frac{\alpha \beta }{1-(1-\alpha \beta )\gamma _{\tau }^A}$.

Consider the second phase. Equating the two non-binding best replies $a_{\tau } = \gamma ^A_{\tau } (1-b_{\tau })$ and $b_{\tau } = \gamma ^B (1-a_{\tau })$ yields:

$$\begin{aligned} a_{\tau }&= (1\!- \!\gamma ^B(1-a_{\tau }) )\gamma _{\tau }^{A}\Leftrightarrow a_{\tau } = \gamma ^A_{\tau }- \gamma ^A_{\tau } \gamma ^B + a_{\tau } \gamma ^A_{\tau } \gamma ^B \quad \Leftrightarrow \quad a_{\tau } = \frac{ \gamma _{\tau }^{A}(1-\gamma ^B)}{1-\gamma _{\tau }^{A}\gamma ^B} \end{aligned}$$

The case for player B is symmetric. The total extraction rate is $d^{II}_{\tau } = a_{\tau } + b_{\tau } = \frac{ \gamma _{\tau }^{A}(1-\gamma ^B) + \gamma ^B(1-\gamma _{\tau }^{A})}{1-\gamma _{\tau }^{A}\gamma ^B}$. It is increasing with $\tau$ when the following holds:

$$\begin{aligned} d^{II}_{\tau +1} \!- \!d^{II}_{\tau } =&\frac{(\gamma ^A_{\tau +1} \!-\! 2 \gamma ^A_{\tau \!+\!1} \gamma ^B \!+\! \gamma ^B)(1-\gamma _{\tau }^{A}\gamma ^B)-(\gamma _{\tau }^{A}- 2 \gamma _{\tau }^{A}\gamma ^B \!+\!\gamma ^B)(1\!-\!\gamma ^A_{\tau +1}\gamma ^B)}{(1\!-\!\gamma ^A_{\tau +1}\gamma ^B)(1\!-\!\gamma _{\tau }^{A}\gamma ^B)} \!>\! 0 \end{aligned}$$

Because the denominator is positive ($\gamma _{\tau }^{A}< 1$ and $\gamma ^B <1$), this is equivalent to:

$$\begin{aligned}&(\gamma ^A_{\tau +1}-\gamma ^A_{\tau })(1+(\gamma ^B)^2-2\gamma ^B) > 0. \end{aligned}$$

This inequality holds because the first bracket is positive, and since $\gamma ^B=1-\alpha \beta$, the second bracket reduces to $(\alpha \beta )^2$, which is also positive.

Total extraction in the third phase is $d^{III}_{\tau } = s_{\tau } + (1-s_{\tau }) \gamma ^B = 1-\alpha \beta +\alpha \beta s_{\tau }$. This is declining as $\tau \rightarrow T$ simply due to the fact that $s_{\tau } > s_{\tau +1}$.

1.2.2 Proof of Proposition 2

In the following, it will be useful to define the following auxiliary parameters:

$$\begin{aligned} \varphi ^A_{\tau } \equiv qk_{\tau }^A + (1-q)k_{\tau +1}^A \qquad {\text {and}} \qquad \varphi ^B \equiv \frac{1}{1-\alpha \beta } \end{aligned}$$

We start by conjecturing that $V_{\tau }^{\mathrm{coop}, i}$ and $V_{\tau }^{\mathrm{nc}, i}$ are of the usual log-linear form and that the coefficients $k_{\tau }^A$ and $k_{\tau }^B$ take the form as described by Eq. (7). Taking the perspective of player A, her relevant value functions are:

$$\begin{aligned} V_{\tau }^{\mathrm{coop}, A}(\lambda , x_t) =&\ln \lambda (1-\alpha \beta ) + \ln x_t + \\&\ldots \beta \left( q\left[ k_{\tau }^A \ln x_{t+1} + K_{\tau }^{\mathrm{coop}, A} \right] + (1-q)\left[ k_{\tau +1}^A \ln x_{t+1} + K_{\tau +1}^{A} \right] \right) \\ V_{\tau }^{\mathrm{nc}, A}(\lambda , x_t) =&\ln a_{\tau } + \ln x_t +\\&\ldots \beta \left( q\left[ k_{\tau }^A \ln x_{t+1} + K_{\tau }^{\mathrm{nc}, A} \right] + (1-q)\left[ k_{\tau +1}^A \ln x_{t+1} + K_{\tau +1}^{A} \right] \right) \end{aligned}$$

