Adaptation and the Allocation of Pollution Reduction Costs

Abstract

We consider a game of abatement of a transboundary pollutant. We use a time-consistent Shapley value allocation of the cost of pollution reduction, and study the sensitivity of such an allocation to countries’ adaptation to pollution. A country’s adaptation to pollution is captured by a change in its damage function. We show that if there is a reduction in the damage cost of one country only, this can harm the other countries. Some countries may end up worse off even in the case where all countries experience a uniform decrease in their damage from pollution. An important policy implication of our analysis is that the Shapley value approach to the allocation of abatement costs does not necessarily provide the right incentives for all players to act on reducing pollution damage. We determine conditions under which a uniform fall in all countries’ pollution damage benefits all countries.

Appendices

Appendix A: Derivation of Value Function (7)

1.1 A.1 Computation of the Value Function of a Coalition of Size k

Suppose there is a coalition K of size k and that each of the n−k players that are not in the coalition individually chooses its cost-minimizing strategy. Let V(K,S) be the value function for coalition K; and, for each player j who stays out of the coalition, let value function be J(j,S). The Hamilton–Jacobi–Bellman (HJB) equations associated with the problem of coalition K and of each of the other n−k outsiders are, respectively,

$$\begin{aligned} & \rho V(K,S) =\min_{e_{i}}\biggl\{\sum _{i\in K}\biggl[\frac{\gamma }{2}(e_{i}- \bar{e}_i)^{2}+\pi _{i}S \biggr]+V^{\prime }(K,S) \biggl(\sum_{i\in K}e_{i}+ \sum_{l\in I\backslash K}e_{l}-\delta S\biggr)\biggr\}\\ &\quad \text{for all }i\in K,\\ & \rho J(j,S) =\min_{e_{j}}\biggl\{\frac{\gamma }{2}(e_{j}- \bar{e}_{j})^{2}+\pi _{j}S+G^{\prime }(j,S) \biggl(\sum_{v\in K}e_{i}+\sum _{l\in I\backslash K}e_{l}-\delta S\biggr)\biggr\}\quad \text{for }j \notin K \end{aligned}$$

where the ′ refers to the derivative with respect to the stock S. Note that, to economize on notation, we also use i and j to respectively refer to country i and j.

The first-order conditions give, for an interior solution:

$$\begin{aligned} e_{i} =&\bar{e}_{i}-\frac{1}{\gamma }V^{\prime }(K,S), \\ \\ e_{j} =&\bar{e}_{j}-\frac{1}{\gamma }J^{\prime }(j,S). \end{aligned}$$

(36)

It can be shown that V and G of the following form, V(K,S)=A _K S+B _K and J(j,S)=D _j S+E _j , satisfy the above system where the coefficients A _K ,B _K ,D _j and E _j are found using the undetermined coefficients technique (see e.g., [9]). Let $\bar{e}=\sum_{i=1}^{n}\bar{e}_{i}$; we have:

$$\begin{aligned} \rho (A_{K}S+B_{K}) =&\frac{k}{2\gamma }A_{K}{}^{2}+ \sum_{i\in K}\pi _{i}S+A_{K}\biggl[ \bar{e}-\frac{k}{\gamma }A_{K}-\frac{1}{\gamma }\sum _{i\in I\backslash K}D_{j}-\delta S\biggr],\\ \rho (D_{j}S+E_{j}) =&\frac{1}{2\gamma }D_{j}{}^{2}+ \pi _{j}S+D_{j}\biggl[\bar{e}-\frac{k}{\gamma }A_{K}-\frac{1}{\gamma }\sum _{i\in I\backslash K}D_{j}-\delta S\biggr] \end{aligned}$$

for all S≥0, which gives

$$\begin{aligned} A_{K} =&\frac{1}{\rho +\delta }\sum_{i\in K}\pi _{i},\\ D_{j} =&\frac{1}{\rho +\delta }\pi _{j} \end{aligned}$$

and

$$\begin{aligned} \rho B_{K} =&A_{K} \biggl( -\frac{k}{2\gamma }A_{K}{}+ \bar{e}-\frac{1}{\gamma }\sum_{i\in I\backslash K}D_{j} \biggr), \\ \rho E_{j} =&D_{j} \biggl( \frac{1}{2\gamma }D_{j}{}+ \bar{e}-\frac{k}{\gamma }A_{K}-\frac{1}{\gamma }\sum _{i\in I\backslash K}D_{j} \biggr) . \end{aligned}$$

The value of a coalition K of size k is thus given by

$$ V(K,S)=\frac{\sum_{i\in K}\pi _{i}}{\rho (\rho +\delta )} \biggl[ \bar{e}-\frac{1}{2\gamma } \frac{1}{\rho +\delta } \biggl( k\sum_{i\in K}\pi _{i}{}+2\sum_{i\in I\backslash K}\pi _{j} \biggr) +\rho S\bigg], $$

or

$$ V(K,S)=\frac{\sum_{i\in K}\pi _{i}}{\rho (\rho +\delta )}\biggl[\bar{e}-\frac{1}{2\gamma (\rho +\delta )}\biggl(2\pi +(k-2)\sum _{i\in K}\pi _{i}\biggr)+\rho S\biggr] $$

where $\pi =\sum_{i=1}^{n}\pi _{i}$.

Appendix B: Proof of Proposition 1

Using the Shapley value formula (13), we obtain

$$ \frac{\partial \phi _{i}(V,S_{0})}{\partial \pi _{i}}=\sum_{K\subset I\backslash \{i\}}\biggl[ \frac{\partial V(K\cup \{i\})}{\partial \pi _{i}}-\frac{\partial V(K)}{\partial \pi _{i}}\biggr]\frac{k!(n-k-1)!}{n!}. $$

Using (8) and (7), we have, after algebraic manipulations:

For i∉K:

$$ \frac{\partial V(K,S_{0})}{\partial \pi _{i}}=\frac{\sum_{j\in K}\pi _{j}}{\rho (\rho +\delta )} \biggl[ -\frac{1}{\gamma (\rho +\delta )} \biggr] \text{ } $$

and

$$ \frac{\partial V(K\cup \{i\},S_{0})}{\partial \pi _{i}}=\frac{1}{\rho (\rho +\delta )} \biggl[ \bar{e}+\rho S+\frac{1}{\gamma (\rho +\delta )} \biggl( -\pi -k \biggl( \sum_{j\in K}\pi _{j}+ \pi _{i} \biggr) \biggr) \biggr]. $$

