Leadership and Free-Riding: Decomposing and Explaining the Paradox of Cooperation in International Environmental Agreements

Abstract

This paper decomposes the canonical model of International Environmental Agreements (Barrett in Oxf Econ Pap 46:878–894, 1994) into three effects: externality, cost-effectiveness and timing. The externality and timing effects are countervailing forces on abatement levels of greenhouse gases. The Paradox of Cooperation in the three-stage Stackelberg game is explained by showing that when the gains to cooperation are small the timing effect dominates the externality effect and large coalitions are stable. The timing effect has the greatest impact when the high benefit nations have low abatement cost. The cost-effectiveness effect arises from asymmetry and generates the need for an agreement with transfers. The cost-effectiveness effect is largest when the high benefit nations are high cost. This creates a larger difference in the marginal abatement cost of the last unit of abatement in the absence of an agreement. Numerical examples illustrate how the parameters and effects interact to result in outcomes ranging from no to full participation.

Appendix

A1: This appendix reproduces the asymmetric Stackelberg coalition results from McGinty (2007) to compare with the new results.

The Stackelberg equilibrium signatory abatement from McGinty (2007) is

$$\begin{aligned} Q^{L}(S^{L}) & = \frac{ab\sum \nolimits _{i\in S}\frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}}{\left[ 1+b\sum \nolimits _{j\in T}\theta _{j} \right] ^{2}+b\sum \nolimits _{i\in S}\frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}} \nonumber \\ q_{i}^{l}(S^{L}) & = \frac{ab\sum \nolimits _{j\in S}\alpha _{j}}{c_{i}\left[ \left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) ^{2}+b\sum \nolimits _{j\in S} \frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}\right] } . \end{aligned}$$

(38)

The non-signatory level of abatement is

$$\begin{aligned} Q^{F}(S^{L}) & = \frac{ab\sum \nolimits _{j\in T}\theta _{j}\left[ 1+b\sum \nolimits _{j\in T}\theta _{j}\right] }{\left[ 1+b\sum \nolimits _{j\in T}\theta _{j}\right] ^{2}+b\sum \nolimits _{i\in S}\frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}} \nonumber \\ q_{i}^{f}(S^{L}) & = \frac{ab\alpha _{i}\left[ 1+b\sum \nolimits _{j\in T}\theta _{j}\right] }{c_{i}\left[ \left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) ^{2}+b\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j} \right] } . \end{aligned}$$

(39)

The global quantity of Stackelberg abatement is $Q(S^{L})=Q^{L}(S^{L})+Q^{F}(S^{L})$.

$$\begin{aligned} Q(S^{L})=\frac{ab\left[ \sum \nolimits _{j\in T}\theta _{j}\left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) +\sum \nolimits _{i\in S}\frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}\right] }{\left[ 1+b\sum \nolimits _{j\in T}\theta _{j}\right] ^{2}+b\sum \nolimits _{i\in S}\frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}} \end{aligned}$$

(40)

A2: Proofs

Proposition 1

With non-trivial $(s>1)$ Cournot coalitions of symmetric nations, joining the IEA always means increasing abatement, i.e., the externality effect is strictly positive.

Proof

A non-signatory outside of an agreement with $s-1$ signatories abates $q_{i}^{t}(s-1)$ from (20). After joining the agreement there are s members each of which abates $q_{i}^{s}(s)$ from (19).