Note that $x_{t+1} = M\left( \alpha \beta x_t\right) ^{\alpha }$ under cooperation but $x_{t+1} = M\left( (1-a_{\tau }-b_{\tau }) x_t\right) ^{\alpha }$ under non-cooperation. Equating coefficients therefore shows that indeed $k_{\tau }^A = 1 + \alpha \beta \varphi ^A_{\tau }$ as in the non-cooperative case:

$$\begin{aligned} k_{\tau }^A \ln x_{t} + K_{\tau }^{\mathrm{coop}, A} =&\ln \lambda (1-\alpha \beta ) + \ln x_t +\\&\ldots \beta \left( q\left[ k_{\tau }^A (\ln M + \alpha \ln \alpha \beta + \alpha \ln x_t) + K_{\tau }^{\mathrm{coop}, A} \right] + \right. \\&\ldots \left. (1-q)\left[ k_{\tau +1}^A (\ln M + \alpha \ln \alpha \beta + \alpha \ln x_t) + K_{\tau +1}^{A} \right] \right) \end{aligned}$$

We have also:

$$\begin{aligned} (1-\beta q)K_{\tau }^{\mathrm{coop}, A}&= \ln \lambda (1-\alpha \beta ) + \beta \left( (1-q)K^A_{\tau +1} + \varphi ^A_{\tau }(\ln M + \alpha \ln \alpha \beta )\right) \end{aligned}$$

(20)

For the non-cooperative case, equating coefficients yields:

$$\begin{aligned} (1-\beta q)K_{\tau }^{\mathrm{nc}, A}&= \ln a_{\tau } + \beta \left( (1-q)K^A_{\tau +1} + \varphi ^A_{\tau }(\ln M + \alpha \ln (1-a_{\tau }-b_{\tau })\right) \end{aligned}$$

(21)

Player A’s participation constraint is satisfied when:

$$\begin{aligned} V_{\tau }^{\mathrm{coop}, A}(\lambda , x_t) - V_{\tau }^{\mathrm{nc}, A}(x_t)&= k_{\tau }^A \ln x_{t} + K_{\tau }^{\mathrm{coop}, A} - \left( k_{\tau }^A \ln x_{t} + K_{\tau }^{\mathrm{nc}, A} \right) \ge 0 \end{aligned}$$

Clearly, the term $k_{\tau }^A \ln x_{t}$ cancels. Similarly, inspecting (20) and (21), we see that the term $K_{\tau +1}^A$, capturing the further development of the game, enters in the same manner in those two equations. Thus, it cancels as well. Therefore, we can define player A’s gain from cooperation by $g^A$:

$$\begin{aligned} g^A(\lambda )&= \frac{1}{1-\beta q} \left( \ln \lambda + \ln \left( \frac{1-\alpha \beta }{a_{\tau }}\right) + \alpha \beta \varphi ^A_{\tau }\ln \left( \frac{\alpha \beta }{1-a_{\tau }-b_{\tau }}\right) \right) \end{aligned}$$

(22)

Whether this $g^A(\lambda ) > 0$ is in fact independent of the current stock level and the future development of the game.

1.2.3 There is Not Always Scope for Cooperation

To see that there cannot always be scope for cooperation, note that $g^A(\lambda )$ is increasing in $\lambda$ and reaches its maximum at $\lambda = 1$. We now show that $g^A(1) < 0$ as $\alpha \beta \rightarrow 1$ or $q \rightarrow 0$, so that player A would not join an agreement, even if she were offered the entire harvest.

To show that $g^A(1) < 0$ as $q \rightarrow 0$, consider A’s gain from cooperation at stage $T-1$. At this stage we have $\lim _{q \rightarrow 0} \varphi _{T-1}^A = 0$ and consequently $a_{T-1} = s_{T-1}$ (Player A will harvest all she can before she loses the resource for sure). We have $g^A(1) < 0 \Leftrightarrow \ln \left( \frac{1-\alpha \beta }{a_{\tau }}\right) < 0 \Leftrightarrow 1-\alpha \beta < a_{T-1}$. When $s_{T-1} < 1-\alpha \beta$, we need to consider the game at stage $\tau < T-1$. But from backward induction, it follows that $a_{\tau } = s_{\tau }$ for all $\tau$ and consequently $b_{\tau } = (1-s_{\tau })(1-\alpha \beta )$. Now there will be some $\tau$ at which $s_{\tau } > 1-\alpha \beta$ and since $\varphi ^A_{\tau }$ is bounded above by $\frac{1}{1-\alpha \beta }$ we know that at some stage (as $s_{\tau } \rightarrow 1$) we have $\ln (1-\alpha \beta ) - (1+\alpha \beta \varphi ^A_{\tau })\ln s_{\tau } < 0$ and thus $g^A(1) < 0$ .

