Partial compensation/responsibility

Abstract

Many theories of fairness distinguish between compensation factors (‘luck’) and responsibility factors (‘effort’). Whereas the distinction between both types of factors is a matter of definition in theory, empirical work usually requires a sharp cut. All determinants of the outcome of interest have to be classified as either a compensation factor or a responsibility factor. We argue that the determinants are often hard to classify. A pragmatic solution to the problem at hand is to introduce a more general soft cut: determinants can be partly compensation, partly responsibility. Still, in a first-best income tax framework, such a soft cut is possible only if the gross income function is additively separable. In case separability fits the data, a simple partial sharing rule emerges as a natural candidate for partial redistribution. This rule can be characterized on the basis of two simple properties, equal treatment of equals and partial solidarity. In case additive separability is rejected by the data, we propose two alternative solutions.

Appendices

Appendix: Proof of Lemma 1

Consider a redistribution scheme $N:\mathbb {D}\rightarrow \mathbb {R}^{I}: \mathbf {x}\mapsto N\left( \mathbf {x}\right)$ that satisfies partial solidarity and equal treatment of equals. We must show that also partial compensation is satisfied. More precisely, for each $\mathbf {x}\in \mathbb {D}$, for each individual $i,j\in \mathbb {I}$ and for each compensation group $\ell \in \mathbb {P}$, if $x_{i}^{k}=x_{j}^{k}$, for each $k\in \mathbb {P}\backslash \left\{ \ell \right\}$ is true, then $N_{i}\left( \mathbf {x}\right) -N_{j}\left( \mathbf {x}\right) =(1-\gamma ^{\ell })\left( G\left( x_{i}\right) -G\left( x_{j}\right) \right)$ must result by combining partial solidarity and equal treatment of equals.

Construct a profile $\mathbf {x}^{\prime }$ with (1) $x_{l}=x_{l}^{\prime }$ for $l\in \mathbb {I}\backslash \left\{ j\right\}$, and (2) $x_{j}^{\prime }=x_{i}$. In words, the transition from $\mathbf {x}$ to $\mathbf {x}^{\prime }$ is such that individual $j$ becomes a copy of individual $i$, ceteris paribus. This only requires a change from $x_{j}^{\ell }$ to $x_{j}^{\prime \ell }=x_{i}^{\ell }$, while $x_{j}^{k}=x_{j}^{\prime k}$, for each $k\in \mathbb {P}\backslash \left\{ \ell \right\}$. We can apply partial solidarity to get

$$\begin{aligned} N_{j}(\mathbf {x}^{\prime })-N_{j}(\mathbf {x)}= N_{i}(\mathbf {x}^{\prime })-N_{i}(\mathbf {x)}+(1-\gamma ^{\ell }) (G(x_{j}^{\prime })-G(x_{j})), \end{aligned}$$

(1)

for $i$ and $j$. Now, since $x_{i}^{\prime }=x_{j}^{\prime }$ by construction, equal treatment of equals in profile $\mathbf {x} ^{\prime }$ requires $N_{i}\left( \mathbf {x}^{\prime }\right) =N_{j}\left( \mathbf {x}^{\prime }\right)$. Using $N_{i}\left( \mathbf {x}^{\prime }\right) =N_{j}\left( \mathbf {x}^{\prime }\right)$ and $x_{j}^{\prime }=x_{i}$ in Eq. (1) leads to

$$\begin{aligned} N_{i}(\mathbf {x)}-N_{j}(\mathbf {x)}=(1-\gamma ^{\ell })(G(x_{j}^{\prime })-G(x_{j}))=(1-\gamma ^{\ell }) (G(x_{i})-G(x_{j})), \end{aligned}$$

as required.

Proof of Lemma 2

Consider a redistribution scheme $N$ that satisfies partial compensation. We must show that there exist functions $G^{1},G^{2},\ldots ,G^{P}$, one function for each compensation group in $\mathbb {P=}\left\{ 1,2,\ldots ,P\right\}$, such that $G\left( x\right) =\sum \nolimits _{k\in \mathbb {P}}G^{k}\left( x^{k}\right)$ for each $x\in \mathbb {R}^{J}$. In case $P=1$, the separability condition is obvious, and so, we focus on $J\ge 2$ and $2\le P\le J$ in the sequel.

