Social Creation of Pro-social Preferences for Collective Action

Abstract

Pro-social preferences are thought to play a significant role in solving society’s collective action problems of providing public goods and reducing public bads. Societies can benefit by deliberately instilling and sustaining such preferences in their members. We construct a theoretical model to examine an intergenerational education process for this. We consider both a one-time action of this kind and a constitution that establishes a steady state, and compare the two.

Appendix: Mathematical Derivations

1.1 Nash Equilibrium with Pro-sociality

Begin with the purely selfish case, set out in Sect. 2, Eqs. (1)–(3). The Kuhn-Tucker conditions for individual i’s choice of (x _i , z _i ) to maximize u _i are

$$ \displaystyle\begin{array}{rcl} \frac{\partial u_{i}} {\partial x_{i}}& =& 1 + \overline{z} -{ 2 \over 3}\,(x_{i} + z_{i}) \leq 0,\quad x_{i} \geq 0\,,{}\end{array}$$

(20)

$$\displaystyle\begin{array}{rcl} \frac{\partial u_{i}} {\partial z_{i}}& =&{ 1 \over n}\ x_{i} -{ 2 \over 3}\,(x_{i} + z_{i}) \leq 0,\quad z_{i} \geq 0\,,{}\end{array}$$

(21)

with complementary slackness in each equation. Note that in (21) we have used $\partial \overline{z}/\partial z_{i} = 1/n$. The matrix of second-order partials is

$$\displaystyle{\left (\begin{array}{cc} -{ 2 \over 3} & -{ 2 \over 3} +{ 1 \over n} \\ -{ 2 \over 3} +{ 1 \over n} & -{ 2 \over 3} \end{array} \right )\,.}$$

The diagonal elements of this are negative, and the determinant is

$$\displaystyle{\left ({2 \over 3}\right )^{2} -\left ({2 \over 3} -{ 1 \over n}\right )^{2}> 0\,.}$$

Therefore the matrix is negative definite, so the second-order sufficient conditions are met and the Kuhn-Tucker conditions yield the global maximum of u _i . (This will continue to be so in all the variants of the model considered here, and will not be mentioned further.)

First try the purely selfish solution where x _i  > 0 and z _i  = 0 for all i. Then $\overline{z} = 0$ also, and the conditions (20) and (21) are

$$\displaystyle{1 -{ 2 \over 3}\,x_{i} = 0,\qquad \left ({ 1 \over n} -{ 2 \over 3}\right )\,x_{i} \leq 0\,,}$$

or

$$\displaystyle{x_{i} ={ 3 \over 2},\qquad { 1 \over n} \leq { 2 \over 3}\,.}$$

The inequality is true for n ≥ 2. Therefore the solution x _i  = 3∕2, z _i  = 0 in (4) is verified. The y _i and u _i are easily computed.

Next consider the symmetric social optimum. Recall that the common private and public effort levels x _i  = x and z _i  = z for all i, are chosen to maximize the common utility level

$$\displaystyle{u = x\,(1 + z) -{ 1 \over 3}\ (x + z)^{2}\,.}$$

The first-order conditions are

$$\displaystyle\begin{array}{rcl} \frac{\partial u} {\partial x}& =& 1 + z -{ 2 \over 3}\,(x + z) = 0\,, {}\\ \frac{\partial u} {\partial z}& =& x -{ 2 \over 3}\,(x + z) = 0\,. {}\\ \end{array}$$

These yield the solution x = 2, z = 1 in (6). The resulting y and u are easily found.

Next consider equilibria where people have the pro-social utility (7), with the same γ for all. The Kuhn-Tucker conditions for person 1 are:

$$\displaystyle\begin{array}{rcl} \frac{\partial v_{1}} {\partial x_{1}}& =& 1 + \overline{z} -{ 2 \over 3}\,(x_{1} + z_{1}) \leq 0,\quad x_{1} \geq 0\,,{}\end{array}$$

(22)

$$\displaystyle\begin{array}{rcl} \frac{\partial v_{1}} {\partial z_{1}}& =&{ 1 \over n}\ x_{1} -{ 2 \over 3}\,(x_{1} + z_{1}) +\gamma \ \sum _{ j=2}^{n}\,{ 1 \over n}\,x_{j} \leq 0,\quad z_{i} \geq 0\,,{}\end{array}$$

(23)

with complementary slackness in each. Similar conditions obtain for the other individuals.