Consider now the case when $\alpha \beta \rightarrow 1$. We have that $g^A(1) < 0$ whenever:

$$\begin{aligned} \lim _{\alpha \beta \rightarrow 1} \ln \left( \frac{1-\alpha \beta }{a_{\tau }}\right) + \alpha \beta \varphi ^A_{\tau }\ln \left( \frac{\alpha \beta }{1-a_{\tau }-b_{\tau }}\right)&< 0 \end{aligned}$$

To evaluate this limit, we need a few building stones. Consider first $k^{A}_{\tau }, \varphi ^A_{\tau }$ and $\gamma _{\tau }^{A}$, where we use L’Hôpital’s rule to show that these terms converge to some constant $\kappa$:

$$\begin{aligned} \lim _{\alpha \beta \rightarrow 1} k^{A}_{\tau }&= \lim _{\alpha \beta \rightarrow 1} \left[ \frac{1}{1-\alpha \beta }\left( 1-\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{T-\tau }\right) \right] \\&= \frac{ \lim _{\alpha \beta \rightarrow 1} \left[ -(T-\tau )\left( \frac{\alpha \beta (1- q)}{1- \alpha \beta q}\right) ^{T-\tau -1}\left( \frac{1-q}{(1-\alpha \beta q)^2}\right) \right] }{ \lim _{\alpha \beta \rightarrow 1} - 1} \\&= (T\!-\!\tau ) \lim _{\alpha \beta \rightarrow 1} \left[ \left( \frac{\alpha \beta (1\!-\! q)}{1\!-\! \alpha \beta q}\right) ^{T-\tau -1} \right] \lim _{\alpha \beta \rightarrow 1} \left[ \frac{1\!-\!q}{(1\!-\!\alpha \beta q)^2} \right] \!=\! \frac{ T-\tau }{1-q} = \kappa _1 > 0 \\ \lim _{\alpha \beta \rightarrow 1} \varphi ^A_{\tau }&= \lim _{\alpha \beta \rightarrow 1} q k^{A}_{\tau }+(1-q)k^{A}_{\tau +1} = q\frac{(T-\tau )}{(1-q)} + T - \tau -1 = \kappa _2 > 0 \\ \lim _{\alpha \beta \rightarrow 1} \gamma _{\tau }^{A}&= \lim _{\alpha \beta \rightarrow 1} \frac{1}{1+\alpha \beta \varphi ^A} = \kappa _3 > 0 \\ \lim _{\alpha \beta \rightarrow 1} \gamma ^B&= \lim _{\alpha \beta \rightarrow 1} (1-\alpha \beta ) = 0 \end{aligned}$$

And therefore:

$$\begin{aligned}&\lim _{\alpha \beta \rightarrow 1} a_{\tau } = \left\{ \begin{array}{lll} \lim _{\alpha \beta \rightarrow 1} s_{\tau }\gamma _{\tau }^{A}&{} = s_{\tau }\kappa _3 &{} \quad {\text {Phase I}}\\ \lim _{\alpha \beta \rightarrow 1} \frac{\gamma _{\tau }^{A}(1-\gamma ^B)}{1-\gamma _{\tau }^{A}\gamma ^B} &{} = \kappa _3 &{} \quad {\text {Phase II}} \\ \lim _{\alpha \beta \rightarrow 1} s_{\tau } &{} = s_{\tau } &{} \quad {\text {Phase III}} \end{array} \right. \quad \lim _{\alpha \beta \rightarrow 1} b_{\tau } = \left\{ \begin{array}{ll} \lim _{\alpha \beta \rightarrow 1} 1-s_{\tau } &{} = 1-s_{\tau } \\ \lim _{\alpha \beta \rightarrow 1} \frac{\gamma ^B(1-\gamma _{\tau }^{A})}{1-\gamma _{\tau }^{A}\gamma ^B} &{} = 0 \\ \lim _{\alpha \beta \rightarrow 1} (1-s_{\tau })\gamma ^B &{} = 0 \end{array} \right. \end{aligned}$$