Therefore,

$$\begin{aligned} \frac{\partial \phi _{i}(V,S_{0})}{\partial \pi _{i}} =&\frac{1}{\rho (\rho +\delta )}\sum_{K\subset I\backslash \{i\}} \biggl( \bar{e}+\rho S+\frac{1}{\gamma (\rho +\delta )} \biggl[ -\pi + ( -k+1 ) \sum _{j\in K}\pi _{j}-k\pi _{i} \biggr] \biggr) \\ &{}\times \frac{k!(n-k-1)!}{n!} \\ >&\frac{1}{\rho (\rho +\delta )} \biggl(\bar{e}+\rho S+\frac{-(k+1)\pi }{\gamma (\rho +\delta )} \biggr) >0 \end{aligned}$$

since: $\bar{e}+\rho S>\frac{n\pi }{\gamma (\rho +\delta )}$ by (11).  □

Appendix C: Proof of Proposition 2

Using the Shapley value formula (13), we obtain

$$\begin{aligned} \frac{\partial \phi _{m}(V,S_{0})}{\partial \pi _{j}} =&\sum_{K\subset I\backslash \{m,j\}}\biggl[ \frac{\partial V(K\cup \{m\})}{\partial \pi _{j}}-\frac{\partial V(K)}{\partial \pi _{j}}\biggr]\frac{k!(n-k-1)!}{n!} \\ &{}+\sum_{\substack{ K\subset I\backslash \{m\} \\ ~j\in K}}\biggl[\frac{\partial V(K\cup \{m\})}{\partial \pi _{j}}- \frac{\partial V(K)}{\partial \pi _{j}}\biggr]\frac{k!(n-k-1)!}{n!}. \end{aligned}$$

(37)

For i∈K, using (8) and (7), we have, after algebraic manipulations:

$$ \frac{\partial V(K,S_{0})}{\partial \pi _{i}}=\frac{\bar{e}+\rho S_{0}}{\rho (\rho +\delta )}-\frac{\pi }{\rho \gamma (\rho +\delta )^{2}}+\frac{-k+1}{\rho \gamma (\rho +\delta )^{2}}\sum _{i\in K}\pi _{i}.\text{ } $$

(38)

For j∉K,

$$ \frac{\partial V(K,S_{0})}{\partial \pi _{j}}=\frac{-1}{\rho \gamma (\rho +\delta )^{2}}\sum_{i\in K}\pi _{i}. $$

(39)

Substituting (38) and (39) into (37) yields

$$\begin{aligned} \frac{\partial \phi _{m}(V,S_{0})}{\partial \pi _{j}} =&\sum_{\substack{ K\subset I\backslash \{m\} \\ j\in K}}\biggl[ \frac{-k}{\rho \gamma (\rho +\delta )^{2}}\biggl(\sum_{i\in K}\pi _{i}+\pi _{m}\biggr)-\frac{-k+1}{\rho \gamma (\rho +\delta )^{2}}\sum _{i\in K}\pi _{i}\biggr]\frac{k!(n-k-1)!}{n!} \\ &{}+\sum_{K\subset I\backslash \{m,j\}}\biggl[\frac{-1}{\rho \gamma (\rho +\delta )^{2}}\biggl(\sum _{i\in K}\pi _{i}+\pi _{m}\biggr)- \frac{-1}{\rho \gamma (\rho +\delta )^{2}}\sum_{i\in K}\pi _{i}\biggr] \\ &{}\times\frac{k!(n-k-1)!}{n!} \\ =&\sum_{\substack{ K\subset I\backslash \{m\} \\ j\in K}}\biggl[\frac{-k}{\rho \gamma (\rho +\delta )^{2}}\pi _{m}+\frac{-1}{\rho \gamma (\rho +\delta )^{2}}\sum_{i\in K}\pi _{i}\biggr]\frac{k!(n-k-1)!}{n!} \\ &{}+\sum_{K\subset I\backslash \{m,j\}}\biggl[\frac{-1}{\rho \gamma (\rho +\delta )^{2}}\pi _{m}\biggr]\frac{k!(n-k-1)!}{n!} \\ <&0. \end{aligned}$$

 □

Appendix D: Proof of Proposition 5

We seek to determine if the Shapley value of a player m can be larger when α<1 than when α=1:

$$ \sum_{K\subset I\backslash \{m\}}\bigl(\bigl[V_{\alpha }\bigl(K\cup \{m\}\bigr)-V_{\alpha }(K)\bigr]- \bigl[ V\bigl(K\cup \{m\}\bigr)-V(K) \bigr] \bigr)\frac{k!(n-k-1)!}{n!}>0 $$

where V _α is the value function when the marginal damage from pollution is απ _i for all i=1,…,n.

Substituting π _i by απ _i for i=1,…,n in (14) yields

$$\begin{aligned} &\bigl[V_{\alpha }\bigl(K\cup \{m\}\bigr)-V_{\alpha }(K)\bigr]- \bigl[ V\bigl(K\cup \{m\}\bigr)-V(K) \bigr] ~ \\ &\quad =\frac{ ( \alpha -1 ) \pi _{m}}{\rho (\rho +\delta )}(\bar{e}+\rho S_{0})- \frac{ ( \alpha ^{2}-1 ) }{2\gamma \rho (\rho +\delta )^{2}}\bigl[(k+1)\pi _{m}^{2}+2 \bigl( n-1+k^{2}-k \bigr) \pi _{o}\pi _{m}+k^{2} \pi _{o}{}^{2}\bigr] \\ &\quad =\frac{ ( \alpha -1 ) }{2\gamma \rho (\rho +\delta )^{2}} \bigl( 2\gamma (\rho +\delta ) (\bar{e}+\rho S_{0})\pi _{m} \\ &\qquad {}- ( \alpha +1 ) \bigl[ (k+1)\pi _{m}^{2}+2 \bigl( n-1+k^{2}-k \bigr) \pi _{o}\pi _{m}+k^{2}\pi _{o}{}^{2} \bigr] \bigr) , \end{aligned}$$

which is positive if

$$ 2\gamma (\rho +\delta ) (\bar{e}+\rho S_{0})\pi _{m}- ( \alpha +1 ) \bigl[ (k+1)\pi _{m}^{2}+2 \bigl( n-1+k^{2}-k \bigr) \pi _{o}\pi _{m}+k^{2} \pi _{o}{}^{2} \bigr] >0. $$

(40)

The left-hand side of (40) is decreasing in k. Therefore if it is positive for k=n−1, it’s positive for all k≤n−1.