See if the selfish solution with x _i  > 0, z _i  = 0 still works. The conditions (22) and (23) become

$$\displaystyle{1 -{ 2 \over 3}\,x_{i} = 0,\qquad { 1 \over n}\ x_{1} -{ 2 \over 3}\,x_{1} +\gamma \ \sum _{ j=2}^{n}\,{ 1 \over n}\,x_{j} \leq 0\,,}$$

or

$$\displaystyle{x_{i} ={ 3 \over 2},\qquad {3 \over 2}\,\left ({ 1 \over n} -{ 2 \over 3} +\gamma { n-1 \over n} \right ) \leq 0\,.}$$

The inequality becomes

$$\displaystyle{\phi = \frac{1 + (n - 1)\,\gamma } {n} \leq \frac{2} {3}\,,}$$

which is equivalent to (8) in the text.

When this condition is not met, look for a symmetric Nash equilibrium with x _i  = x > 0 and z _i  = z > 0 for all i. The conditions (22) and (23) become

$$\displaystyle\begin{array}{rcl} 1 + z& =&{ 2 \over 3}\,(x + z)\,, {}\\ \phi \,x& =&{ 2 \over 3}\,(x + z)\,. {}\\ \end{array}$$

These yield the solution (10) in the text.

Writing u for the common level of selfish utility, it is then mechanical to verify

$$\displaystyle{\frac{du} {d\phi } = \frac{8\,(1-\phi )} {(2-\phi )^{2}} \,.}$$

Therefore u is an increasing function of ϕ over the range $({2 \over 3},1)$. Thus more pro-socialness achieves higher selfish utilities all round.

1.2 One-Time Education for Pro-socialness

In the range 0 ≤ ϕ ≤ 2∕3, it is obviously best to set ϕ = 0 and get U ^a = 0. 75 (1 +δ). In the range 2∕3 ≤ ϕ ≤ 1, use θ = (k∕δ)^1∕3, to substitute k = δ θ ³, and write the formula defining the function for $\phi \geq { 2 \over 3}$ as

$$\displaystyle{ f(\phi ) = 0.75 +\delta \ \left [\,\frac{\phi \,(4 - 3\phi )} {(2-\phi )^{2}} - \frac{\theta ^{3}} {1-\phi }\,\right ]\,. }$$

(24)

It is then mechanical to verify

$$\displaystyle{f^{{\prime}}(\phi ) = \frac{\delta } {(1-\phi )^{2}}\ \left [\ 8\,\left (\frac{1-\phi } {2-\phi }\right )^{3} -\theta ^{3}\ \right ]\,.}$$

Therefore

$$\displaystyle{ f^{{\prime}}(\phi )> 0\quad \mbox{ iff}\quad 2\,\frac{1-\phi } {2-\phi }>\theta,\quad \mbox{ i.e.}\quad \phi <\phi ^{{\ast}} = \frac{2\,(1-\theta )} {2-\theta } \,. }$$

(25)

Therefore f(ϕ) is single-peaked, and its maximum occurs where f ^′(ϕ) = 0, that is, at ϕ = ϕ ^∗. Substituting and simplifying, the maximum value is

$$\displaystyle{f(\phi ^{{\ast}}) = 0.75 +\delta \, \left (\ \theta ^{3} - 3\,\theta ^{2} + 1\ \right )\,.}$$

If this exceeds 0. 75 (1 +δ), then ϕ ^∗ maximizes f(ϕ) in (14); otherwise ϕ = 0 yields the maximum of f(ϕ).