$$\begin{aligned} q_{i}^{s}(s)-q_{i}^{t}(s-1) & = \frac{as}{n\gamma +n-s+s^{2}}-\frac{a}{ n\gamma +n-(s-1)+(s-1)^{2}} \nonumber \\ & = \frac{as\left[ n\gamma +n-(s-1)+(s-1)^{2}\right] -a\left[ n\gamma +n-s+s^{2}\right] }{\left[ n\gamma +n-(s-1)+(s-1)^{2}\right] \left[ n\gamma +n-s+s^{2}\right] } \nonumber \\ & = \frac{a\left\{ s\left[ n\gamma +n-(s-1)+(s-1)^{2}\right] -\left[ n\gamma +n-s+s^{2}\right] \right\} }{\Delta _{1}} \nonumber \\ & = \frac{a\left\{ (s-1)\left( n\gamma +n\right) +s\left[ -(s-1)+(s-1)^{2}+1-s\right] \right\} }{\Delta _{1}} \nonumber \\ & = \frac{a\left\{ n(s-1)\left( \gamma +1\right) +s\left[ s^{2}-2s+1-2s+2 \right] \right\} }{\Delta _{1}} \nonumber \\ & = \frac{a\left\{ n(s-1)\left( \gamma +1\right) +s\left[ s^{2}-4s+3\right] \right\} }{\Delta _{1}}>0\text { for all} \, s>1 \end{aligned}$$

(41)

where the denominator $\Delta _{1}\equiv$$\left[ n\gamma +n-(s-1)+(s-1)^{2} \right] \left[ n\gamma +n-s+s^{2}\right]$ is positive for all $n\ge s\ge 1$. $\square$

Proposition 2

The timing effect in Stackelberg coalitions implies that a single signatory exploits the leadership advantage by (i) reducing abatement relative to the Nash equilibrium, which (ii) induces followers to increase abatement relative to the Nash equilibrium.

Proof

(i) Using (12), the definition $\gamma \equiv \frac{c}{ b}$ and (24), $q^{ne}-q_{|_{s=1}}^{l}$ is

$$\begin{aligned} q^{ne}-q_{|_{s=1}}^{l} & = \frac{ab}{n\left( b+c\right) }-\frac{abcn}{\left[ cn+b(n-1)\right] ^{2}+bcn} \nonumber \\ & = ab\left[ \frac{1}{n\left( b+c\right) } -\frac{cn}{\left[ cn+b(n-1)\right] ^{2}+bcn}\right] \nonumber \\ & = \frac{ab}{\Delta _{2}}\left[ \left[ cn+b(n-1)\right] ^{2}+bcn-cn^{2} \left( b+c\right) \right] \nonumber \\ & = \frac{ab}{\Delta _{2}}\left[ c^{2}n^{2}+2bcn(n-1)+b^{2}(n-1)^{2}+bcn-bcn^{2}-c^{2}n^{2}\right] \nonumber \\ & = \frac{ab}{\Delta _{2}}\left[ 2bcn^{2}-bcn+b^{2}(n-1)^{2}-bcn^{2}\right] \nonumber \\ & = \frac{ab^{2}}{\Delta _{2}}\left[ cn^{2}-cn+b(n-1)^{2}\right] \nonumber \\ & = \frac{ab^{2}}{\Delta _{2}}\left[ cn(n-1)+b(n-1)^{2}\right] \nonumber \\ & = \frac{ab^{2}(n-1)}{\Delta _{2}}\left[ cn+b(n-1)\right]>0\text { for all} \, n>1 \end{aligned}$$

(42)

where $\Delta _{2}\equiv n\left( b+c\right) \left[ \left[ cn+b(n-1)\right] ^{2}+bcn\right] >0$ for all $n>1$.

Then (ii), using (12) with $\gamma \equiv \frac{c}{b}$ and (24), evaluating $q_{|_{s=1}}^{f}-q^{ne}$ results in