For k=n−1, condition (40) gives

$$ 2\gamma (\rho +\delta ) (\bar{e}+\rho S_{0})\pi _{m}- ( \alpha +1 ) \bigl[ n\pi _{m}^{2}+2 ( n-1 ) ^{2}\pi _{o}\pi _{m}+ ( n-1 ) ^{2}\pi _{o}{}^{2} \bigr] >0.\text{ } $$

By (11) and the fact that S ₀≥0, a sufficient condition for (40) to hold is that

$$ 2n \bigl( ( n-1 ) \pi _{o}+\pi _{m} \bigr) \pi _{m}- ( \alpha +1 ) \bigl( n\pi _{m}^{2}+2 ( n-1 ) ^{2}\pi _{o}\pi _{m}+ ( n-1 ) ^{2}\pi _{o}{}^{2} \bigr) >0 $$

or

$$ 2n \bigl( ( n-1 ) r+r^{2} \bigr) - ( \alpha +1 ) \bigl( nr^{2}+2 ( n-1 ) ^{2}r+ ( n-1 ) ^{2} \bigr) >0. $$

Therefore, we can infer that

$$ \alpha <\frac{nr^{2}+2r ( n-1 ) - ( n-1 ) ^{2}}{nr^{2}+2 ( n-1 ) ^{2}r+ ( n-1 ) ^{2}}. $$

For the right-hand side to be positive, we need

$$ nr^{2}+2r ( n-1 ) - ( n-1 ) ^{2}>0 $$

or

$$ r>\frac{1}{n} \bigl( \sqrt{n+1}-1 \bigr) ( n-1 ) . $$

 □

Appendix E: Proof of Proposition 6

In this appendix we provide sufficient conditions to have

$$ \sum_{K\subset I\backslash \{m\}}\bigl(\bigl[V\bigl(K\cup \{m\}\bigr)-V(K) \bigr]-\bigl[V_{\alpha }\bigl(K\cup \{m\}\bigr)-V_{\alpha }(K)\bigr] \bigr)\frac{k!(n-k-1)!}{n!}>0 $$

(41)

where V _α is the value function corresponding to {απ _i ,i∈I} and V is the value function corresponding to {π _i ,i∈I}. From (14), after substituting π _i by απ _i for all i=1,…,n, we have

$$\begin{aligned} V_{\alpha }\bigl(K\cup \{m\}\bigr)-V_{\alpha }(K) =&\frac{\alpha \pi _{m}}{\rho (\rho +\delta )}( \bar{e}+\rho S_{0})-\frac{\alpha ^{2}}{2\gamma \rho (\rho +\delta )^{2}} \\ &{}\times\biggl[2\pi \pi _{m}+2(k-1)\pi _{m}\sum_{i\in K} \pi _{i}+(k-1)\pi _{m}^{2}+\biggl(\sum _{i\in K}\pi _{i}\biggr)^{2}\biggr]. \end{aligned}$$

(42)

We first show that

$$ \frac{\partial }{\partial \alpha } \bigl[ V_{\alpha }\bigl(K\cup \{m\}\bigr)-V_{\alpha }(K) \bigr] >0, $$

(43)

which implies that V _α (K∪{m})−V _α (K), and consequently, that the Shapley value for player m is increasing in α. In particular, the Shapley value will be highest for α=1; thus, we can conclude that the Shapley value for player m when α<1 is strictly smaller than when α=1.

Using (42), condition (43) holds iff

$$\begin{aligned} &\frac{\pi _{m}}{\rho (\rho +\delta )}(\bar{e}+\rho S_{0})-\frac{\alpha }{\gamma \rho (\rho +\delta )^{2}}\biggl[2 \pi \pi _{m}+2(k-1)\pi _{m}\sum _{i\in K}\pi _{i} \\ &\quad {}+(k-1)\pi _{m}^{2}+ \biggl(\sum_{i\in K}\pi _{i} \biggr)^{2}\biggr] >0, \end{aligned}$$

(44)

$$\begin{aligned} &\quad \Leftrightarrow \quad \alpha \biggl[ 2\pi \pi _{m}+2(k-1)\pi _{m}\sum _{i\in K}\pi _{i}+(k-1)\pi _{m}^{2}+ \biggl(\sum_{i\in K}\pi _{i} \biggr)^{2}\biggr] \\ &\hphantom{\quad \Leftrightarrow \quad}\quad <\pi _{m}\gamma (\rho +\delta ) ( \bar{e}+\rho S_{0}). \end{aligned}$$

(45)

In the case π _k =π _o for all k≠m, this condition becomes

$$ \alpha \bigl[ 2 \bigl( ( n-1 ) \pi _{o}+\pi _{m} \bigr) \pi _{m}+2(k-1)\pi _{m}k\pi _{o}+(k-1)\pi _{m}^{2}+(k\pi _{o})^{2}\bigr]<\pi _{m}\gamma (\rho +\delta ) (\bar{e}+\rho S_{0}). $$

(46)

The left-hand side of this inequality is a strictly increasing function of k, and therefore, if the inequality holds for k=n−1, then it holds for all k≤n−1:

$$\begin{aligned} & \alpha \bigl[ 2 \bigl( ( n-1 ) \pi _{o}+\pi _{m} \bigr) \pi _{m}+2(n-2) ( n-1 ) \pi _{m}\pi _{o}+(n-2)\pi _{m}^{2}+ \bigl( ( n-1 ) \pi _{o} \bigr)^{2}\bigr] \\ &\quad <\pi _{m}\gamma (\rho +\delta ) (\bar{e}+\rho S_{0}) \end{aligned}$$

or

$$ \alpha \bigl[ 2 ( n-1 ) ^{2}\pi _{o}\pi _{m}+2 \pi _{m}^{2}+(n-2)\pi _{m}^{2}+\bigl( ( n-1 ) \pi _{o}\bigr)^{2}\bigr]<\pi _{m}\gamma (\rho + \delta ) (\bar{e}+\rho S_{0}). $$

A sufficient condition is

$$ \alpha \bigl[ 2 ( n-1 ) ^{2}\pi _{o}\pi _{m}+n \pi _{m}^{2}+\bigl( ( n-1 ) \pi _{o} \bigr)^{2}\bigr]<\pi _{m}\gamma (\rho +\delta )\bar{e}. $$