Step 1 Recall that $J^{k}$ is the cardinality of $\mathbb {J}^{k}$. For any two compensation groups $k$ and $\ell$, with $\ell >k$, we show that there must exist functions $G_{k\ell }^{-\ell }: \mathbb {R}^{J-J_{\ell }}\rightarrow \mathbb {R}$ and $G_{k\ell }^{-k}:\mathbb { R}^{J-J_{k}}\rightarrow \mathbb {R}$ such that

$$\begin{aligned} G\left( x\right) =G_{k\ell }^{-\ell }\left( x^{1},\ldots ,x^{\ell -1},x^{\ell +1},\ldots ,x^{P}\right) +G_{k\ell }^{-k}\left( x^{1},\ldots , x^{k-1},x^{k+1},\ldots ,x^{P}\right) \end{aligned}$$

for each $x\in \mathbb {R}^{J}$. Consider two compensation groups $k$ and $\ell$ with $\ell >k$ and consider four individuals (1, 2, 3 and 4) with types

$$\begin{aligned} x_{1}&= \left( x^{1},\ldots ,x^{k-1},x^{k},x^{k+1},\ldots ,x^{\ell -1},x^{\ell },x^{\ell +1},\ldots ,x^{P}\right) \equiv x,\\ x_{2}&= \left( x^{1},\ldots ,x^{k-1},x^{k},x^{k+1},\ldots ,x^{\ell -1},b,x^{\ell +1},\ldots ,x^{P}\right) ,\\ x_{3}&= \left( x^{1},\ldots ,x^{k-1},a,x^{k+1},\ldots ,x^{\ell -1},x^{\ell }, x^{\ell +1},\ldots ,x^{P}\right) ,\\ x_{4}&= \left( x^{1},\ldots ,x^{k-1},a,x^{k+1},\ldots ,x^{\ell -1},b,x^{\ell +1},\ldots ,x^{P}\right) . \end{aligned}$$

for arbitrary vectors $x\in \mathbb {R}^{J}$, $a\in \mathbb {R}^{J^{k}}$ and $b\in \mathbb {R}^{J^{\ell }}$. Partial compensation requires

$$\begin{aligned} N_{1}\left( \mathbf {x}\right) -N_{2}\left( \mathbf {x}\right)&= (1-\gamma ^{\ell })\left( G\left( x_{1}\right) -G\left( x_{2}\right) \right) , \end{aligned}$$

(2)

$$\begin{aligned} N_{3}\left( \mathbf {x}\right) -N_{4}\left( \mathbf {x}\right)&= (1-\gamma ^{\ell })\left( G\left( x_{3}\right) -G\left( x_{4}\right) \right) , \end{aligned}$$

(3)

$$\begin{aligned} N_{1}\left( \mathbf {x}\right) -N_{3}\left( \mathbf {x}\right)&= (1-\gamma ^{k})\left( G\left( x_{1}\right) -G\left( x_{3}\right) \right) , \end{aligned}$$

(4)

$$\begin{aligned} N_{2}\left( \mathbf {x}\right) -N_{4}\left( \mathbf {x}\right)&= (1-\gamma ^{k})\left( G\left( x_{2}\right) -G\left( x_{4}\right) \right) . \end{aligned}$$

(5)

Subtracting (3) from (2) and (5) from (4), and noting that both differences have to be the same, we get:

$$\begin{aligned}&(1-\gamma ^{\ell })\left( G\left( x_{1}\right) -G\left( x_{2}\right) -G\left( x_{3}\right) +G\left( x_{4}\right) \right) \\&\qquad \qquad \qquad \qquad \parallel \\&(1-\gamma ^{k})\left( G\left( x_{1}\right) -G\left( x_{3}\right) -G\left( x_{2}\right) +G\left( x_{4}\right) \right) . \end{aligned}$$

Given $\gamma _{k}\ne \gamma _{\ell }$ and $G\left( x_{1}\right) -G\left( x_{2}\right) -G\left( x_{3}\right) +G\left( x_{4}\right) =G\left( x_{1}\right) -G\left( x_{3}\right) -G\left( x_{2}\right) +G\left( x_{4}\right)$, this is only possible if $G\left( x_{1}\right) -G\left( x_{2}\right) -G\left( x_{3}\right) +G\left( x_{4}\right) =0$, or (using $x_{1}=x$)

$$\begin{aligned} G\left( x\right) =G\left( x_{2}\right) +\left( G\left( x_{3}\right) -G\left( x_{4}\right) \right) , \end{aligned}$$

