Which Ethics Will Make us Individually and Socially Happier? A Cross-Culture and Cross-Development Analytical Model

Abstract

This paper provides an analytical model representing four polar ethical approaches drawn from the main ethical positions suggested by the philosophical, psychological, and socio-economic literature. Moreover, it develops the model in order to obtain rankings of the four approaches in terms of happiness and, consequently, to provide insights into which ethical approach should best be adopted by each individual, according to his characteristics (income level, in developed countries (DCs) or in less developed countries (LDCs), aspiration level): some dynamics are also predicted, if the Golden and the Copper Rules are applied. Finally, this paper provides insights into which ethical approach should best be adopted by each society, according to its characteristics (DCs or LDCs, social distribution of aspiration levels), by predicting happiness levels in alternative countries, according to the prevailing ethics, and by comparing these predictions with the observed happiness levels, thus providing an empirical test of the analytical model: some dynamics are again predicted, with non-Protestant DCs moving to higher, and Protestant DCs towards lower, happiness levels (conditioned to the per capita income), due to the increasing and decreasing rejection of the Golden and Copper Rules, respectively, and with LDCs moving to lower (conditioned to the increasing per capita income) in the short-run and higher happiness levels in the long run, by establishing and entertaining conditions that set clear incentives for moral behaviour, in order to increase and decrease the adoption of the Golden and Copper Rules, respectively.

Appendices

Appendix I

Case 1.

$$ a_{11} = (1/a)^{(1/(a - 1))} \quad a_{12} = (1/b) \, ^{(1/(b - 1))} $$

$$ a_{22} = { \max }\,a_{22} - (1 /b)^{(b/(b - 1))} \quad a_{21} = \left\{ {\max a_{21} - (1/a)^{(a/(a - 1))} + (1/a)^{(1/(a - 1))} } \right\} - (1/a)^{(1/(a - 1))} $$

$$ \lambda_{1} = f_{1} \quad \lambda_{2} = f_{2} $$

Case 2.

$$ a_{11} = (1/a)^{(1/(a - 1))} \quad a_{12} = (1/b)^{(1/(b - 1))} $$

$$ a_{22} = \max \,a_{22} - (1/b)^{(b/(b - 1))} \quad a_{21} = \left\{ {\max a_{21} - (1/a)^{(a/(a - 1))} + (1/a)^{(1/(a - 1))} } \right\} - (1/a)^{(1/(a - 1))} $$

$$ \lambda_{2} = f_{2} = \max f_{2} - \left\{ {(1/c){{\left.\left[ {\max a_{21} - (1/a)^{(a/(a - 1))} + (1/a)^{(1/(a - 1))} } \right]\right/} {\left[ {\max a_{22} - \left( {1/b} \right)^{(b/(b - 1))} + \left( {1/b} \right)^{(1/(b - 1))} } \right]}}} \right\}^{(c/(c - 1))} $$

$$ \lambda_{1} \, = f_{1} \, = \, \left\{ {(1/c){{\left.\left[ {\max a_{21} - \left( {1/a} \right)^{(a/(a - 1))} + \left( {1/a} \right)^{(1/(a - 1))} } \right]\right/} {\left[ {\max a_{22} - \left( {1/b} \right)^{(b/(b - 1))} + \left( {1/b} \right)^{(1/(b - 1))} } \right]}}} \right\}^{(1/(c - 1))} $$

$$ \mu = \max a_{22} - (1/b)^{(b/(b - 1))} + (1/b)^{(1/(b - 1))} $$

Case I.

$$ a_{11} = (1/a)^{(1/(a - 1))} \;a_{21} = \max a_{21} - (1/a)^{(a/(a - 1))} - \varepsilon $$

$$ a_{12} = (1/b)^{(1/(b - 1))} \;a_{22} = \max a_{22} - (1/b)^{(b/(b - 1))} - \varepsilon $$

Case II.

$$ a_{11} = (1 /a)^{(1/(a - 1))} \;a_{21} = \max a_{21} - (1/a)^{(a/(a - 1))} - \varepsilon $$

$$ a_{12} = (1/b)^{(1/(b - 1))} \;a_{22} = \max a_{22} - (1/b)^{(b/(b - 1))} - \varepsilon $$

$$ f_{2} = \max f_{2 } - f_{1}^{c} - \varepsilon $$

Notice that λ₁ = f ₁ and λ₂ = f ₂ obtained in Case 1 (where goals are given) with respect to Case 2 (where goals are chosen) confirm that Lagrange multipliers measure the lack of freedom to choose goals.

Let us fix a = b = c = 2, max a ₂₂ = max a ₂₁ = 1, and max f ₂ = 3 so that the Lagrangian becomes:

$$ L = f_{1} (a_{11} + a_{21} ) + f_{2} (a_{12} + a_{22} ) - \lambda_{1} \left[ {a_{21} - \left( {1 - a_{11}^{2} } \right)} \right] - \lambda_{2} \left[ {a_{22} - \left( {1 - a_{12}^{2} } \right)} \right] - \mu \left[ {f_{2} - \left( {3 - f_{1}^{2} } \right)} \right] $$

Thus, numerical results for ε = 1/5 are given by:

Case 1

$$a_{11} = 1 /2,a_{21} = 7 /4,a_{12} = 1 /2,a_{22} = 3 /4,\lambda_{1} = f_{1} ,\lambda_{2} = f_{2} ,V(1) = 1 /4(9 \, f_{1} + 5 \, f_{2} ) - f_{1} - f_{2} $$