From (11), it is sufficient that

$$ \alpha \bigl[ 2 ( n-1 ) ^{2}\pi _{o}\pi _{m}+n \pi _{m}^{2}+\bigl( ( n-1 ) \pi _{o} \bigr)^{2}\bigr]<\pi _{m}n\pi \text{,} $$

which, after substitution of π=(n−1)π _o +π _m , gives

$$ h ( \alpha ) \equiv \alpha \bigl( 2 ( n-1 ) ^{2}r+nr^{2}+(n-1)^{2} \bigr) -nr \bigl( ( n-1 ) +r \bigr) <0, $$

where $r=\frac{\pi _{m}}{\pi _{o}}$. Thus it is sufficient to have h(α)<0 for (41) to be satisfied. The function h is strictly increasing in α with h(0)<0 and h(1)>0. We can therefore conclude that there exists $\tilde{\alpha}=\frac{nr ( ( n-1 ) +r ) }{2 ( n-1 ) ^{2}r+nr^{2}+(n-1)^{2}}<1$ such that, for all $\alpha <\tilde{\alpha}$, we have, for all $( \bar{e},S_{0} )$: $\frac{\partial }{\partial \alpha } [ V_{\alpha }(K\cup \{m\})-V_{\alpha }(K) ] >0$ for all K⊂I∖{m} and therefore (41) is satisfied.  □

Appendix F: Proof of Proposition 7a

We have

$$ \frac{\partial }{\partial \pi _{m}}\beta _{m}(t)=\rho \frac{\partial }{\partial \pi _{m}}\phi _{m}\bigl(V,S^{I}(t)\bigr)-\frac{\partial }{\partial \pi _{m}}\frac{d}{dt}\phi _{m}\bigl(V,S^{I}(t)\bigr). $$

To study the sign of $\frac{\partial }{\partial \pi _{m}}\beta _{m}(t)$, we separately compute $\frac{\partial }{\partial \pi _{m}}\phi _{m}(V,S^{I}(t))$ and $\frac{\partial }{\partial \pi _{m}}\frac{d}{dt}\phi _{m}(V, S^{I}(t))$.

Using (13), we get, after simplification,

$$ \frac{\partial }{\partial \pi _{m}}\phi _{m}\bigl(V,S^{I} ( t ) \bigr)= \frac{\bar{e}}{\rho (\rho +\delta )}+\frac{S^{I} ( t ) }{(\rho +\delta )}+\frac{\pi _{m}}{(\rho +\delta )} \frac{\partial S^{I} ( t ) }{\partial \pi _{m}}-\frac{1}{\rho \gamma (\rho +\delta )^{2}}\varOmega $$

(47)

where

$$ \varOmega \equiv \pi +\sum_{K\subset I\backslash \{m\}} \biggl( \pi _{m}k+\sum_{i\in K}\pi _{i} ( k-1 ) \biggr) \frac{k!(n-k-1)!}{n!}. $$

When π _j =π _o for all j≠m, we have

$$ \varOmega =\pi +\frac{1}{n} \biggl( \frac{1}{2}n ( n-1 ) \pi _{m}+\pi _{o} \biggl( \frac{1}{3}n \bigl( n^{2}-3n+2 \bigr) \biggr) \biggr). $$

Moreover, differentiating (12) with respect to π _m gives

$$ \frac{\partial }{\partial \pi _{m}}S^{I}(t)=\frac{1}{\delta }\biggl(- \frac{n}{\gamma }\frac{1}{\rho +\delta }\biggr) \bigl(1-e^{-\delta t}\bigr)\text{,} $$

and therefore,

$$ \frac{S^{I} ( t ) }{(\rho +\delta )}+\frac{\pi _{m}}{(\rho +\delta )}\frac{\partial S^{I} ( t ) }{\partial \pi _{m}}=\frac{1}{(\rho +\delta )} \biggl( S_{0}e^{-\delta t}+\frac{1}{\delta }\biggl(\bar{e}- \frac{n}{\gamma }\frac{1}{\rho +\delta } ( \pi +\pi _{m} ) \biggr) \bigl(1-e^{-\delta t}\bigr) \biggr) . $$

(48)

Replacing (48) into (47) and π by π _m +(n−1)π _o gives

$$\begin{aligned} \frac{\partial }{\partial \pi _{m}}\phi _{m}(V,S) =&\frac{\bar{e}}{\rho (\rho +\delta )}+ \frac{1}{(\rho +\delta )} \biggl( S_{0}e^{-\delta t}+\frac{1}{\delta } \biggl(\bar{e}-\frac{n}{\gamma }\frac{1}{\rho +\delta } \bigl( ( n-1 ) \pi _{o}+2\pi _{m} \bigr) \biggr) \\ &{}\times \bigl(1-e^{-\delta t}\bigr) \biggr) \\ &{}-\frac{ 1 }{\rho \gamma (\rho +\delta )^{2}} \biggl( \frac{1}{2} ( n+1 ) \pi _{m}+ \pi _{o}\frac{1}{3} \bigl( n^{2}-1 \bigr) \biggr) . \end{aligned}$$

(49)

We now compute $\frac{\partial }{\partial \pi _{m}}\frac{d}{dt}\phi _{m}(V,S^{I}(t))$:

$$\begin{aligned} \frac{\partial }{\partial \pi _{m}}\frac{d}{dt}\phi _{m}\bigl(V,S^{I}(t) \bigr) =&\frac{\partial }{\partial \pi _{m}} \biggl( \sum _{K\subset I\backslash \{m\}}\biggl[\frac{dV(K\cup \{m\})}{dt}- \frac{dV(K)}{dt}\biggr]\frac{k!(n-k-1)!}{n!} \biggr) \notag \\ =&\frac{\partial }{\partial \pi _{m}} \biggl( \sum_{K\subset I\backslash \{m\}} \biggl[ \frac{\pi _{m}}{\rho (\rho +\delta )}\biggl[\rho \frac{d}{dt}S^{I} ( t ) \biggr] \biggr] \frac{k!(n-k-1)!}{n!} \biggr) . \end{aligned}$$

(50)

Using (12), we compute $\frac{d}{dt}S^{I} ( t )$ and substitute it into (50) to obtain

$$ \frac{\partial }{\partial \pi _{m}}\frac{d}{dt}\phi _{m}(V,S)= \frac{1}{(\rho +\delta )} \biggl( -\delta S_{0}+\bar{e}-\frac{n}{\gamma } \biggl( \frac{ ( n-1 ) \pi _{o}+2\pi _{m}}{\rho +\delta } \biggr) \biggr) e^{-\delta t}. $$

(51)

Proof of Part (i)

By (49) and (51):

$$\begin{aligned} \lim_{t\rightarrow 0}\frac{\partial }{\partial \pi _{m}}\beta _{m}(t) =&\lim _{t\rightarrow 0} \biggl( \rho \frac{\partial }{\partial \pi _{m}}\phi _{m} \bigl(V,S^{I}(t)\bigr)-\frac{\partial }{\partial \pi _{m}}\frac{d}{dt}\phi _{m}\bigl(V,S^{I}(t)\bigr) \biggr) \notag \\ >&\frac{\bar{e}}{(\rho +\delta )}+\frac{\rho }{(\rho +\delta )}S_{0}e^{-\delta t}- \frac{ ( n+3 ) \pi }{2\gamma (\rho +\delta )^{2}} \\ &{}-\frac{1}{(\rho +\delta )} \biggl( -\delta S_{0}+\bar{e}-\frac{n}{\gamma }\biggl( \frac{\pi +\pi _{m}}{\rho +\delta } \biggr) \biggr) \end{aligned}$$

