Relinquishing power, exploitation and political unemployment in democratic organizations

Abstract

We analyze the evolution of organizations which take decisions on whom to hire and how to share the output by plurality voting. Agents are grouped in three classes, high, medium and low productivity. We study the evolution of political power and show that in some cases, rational agents who value the future may yield political power to another class. This is what we call the relinquish effect. We show that high productivity agents may receive less than their individual output, i.e. exploitation is possible. We also show that high productivity agents may be left out in the cold because their entrance in an organization may threaten the dominance of other classes. We call this political unemployment.

Appendix

First, let us introduce some notation. Let $V_{J}(T)$ be the continuation payoff for a type J agent when hiring a type T agent. Note first that $V_{L}(T)\le V_{M}(T)\le V_{H}(T)$. This is because in each period, the rule is either meritocratic or egalitarian. If it is egalitarian, everyone receives the same, if it is meritocratic the higher the type the higher the payoff obtained.

Proof of Lemma 1

Given A5, a high type agent will always vote for a meritocratic sharing rule and a low type agent for an egalitarian rule independently of the hiring decision. Thus, when there is no dominant class, when deciding on the sharing rule, the pivotal voter is a medium type agent. Let us see that for $\delta$ sufficiently high, the medium type is also pivotal when voting on the hiring. First note that, under A4 and A5, an agent will always vote for hiring another agent oh her type or of a higher type. Thus, if a medium type prefers to hire a high type, a high type will be hired. Furthermore, given A4, and A5, in all $\tau$ such that $n_{H}^{\tau -1}+1>n_{M}^{\tau -1}+n_{L}^{\tau -1}$, a medium type will always prefer to hire a high type because her pivotal vote on the sharing rule for the next period is not at risk. If $n_{H}^{\tau -1}+1=n_{M}^{\tau -1}+n_{L}^{\tau -1}$, there is a trade off because by hiring a high type, the medium class will lose its pivotal vote on the sharing rule in the next period in favor of the high type, and by part (i) in Proposition 1 we will get meritocracy from there on. To discuss the implications of the voting behavior we distinguish three cases.

(i) Suppose that at $\tau$, $(X^{\tau -1}+x_{H})/n^{\tau }\le x_{M}$.

Since the pivotal voter on the sharing rule is the medium type, she will vote for a meritocratic sharing rule independently of the new hiring. Suppose that the medium type prefers to hire another medium type. Thus,

$$\begin{aligned} x_{M}+\delta V_{M}(M)>x_{M}+\delta V_{M}(H)=x_{M}+\frac{\delta }{1-\delta }x_{M}. \end{aligned}$$

(6.1)

Which implies that some of the terms in $V_{M}(M)$ are the result of an egalitarian rule with a payoff greater than $x_{M}$. Given that $V_{L}(T)\ge (1/(1-\delta ))x_{L}$ for all $T\in \{H,M,L\}$, $V_{L}(M)>(1/(1-\delta ))x_{L}$. Thus, in this situation a low type agent will also prefer to hire a medium type. As a result, a medium type will be hired.

(ii) Suppose that at $\tau$, $x_{M}\le X^{\tau -1}/n^{\tau -1}<(X^{\tau -1}+x_{H})/n^{\tau }$.

Since the pivotal voter on the sharing rule is the medium type, she will vote for an egalitarian sharing rule independently of hiring a high or a medium type. Suppose that the medium type prefers to hire another medium type. Thus,

$$\begin{aligned} \frac{X^{\tau -1}+x_{M}}{n^{\tau }}+\delta V_{M}(M)>\frac{X^{\tau -1}+x_{H}}{n^{\tau }}+\frac{\delta }{1-\delta }x_{M}. \end{aligned}$$

(6.2)

Or equivalently

$$\begin{aligned} \frac{X^{\tau -1}+x_{H}}{n^{\tau }}-\frac{X^{\tau -1}+x_{M}}{n^{\tau }}<\delta \left( V_{M}(M)-\frac{1}{1-\delta }x_{M}\right) . \end{aligned}$$

(6.3)

Let us see that the low type also prefers to vote for a medium type. Note that $V_{M}(M)$ and $V_{L}(M)$ only differ in the periods where meritocracy is the resulting sharing rule (if any of these periods exist). Thus,

$$\begin{aligned} V_{M}(M)-V_{L}(M)<\frac{1}{1-\delta }(x_{M}-x_{L}). \end{aligned}$$

(6.4)

Therefore, if (6.3) holds, then

$$\begin{aligned} \frac{X^{\tau -1}+x_{H}}{n^{\tau }}-\frac{X^{\tau -1}+x_{M}}{n^{\tau }}<\delta \left( V_{L}(M)-\frac{1}{1-\delta }x_{L}\right) \end{aligned}$$

(6.5)

also holds and the low type will also prefer to hire a medium type.

(iii) Suppose that at $\tau$, $X^{\tau -1}/n^{\tau -1}\le$ $x_{M}<(X^{\tau -1}+x_{H})/n^{\tau }$.