Hence, $\lim _{\alpha \beta \rightarrow 1} \ln \left( \frac{1-\alpha \beta }{a_{\tau }}\right) = -\infty$, but $\lim _{\alpha \beta \rightarrow 1} \alpha \beta \varphi ^A_{\tau }\ln \left( \frac{\alpha \beta }{1-a_{\tau }-b_{\tau }}\right) = \kappa _2 \ln \left( \frac{1}{1-s_{\tau }\kappa _3}\right)$ in Phase I, $= \kappa _2 \ln \left( \frac{1}{1-\kappa _3}\right)$ in Phase II, and $= \kappa _2 \ln \left( \frac{1}{1-s_{\tau }}\right)$ in Phase III. In any case, it is bounded above so that the entire term $\left[ \ln \left( \frac{1-\alpha \beta }{a_{\tau }}\right) + \alpha \beta \varphi ^A_{\tau }\ln \left( \frac{\alpha \beta }{1-a_{\tau }-b_{\tau }}\right) \right]$ must be smaller than zero as claimed.

1.2.4 Derivation of $\lambda _\mathrm{min}^A$ and $\lambda _\mathrm{max}^B$

Now consider the situation where $\alpha \beta$ take values such that $g^A(\lambda _\mathrm{min}^A) = 0$ exists. By rewriting Eq. (22) and taking exponents on both sides, we get:

$$\begin{aligned} \lambda _\mathrm{min}^A&= \frac{a_{\tau }}{1-\alpha \beta } \left( \frac{1-a_{\tau }-b_{\tau }}{\alpha \beta } \right) ^{\alpha \beta \varphi ^A_{\tau }} \end{aligned}$$

(23)

By parallel reasoning, player B’s participation constraint is satisfied when $V_{\tau }^{\mathrm{coop}, B} - V_{\tau }^{\mathrm{nc}, B} \ge 0$ and $\lambda _\mathrm{max}^B$, the maximum share that player B would be willing to give to player A and still harvest cooperatively is—if it exists —defined by:

$$\begin{aligned} \lambda _\mathrm{max}^B&= 1 - \frac{b_{\tau }}{1-\alpha \beta } \left( \frac{1-a_{\tau }-b_{\tau }}{\alpha \beta } \right) ^{\alpha \beta \varphi ^B} \end{aligned}$$

(24)

1.2.5 Cooperation Possibilities are not Constrained by the Available Harvest Shares

At any stage $\tau$, player A needs to harvest at least $(1-\alpha \beta )\lambda _\mathrm{min}^A$ to join the agreement. This is less than he would harvest under non-cooperation:

$$\begin{aligned} (1-\alpha \beta )\lambda _\mathrm{min}^A&< a_{\tau } \quad \Leftrightarrow \quad a_{\tau }\left( \frac{1-a_{\tau }-b_{\tau }}{\alpha \beta } \right) ^{\alpha \beta \varphi ^A_{\tau }} < a_{\tau } \quad \Leftrightarrow \quad \left( \frac{1-a_{\tau }-b_{\tau }}{\alpha \beta } \right) ^{\alpha \beta \varphi ^A_{\tau }} < 1 \end{aligned}$$

The last statement is true because $1-a_{\tau }-b_{\tau } = 1-d^{\mathrm{nc}} < 1-d^* = \alpha \beta$ and $\alpha \beta \varphi > 0$ and a function $x^y < 1$ when $x<1$ and $y>0$. For player B, the argumentation is parallel:

$$\begin{aligned} (1\!-\!\alpha \beta )\left( 1\!-\!\lambda _\mathrm{max}^B\right)&< \!b_{\tau } \!\quad \Leftrightarrow \quad \! b_{\tau }\left( \frac{1-a_{\tau }-b_{\tau }}{\alpha \beta } \right) ^{\alpha \beta \varphi ^B} \!<\! b_{\tau } \!\quad \Leftrightarrow \quad \! \left( \frac{1\!-\!a_{\tau }\!-\!b_{\tau }}{\alpha \beta } \right) ^{\alpha \beta \varphi ^B} \!<\! 1 \end{aligned}$$

As $a_{\tau } \le s_{\tau }$ and $b_{\tau } \le 1-s_{\tau }$ by construction, it follows immediately that cooperation possibilities are not constrained by the accessible harvest shares.

This completes the proof of Proposition 2.

