Lawyers and politicians: the impact of organized legal professions on institutional reforms

Abstract

Organized legal professions often play a central role in successful institutional development. The paper’s model examines how legal professions affect institutional reform. Professional review of reform proposals solves a politician’s informational problem in a way that makes democracy, political stability, and professional power substitutes. The model’s applicability is examined by showing that its predictions track the fortunes of lawyers in the USSR and early transition and are consistent with events in 1688 in England and 1789 in France, indicating why these two revolutions had different consequences. The model suggests why and when civil law and common law systems differ.

Appendix Proofs of Propositions

Proposition 1

When $\lambda<\bar{\lambda}$ the politician accepts all reform proposals. Consider reform j∈{2,3}. β will never want to lobby for j∈{2,3}, since B _βj  < 0 for j∈{2,3}. There are four possible sub-games at the beginning of period 2, depending on whether α chose to lobby for reform j∈{2,3} in period 1 or not and whether A or B is in power in period 2.

• 1. If α lobbied for j∈{2,3} in period 1 and A is in power in period 2, then the payoff to α when it lobbies for reform j∈{2,3} in period 2 is B _αj  − Ψ_αA and B _αj when it does not lobby. Therefore, α does not lobby.

• 2. If α lobbied for j∈{2,3} in period 1 and B is in power in period 2, then the payoff to α when it lobbies for reform j∈{2,3} in period 2 is B _αj  − Ψ _αB  < 0 and 0 when it does not lobby. Therefore, α does not lobby.

• 3. If α did not lobby for j∈{2,3} in period 1 and politician A is in power in period 2, then the payoff to α when it lobbies for reform j∈{2,3} in period 2 is − Ψ_αA  < 0 and 0 when it does not lobby. Therefore, α does not lobby.

• 4. If α did not lobby for j∈{2,3} in period 1 and politician B is in power in period 2, then the payoff to interest group α when it lobbies for reform j∈{2,3} in period 2 is − Ψ_αB  < 0 and 0 when it does not lobby. Therefore, α does not lobby.

If α chooses to lobby for j∈{2,3} in period 1, its payoff is pB _αj  − Ψ_αA . If α chooses not to lobby for j∈{2,3} in period 1, its payoff is 0. Therefore, α chooses to lobby for j∈{2,3} in period 1 if and only if p ≥ Ψ_αA /B _αj .

Consider now reform j∈{4,5}. α will never lobby for j∈{4,5}, since B _αj  < 0 for j∈{4,5}. There are four possible sub-games at the beginning of period 2, depending on whether β chose to lobby for j∈{4,5} in period 1 or not and whether politician A or politician B is in power in period 2.

• 1. If β lobbied for j∈{4,5} in period 1 and A is in power in period 2, then the payoff to β when it lobbies for j∈{4,5} in period 2 is B _βj  − Ψ_βA  < 0 and B _βj when it does not lobby. Therefore, β does not lobby.

• 2. If β lobbied for j∈{4,5} in period 1 and B is in power in period 2, then the payoff to β when it lobbies for j∈{4,5} in period 2 is B _βj  − Ψ_βB  > 0 and 0 when it does not lobby. Therefore, β lobbies.

• 3. If β did not lobby for j∈{4,5} in period 1 and B is in power in period 2, then the payoff to β when it lobbies for j∈{4,5} in period 2 is − Ψ_βB  < 0 and 0 when it does not lobby. Therefore, β does not lobby.

• 4. If β did not lobby for j∈{4,5} in period 1 and politician A is in power in period 2, then the payoff to β when it lobbies for j∈{4,5} in period 2 is − Ψ _βA  < 0 and 0 when it does not lobby. Therefore, β does not lobby.

If β chooses to lobby for j∈{4,5} in period 1, its payoff is $pB_{\beta j}+(1-p)(B_{\beta j}-\Psi_{\beta B})- \Psi_{\beta A}<0$. If β chooses not to lobby for j∈{4,5} in period 1, its payoff is 0. Therefore, β chooses not to lobby for reform j∈{4,5} in period 1.