$$\begin{aligned}&q_{|_{s=1}}^{f}-q^{ne} \nonumber \\&\quad =\frac{ab\left[ cn+b(n-1)\right] }{\left[ cn+b(n-1) \right] ^{2}+bcn}-\frac{ab}{n\left( b+c\right) } \nonumber \\&\quad =ab\left[ \frac{cn+b(n-1)}{\left[ cn+b(n-1)\right] ^{2}+bcn}-\frac{1}{ n\left( b+c\right) } \right] \nonumber \\&\quad =\frac{ab}{\Delta _{2}}\left[ n\left( b+c\right) \left[ cn+b(n-1)\right] - \left[ cn+b(n-1)\right] ^{2}-bcn\right] \nonumber \\&\quad =\frac{ab}{\Delta _{2}}\left[ bcn^{2}+c^{2}n^{2}+b^{2}n(n-1)+bcn(n-1)-c^{2}n^{2}-2bcn(n-1)-b^{2}(n-1)^{2}-bcn \right] \nonumber \\&\quad =\frac{ab}{\Delta _{2}}\left[ bcn^{2}+b^{2}n(n-1)+bcn(n-1)-2bcn(n-1)-b^{2}(n-1)^{2}-bcn\right] \nonumber \\&\quad =\frac{ab}{\Delta _{2}}\left[ b^{2}n(n-1)-bcn+2bcn-b^{2}(n-1)^{2}-bcn \right] \nonumber \\&\quad =\frac{ab}{\Delta _{2}}\left[ b^{2}n(n-1)-b^{2}(n-1)^{2}\right] \nonumber \\&\quad =\frac{ab^{3}(n-1)}{\Delta _{2}}\left[ n-n+1\right] \nonumber \\&\quad =\frac{ab^{3}(n-1)}{\Delta _{2}}>0\text { for all} \, n>1. \end{aligned}$$

(43)

$\square$

Proposition 3

The timing and externality effects exactly offset when $s={\hat{s}}$. When $s<{\hat{s}}$the timing effect dominates and when $s>{\hat{s}}$the externality effect dominates, i.e., for all coalitions with s members (i) $Q(S^{L})\lesseqqgtr Q^{ne}$(ii) $q^{f}\gtreqless q^{ne}$(iii) $q^{l}\lesseqgtr q^{ne}$if and only if $s\lesseqgtr {\hat{s}}$.

Proof

Proof of (i) $Q(S^{L})\lesseqqgtr Q^{ne}$ if and only if $s\lesseqgtr \hat{s}$. Using (23) and (12) with $\gamma \equiv \frac{c}{b}$.

$$\begin{aligned}&Q^{ne}-Q(S^{L}) \nonumber \\&\quad =\frac{ab}{b+c}-\frac{ab\left[ cn\left( s^{2}+n-s\right) +b(n-s)^{2}\right] }{\left[ cn+b(n-s)\right] ^{2}+bcns^{2}} \nonumber \\&\quad =\frac{ab}{\Delta _{3}}\left[ \left[ cn+b(n-s)\right] ^{2}+bcns^{2}-\left( b+c\right) \left[ cn\left( s^{2}+n-s\right) +b(n-s)^{2}\right] \right] \nonumber \\&\quad =\frac{ab}{\Delta _{3}}\left[ \begin{array}{c} c^{2}n^{2}+2bcn(n-s)+b^{2}(n-s)^{2}+bcns^{2} \\ -\left( b+c\right) \left[ cn\left( s^{2}+n-s\right) +b(n-s)^{2}\right] \end{array} \right] \nonumber \\&\quad =\frac{ab}{\Delta _{3}}\left[ \begin{array}{c} c^{2}n^{2}+2bcn(n-s)+b^{2}(n-s)^{2}+bcns^{2}-bcn\left( s^{2}+n-s\right) -b^{2}(n-s)^{2} \\ -c^{2}n\left( s^{2}+n-s\right) -bc(n-s)^{2} \end{array} \right] \nonumber \\&\quad =\frac{ab}{\Delta _{3}}\left[ 2bcn(n-s)+bcns^{2}-bcn\left( s^{2}+n-s\right) -c^{2}n\left( s^{2}-s\right) -bc(n-s)^{2}\right] \nonumber \\&\quad =\frac{ab}{\Delta _{3}}\left[ 2bcn(n-s)-bcn\left( n-s\right) -c^{2}ns\left( s-1\right) -bc(n^{2}-2ns+s^{2})\right] \nonumber \\&\quad =\frac{abc}{\Delta _{3}}\left[ bn(n-s)-cns\left( s-1\right) -b(n^{2}-2ns+s^{2})\right] \nonumber \\&\quad =\frac{abc}{\Delta _{3}}\left[ -bns-cns^{2}+cns+2bns-bs^{2}\right] \nonumber \\&\quad =\frac{abcs}{\Delta _{3}}\left[ -cns+cn+bn-bs\right] \nonumber \\&\quad =\frac{abcs}{\Delta _{3}}\left[ -s(b+cn)+n(b+c)\right] \gtreqqless 0\Longleftrightarrow s\lesseqgtr {\hat{s}}=\frac{n(b+c)}{b+cn}=\frac{ n(1+\gamma )}{n\gamma +1} \end{aligned}$$

(44)

where $\Delta _{3}\equiv \left( b+c\right) \left[ \left[ cn+b(n-s)\right] ^{2}+bcns^{2}\right] >0$.