for each vector $x\in \mathbb {R}^{J}$, $a\in \mathbb {R}^{J^{k}}$ and $b\in \mathbb {R}^{J^{\ell }}$. Arbitrarily fixing $a$ and $b$, we can define

$$\begin{aligned} G_{k\ell }^{-\ell }\left( \ldots ,x^{\ell -1},x^{\ell +1},\ldots \right)&\equiv G\left( x_{2}\right) \\&\equiv G\left( \ldots ,x^{k-1},x^{k},x^{k+1},\ldots ,x^{\ell -1},b,x^{\ell +1},\ldots \right) \end{aligned}$$

and

$$\begin{aligned} G_{k\ell }^{-k}\left( \ldots ,x^{k-1},x^{k+1},\ldots \right)&\equiv G\left( x_{3}\right) -G\left( x_{4}\right) \\&\equiv G\left( \ldots ,x^{k-1},a,x^{k+1},\ldots ,x^{\ell -1},x^{\ell },x^{\ell +1},\ldots \right) \\&-G\left( \ldots ,x^{k-1},a,x^{k+1},\ldots ,x^{\ell -1},b,x^{\ell +1},\ldots \right) , \end{aligned}$$

which leads to the desired result.

Step 2 We show that there must exist a list of functions $G^{1},G^{2},\ldots ,G^{P}$ s.t. $G\left( x\right) =\sum \nolimits _{k\in \mathbb {P}}G^{k}\left( x^{k}\right)$ for each $x\in \mathbb {R}^{J}$.

If $P=2$, the representation follows directly from step 1. We proceed by induction. Consider $P$ compensation groups, with $2\le P<J$, and suppose that the existence of functions $G_{k\ell }^{-\ell }:\mathbb {R} ^{J-J_{\ell }}\rightarrow \mathbb {R}$ and $G_{k\ell }^{-k}:\mathbb {R} ^{J-J_{k}}\rightarrow \mathbb {R}$ for any two compensation groups $k$ and $\ell$, with $k<\ell \le P$, such that

$$\begin{aligned} G\left( x\right) =G_{k\ell }^{-\ell }\left( x^{1},\ldots ,x^{\ell -1},x^{\ell +1},\ldots ,x^{P}\right) +G_{k\ell }^{-k}\left( x^{1},\ldots , x^{k-1},x^{k+1},\ldots ,x^{P}\right) \end{aligned}$$

holds for each $x\in \mathbb {R}^{J}$, implies additive separability of $G$ (induction hypothesis). We show next that it also holds for $P+1$ groups. Consider a function with $P+1$ groups. From step 1, we know that, for each two compensation groups $k$ and $\ell$, with $k<\ell \le P+1$, there exist functions $G_{k\ell }^{-\ell }:\mathbb {R}^{J-J_{\ell }}\rightarrow \mathbb {R}$ and $G_{k\ell }^{-k}:\mathbb {R}^{J-J_{k}}\rightarrow \mathbb {R}$ such that

$$\begin{aligned} G\left( x\right)&= G_{k\ell }^{-\ell }\left( x^{1},\ldots ,x^{\ell -1},x^{\ell +1},\ldots ,x^{P+1}\right) \\&\quad +G_{k\ell }^{-k}\left( x^{1},\ldots , x^{k-1},x^{k+1},\ldots ,x^{P+1}\right) \end{aligned}$$

for each $x\in \mathbb {R}^{J}$. Using these conditions for arbitrary $k<\ell \le P$, and using the induction hypothesis, there must exist functions $\overline{G}^{k}\left( \cdot ,x^{P+1}\right)$ for $k=1,\ldots ,P$, such that

$$\begin{aligned} G\left( x^{1},\ldots ,x^{P},x^{P+1}\right) =\sum \limits _{k=1}^{P}\overline{G}^{k}\left( x^{k},x^{P+1}\right) , \end{aligned}$$