Proposition 2

If $p<\Psi_{\alpha A}/B_{\alpha3}, \alpha$ chooses not to lobby, so W ⁺ = 0 > W ⁻. If $\Psi _{\alpha A}/B_{\alpha 3}\leq p<\Psi_{\alpha A}/B_{\alpha 2}, \alpha$ lobbies for reform 3, so W ⁺ = p(B _α3 + B _β3) > W ⁻. If p ≥ Ψ_αA /B _α2, α lobbies for reforms 2 and 3, so $W^{+}=p(B_{\alpha 2} +B_{\beta 2}+B_{\alpha 3}+B_{\beta 3})> W^{-}$.

Proposition 3

If p < Ψ_αA /B _α3, α chooses not to lobby in period 1, so V ⁺ = 0 < V ⁻.

If p ≥ Ψ_αA /B _α2, α lobbies for 2 and 3 in period 1, so

$$V^{+}=\lambda p(B_{\alpha 2}+B_\beta 2+B_{\alpha 3}+B_{\beta 3})+(1-\lambda)2\Psi_{\alpha A}> V^{-}. $$

If $\Psi_{\alpha A}/B_{\alpha 3}\leq p<\Psi_{\alpha A} /B_{\alpha 2}$, α lobbies for 3 in period 1, so $V^{+}=\lambda p(B_{\alpha 3}+ B_{\beta 3})+\Psi_{\alpha A}$. Then,

$$ V^{+}>(<)V^{-}\Leftrightarrow p>(<) p^{\ast}(\lambda)\equiv1+[\lambda(B_{\alpha 2}+B_{\beta 2})+(1-\lambda)\Psi_{\alpha A}] /[\lambda(B_{\alpha 3}+B_{\beta 3})]. $$

The following is useful for the purposes of constructing a diagram to examine the trade-offs between λ and p. Observe that p ^*(λ)<(>)1 if and only if $\lambda>(<) \underline{\lambda}$. It can be easily shown that $\partial p^{\ast}(\lambda)/\partial\lambda< 0$ and $\partial^{2}p^{*}(\lambda)/\partial \lambda^{2}>0$. Define

$$ \begin{aligned} \lambda_{L}&=\Psi_{\alpha A}/[(\Psi_{\alpha A} /B_{\alpha 2}-1)(B_{\alpha 3} +B_{\beta 3})+|B_{\alpha 2}+B_{\beta2}|+\Psi_{\alpha A}]\hbox{ and}\\ \lambda_{H}&=\Psi _{\alpha A}/[(\Psi_{\alpha A}/B_{\alpha 3} -1)(B_{\alpha 3}+B_{\beta 3})+|B_{\alpha 2}+B_{\beta 2}| +\Psi_{\alpha A}]. \end{aligned} $$

Then $\underline{\lambda}<\lambda_{L}<\lambda _{H}<\bar{\lambda}, p^{\ast}(\lambda_{H}) =\Psi_{\alpha A}/B_{\alpha 3}$ and p ^* (λ _L ) = Ψ_αA /B _α2.

Proposition 4

Define C ^m(p,λ) as the maximum political cost that the politician is willing to incur to remove the profession from its advisory role. If $V^{+}< V^{-},C^{m}(p,\lambda)=V^{-}(p, \lambda)-V^{+}(p,\lambda)$ and if V ⁺ ≥ V ⁻,C ^m(p,λ) = 0. More precisely,

$$ \begin{aligned} C^{m}(p,\lambda)=\left\{\begin{array}{ll} \lambda[(B_{\alpha 2}+B_{\beta 2})+(B_{\alpha 3}+B_{\beta 3})]+(1-\lambda)2\Psi_{\alpha A},& \hbox{when} p<\Psi _{\alpha A}/B_{\alpha 3}\hbox{and}\lambda<\bar{\lambda}\\ \lambda[(B_{\alpha 2}+B_{\beta2}) +(B_{\alpha 3}+B_{\beta 3})]-p\lambda(B_{\alpha3} +B_{\beta 3})+(1-\lambda)2\Psi_{\alpha A},& \hbox{when} p< p^{*}(\lambda) \hbox{ and} p\in[\Psi_{\alpha A}/B_{\alpha 3},\Psi_{\alpha A}/B_{\alpha 2})\\ 0,&\hbox{elsewhere}.\end{array} \right. \end{aligned} $$