Since the pivotal voter on the sharing rule is the medium type, she will vote for a meritocratic sharing rule if a medium type is hired, and for an egalitarian one if a high type is hired. Suppose that the medium type prefers to hire another medium type. Thus,

$$\begin{aligned} x_{M}+\delta V_{M}(M)>\frac{X^{\tau -1}+x_{H}}{n^{\tau }}+\frac{\delta }{1-\delta }x_{M}. \end{aligned}$$

(6.6)

Or equivalently

$$\begin{aligned} \frac{1-2\delta }{1-\delta }x_{M}+\delta V_{M}(M)>\frac{X^{\tau -1}+x_{H}}{n^{\tau }}. \end{aligned}$$

(6.7)

Because in all periods where the pivotal vote of the medium type on the sharing rule is not at risk a new high type is hired, from $\tau +1$ on, the sharing rule will be egalitarian. This is because once we hire a high type, the average surplus is above $x_{M}$ and from there on, regardless of whether a high or a medium type is hired, the average surplus will remain above $x_{M}$. Thus, $V_{M}(M)=V_{L}(M)$. Furthermore, let us see that the left hand side of (6.7) is increasing in $\delta$. Let $y_{t}$ denote the t-term in $V_{M}(M)$, recall that $y_{t}>x_{M}$ for all t. The derivative with respect to $\delta$ of the left hand side of (6.7) is

$$\begin{aligned} \frac{-1}{(1-\delta )^{2}}x_{M}+V_{M}(M)+\delta \frac{\partial V_{M}(M)}{\partial \delta }. \end{aligned}$$

(6.8)

Given that $V_{M}(M)=\sum _{t=0}^{\infty }\delta ^{t}y_{t}>\sum _{t=0}^{\infty }\delta ^{t}x_{M}=(1/(1-\delta ))x_{M}$, and

$$\begin{aligned} \frac{\partial V_{M}(M)}{\partial \delta }=\sum _{t=1}^{\infty }t\delta ^{t-1}y_{t}>~\ \sum _{t=1}^{\infty }t\delta ^{t-1}x_{M}= (1/(1-\delta )^{2})x_{M}, \end{aligned}$$

(6.9)

then (6.8) is positive. Thus, the left hand side of (6.7) is increasing in $\delta$. Therefore, if (6.7) holds for some $\delta$, it holds for all $\delta ^{\prime }>\delta$. Finally note that for $\delta >0.5$, whenever (6.7) holds,

$$\begin{aligned} \frac{1-2\delta }{1-\delta }x_{L}+\delta V_{L}(M)>\frac{X^{\tau -1}+x_{H}}{n^{\tau }} \end{aligned}$$

(6.10)

also holds and thus low types will also prefer to hire a medium type.

Summarizing, there exist $\bar{\delta }_{0}>0.5$ such that for all $\delta \ge \bar{\delta }_{0}$ in any MPE, if at period $\tau$ there is no a dominant class, a type M agent is the pivotal voter. $\square$

Proof of Lemma 2

First of all note that $\delta V_{E}^{1}(\delta )$ is

$$\begin{aligned} \delta V_{E}^{1}(\delta )= & {} \sum _{t=0}^{\infty }\delta ^{2t+1}\left[ \frac{X^{\tau -1}+(t+1)x_{M}+(t+1)x_{H}}{n^{\tau }+2t+1}\right] \nonumber \\&+\sum _{t=1}^{\infty }\delta ^{2t}\left[ \frac{X^{\tau -1}+(t+1)x_{M}+tx_{H}}{n^{\tau }+2t}\right] . \end{aligned}$$

(6.11)

The first term corresponds to all the periods where a high type is hired and the second term to the periods where a medium type is hired. First, note that since $\delta <1$, all power series in $\delta V_{E}^{1}(\delta )$ are convergent. Note also that all terms in $\delta V_{E}^{1}(\delta )$ are larger than $x_{M}$. Let us see that there exist $\bar{\delta }_{1}>0.5$ such that the first three terms in $x_{M}+\delta V_{E}^{1}(\delta )$ are bigger than $(X^{\tau -1}+x_{H})/n^{\tau }+\delta x_{M}+\delta ^{2}x_{M}$. Formally, let us see that there is $\bar{\delta }_{1}>0.5$ such that for all $\delta >\bar{\delta }_{1}$,

$$\begin{aligned}&x_{M}+\delta \left( \frac{X^{\tau -1}+x_{M}+x_{H}}{n^{\tau }+1}\right) +\delta ^{2}\left( \frac{X^{\tau -1}+2x_{M}+x_{H}}{n^{\tau }+2}\right) \nonumber \\&\quad >\frac{X^{\tau -1}+x_{H}}{n^{\tau }}+\delta x_{M}+\delta ^{2}x_{M}. \end{aligned}$$

(6.12)