(52)

which gives, after substitution of π by (n−1)π _o +π _m and simplifications,

$$ \lim_{t\rightarrow 0}\frac{\partial }{\partial \pi _{m}}\beta _{m}(t)>S_{0}+\frac{ ( ( n-3 ) \pi _{o}+3\pi _{m} ) (n-1)}{2\gamma (\rho +\delta )^{2}}, $$

(53)

which is clearly positive for n≥3. For n=2 from (49) and (51) we have

$$ \frac{\partial }{\partial \pi _{m}}\phi _{m}(V,S)=\frac{\bar{e}}{\rho (\rho +\delta )}+ \frac{S^{I} ( t ) }{(\rho +\delta )}+\frac{\pi _{m}}{(\rho +\delta )}\frac{\partial S ( t ) }{\partial \pi _{m}}-\frac{\pi +\pi _{m}}{\rho \gamma (\rho +\delta )^{2}} $$

and

$$ \lim_{t\rightarrow 0} \biggl( \frac{\partial }{\partial \pi _{m}}\frac{d}{dt}\phi _{m}(V,S) \biggr) =\frac{1}{(\rho +\delta )} \biggl( -\delta S_{0}+\bar{e}-\frac{2}{\gamma } \biggl( \frac{\pi +\pi _{m}}{\rho +\delta } \biggr) \biggr) . $$

Therefore, using (48),

$$\begin{aligned} \lim_{t\rightarrow 0}\frac{\partial }{\partial \pi _{m}}\beta _{m}(t) =&\lim _{t\rightarrow 0} \biggl( \frac{\bar{e}}{(\rho +\delta )}+\frac{\rho S^{I} ( t ) }{(\rho +\delta )}+ \frac{\rho \pi _{m}}{(\rho +\delta )}\frac{\partial S ( t ) }{\partial \pi _{m}}-\frac{\pi +\pi _{m}}{\gamma (\rho +\delta )^{2}} \biggr) \\ &-\lim_{t\rightarrow 0} \biggl( \frac{1}{(\rho +\delta )} \biggl( \bar{e}-\delta S_{0}-\frac{2}{\gamma } \biggl( \frac{\pi +\pi _{m}}{\rho +\delta }\biggr) \biggr) \biggr) \\ =&S_{0}+\frac{ ( \pi +\pi _{m} ) }{\gamma (\rho +\delta )^{2}}>0. \end{aligned}$$

(54)

Proof of Part (ii)

Differentiating (13) with respect to π _j yields

$$\begin{aligned} \frac{\partial }{\partial \pi _{j}}\phi _{m}(V,S) =&\sum _{\substack{ K\subset I\backslash \{m\} \\ ~j\in K}}\biggl[\frac{-k}{\rho \gamma (\rho +\delta )^{2}}\pi _{m}+ \frac{-1}{\rho \gamma (\rho +\delta )^{2}}\sum_{i\in K}\pi _{i} \biggr]\frac{k!(n-k-1)!}{n!} \\ &{}+\sum_{K\subset I\backslash \{m,j\}}\biggl[\frac{-1}{\rho \gamma (\rho +\delta )^{2}}\pi _{m}\biggr]\frac{k!(n-k-1)!}{n!} \\ &{}+\frac{-n\pi _{m}}{\delta \gamma ( \rho +\delta ) ^{2}}\bigl(1-e^{-\delta t}\bigr) \end{aligned}$$

(55)

and

$$\begin{aligned} & \frac{d}{dt}\phi _{m}(V,S) =\frac{\pi _{m}}{(\rho +\delta )} \biggl( - \delta S_{0}+\bar{e}-\frac{n}{\gamma }\frac{\pi }{\rho +\delta } \biggr) e^{-\delta t}, \\ \\ &\frac{\partial }{\partial \pi _{j}} \biggl( \frac{d}{dt}\phi _{m} \bigl(V,S^{I}(t)\bigr) \biggr) =-\frac{n\pi _{m}}{\gamma (\rho +\delta )^{2}}e^{-\delta t}. \end{aligned}$$

(56)

Therefore, we can obtain the following:

$$\begin{aligned} \lim_{t\rightarrow 0}\frac{\partial }{\partial \pi _{j}}\beta _{m}(t) =&\lim _{t\rightarrow 0} \biggl( \rho \frac{\partial }{\partial \pi _{j}}\phi _{m} \bigl(V,S^{I}(t)\bigr)-\frac{\partial }{\partial \pi _{j}}\frac{d}{dt}\phi _{m}\bigl(V,S^{I}(t)\bigr) \biggr) \\ =&\sum_{\substack{ K\subset I\backslash \{m\} \\ ~j\in K}}\biggl[\frac{-k}{\gamma (\rho +\delta )^{2}}\pi _{m}+\frac{-k}{\gamma (\rho +\delta )^{2}}\pi _{o}\biggr]\frac{k!(n-k-1)!}{n!} \\ &{}+\sum_{K\subset I\backslash \{m,j\}}\biggl[\frac{-1}{\gamma (\rho +\delta )^{2}}\pi _{m}\biggr]\frac{k!(n-k-1)!}{n!}+\frac{n\pi _{m}}{\gamma (\rho +\delta )^{2}} \\ =&-\frac{ ( n-2 ) ( 2n-3 ) }{6n} \biggl( \frac{\pi _{m}+\pi _{o}}{\gamma (\rho +\delta )^{2}} \biggr) -\frac{1}{\gamma (\rho +\delta )^{2}} \pi _{m} \biggl( \frac{1}{2} \biggr) +\frac{n\pi _{m}}{\gamma (\rho +\delta )^{2}} \\ =& \biggl( \frac{1}{6n} \bigl( \bigl( 4n^2+4n-6 \bigr) \pi _{m}- \bigl( 2n^{2}-7n+6 \bigr) \pi _{o} \bigr) \biggr) \frac{1}{\gamma (\rho +\delta )^{2}} \end{aligned}$$

(57)

Therefore $\lim_{t\rightarrow 0}\frac{\partial }{\partial \pi _{j}}\beta _{m}(t)|_{\pi_{o}=0}>0$.  □

Appendix G: Proof of Proposition 7b

Proof of Part (i)