1.2.6 Proof of Proposition 3

To show how the extraction pattern changes with, we need to derive $\frac{\partial \gamma _{\tau }^A}{\partial q}$($k^B$ and thus $\gamma ^B$ do not depend on q). In this derivation, we will again employ the following auxiliary parameters to make the derivations more concise.

$$\begin{aligned} \varphi ^A_{\tau } \equiv qk_{\tau }^A + (1-q)k_{\tau +1}^A \qquad {\text {and}} \qquad \varphi ^B \equiv \frac{1}{1-\alpha \beta } \end{aligned}$$

To start, consider $\frac{\partial k_{\tau }^A}{\partial q}$ :

$$\begin{aligned} \frac{\partial k_{\tau }^A}{\partial q}&= \frac{-(T-\tau )\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{T-\tau -1}\left( \frac{-\alpha \beta (1-\alpha \beta q)-\alpha \beta (1-q)(-\alpha \beta )}{(1-\alpha \beta q)^2}\right) }{1-\alpha \beta } \nonumber \\&= -(T-\tau )\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{T-\tau -1}\left( \frac{ - \alpha \beta }{(1-\alpha \beta q)^2 }\right) > 0 \end{aligned}$$

(25)

The above follows from $T>\tau$ and because the term $\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) > 0$ as $\alpha , \beta$ and $q \in (0,1)$. As a consequence:

$$\begin{aligned} \frac{\partial \varphi _{\tau }^A}{\partial q}&= k_{\tau }^A + q\frac{\partial k_{\tau }^A}{\partial q} - k_{\tau +1}^A + (1-q) \frac{\partial k_{\tau +1}^A}{\partial q} \nonumber \\&= k_{\tau }^A - k_{\tau +1}^A + q \left( \frac{\partial k_{\tau }^A}{\partial q} -\frac{\partial k_{\tau +1}^A}{\partial q} \right) + \frac{\partial k_{\tau +1}^A}{\partial q} > 0 \end{aligned}$$

(26)

Here, we have to show that $\frac{\partial k_{\tau }^A}{\partial q} > \frac{\partial k_{\tau +1}^A}{\partial q}$ for all $k_{\tau }^A > k_{\tau +1}^A$ and $q>0$ and $\frac{\partial k_{\tau +1}^A}{\partial q} > 0$. From (25) we have:

$$\begin{aligned} \frac{\partial k_{\tau }^A}{\partial q} \!- \!\frac{\partial k_{\tau +1}^A}{\partial q} \!>\! 0&\Leftrightarrow \! (T-\tau -1)\left( \frac{\alpha \beta (1-q)}{1\!-\!\alpha \beta q}\right) ^{T-\tau -2}-(T-\tau )\left( \frac{\alpha \beta (1-q)}{1\!-\!\alpha \beta q}\right) ^{T-\tau -1} \!>\! 0 \\&\Leftrightarrow (T-\tau -1)\left( \frac{1-\alpha \beta q}{\alpha \beta (1-q)}\right) > T-\tau \\&\Leftrightarrow T-\tau -1 > \alpha \beta (T-\tau - q) \end{aligned}$$

This is true because $\alpha , \beta$ and $q \in (0,1)$. This implies:

$$\begin{aligned} \frac{\partial \gamma _{\tau }^A}{\partial q}&= \frac{-\alpha \beta \frac{\partial \varphi _{\tau }^A}{\partial q}}{(1+\alpha \beta \varphi ^A)^{2}} < 0 \end{aligned}$$

(27)

Furthermore, we need to derive $\frac{\partial \gamma _{\tau }^A}{\partial T}$ ($k^B$ and thus $\varphi ^B$ and $\gamma ^B$ do not depend on T). Again, consider first $\frac{\partial k_{\tau }^A}{\partial T}$ :

$$\begin{aligned} \frac{\partial k_{\tau }^A}{\partial T}&= \frac{-1}{1-\alpha \beta }\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{T-\tau }\ln \left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q} \right) > 0 \end{aligned}$$

(28)

Since $\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q} \right) < 1$, the logarithm yields a negative number, so that the entire term is positive. Consequently:

$$\begin{aligned}&\frac{\partial \varphi _{\tau }^A}{\partial T} = q \frac{\partial k_{\tau }^A}{\partial T} + (1-q) \frac{\partial k_{\tau +1}^A}{\partial T} > 0 \end{aligned}$$