For $\delta =0$, the left hand side of (6.12) is $x_{M}$ which is smaller than $(X^{\tau -1}+x_{H})/n^{\tau }$. As $\delta$ converges to 1, the left hand side converges to

$$\begin{aligned} x_{M}+\frac{X^{\tau -1}+x_{M}+x_{H}}{n^{\tau }+1}+\frac{X^{\tau -1}+2x_{M}+x_{H}}{n^{\tau }+2}, \end{aligned}$$

(6.13)

and the right hand side converges to

$$\begin{aligned} \frac{X^{\tau -1}+x_{H}}{n^{\tau }}+x_{M}+x_{M}. \end{aligned}$$

(6.14)

Let us see that

$$\begin{aligned} \frac{X^{\tau -1}+x_{M}+x_{H}}{n^{\tau }+1}+\frac{X^{\tau -1}+2x_{M}+x_{H}}{n^{\tau }+2}>\frac{X^{\tau -1}+x_{H}}{n^{\tau }}+x_{M}, \end{aligned}$$

(6.15)

or equivalently,

$$\begin{aligned} \frac{X^{\tau -1}+2x_{M}+x_{H}}{n^{\tau }+2}-x_{M}> & {} \frac{X^{\tau -1}+x_{H}}{n^{\tau }}-\frac{X^{\tau -1}+x_{M}+x_{H}}{n^{\tau }+1}, \end{aligned}$$

(6.16)

$$\begin{aligned} \frac{X^{\tau -1}+x_{H}-n^{\tau }x_{M}}{n^{\tau }+2}> & {} \frac{X^{\tau -1}+x_{H}-n^{\tau }x_{M}}{n^{\tau }(n^{\tau }+1)} \end{aligned}$$

(6.17)

Since $n^{\tau }\ge 2$, $n^{\tau }(n^{\tau }+1)$ is larger than $(n^{\tau }+2)$, (6.15) holds.

Let

$$\begin{aligned} F(\delta )=(1-\delta -\delta ^{2})x_{M}+\delta \left( \frac{X^{\tau -1}+x_{M}+x_{H}}{n^{\tau }+1}\right) +\delta ^{2}\left( \frac{X^{\tau -1}+2x_{M}+x_{H}}{n^{\tau }+2}\right) . \nonumber \\ \end{aligned}$$

(6.18)

We have seen that $F(0)<(X^{\tau -1}+x_{H})/n^{\tau }$, and that $\lim _{\delta \rightarrow 1}F(\delta )>(X^{\tau -1}+x_{H})/n^{\tau }$, note also that since $(X^{\tau -1}+x_{M}+x_{H})/(n^{\tau }+1)>x_{M},~$and $(X^{\tau -1}+2x_{M}+x_{H})/(n^{\tau }+2)>x_{M}$, $F(\delta )$ is increasing in $\delta$. Thus, we can find $\bar{\delta }_{1}>0.5$ such that for all $\delta >\bar{\delta }_{1}$, $F(\delta )>(X^{\tau -1}+x_{H})/n^{\tau }$, as we wanted to prove.

Finally, note that for $\delta >0.5$, if (3.1) holds, then (3.2) also holds. $\square$

Proof of Lemma 3

Note first that if (3.3) holds, given that $x_{M}>x_{L}$, (3.4) also holds. Also note that $\delta V_{E}^{2}(\delta )$ can be written as (6.11) in Lemma 2. As in Lemma 2, all terms in $\delta V_{E}^{2}(\delta )$ are larger than $x_{M}$. Let us see that there exist $\bar{\delta }_{2}$ such that the first three terms in $(X^{\tau -1}+x_{M})/n^{\tau }+\delta V_{E}^{2}(\delta )$ are larger than $(X^{\tau -1}+x_{H})/n^{\tau }+\delta x_{M}+\delta ^{2}x_{M}$. That is, there is $\bar{\delta }_{2}$ such that in three periods, we compensate the initial loss of getting $(X^{\tau -1}+x_{M})/n^{\tau }$ compared with obtaining $(X^{\tau -1}+x_{H})/n^{\tau }$. Formally, let us see that there is $\bar{\delta }_{2}~$such that for all $\delta >\bar{\delta }_{2}$,

$$\begin{aligned} (X^{\tau -1}+x_{M})/n^{\tau }+ & {} \delta \left( \frac{X^{\tau -1}+x_{M}+x_{H}}{n^{\tau }+1}\right) \nonumber \\+ & {} \delta ^{2}\left( \frac{X^{\tau -1}+2x_{M}+x_{H}}{n^{\tau }+2}\right) \quad > \frac{X^{\tau -1}+x_{H}}{n^{\tau }}+\delta x_{M}+\delta ^{2}x_{M}. \end{aligned}$$

(6.19)

Given that $(X^{\tau -1}+x_{M})/n^{\tau }>x_{M}$, we can apply the argument in Lemma 2 to ensure the existence of $\bar{\delta }_{2}$. $\square$