When t→∞ by (51) $\lim_{t\rightarrow \infty } ( \frac{\partial }{\partial \pi _{m}}\frac{d}{dt}\phi _{m}(V,S^{I} ( t ) ) ) =0$, so using (49), at π _o =0, we have

$$ \lim_{t\rightarrow \infty } \frac{\partial }{\partial \pi _{m}}\beta _{m}(t) \vert _{\pi _{o}=0}=\frac{1}{\delta }\bar{e}-\frac{1}{\delta \gamma (\rho +\delta )^{2}} \bigl( \bigl(4\rho n+(n+1)\delta \bigr)\bigl((\pi _{m})/2\bigr)\bigr) $$

(58)

therefore

$$ \lim_{t\rightarrow \infty } \frac{\partial }{\partial \pi _{m}}\beta _{m}(t) \bigg \vert _{\pi _{o}=0}<0 $$

if and only if

$$ \bar{e}<\frac{1}{\gamma (\rho +\delta )^{2}} \bigl( \bigl(4\rho n+(n+1)\delta \bigr)\bigl((\pi _{m})/2\bigr) \bigr) . $$

Therefore if (2n−1)ρ−δ>0, then for $\bar{e}$, such that

$$ \frac{1}{\delta }\frac{\pi _{m}}{\gamma (\rho +\delta )}\leq \frac{1}{\delta }\bar{e}< \frac{1}{\delta \gamma (\rho +\delta )^{2}} \bigl( \bigl(4\rho n+(n+1)\delta \bigr)\bigl((\pi _{m})/2\bigr) \bigr), $$

(59)

we have $\lim_{t\rightarrow \infty }\frac{\partial }{\partial \pi _{m}}\beta _{m}(t)=\frac{1}{\delta }\bar{e}-\frac{1}{\delta \gamma (\rho +\delta )^{2}}( (4\rho n+(n+1)\delta )((\pi _{m})/2) ) <0$. Condition (59), after simple manipulations, can be rewritten as (24).

Proof of Part (ii)

Using (55) and (56), we have

$$\begin{aligned} \lim_{t\rightarrow \infty }\frac{\partial }{\partial \pi _{j}}\beta _{m}(t) =&\lim _{t\rightarrow \infty } \biggl( \rho \frac{\partial }{\partial \pi _{j}}\phi _{m}\bigl(V,S^{I}(t)\bigr)-\frac{\partial }{\partial \pi _{j}} \frac{d}{dt}\phi _{m}\bigl(V,S^{I}(t)\bigr) \biggr) \\ =&\sum_{\substack{ K\subset I\backslash \{m\} \\ ~j\in K}}\biggl[\frac{-k}{\gamma (\rho +\delta )^{2}}\pi _{m}+\frac{-k}{\gamma (\rho +\delta )^{2}}\pi _{o}\biggr]\frac{k!(n-k-1)!}{n!} \\ &{}+\sum_{K\subset I\backslash \{m,j\}}\biggl[\frac{-1}{\gamma (\rho +\delta )^{2}}\pi _{m}\biggr]\frac{k!(n-k-1)!}{n!}+\frac{-\rho n\pi _{m}}{\delta \gamma ( \rho +\delta ) } \\ <&0. \end{aligned}$$

 □

Appendix H: Derivation of the Value Functions for Quadratic Damage Costs

Let W(I,S) be the value function for the grand coalition. The HJB equation associated with the grand coalition reads:

$$ \rho W(I,S)=\min \Biggl\{\sum_{i=1}^{3} \biggl[\frac{1}{2}(e_{i}-\bar{e}_{i})^{2} \biggr]+\frac{1}{2}\eta _{1}S^{2}+W^{\prime}(I,S) \Biggl(\sum_{i=1}^{3}e_{i}- \delta S\Biggr)\Biggr\} $$

where the ′ refers to the derivative with respect to the stock S.

The first order conditions give for an interior solution:

$$ e_{i}=\bar{e}_{i}-W^{\prime }(I,S). $$

A quadratic value function, W(I,S)=Z _I S ²+A _I S+B _I satisfies the above equation and the coefficients Z _I , A _I , B _I are found using undetermined coefficient technique. Let $\bar{e}=\sum_{i\in I}\bar{e}_{i}$; we have:

$$\begin{aligned} \rho \bigl(Z_{I}S^{2}+A_{I}S+B_{I} \bigr) =&\sum_{i=1}^{3}\biggl[ \frac{1}{2}(2Z_{I}S+A_{I})^{2} \biggr]+\frac{1}{2}\eta _{1}S^{2}\\ &{}+(2Z_{I}S+A_{I}) \Biggl(\bar{e}-\sum_{j=1}^{3}(2Z_{I}S+A_{I})- \delta S\Biggr). \end{aligned}$$

Equating the coefficients on both sides of the equality provides two sets of solutions:

$$\begin{aligned} Z_{I} =&\frac{1}{12} \Bigl( - ( 2\delta +\rho ) \pm \sqrt{ \bigl( 12\eta _{1}+4\delta ^{2}+\rho ^{2}+4\delta \rho \bigr) } \,\Bigr), \\ A_{I} =&2\bar{e}\frac{Z_{I}}{\delta +\rho +6Z_{I}}, \\ B_{I} =&\frac{1}{2\rho }A_{I}(2\bar{e}-3A_{I}). \end{aligned}$$

Evolution of stock of pollution under full cooperation is given by:

$$ S^{C}(t)=\frac{\bar{e}-3A_{I}}{6Z_{I}+\delta }+ \biggl( S_{0}- \frac{\bar{e}-3A_{I}}{6Z_{I}+\delta } \biggr) e^{-(6Z_{I}+\delta )t} $$

where we should have:

$$ 6Z_{I}+\delta \geq 0. $$

Therefore, the coefficient of the value function for the grand coalition are:

$$\begin{aligned} Z_{I} =&\frac{1}{12} \bigl( - ( 2\delta +\rho ) +\sqrt{12\eta _{1}+(2\delta +\rho )^{2}}\, \bigr), \\ A_{I} =&2\bar{e}\frac{Z_{I}}{\delta +\rho +6Z_{I}}, \\ B_{I} =&\frac{1}{2\rho }A_{I}(2\bar{e}-3A_{I}), \end{aligned}$$

which can be written as:

$$ Z_{I}=\frac{1}{12}f ( 12 ) ,\qquad A_{I}=g ( 12,6 ) ,\qquad B_{I}=u ( 12,6,3 ) , $$

where

$$\begin{aligned} f ( x ) =&\sqrt{(2\delta +\rho )^{2}+\eta _{1}x}- ( 2 \delta +\rho ), \\ g ( x,q ) =&\frac{2\bar{e}f ( x ) }{x ( \delta +\rho ) +qf ( x ) }, \\ u ( x,q,\kappa ) =&\frac{1}{2\rho }g ( x,q ) \bigl(2\bar{e}- \kappa g ( x,q ) \bigr). \end{aligned}$$