The politician chooses not to use the profession for all values of λ and p such that C ^m(p,λ) > cλ. C ^m(p,λ) has the following properties:

• 1. When p < Ψ_αA /B _α3 and $\lambda <\bar{\lambda}$

  $$ \begin{aligned} \partial C^{m}(p,\lambda)/\partial p&=0,\\ \partial C^{m}(p,\lambda)/\partial\lambda&=[(B_{\alpha 2}+B_{\beta 2})+(B_{\alpha 3} +B_{\beta 3})]-2\Psi_{\alpha A}< 0 \end{aligned} $$

• 2. When $p\in[\Psi_{\alpha A}/B_{\alpha 3},\Psi_{\alpha A}/B_{\alpha 2})$ and p < p ^*(λ),

  $$ \begin{aligned} \partial C^{m}(p,\lambda)/\partial p&=\lambda (B_{\alpha 2}+B_{\beta 2})< 0,\\ \partial C^{m}(p,\lambda)/\partial \lambda &=[(B_{\alpha 2}+B_{\beta 2})+(B_{\alpha 3} +B_{\beta 3})]- p(B_{\alpha 3}+B_{\beta 3}) -\Psi_{\alpha A}< 0,\\ \partial^{2}C^{m}(p,\lambda)/\partial \lambda\partial p&=-(B_{\alpha 2}+B_{\beta2})< 0. \end{aligned} $$

• 3. Elsewhere, C ^m(p,λ) = 0.

• 4. Two additional properties that enable us to construct Fig. 6 are:

  $$ \begin{aligned} &\left.{\partial C^m(p,\lambda^0)/\partial\lambda}\right|_{p<\Psi _{\alpha A}/B_{\alpha 3}} <\left.{\partial C^m(p,\lambda ^0)/\partial \lambda}\right|_{p=\Psi_{\alpha A}/B_{\alpha 3}} \ \hbox{for all} \ \lambda ^0<\bar{\lambda}\\ &\left.{C^m(p,\lambda^0)}\right|_{p<\Psi_{\alpha A} /B_{\alpha 3}} >\left.{C^m(p,\lambda^0)}\right|_{p\in [\Psi_{\alpha A} /B_{\alpha 3},\Psi_{\alpha A}/B_{\alpha 3}) \ \hbox{and} \ p< p\ast(\lambda^0)} \ \hbox{ for all} \ \lambda ^0< \bar {\lambda} \end{aligned} $$

Fig. 6

The political cost of removing the legal profession

The properties of C ^m(p,λ) summarized above imply then for any pair (p,c), there exists a unique value of λ^T(p,c) > 0 such that C ^m(p,λ^T(p,c)) = cλ ^T(p,c) if V ⁺ < V ⁻. On the other hand, C ^m(p,λ) = 0 if V ⁺ ≥ V ⁻ and hence λ^T(p,c) = 0. This proves that λ^T(p,c) is a function. To prove that λ^T(p,c) is discontinuous at p = Ψ_αA /B _α2 and at p = Ψ_αA /B _α3, and points (ii)–(iv), note also the following:

• 1. When p < Ψ_αA /B _α3, λ ^T(p,c) is defined by

  $$ \begin{aligned} c\lambda^{T}(p,c)&=\lambda^{T}(p,c)(B_{\alpha 2}+B_{\beta 2} +B_{\alpha 3}+B_{\beta 3})+(1-\lambda^{T}(p,c))2\Psi _{\alpha A}, \hbox{so that }\\ \lambda^{T}(p,c)&=2\Psi_{\alpha A}/[ c-(B_{\alpha 2}+B_{\beta 2}+B_{\alpha 3}+B_{\beta 3}) +2\Psi_{\alpha A}].\hbox{ Then}, \partial \lambda^{T}(p,c)/\partial p=0. \end{aligned} $$

