To be fair: claims have amounts and strengths

Abstract

John Broome (Proc Aristot Soc 91:87–101, 1990) has developed an influential theory of fairness, which has generated a thriving debate about the nature of fairness. In its initial conception, Broomean fairness is limited to a comparative notion. More recent commentators such as Hooker (Ethical Theory Moral Pract 8:329–52, 2005), Saunders (Res Publica 16:41–55, 2010), Lazenby (Utilitas 26:331–345, 2014), Curtis (Analysis 74:47–57, 2014) have advocated, for different reasons, to also take into account non-comparative fairness. Curtis’ (Analysis 74:47-57, 2014) theory does just that. He also claims that he furthers Broome’s theory by saying precisely what one must do in order to be fair. However, Curtis departs from Broome’s (Proc Aristot Soc 91:87-101, 1990) requirement that claims are satisfied in proportion to their strength. He neglects claim-strengths altogether and identifies claims with their amount. As a result, the theory of Curtis has limited scope. I present a theory of fairness that fulfils all three desiderata: it incorporates non-comparative fairness, it recognizes that claims have both amounts and strengths, and it tells us precisely what one must do in order to be fair.

Appendix

Appendix I shows that Curtis’s argument for $L^\dag$ being well-defined is flawed. Appendix II presents the proof of the Lottery Theorem.

1.1 I. Curtis’s flawed argument

Remember from Sect. 2.3 that method $L^\dag$ constructs a lottery in terms of a probability vector that solves equation

$$\begin{aligned} Ap = s^\star \end{aligned}$$

(⋆)

Also, remember that Curtis argued that the construction of $L^\dag$ is well-defined:

The number of equations in the system generated will be equal to the number of receiving agents, and the number of variables equal to the number of allocations. Moreover, the only possible coefficients in any equation generated are 0 and 1, and there can never be more receiving agents than there are allocations. These two latter facts are enough (due to the Rouché-Capelli theorem) to guarantee that every system of linear equations will have at least one solution.

Curtis (2014: 53)

Indeed, “the number of equations in the system” equals the number $|N^\star |$ of receiving agents and the number of variables equals the number m of integer allocations of $\textbf{MaxSat}({\mathcal {B}}^\star )$, which is to say that, in ($\star$), A is an $|N^\star | \times m$ matrix. Also, it is immediately clear that the only possible entries of A are 0 and 1 and that $|N^\star | \le m$. But from these observations it does not follow, pace Curtis, that ($\star$) will have at least one solution. To see this, consider the following system of equations of form $Ap = s$:

$$\begin{aligned} \begin{pmatrix} 1 &{} 1 &{} 0 \\ 1 &{} 1 &{} 1 \\ 0 &{} 0 &{} 1 \end{pmatrix} \begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix} = \begin{pmatrix} 0.8 \\ 0.6 \\ 0.6 \end{pmatrix} \end{aligned}$$

(4)

Matrix Eq. (4) involves a matrix A which fully respects the conditions cited by Curtis. And yet, as one readily verifies, the equation has no solution. Further, the matrix of the following equation also fully respects the conditions cited by Curtis:

$$\begin{aligned} \begin{pmatrix} 1 &{} 1 &{} 0 \\ 1 &{} 0 &{} 1 \\ 0 &{} 1 &{} 1 \end{pmatrix} \begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix} = \begin{pmatrix} 0.35 \\ 0.05 \\ 0.6 \end{pmatrix} \end{aligned}$$

(5)

Now, (5) has a unique solution. However, this solution is $(-0.1, 0.45, 0.15)$, which is not a probability vector and so does not induce a lottery.