(29)

$$\begin{aligned}&\quad \frac{\partial \gamma _{\tau }^A}{\partial T} = \frac{-\alpha \beta \frac{\partial \varphi _{\tau }^A}{\partial T}}{(1+\alpha \beta \varphi ^A)^{2}} < 0 \end{aligned}$$

(30)

In the first phase, player B is harvesting the entire share available to him ($b^{I}_{\tau } = 1-s_{\tau }$). Hence his extraction rate does not depend on q. It does depend on T, however, since $s_{\tau } = \frac{T- \tau }{T}$. $s_{\tau }$ increases with T so that $b^{I}_{\tau }$ decreases. From (27), it follows that player A’s and thus the total extraction rate is decreasing in q. Regarding an increase in T, we see from (30) that $\gamma _{\tau }^{A}$ decreases, but $s_{\tau }$ increases. The total effect is still negative:

$$\begin{aligned} \frac{\partial d_{\tau }^{I} }{\partial T} =&\frac{\partial s_{\tau }}{\partial T} \gamma _{\tau }^{A}+ s_{\tau } \frac{\partial \gamma _{\tau }^{A}}{\partial T} - \frac{\partial s_{\tau }}{\partial T} < 0 \\&\Leftrightarrow \frac{\partial s_{\tau }}{\partial T} \gamma _{\tau }^{A}+ s_{\tau } \frac{\partial \gamma _{\tau }^{A}}{\partial T} - \frac{\partial s_{\tau }}{\partial T} <0 \\&\Leftrightarrow s_{\tau } \frac{\partial \gamma _{\tau }^{A}}{\partial T} < (1-\gamma _{\tau }^{A})\frac{\partial s_{\tau }}{\partial T} \end{aligned}$$

This is true because $\frac{\partial \gamma _{\tau }^{A}}{\partial T} <0$ and all other terms are positive (and $\gamma _{\tau }^{A}<1$).

In the second phase, we have $d^{II} =\frac{\gamma _{\tau }^{A}+\gamma ^B-2\gamma _{\tau }^{A}\gamma ^B}{1-\gamma _{\tau }^{A}\gamma ^B}$. Consequently:

$$\begin{aligned} \frac{\partial d_{\tau }^{II} }{\partial q}&= \frac{\left( \frac{ \partial \gamma _{\tau }^A}{\partial q}-\frac{ \partial 2\gamma _{\tau }^{A}\gamma ^B}{\partial q}\right) \left( 1-\gamma _{\tau }^{A}\gamma ^B\right) - \frac{ \partial (1-\gamma _{\tau }^{A}\gamma ^B)}{\partial q}\left( \gamma _{\tau }^{A}+\gamma ^B-2\gamma _{\tau }^{A}\gamma ^B\right) }{\left( 1-\gamma _{\tau }^{A}\gamma ^B\right) ^2} \\&= \frac{\left( \frac{ \partial \gamma _{\tau }^A}{\partial q}-2\gamma ^B\frac{ \partial \gamma _{\tau }^A}{\partial q} \right) \left( 1-\gamma _{\tau }^{A}\gamma ^B \right) + \gamma ^B\frac{ \partial \gamma _{\tau }^A}{\partial q}\left( \gamma _{\tau }^{A}+\gamma ^B-2\gamma _{\tau }^{A}\gamma ^B\right) }{(1-\gamma _{\tau }^{A}\gamma ^B)^2} \\&= \frac{(1-\gamma ^B)^2\frac{\partial \gamma _{\tau }^{A}}{\partial q}}{\left( 1-\gamma _{\tau }^{A}\gamma ^B\right) ^2} =\frac{(\alpha \beta )^2\frac{ \partial \gamma _{\tau }^A}{\partial q}}{\left( 1-\gamma _{\tau }^{A}\gamma ^B\right) ^2} \end{aligned}$$

As we know, $(\alpha \beta )^2>0$ and $\frac{ \partial \gamma _{\tau }^A}{\partial q}<0$ (confer (27)), which implies a negative nominator and a positive denominator. Consequently, the total extraction is decreasing in q.