Proof of (ii) $q^{f}\gtreqless q^{ne}$ if and only if $s\lesseqgtr {\hat{s}}$ using (22) and (12).

$$\begin{aligned} q^{f}-q^{ne} & = \frac{ab\left[ cn+b(n-s)\right] }{\left[ cn+b(n-s)\right] ^{2}+bcns^{2}}-\frac{ab}{n(b+c)} \nonumber \\ & = \frac{ab}{n\Delta _{3}}\left[ n(b+c)\left[ cn+b(n-s)\right] -\left[ cn+b(n-s)\right] ^{2}-bcns^{2}\right] \nonumber \\ & = \frac{ab}{n\Delta _{3}}\left[ \begin{array}{c} bcn^{2}+c^{2}n^{2}+b^{2}n(n-s)+bcn(n-s) \\ -c^{2}n^{2}-2bcn(n-s)-b^{2}(n-s)^{2}-bcns^{2} \end{array} \right] \nonumber \\ & = \frac{ab}{n\Delta _{3}}\left[ bcn^{2}+b^{2}n^{2}-b^{2}ns-bcn(n-s)-b^{2}(n-s)^{2}-bcns^{2}\right] \nonumber \\ & = \frac{ab^{2}}{n\Delta _{3}}\left[ cn^{2}+bn^{2}-bns-cn^{2}+cns-b(n^{2}-2ns+s^{2})-cns^{2}\right] \nonumber \\ & = \frac{ab^{2}}{n\Delta _{3}}\left[ -bns+cns-b(-2ns+s^{2})-cns^{2}\right] \nonumber \\ & = \frac{ab^{2}}{n\Delta _{3}}\left[ -cns(s-1)+b(ns-s^{2})\right] \nonumber \\ & = \frac{ab^{2}s}{n\Delta _{3}}\left[ -cn(s-1)+b(n-s)\right] \nonumber \\ & = \frac{ab^{2}s}{n\Delta _{3}}\left[ -s(cn+b)+n(b+c)\right] \gtreqqless 0\Longleftrightarrow \nonumber \\ s\lesseqgtr {\hat{s}} & = \frac{n(b+c)}{b+cn} =\frac{ n(1+\gamma )}{n\gamma +1}. \end{aligned}$$

(45)

Proof of (iii) $q^{l}\lesseqgtr q^{ne}$ if and only if $s\lesseqgtr {\hat{s}}$ .

This follows directly from (i) and (ii). If $Q(S^{L})\lesseqqgtr Q^{ne}$ and $q^{f}\gtreqless q^{ne}$ if and only if $s\lesseqgtr {\hat{s}}$ then $q^{l}\lesseqgtr q^{ne}$ when $s\lesseqgtr {\hat{s}}$.

$$\begin{aligned} q^{l} & = q^{f} \nonumber \\ \frac{abcns}{\left[ cn+b(n-s)\right] ^{2}+bcns^{2}} & = \frac{ab\left[ cn+b(n-s)\right] }{\left[ cn+b(n-s)\right] ^{2}+bcns^{2}} \nonumber \\ cns & = cn+b(n-s) \nonumber \\ s(b+cn) & = n(b+c) \nonumber \\ {\hat{s}} & = \frac{n(b+c)}{b+cn}<n\,\,{\text {for}}\,\, n>1 \end{aligned}$$

(46)

Of course if $q^{l}=q^{f}$ then they must both equal $q^{ne}$ since this is where the positive externality effect and the negative timing effect exactly offset. It is easily verified that $q^{l}({\hat{s}})=q^{f}({\hat{s}})=q^{ne}$. $\square$

Proposition 4

For all possible subcoalitions (i.e., partial participation) with $1\le s<n$members Cournot coalitions compared to Stackelberg coalitions result in (i) higher global abatement, $Q(S)>Q(S^{L})$, (ii) higher signatory abatement $q_{i}^{s}(S)>q_{i}^{l}(S^{L})$, and (iii) lower non-signatory abatement, $q_{i}^{t}(S)<q_{i}^{f}(S^{L})$.