We also need to restrict ϕ ^∗ > 2∕3 to have an equilibrium that results in the utilities that enter the construction of f(ϕ). From the definition in (7), we see that 1 ≥ ϕ ^∗ > 2∕3 corresponds to 0 ≤ θ < 1∕2. Now define

$$\displaystyle{h(\theta ) =\theta ^{3} - 3\,\theta ^{2} + 1 - 0.75 =\theta ^{3} - 3\,\theta ^{2} + 0.25\,.}$$

We have

$$\displaystyle{h^{{\prime}}(\theta ) = 3\ \theta ^{2} - 6\,\theta = 3\,\theta \,(\theta -2)\,,}$$

which is negative over the interval $(0,{ 1 \over 2})$. Therefore h(θ) is a decreasing function throughout this range. Also h(0) = 1∕4 > 0 and $h({1 \over 2}) = -3/8 <0$. Therefore there is a unique θ ^∗ in the interval such that h(θ) > 0 for θ < θ ^∗ and h(θ) < 0 for θ > θ ^∗. Numerical calculation shows that θ ^∗ ≈ 0. 305. This completes the proof of the statements in the text leading to (15).

1.3 Education for Pro-socialness in Steady State

Write the objective function in (26) as v = n g(ϕ) where

$$\displaystyle{ g(\phi ) = \left \{\begin{array}{ll} 0.75\ \phi - \frac{k\,\phi } {1-\phi } &\mbox{ for }\phi <2/3 \\ \frac{\phi ^{2}\,(4-3\phi )} {(2-\phi )^{2}} - \frac{k\,\phi } {1-\phi }&\mbox{ for }2/3 <\phi <1 \end{array} \right. }$$

(26)

For 0 ≤ ϕ < 2∕3 we have

$$\displaystyle{g^{{\prime}}(\phi ) = 0.75 - \frac{k} {(1-\phi )^{2}}}$$

and

$$\displaystyle{g^{{\prime\prime}}(\phi ) = \frac{-2\,k} {(1-\phi )^{3}} <0}$$

Therefore g(ϕ) is concave in this range. Also

$$\displaystyle{g^{{\prime}}(0) = 0.75 - k> 0\quad \mbox{ if }k <0.75}$$

and

$$\displaystyle{g^{{\prime}}(2/3) = 0.75 - 9\,k <0\quad \mbox{ if }k> 1/12}$$

For 2∕3 < ϕ < 1 we have, after some tedious algebra,

$$\displaystyle{g^{{\prime}}(\phi ) = \frac{\phi \,(16 - 18\ \phi + 3\ \phi ^{2})} {(2-\phi )^{3}} - \frac{k} {(1-\phi )^{2}}}$$

Then

$$\displaystyle{g^{{\prime}}(2/3) = 3/2 - 9\,k> 0\quad \mbox{ if }k <1/6}$$

and

$$\displaystyle{g^{{\prime}}(\phi ) \rightarrow -\infty \quad \mbox{ as}\quad \phi \rightarrow 1}$$

In general, g(ϕ) is not concave throughout the range 2∕3 < ϕ < 1. (Specifically, the first term on the right hand side of (26) is convex for 2∕3 < ϕ < 4∕5 and concave for 4∕5 < ϕ < 1, so g(ϕ) can be convex for a sub-range to the right of 2/3 if k is small.) But numerical calculation shows that

$$\displaystyle{(1-\phi )^{2}\ g^{{\prime}}(\phi ) = \frac{\phi \,(1-\phi )^{2}\,(16 - 18\ \phi + 3\ \phi ^{2})} {(2-\phi )^{3}} - k}$$

is a decreasing function of ϕ. It equals 1∕6 − k for ϕ = 2∕3 and − k for ϕ = 1. Therefore, if k < 1∕6, g ^′(ϕ) is positive for ϕ < a critical value ϕ ^∗ (which is uniquely defined for each k and of course depends on k), and negative for ϕ > ϕ ^∗. Then g(ϕ) is increasing for 2∕3 < ϕ < ϕ ^∗ and decreasing for ϕ ^∗ < ϕ < 1, i.e. g(ϕ) has a unique interior maximum at ϕ ^∗ in the range 2∕3 < ϕ < 1.

Putting all this information together gives the statements in the text.