Similarly:

$$\begin{aligned} \frac{\partial d_{\tau }^{II} }{\partial T}&= \frac{\left( \frac{ \partial \gamma _{\tau }^A}{\partial T}-\frac{\partial 2\gamma _{\tau }^{A}\gamma ^B}{\partial T} \right) \left( 1-\gamma _{\tau }^{A}\gamma ^B\right) - \frac{\partial (1-\gamma _{\tau }^{A}\gamma ^B)}{\partial T} \left( \gamma _{\tau }^{A}+\gamma ^B-2\gamma _{\tau }^{A}\gamma ^B\right) }{\left( 1-\gamma _{\tau }^{A}\gamma ^B\right) ^2} \\&=\frac{(\alpha \beta )^2\frac{ \partial \gamma _{\tau }^A}{\partial T}}{\left( 1-\gamma _{\tau }^{A}\gamma ^B\right) ^2} \end{aligned}$$

Again, $(\alpha \beta )^2>0$ and $\frac{ \partial \gamma _{\tau }^A}{\partial T}<0$ (following from (30)) , which implies a negative nominator and a positive denominator. Consequently, the total extraction is decreasing in T in the second phase.

In the third phase, neither player’s extraction rate reacts to changes in q. Instead, both player’s extraction rates depend on T as $a^{III}_{\tau }=s_{\tau }$ and $b^{III}_{\tau }=(1-s_{\tau })\gamma ^B$. As an increase in T means a larger $s_{\tau }$, the extraction rate of player A increases in T, and consequently player B’s extraction rate decreases. The total effect is increasing, as player B’s declining effect is smaller than the effect of player A ($1-\alpha \beta <1$).

1.2.7 Proof of Proposition 4

To see that an increase in $\alpha$ or $\beta$ (and hence $\alpha \beta$) leads to a lower total extraction rate in all phases, we first need to establish $\frac{\partial \gamma _{\tau }^A}{\partial \alpha \beta } < 0$ and $\frac{\partial \gamma ^B}{\partial \alpha \beta } < 0$. As $\gamma ^B = 1-\alpha \beta$, the sign of the latter derivative is obvious. For $\gamma _{\tau }^A = (1+\alpha \beta (qk_{\tau }^A + (1-q)k_{\tau +1}^A))^{-1}$, it is more involved. We first need to show that $\frac{\partial k_{\tau }^A}{\partial \alpha \beta } > 0$:

$$\begin{aligned} \frac{\partial k_{\tau }^A}{\partial \alpha \beta }&= \frac{\left( 1-\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{T-\tau }\right) -(T-\tau )\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{T-\tau -1}\left( \frac{(1-q)(1-\alpha \beta q) + \alpha \beta q(1-q)}{(1-\alpha \beta q)^2}\right) (1-\alpha \beta ) }{(1-\alpha \beta )^2} > 0 \\&\Leftrightarrow \left( 1-\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{T-\tau }\right) > (T-\tau )\left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{T-\tau -1}\left( \frac{(1-q)(1-\alpha \beta )}{(1-\alpha \beta q)^2}\right) \\&\Leftrightarrow \left( \frac{\alpha \beta (1-q)}{1-\alpha \beta q}\right) ^{\tau -T} - 1 > (T-\tau )\left( \frac{1-\alpha \beta q}{\alpha \beta (1-q)}\right) \frac{(1-q)(1-\alpha \beta )}{(1-\alpha \beta q)^2} \\&\Leftrightarrow \left( \frac{1-\alpha \beta q}{\alpha \beta - \alpha \beta q}\right) ^{T-\tau }\!- \!1\! >\! (T\!-\!\tau ) \frac{1-\alpha \beta }{\alpha \beta (1\!-\!\alpha \beta q)} \end{aligned}$$

We know that this holds because at $\tau = T-1$ (the largest value that $\tau$ can take at which $k_{\tau }^A$ is still strictly positive), the last line reads: $\frac{1-\alpha \beta q}{\alpha \beta - \alpha \beta q}- 1 > \frac{1-\alpha \beta }{\alpha \beta (1-\alpha \beta q)}$. This can be transformed to $q(1-\alpha \beta )^2>0$ which is true because $q < 1$ and $\alpha \beta < 1$. For $\tau > T-1$ the left-hand side (LHS) of the above inequality is growing exponentially, while the right-hand side (RHS) is growing linearly.