Proof

For a given set of coalition members, $S=S^{L}$, and (i) using (34) and (40),

$$\begin{aligned}&Q(S)-Q(S^{L}) \nonumber \\&\quad =\frac{ab\left( \sum \nolimits _{j\in T}\theta _{j}+\sum \nolimits _{i\in S}\frac{1}{ci}\sum \nolimits _{i\in S}\alpha _{i}\right) }{ 1+b\left( \sum \nolimits _{j\in T}\theta _{j}+\sum \nolimits _{i\in S}\frac{1}{ci} \sum \nolimits _{i\in S}\alpha _{i}\right) } -\frac{ab\left[ \sum \nolimits _{j\in T}\theta _{j}\left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) +\sum \nolimits _{i\in S}\frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}\right] }{ \left[ 1+b\sum \nolimits _{j\in T}\theta _{j}\right] ^{2}+b\sum \nolimits _{i\in S} \frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}} \nonumber \\&\quad =\frac{ab}{\Delta _{4}}\left[ \begin{array}{l} \sum \limits _{j\in T}\theta _{j}\left[ 1+b\sum \limits _{j\in T}\theta _{j} \right] ^{2}+b\sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{ ci}\sum \limits _{i\in S}\alpha _{i} \\ +\left[ 1+b\sum \limits _{j\in T}\theta _{j}\right] ^{2}\sum \limits _{i\in S} \frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}+b\left( \sum \limits _{i\in S} \frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}\right) ^{2} \\ -\sum \limits _{j\in T}\theta _{j}\left( 1+b\sum \limits _{j\in T}\theta _{j}\right) -\sum \limits _{i\in S}\frac{1}{ci}\sum \limits _{i\in S}\alpha _{i}-b\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}\left( 1+b\sum \limits _{j\in T}\theta _{j}\right) \\ -b\sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}-b\sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}\left( 1+b\sum \limits _{j\in T}\theta _{j}\right) \\ -b\left( \sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}\right) ^{2} \end{array} \right] \nonumber \\&\quad =\frac{ab}{\Delta _{4}}\left[ \begin{array}{l} \sum \limits _{j\in T}\theta _{j}\left[ 1+b\sum \limits _{j\in T}\theta _{j} \right] ^{2}+\left[ 1+b\sum \limits _{j\in T}\theta _{j}\right] ^{2}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i} \\ -\sum \limits _{j\in T}\theta _{j}\left( 1+b\sum \limits _{j\in T}\theta _{j}\right) -\sum \limits _{i\in S}\frac{1}{ci}\sum \limits _{i\in S}\alpha _{i}-b\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}\left( 1+b\sum \limits _{j\in T}\theta _{j}\right) \\ -b\sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}\left( 1+b\sum \limits _{j\in T}\theta _{j}\right) \end{array} \right] \nonumber \\&\quad =\frac{ab}{\Delta _{4}}\left[ \begin{array}{l} \left[ 1+2b\sum \limits _{j\in T}\theta _{j}+b^{2}\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}\right] \left( \sum \limits _{j\in T}\theta _{j}+\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}\right) \\ -\sum \limits _{j\in T}\theta _{j}-b\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}-\sum \limits _{i\in S}\frac{1}{ci}\sum \limits _{i\in S}\alpha _{i}-b\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}-b^{2}\left( \sum \limits _{j\in T}\theta _{j}\right) ^{3} \\ -b\sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}-b^{2}\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i} \end{array} \right] \nonumber \\&\quad =\frac{ab}{\Delta _{4}}\left[ \begin{array}{l} \sum \limits _{j\in T}\theta _{j}+\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}+2b\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}+2b\sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{c_{i}} \sum \limits _{i\in S}\alpha _{i} \\ +b^{2}\left( \sum \limits _{j\in T}\theta _{j}\right) ^{3}+b^{2}\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}\sum \limits _{i\in S}\frac{1}{c_{i}} \sum \limits _{i\in S}\alpha _{i} \\ -\sum \limits _{j\in T}\theta _{j}-b\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}-\sum \limits _{i\in S}\frac{1}{ci}\sum \limits _{i\in S}\alpha _{i}-b\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}-b^{2}\left( \sum \limits _{j\in T}\theta _{j}\right) ^{3} \\ -b\sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}-b^{2}\left( \sum \limits _{j\in T}\theta _{j}\right) ^{2}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i} \end{array} \right] \nonumber \\&\quad =\frac{ab}{\Delta _{4}}\left[ b\sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}\right] \nonumber \\&\quad =\frac{ab^{2}}{\Delta _{4}}\left[ \sum \limits _{j\in T}\theta _{j}\sum \limits _{i\in S}\frac{1}{c_{i}}\sum \limits _{i\in S}\alpha _{i}\right] >0 \end{aligned}$$