(6)

for each $x^{1},\ldots ,x^{P},x^{P+1}$. Now, consider an arbitrary compensation group $k<P+1$. Step 1 applied to $k$ and $P+1$ gives us a representation

$$\begin{aligned} G\left( x^{1},\ldots ,x^{P},x^{P+1}\right)&= G_{k\left( P+1\right) }^{-k}\left( x^{1},\ldots ,x^{k-1},x^{k+1},\ldots ,x^{P+1}\right) \\&+G_{k\left( P+1\right) }^{-\left( P+1\right) }\left( x^{1},\ldots , x^{P}\right) , \end{aligned}$$

which can be combined with (6) to obtain

$$\begin{aligned} \sum \limits _{k=1}^{P}\overline{G}\left( x^{k},x^{P+1}\right)&= G_{k\left( P+1\right) }^{-k}\left( x^{1},\ldots ,x^{k-1},x^{k+1},\ldots , x^{P+1}\right) \\&+G_{k\left( P+1\right) }^{-\left( P+1\right) }\left( x^{1},\ldots , x^{P}\right) \end{aligned}$$

or equivalently,

$$\begin{aligned} \overline{G}\left( x^{k},x^{P+1}\right)&= G_{k\left( P+1\right) }^{-k}\left( x^{1},\ldots ,x^{k-1},x^{k+1},\ldots ,x^{P+1}\right) \\&+G_{k\left( P+1\right) }^{-\left( P+1\right) }\left( x^{1}, \ldots , x^{P}\right) -\sum \limits _{\ell \ne k}\overline{G}\left( x^{\ell }, x^{P+1}\right) , \end{aligned}$$

for each $x^{1},\ldots ,x^{P},x^{P+1}$. Fixing all variables, except $x^{k}$ and $x^{P+1}$, we get a representation of $\overline{G}\left( x^{k},x^{P+1}\right)$ as

$$\begin{aligned} \overline{G}\left( x^{k},x^{P+1}\right) =G^{k}\left( x^{k}\right) +\widetilde{G}_{k}^{P+1}\left( x^{P+1}\right) , \end{aligned}$$

with

$$\begin{aligned} G^{k}\left( x^{k}\right)&\equiv G_{k\left( P+1\right) }^{-\left( P+1\right) }\left( \overline{x}^{1},\ldots ,\overline{x}^{k-1},x^{k},\overline{x}^{k+1}\ldots ,\overline{x}^{P}\right) ,\\ \widetilde{G}_{k}^{P+1}\left( x^{P+1}\right)&\equiv G_{k\left( P+1\right) }^{-k}\left( \overline{x}^{1},\ldots ,\overline{x}^{k-1},\overline{x}^{k+1},\ldots ,\overline{x}^{P},x^{P+1}\right) \\&-\sum \limits _{\ell \ne k}\overline{G}\left( \overline{x}^{\ell }, x^{P+1}\right) . \end{aligned}$$

Since this holds for any compensation group $k<P+1$, we can plug it in Eq. (6) to obtain the desired result, i.e. the existence of functions $G^{k}$ for $k=1,\ldots ,P+1$ such that

$$\begin{aligned} G\left( x^{1},\ldots ,x^{P},x^{P+1}\right) =\sum \limits _{k=1}^{P+1}G^{k}\left( x^{k}\right) , \end{aligned}$$

for each $x^{1},\ldots ,x^{P},x^{P+1}$, with $G^{P+1}\left( x^{P+1}\right)$ equal to $\sum _{k=1}^{P}\widetilde{G}_{k}^{P+1}\left( x^{P+1}\right)$.

Proof of proposition 1

Proposition 1a follows directly from Lemmas 1 and 2. We prove proposition 1b, i.e. given separability of G, partial solidarity and equal treatment of equals lead to the partial sharing rule.

Suppose by contradiction that the partial sharing rule does not follow, i.e. that there exist $\mathbf {x}\in \mathbb {D}$ and $j\in \mathbb {I}$ such that

$$\begin{aligned} N_{j}\left( \mathbf {x}\right) \ne \sum \limits _{k\in \mathbb {P}}(1-\gamma ^{k})G^{k}(x_{j}^{k})+\frac{1}{I}\sum \limits _{i\in \mathbb {I}}\sum \limits _{k\in \mathbb {P}}G^{k}(x_{i}^{k}). \end{aligned}$$

Without loss of generality, we assume

$$\begin{aligned} N_{j}\left( \mathbf {x}\right) >\sum \limits _{k\in \mathbb {P}}(1-\gamma ^{k})G^{k}(x_{j}^{k})+\frac{1}{I}\sum \limits _{i\in \mathbb {I}}\sum \limits _{k\in \mathbb {P}}G^{k}(x_{i}^{k}). \end{aligned}$$

Since the budget must be balanced in a redistribution scheme, there must exist some $m\in \mathbb {I}$ such that

$$\begin{aligned} N_{m}\left( \mathbf {x}\right) <\sum \limits _{k\in \mathbb {P}}(1-\gamma ^{k})G^{k}\left( x_{m}^{k}\right) +\frac{1}{I}\sum \limits _{i\in \mathbb {I}}\sum \limits _{k\in \mathbb {P}}G^{k}\left( x_{i}^{k}\right) . \end{aligned}$$