Let W({1},S) be the value function for player i corresponding to the Nash equilibrium. The HJB equation for player 1 reads:

$$ \rho W\bigl(\{1\},S\bigr)=\min \Biggl\{\frac{1}{2}(e_{1}- \bar{e}_{1})^{2}+\frac{1}{2}\eta _{1}S^{2}+W^{\prime } \bigl( \{ 1 \} ,S \bigr) \Biggl(\sum_{j=1}^{3}e_{j}- \delta S\Biggr)\Biggr\}. $$

In the case where players 2 and 3 have no damage cost, their emission strategy is constant and equal to $\bar{e}_{i}$. Therefore, their value function is zero.

In the same manner as computing the value function for the grand coalition, we can find the coefficient of player 1’s value function for the Nash equilibrium W({1},S)=Z ₁ S ²+A ₁ S+B ₁:

$$\begin{aligned} Z_{1} =&\frac{1}{4}\bigl(-(2\delta +\rho )+\sqrt{(2\delta + \rho )^{2}+4\eta _{1}}\bigr)=\frac{1}{4}f ( 4 ), \\ A_{1} =&2\bar{e}\frac{Z_{1}}{\delta +\rho +2Z_{1}}=g ( 4,2 ), \\ B_{1} =&\frac{1}{2\rho }A_{1}(2\bar{e}-A_{1}) =u ( 4,2,1 ) . \end{aligned}$$

Computing the value function for coalition of Countries 1 and 2:

Denote this coalition by W(K,S)=W({1,2},S), and the value function for player 3 by G(j,S). HJB equation for this coalition reads:

$$\begin{aligned} \rho W(K,S) =&\sum_{i\in \{1,2\}}\biggl[\frac{1}{2} \bigl(W^{\prime }(K,S)\bigr)^{2}\biggr]+\frac{1}{2}\eta _{1}S^{2} \\ &{}+W^{\prime }(K,S)\biggl[\sum _{i\in \{1,2\}}\bigl(\bar{e}_{i}-W^{\prime }(K,S)\bigr)+ \sum_{i=3}\bigl(\bar{e}_{i}-G^{\prime }(i,S) \bigr)-\delta S\biggr]. \end{aligned}$$

(60)

Emission strategy for player 3 is to choose $e=\bar{e}_{3}$, so G=0. A quadratic value function, W({1,2},S)=Z ₁₂ S ²+A ₁₂ S+B ₁₂ satisfies equality (60). The coefficients of this value function are as follows:

$$\begin{aligned} Z_{12} =&Z_{13}=\frac{1}{8}\bigl(-(2\delta +\rho )+ \sqrt{(2\delta +\rho )^{2}+8\eta _{1}}\bigr), \\ A_{12} =&A_{13}=2\bar{e}\frac{Z_{12}}{\delta +\rho +4Z_{12}}, \\ B_{12} =&B_{13}=\frac{1}{\rho }A_{12}( \bar{e}-A_{12}), \\ Z_{23} =&A_{23}=B_{23}=0. \end{aligned}$$

Finally, the Shapley values can be obtained as follows:

$$\begin{aligned} \phi _{1} =&\frac{1}{3}\bigl[(Z_{I}+Z_{12}+Z_{1})S^{2}+(A_{I}+A_{12}+A_{1})S+(B_{I}+B_{12}+B_{1}) \bigr], \\ \phi _{2} =&\phi _{3} \\ =&\frac{1}{2}Z_{I}S^{2}- \frac{1}{6}(Z_{I}+Z_{12}+Z_{1})S^{2}+ \frac{1}{2}A_{I}S-\frac{1}{6}(A_{I}+A_{12}+A_{1})S+ \frac{1}{2}B_{I}\\ &{}- \frac{1}{6}(B_{I}+B_{12}+B_{1}) \\ =&\frac{1}{6}(2Z_{I}-Z_{12}-Z_{1})S^{2}+ \frac{1}{6}(2A_{I}-A_{12}-A_{1})S+\frac{1}{6}(2B_{I}-B_{12}-B_{1}). \end{aligned}$$

Appendix I: Proof of Proposition 9

Denote by $\phi _{i,\eta _{1}}^{\prime }$ the derivative of ϕ _i (i=1,2,3) with respect to η ₁.

$$ \phi _{2,\eta _{1}}^{\prime }=\frac{1}{6}\bigl(2Z_{I}^{\prime }-Z_{12}^{\prime }-Z_{1}^{\prime } \bigr)S^{2}+\frac{1}{6}\bigl(2A_{I}^{\prime }-A_{12}^{\prime }-A_{1}^{\prime } \bigr)S+\frac{1}{6}\bigl(2B_{I}^{\prime }-B_{12}^{\prime }-B_{1}^{\prime } \bigr). $$

Here ′ refers to the derivative with respect to η ₁.

Let

$$\begin{aligned} \mathit{ZZ} =&2Z_{I}^{\prime }-Z_{12}^{\prime }-Z_{1}^{\prime },\\ \mathit{AA} =&2A_{I}^{\prime }-A_{12}^{\prime }-A_{1}^{\prime },\\ \mathit{BB} =&2B_{I}^{\prime }-B_{12}^{\prime }-B_{1}^{\prime }. \end{aligned}$$

Below we show that ZZ,AA and BB are negative.

Using (29) and (30) we can present ZZ,AA,BB in the following forms:

$$\begin{aligned} \mathit{ZZ} =&2h ( 12 ) -h ( 8 ) -h ( 4 ), \\ \mathit{AA} =&2w ( 12 ) -w ( 8 ) -w ( 4 ), \\ \mathit{BB} =&\bar{e} \bigl( 2w ( 12 ) -w ( 8 ) -w ( 4 ) \bigr) +2m ( 12 ) -m ( 8 ) -m ( 4 ), \end{aligned}$$

where $h ( x ) =f_{x}^{\prime } ( x )$, and w(x) and m(x) are given by:

$$\begin{aligned} w ( x ) =&\frac{8\bar{e}(\delta +\rho )}{(2 ( \delta +\rho ) +f ( x ) )^{2}}h ( x ),\\ m ( x ) =&\frac{-\bar{e}~f ( x ) }{2 ( \delta +\rho ) +f ( x ) }w ( x ). \end{aligned}$$

The h(x), w(x) and m(x) are decreasing functions of x. Therefore ZZ, AA and BB are negative, and the Shapley values for players 2 and 3 are decreasing in η ₁. □

Appendix J: Proof of Proposition 10

Proof of Part (i)

According to (15), it is enough to show that $\frac{\partial }{\partial \eta _{1}} \phi _{1} ( t ) \vert _{t=0}$ is positive and $\frac{\partial }{\partial \eta _{1}} \frac{\partial }{\partial t}\phi _{1} ( t ) \vert _{t =0}$ is negative.