• 2. When $p\in[\Psi_{\alpha A}/B_{\alpha 3},\Psi_{\alpha A}/B_{\alpha 2}), \lambda^{T}(p,c)$ is defined by

  $$ \begin{aligned} &c\lambda^{T}(p,c)=\lambda ^{T}(p,c)(B_{\alpha 2}+B_{\beta 2} +B_{\alpha 3}+B_{\beta 3})-p\lambda^{T}(p,c)(B_{\alpha 3}+B_{\beta 3})+(1 -\lambda^{T}(p,c))\Psi_{\alpha A},\hbox{ so that}\\ &\lambda ^{T}(p,c)=\Psi_{\alpha A}/[c-(B_{\alpha 2}+B_{\beta 2}+B_{\alpha 3} +B_{\beta 3})+p(B_{\alpha 3}+B_{\beta 3})+\Psi_{\alpha A}]. \end{aligned} $$
  $$ \begin{aligned} &\partial \lambda^{T}(p,c)/\partial p=-(B_{\alpha 3}+B_{\beta 3})\lambda^{T}(p,c)/[c-(B_{\alpha 2} +B_{\beta 2}+B_{\alpha 3}+B_{\beta 3}) +(B_{\alpha 3}+B_{\beta 3})p+\Psi_{\alpha A}]< 0\hbox{ and}\\ &\partial^{2}\lambda^{T}(p,c)/\partial p^{2}> 0. \end{aligned} $$

• 3. When p ≥ Ψ_αA /B _α3, λ^T(p,c) = 0.

• 4. $\left.\lambda^T(p,c^0)\right|_{p<\Psi_{\alpha A}/B_{\alpha 3}}>\left.\lambda^T(p,c^0)\right|_{p\in [\Psi_{\alpha A}/B_{\alpha 3},\Psi_{\alpha A} /B_{\alpha 2})}>\left.\lambda^T(p,c^0) \right|_{p\geq\Psi_{\alpha A}/B_{\alpha 2}}=0 \quad\hbox{for all} \quad c^0>0.$

Proposition 5

Let c ¹ > c ² ≥ 0. Define $\Delta ^{T}(p,c^{1},c^{2})= \lambda^{T}(p,c^{2})-\lambda^{T}(p,c^{1})$.

• 1. When $p<\Psi_{\alpha A}/B_{\alpha 3},\Delta ^{T}(p,c^{1},c^{2})=2\Psi_{\alpha A}/[c^{2}-(B_{\alpha 2}+B_{\beta 2} +B_{\alpha 3}+B_{\beta 3})+2\Psi_{\alpha A}] -2\Psi_{\alpha A}/[c^{1}-(B_{\alpha 2}+B_{\beta 2}+B_{\alpha 3}+B_{\beta 3}) +2\Psi_{\alpha A}]>0\hbox{ and }\partial\Delta ^{T}(p,c^{1},c^{2})/\partial p=0$.

• 2. When, $p\in[\Psi_{\alpha A}/B_{\alpha 3}, \Psi_{\alpha A}/B_{\alpha 2})$.

  $$\begin{aligned} &\Delta^{T}(p,c^{1},c^{2})=\Psi_{\alpha A}/[c^{2} -(B_{\alpha 2}+B_{\beta 2}+B_{\alpha 3} +B_{\beta 3})+p(B_{\alpha 3}+B_{\beta 3}) +\Psi_{\alpha A}]-\\ &\Psi_{\alpha A}/[c^{1} -(B_{\alpha 2}+B_{\beta 2}+B_{\alpha 3} +B_{\beta 3})+p(B_{\alpha 3}+B_{\beta 3}) +\Psi_{\alpha A}]>0\hbox{ and }\partial\Delta ^{T}(p,c^{1},c^{2})/\partial p<0 \end{aligned} $$

• 3. When $p\geq\Psi_{\alpha A}/B_{\alpha 2}, \lambda^{T}( p,c^{2})=\lambda^{T}(p,c^{1})=0,\hbox{ so }\Delta^{T}(p,c^{1} ,c^{2})=0$.

Note that when $p<\Psi_{\alpha A}/B_{\alpha 3}, \lambda^{T}(p,0)=\bar{\lambda}$ and when $p\in[\Psi_{\alpha A}/B_{\alpha 3},\Psi_{\alpha A}/B_{\alpha 2}),\lambda^{T}(p,0) =p{^\ast -1}(\lambda)$.