The subsequent comparative static results follow immediately from $\frac{\partial k_{\tau }^A}{\partial \alpha \beta } > 0$:

$$\begin{aligned}&\frac{\partial \left[ qk_{\tau }^A + (1-q)k_{\tau +1}^A\right] }{\partial \alpha \beta } = q \frac{\partial k_{\tau }^A}{\partial \alpha \beta } + (1-q) \frac{\partial k_{\tau +1}^A}{\partial \alpha \beta } > 0 \nonumber \\&\quad \frac{\partial \gamma _{\tau }^A}{\partial \alpha \beta } = \frac{-\left[ qk_{\tau }^A + (1-q)k_{\tau +1}^A\right] - \alpha \beta \frac{\partial \left[ qk_{\tau }^A + (1-q)k_{\tau +1}^A\right] }{\partial \alpha \beta } }{\left( 1+\alpha \beta [qk_{\tau }^A + (1-q)k_{\tau +1}^A]\right) ^2} < 0 \end{aligned}$$

(31)

In the first phase, Player B’s extraction rate is given by $b_{\tau }^I = 1-s_{\tau }$ and does not depend on $\alpha \beta$. Player A’s extraction rate in the first phase is given by $a_{\tau }^I = s_{\tau } \gamma _{\tau }^{A}$ and $\frac{\partial a_{\tau }^I }{\partial \alpha \beta } = s_{\tau }\frac{\partial \gamma _{\tau }^A}{\partial \alpha \beta } < 0$, which follows from (31).

In Phase II, total extraction is $d^{II}_\tau =a^{II}_{\tau }+b_{\tau }^{II} =\frac{\gamma _\tau ^A(1-\gamma ^B)}{1-\gamma _\tau ^A\gamma ^B} +\frac{\gamma ^B(1-\gamma _\tau ^A)}{1-\gamma _\tau ^A\gamma ^B}=\frac{\gamma _{\tau }^{A}+\gamma ^B-2\gamma _{\tau }^{A}\gamma ^B}{1-\gamma _{\tau }^{A}\gamma ^B}$. Consequently:

$$\begin{aligned} \frac{\partial d_{\tau }^{II} }{\partial \alpha \beta }&= \frac{\left( \frac{ \partial \gamma _{\tau }^A}{\partial \alpha \beta }+\frac{ \partial \gamma ^B}{\partial \alpha \beta }-\frac{ \partial 2\gamma _{\tau }^{A}\gamma ^B}{\partial \alpha \beta } \right) \left( 1-\gamma _{\tau }^{A}\gamma ^B\right) - \frac{ \partial (1-\gamma _{\tau }^{A}\gamma ^B)}{\partial \alpha \beta }\left( \gamma _{\tau }^{A}+\gamma ^B-2\gamma _{\tau }^{A}\gamma ^B\right) }{\left( 1-\gamma _{\tau }^{A}\gamma ^B\right) ^2} \\&= \frac{\left( \frac{ \partial \gamma _{\tau }^A}{\partial \alpha \beta }\!-\!1\!+\!2\gamma _{\tau }^{A}\!-\!2\gamma ^B\frac{ \partial \gamma _{\tau }^A}{\partial \alpha \beta }\right) \left( 1\!-\!\gamma _{\tau }^{A}\gamma ^B\right) \!-\! \left( \gamma _{\tau }^{A}-\gamma ^B\frac{ \partial \gamma _{\tau }^A}{\partial \alpha \beta }\right) \left( \gamma _{\tau }^{A}+\gamma ^B-2\gamma _{\tau }^{A}\gamma ^B\right) }{\left( 1-\gamma _{\tau }^{A}\gamma ^B\right) ^2} \\&= \frac{(\alpha \beta )^2\frac{\partial \gamma _{\tau }^A}{\partial \alpha \beta }-\left( 1-\gamma _{\tau }^{A}\right) ^2}{\left( 1-\gamma _{\tau }^{A}\gamma ^B\right) ^2} \end{aligned}$$

As $(\alpha \beta )^2>0$, $\frac{\partial \gamma _{\tau }^A}{\partial \alpha \beta }<0$ and $(1-\gamma _{\tau }^{A})^2>0$, the nominator is negative. Additionally, the denominator is $(1-\gamma _{\tau }^{A}\gamma ^B)^2>0$, and consequently the total extraction is decreasing in $\alpha \beta$ in the second phase.

In the third phase, player A’s extraction rate is independent of $\alpha$ and $\beta$, and thus changes in those do not affect her extraction rate. However, player B’s extraction is linearly dependent on both $\alpha$ and $\beta$ as $b_\tau =(1-s_{\tau })1-\alpha \beta$, and an increase in those parameters results in a decrease in player B’s and consequently in the total extraction rate.