(47)

where $\Delta _{4}\equiv \left[ 1+b\left( \sum \nolimits _{i\in T}\theta _{i}+\sum \nolimits _{i\in S}\frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}\right) \right] \left\{ \left[ 1+b\sum \nolimits _{i\in T}\theta _{i} \right] ^{2}+b\sum \nolimits _{i\in S}\frac{1}{c_{i}}\sum \nolimits _{i\in S}\alpha _{i}\right\} >0$.

(ii) Using (32) and (38) $q_{i}^{s}(S)-q_{i}^{l}(S^{L})>0$ since

$$\begin{aligned} q_{i}^{s}(S) & = \frac{ab\sum \nolimits _{j\in S}\alpha _{j}}{c_{i}\left[ 1+b\left( \sum \nolimits _{j\in T}\theta _{j}+\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}\right) \right] }\text { and} \nonumber \\ q_{i}^{l}(S^{L}) & = \frac{ab\sum \nolimits _{j\in S}\alpha _{j}}{c_{i}\left[ \left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) ^{2}+b\sum \nolimits _{j\in S}\frac{1}{c_{j}} \sum \nolimits _{j\in S}\alpha _{j}\right] } . \end{aligned}$$

(48)

The numerators are identical and the denominator of $q_{i}^{l}(S^{L})$ is greater since $\left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) ^{2}>$$1+b\sum \nolimits _{j\in T}\theta _{j}$ for $\left| T\right| >0$.

(iii) Using (33) and (39) $\frac{q_{i}^{t}(S)}{q_{i}^{f}(S^{L})}$ is

$$\begin{aligned}&\frac{q_{i}^{t}(S)}{q_{i}^{f}(S^{L})} \nonumber \\&\quad = \frac{\left( \frac{ab\alpha _{i}}{c_{i}\left[ 1+b\left( \sum \nolimits _{j\in T}\theta _{j}+\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}\right) \right] } \right) }{\left( \frac{ab\alpha _{i}\left[ 1+b\sum \nolimits _{j\in T}\theta _{j}\right] }{c_{i}\left[ \left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) ^{2}+b\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}\right] } \right) } \nonumber \\&\quad = \frac{\left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) ^{2}+b\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}}{\left[ 1+b\sum \nolimits _{j\in T}\theta _{j}\right] \left[ 1+b\left( \sum \nolimits _{j\in T} \theta _{j}+\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}\right) \right] } \nonumber \\&\quad = \frac{\left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) ^{2}+b\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}}{\left( 1+b\sum \nolimits _{j\in T}\theta _{j}\right) ^{2}+b\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}+b^{2}\sum \nolimits _{j\in T}\theta _{j}\sum \nolimits _{j\in S}\frac{1}{c_{j}}\sum \nolimits _{j\in S}\alpha _{j}} <1 \text { for }\, \left| T\right| > 0. \end{aligned}$$

(49)

$\square$