Both inequalities together imply

$$\begin{aligned} N_{j}\left( \mathbf {x}\right) -N_{m}\left( \mathbf {x}\right) >\sum \limits _{k\in \mathbb {P}}(1-\gamma ^{k})(G^{k}(x_{j}^{k})-G^{k}(x_{m}^{k})). \end{aligned}$$

(7)

Define $\mathbf {x}\left( 1\right) \in \mathbb {D}$ to be such that for each $i\in \mathbb {I}\backslash \left\{ j\right\}$, $x(1)_{i}=x_{i}$ and for each $k\in \mathbb {P}\backslash \left\{ 1\right\}$, $x(1)_{j}^{k}=x_{j}^{k}$ and $x(1)_{j}^{1}=x_{m}^{1}$. Using partial solidarity, we get

$$\begin{aligned} N_{j}\left( \mathbf {x}\left( 1\right) \right) -N_{m}\left( \mathbf {x}\left( 1\right) \right)&= N_{j}\left( \mathbf {x}\right) -N_{m}\left( \mathbf {x}\right) +\left( 1-\gamma ^{1}\right) (G(x(1)_{j})-G(x_{j}))\\&= N_{j}\left( \mathbf {x}\right) -N_{m}\left( \mathbf {x}\right) +\left( 1-\gamma ^{1}\right) (G^{1}(x_{m}^{1})-G^{1}(x_{j}^{1})), \end{aligned}$$

where the last step follows using additive separability and the construction of $\mathbf {x}\left( 1\right)$. Combining with (7), we get

$$\begin{aligned} N_{j}\left( \mathbf {x}\left( 1\right) \right) -N_{m}\left( \mathbf {x}\left( 1\right) \right)&> \sum \limits _{k\in \mathbb {P}}(1-\gamma ^{k})(G^{k}(x_{j}^{k})-G^{k}(x_{m}^{k}))\\&+\left( 1-\gamma ^{1}\right) (G^{1}(x_{m}^{1})-G^{1}(x_{j}^{1})).\nonumber \end{aligned}$$

(8)

Now, define $\mathbf {x}\left( 2\right) \in \mathbb {D}$ to be such that for each $i\in \mathbb {I}\backslash \left\{ j\right\}$, $x(2)_{i}=x\left( 1\right) _{i}$ and for each $k\in \mathbb {P}\backslash \left\{ 2\right\}$, $x(2)_{j}^{k}=x(1)_{j}^{k}$ and $x(2)_{j}^{2}=x(1)_{m}^{2}$. Note that by construction $x(2)_{j}^{2}=x_{m}^{2}$ and $x(1)_{j}^{2}=x_{j}^{2}$. Applying the same reasoning as above, using Eq. (8) this time, yields

$$\begin{aligned} N_{j}\left( \mathbf {x}\left( 2\right) \right) -N_{m}\left( \mathbf {x}\left( 2\right) \right)&> \sum \limits _{k\in \mathbb {P}}(1-\gamma ^{k})(G^{k}(x_{j}^{k})-G^{k}(x_{m}^{k}))\\&+\sum \limits _{k=1}^{2}(1-\gamma ^{k})(G^{k}(x_{m}^{k})-G^{k}(x_{j}^{k})). \end{aligned}$$

Proceeding in this way, we end up with a distribution $\mathbf {x}\left( P\right)$ such that

$$\begin{aligned} N_{j}\left( \mathbf {x}\left( P\right) \right) -N_{m}\left( \mathbf {x}\left( P\right) \right)&> \sum \limits _{k\in \mathbb {P}}(1-\gamma ^{k})(G^{k}(x_{j}^{k})-G^{k}(x_{m}^{k}))\\&+\sum \limits _{k=1}^{P}(1-\gamma ^{k})(G^{k}(x_{m}^{k})-G^{k}(x_{j}^{k})). \end{aligned}$$

Note that the right-hand side sums up to zero and that $x\left( P\right) _{j}=x\left( P\right) _{m}$ holds by construction. The inequality $N_{j}\left( \mathbf {x}\left( P\right) \right) >N_{m}\left( \mathbf {x}\left( P\right) \right)$, therefore, violates equal treatment of equals, a contradiction.