Case 2

$$a_{11} = 1 /2,\,a_{21} = 7 /4,\,a_{12} = 1 /2,\,a_{22} = 3 /4,\,\lambda_{1} = f_{1} = 9 /10,\,\lambda_{2} = f_{2} = 219 /100,\,\mu = 5 /4,V(2) = 169 /400 $$

Case I

$${\text{With}}\,{\text{given}}\,a_{11} = 1 /2,a_{12} = 1 /2,V({\text{I}}) = 1 /20(41 \, f_{1} + 21 \, f_{2} ) $$

Case II

$${\text{With}}\,{\text{given}}\,a_{11} = 1 /2,a_{12} = 1 /2,\,V({\text{II}}) = 1 /100(294 + 205 \, f_{1} - 105 \, f_{1}^{2} ) $$

Notice that the obtained values of V(1), V(2), V(I) and V(II) depend on the assumed value of ε, while rankings are independent of it.

Appendix II

Case 1.

$$ a_{11} = (1/a)^{(1/(a - 1))} a_{21} = \max a_{21} - (1/a)^{(a/(a - 1))} $$

$$ a_{12} \, = (1/b)^{(1/(b - 1))} a_{22} = (1/c)\left\{ {\left[ {\max a_{21} - (1/a)^{(a/(a - 1))} + (1/a)^{(1/(a - 1))} } \right]\max f_{2}^{((1 - c)/c)} } \right\} - (1/b)^{(1/(b - 1))} $$

$$ \lambda_{1} = f_{1} \quad \lambda_{2} = f_{2} $$

Case 2.

$$ a_{11} = (1/a)^{(1/(a - 1))} a_{21} = \max a_{21} - (1/a)^{(a/(a - 1))} $$

$$ a_{12} = (1/b)^{(1/(b - 1))} a_{22} = (1/c )\left\{ {\left[ {\max a_{21} - \left( {1/a} \right)^{(a/(a - 1))} + \left( {1/a} \right)^{(1/(a - 1))} } \right]\max f_{2}^{((1 - c)/c)} } \right\} - (1/b)^{(1/(b - 1))} $$

$$ \lambda_{1} = f_{1} = \max f_{2}^{(1/c)} \quad \lambda_{2} = f_{2} = 0 $$

$$ \mu = (1/c)\left\{ {\left[ {\max a_{21} - (1/a )^{(a/(a - 1))} + (1/a)^{(1/(a - 1))} } \right]\max f_{2}^{((1 - c)/c)} } \right\} $$

Case I.

$$ a_{11} = (1 /a)^{(1/(a - 1))} a_{21} = \max a_{21} - (1/a)^{(a/(a - 1))} - \varepsilon $$

$$ a_{12} \, = (1/b)^{(1/(b - 1))} a_{22} = \max a_{22} - (1/b)^{(b/(b - 1))} - \varepsilon $$

Case II.

$$ a_{11} = (1/a)^{(1/(a - 1))} a_{21} = \max a_{21} - (1/a)^{(a/(a - 1))} - \varepsilon $$

$$ a_{12} = (1/b)^{(1/(b - 1))} a_{22} = \max a_{22} - (1/b)^{(b/(b - 1))} - \varepsilon $$

$$ f_{2} = \max f_{2} - f_{1}^{c} - \varepsilon $$

Notice again that λ₁ = f ₁ and λ₂ = f ₂ obtained in Case 1 (where goals are given) with respect to Case 2 (where goals are chosen) confirm that Lagrange multipliers measure the lack of freedom to choose goals.

Let us fix a = b = 2, c = 1/2, max a ₂₂ = max a ₂₁ = 1, and max f ₂ = 3 so that the Lagrangian becomes:

• $$ L = \, f_{1} (a_{11} + a_{21} ) + f_{2} (a_{12} + a_{22} ) - \lambda_{1} \left[ {a_{21} - \left( {1 - a_{11}^{2} } \right)} \right] - \lambda_{2} \left[ {a_{22} - \left( {1 - a_{12}^{2} } \right)} \right] - \mu \left[ {f_{2} - \left( {3 - f_{1}^{1/2} } \right)} \right] $$

Thus, numerical results for ε = 1/5 are given by:

Case 1

$$a_{11} = 1 /2,\,a_{21} = 7 /4,\,a_{12} = 1 /2,\,a_{22} = 13,\lambda_{1} = f_{1} ,\lambda_{2} = f_{2} ,V(1) = 9 /4(f_{1} + 6f_{2} ) - f_{1} - f_{2} $$

Case 2

$$a_{11} = 1/4,a_{21} = 1 /4,a_{12} = 3 /2,a_{22} = 13,\lambda_{1} = f_{1} = 9,\lambda_{2} = f_{2} = 0,\mu = 27 /2,V(2) = - 9 /4 $$

Case I

$${\text{With}}\,{\text{ given}}\,a_{11} = 1 /2,a_{12} = 1 /2,V({\text{I}}) = 1 /20(41 \, f_{1} + 21 \, f_{2} ) $$

Case II

$${\text{With}}\,{\text{given}}\,a_{11} = 1 /2,a_{12} = 1 /2,\,V(II) = 1 /100\left( {294 - 105 \, \sqrt {f_{1} } + 205 \, f_{1} } \right) $$

Notice that numerical results turn out to be consistent with the assumption of a concave and convex relationship between f ₂ and f ₁ in DCs and LDCs, respectively: indeed, in case 2, if f ₂ is concave in f ₁, i.e. in DCs, the total value of lack of freedom amounts to 217/50, while if f ₂ is convex in f ₁, i.e. in LDCs, the total value of lack of freedom amounts to 45/2.