First, note that from (35) we can infer that

$$ \frac{\partial }{\partial \eta _{1}}S ( t ) \bigg \vert _{t=0}=0. $$

(61)

Now:

$$\begin{aligned} \frac{\partial }{\partial \eta _{1}} \phi _{1} ( t ) =&\frac{1}{3} \biggl( \frac{\partial }{\partial \eta _{1}}(Z_{I}+Z_{12}+Z_{1})S^{2} ( t ) +\frac{\partial }{\partial \eta _{1}}(A_{I}+A_{12}+A_{1})S ( t )\\ &{} +\frac{\partial }{\partial \eta _{1}}(B_{I}+B_{12}+B_{1}) \biggr) \\ &{}+\frac{1}{3} \bigl( 2(Z_{I}+Z_{12}+Z_{1})S+ ( A_{I}+A_{12}+A_{1} ) \bigr) \frac{\partial }{\partial \eta _{1}}S ( t) \end{aligned}$$

Derivatives of (29) and (30) with respect to η ₁ are positive. This, together with (61), implies that $\frac{\partial }{\partial \eta _{1}} \phi _{1} ( t ) \vert _{t=0}$ is positive.

On the other hand, around t=0, we can infer that

$$\begin{aligned} \frac{\partial }{\partial \eta _{1}} \frac{\partial }{\partial t}\phi _{1} ( t ) \bigg \vert _{t=0} =&\frac{1}{3} \frac{\partial }{\partial \eta _{1}} \biggl( (Z_{I}+Z_{12}+Z_{1}) \frac{\partial }{\partial t}S^{2} ( t ) \biggr) \bigg \vert _{t=0}\\ &{}+ \frac{1}{3} \frac{\partial }{\partial \eta _{1}} \biggl( (A_{I}+A_{12}+A_{1}) \frac{\partial }{\partial t}S ( t ) \biggr) \bigg \vert _{t=0} \\ =&\frac{1}{3}\frac{\partial }{\partial \eta _{1}} \biggl( (Z_{I}+Z_{12}+Z_{1}) \biggl( -2(6Z_{I}+ \delta ) \biggl( S_{0}-\frac{1}{6Z_{I}+\delta }(\bar{e}-3A_{I}) \biggr) S_{0} \biggr) \biggr) \\ &{}+\frac{1}{3}\frac{\partial }{\partial \eta _{1}} \biggl( (A_{I}+A_{12}+A_{1}) \biggl( -(6Z_{I}+\delta ) \biggl( S_{0}- \frac{1}{6Z_{I}+\delta }(\bar{e}-3A_{I}) \biggr) \biggr) \biggr). \end{aligned}$$

Therefore, by using (28) and (32), we can obtain that $\frac{\partial }{\partial \eta _{1}} \frac{\partial }{\partial t}\phi _{1} ( 0 )$ is negative.

Proof of Part (ii)

From (34) we can infer that

$$\begin{aligned} \frac{\partial }{\partial \eta _{1}} \phi _{2} ( t ) \bigg \vert _{t=0} =&\frac{1}{6} \biggl( \frac{\partial }{\partial \eta _{1}}(2Z_{I}-Z_{12}-Z_{1})S^{2} ( t ) +\frac{\partial }{\partial \eta _{1}}(2A_{I}-A_{12}-A_{1})S ( t )\\ &{} +\frac{\partial }{\partial \eta _{1}}(2B_{I}-B_{12}-B_{1}) \biggr) \\ &{}+\frac{1}{6} \bigl( 2(2Z_{I}-Z_{12}-Z_{1})S ( t ) + ( 2A_{I}-A_{12}-A_{1} ) \bigr) \frac{\partial }{\partial \eta _{1}}S ( t ). \end{aligned}$$

By the proof of Proposition 9, Appendix I, $\frac{\partial }{\partial \eta _{1}}(2Z_{I}-Z_{12}-Z_{1}),~\frac{\partial }{\partial \eta _{1}}(2A_{I}-A_{12}-A_{1})$, and $\frac{\partial }{\partial \eta _{1}}(2B_{I}-B_{12}-B_{1})$ are negative. Implementing (61), we can deduce that $\frac{\partial }{\partial \eta _{1}} \phi _{2} ( t ) \vert _{t=0}<0$.

It remains to show that $\frac{\partial }{\partial \eta _{1}} \frac{\partial }{\partial t}\phi _{2} ( t ) \vert _{t=0}>0$:

$$\begin{aligned} \frac{\partial }{\partial \eta _{1}} \frac{\partial }{\partial t}\phi _{2} ( t ) \bigg \vert _{t=0} =&\frac{1}{6} \frac{\partial }{\partial \eta _{1}} \biggl( (2Z_{I}-Z_{12}-Z_{1}) \frac{\partial }{\partial t}S^{2} ( t ) \biggr)\\ &{} +\frac{1}{6} \frac{\partial }{\partial \eta _{1}} \biggl( (2A_{I}-A_{12}-A_{1}) \frac{\partial }{\partial t}S ( t ) \biggr) \\ =&\frac{1}{6}\frac{\partial }{\partial \eta _{1}} \biggl( (2Z_{I}-Z_{12}-Z_{1}) \\ &{}\times\biggl( -2(6Z_{I}+ \delta )\biggl( S_{0}-\frac{1}{6Z_{I}+\delta }(\bar{e}-3A_{I}) \biggr) S_{0} \biggr) \biggr) \\ &{}+\frac{1}{6}\frac{\partial }{\partial \eta _{1}} \biggl( (2A_{I}-A_{12}-A_{1}) \\ &{}\times\biggl( -(6Z_{I}+\delta )\biggl( S_{0}- \frac{1}{6Z_{I}+\delta}(\bar{e}-3A_{I}) \biggr) \biggr) \biggr). \end{aligned}$$

After some simple manipulations, we can obtain that $\frac{\partial }{\partial \eta _{1}} \frac{\partial }{\partial t}\phi _{2} ( t ) \vert _{t=0}>0$. Therefore β ₂(t)=β ₃(t) is decreasing in η ₁ around zero.  □