Although (4) and (5) both respect Curtis’s conditions, neither of them provides an instantiation of ($\star$). For, instantiations of ($\star$) are induced by remainder problems ${\mathcal {B}}^\star$ and neither (4) nor (5) can be induced as such.²¹ Curtis further mentions the so-called Rouché-Capelli theorem. This theorem states that an arbitrary matrix equation of form $Ap = s$ has a solution if and only if the rank of its coefficient matrix A is equal to the rank of its augmented matrix (A|s). As testified²² by (4), the conditions cited by Curtis do not ensure that the Rouché-Capelli theorem is applicable. Now, for proper instantiations of ($\star$) it can²³ be shown that the rank of A equals that of (A|s). But this only establishes, via the Rouché-Capelli theorem, that instantiations of ($\star$) have solutions, not that they have probability vectors amongst their solutions. And yet clearly, we have to establish the latter in order to show that method $L^\dag$ is well-defined.

1.2 II. Proof of the Lottery Theorem

Theorem 1

(Lottery Theorem) Let ${\mathcal {B}} = (E, N, \textbf{1}, s)$ be a Broomean problem such that $E >0$ is an integer and such that $0< s_i < 1$ for all $i \in N = \{1, \ldots , n \}$ and $\sum _{i \in N} s_i = E$. Let A be an $n \times m$ matrix whose columns $\alpha _1, \ldots , \alpha _m$ are the integer allocations of $\textbf{MaxSat}({\mathcal {B}})$. Then, the matrix equation $Ap = s$ has at least one solution $p \in {\mathbb {R}}^m$ that is a probability vector, i.e. $p_i \ge 0$ for all $i \in N$ and $\sum _{i \in N} p_i = 1$.

Proof

See below. $\square$

We will call an equation of form $Ap = s$ which respects the conditions of the Lottery Theorem a lottery equation. It readily follows that the entries of any solution of a lottery equation must sum to one:

Lemma 1

If ${\bar{p}}$ is a solution to lottery equation $Ap =s$ then $\sum _{i \in N} {\bar{p}}_i = 1$.

Proof

Let lottery equation $Ap =s$ be induced from $(E, N, \textbf{1}, s)$. As the number of 1s and 0s in a column of A equal E and $n-E$ respectively, it follows that, for any $p \in {\mathbb {R}}^m$, $\sum _{i \in N} (Ap)_i = E \cdot (\sum _{i \in N} p_i)$. So, when $A{\bar{p}} =s$ we get $\sum _{i \in N} s_i = E \cdot (\sum _{i \in N} {\bar{p}}_i)$ from which, as $\sum _{i \in N} s_i = E$, it follows that $\sum _{i \in N} {\bar{p}}_i = 1$. $\square$

Thus, showing that lottery equations have non-negative solutions suffices, in virtue of Lemma 1, to establish the Lottery Theorem. In order to show that lottery equations have non-negative solutions we will invoke Farkas’ lemma:

Proposition 2

(Farkas’ lemma) For any $n \times m$ matrix A and $b \in {\mathbb {R}}^n$, exactly one of the following two statements is true.

• 1. There is an $x \in {\mathbb {R}}^m$ such that such that $Ax = b$ and $x \ge 0$

• 2. There is an $y \in {\mathbb {R}}^n$ such that $A^\top y \ge 0$ and $b^\top y < 0$

Proof

See Goldman and Tucker (1956). $\square$

The following lemma shows that, for lottery equations, the second statement of Farkas’ lemma is false.

Lemma 2

Let $Ap =s$ be a lottery equation with $n \times m$ lottery matrix A. There is no $y \in {\mathbb {R}}^n$ such that $A^\top y \ge 0$ and $s^\top y < 0$.

Proof

Let $Ap = s$ be a lottery equation induced by Broomean problem $(E, N, \textbf{1}, s)$ and suppose, for a reductio, that $A^\top y \ge 0$ and $s^\top y < 0$ for some $y \in {\mathbb {R}}^n$ from which, as I will show, it follows that (i) $\sum \nolimits _{i \in X} y_i \ge 0$ for each $X \subseteq N$ with $|X| = E$ and that (ii) $\sum \nolimits _{i \in N} s_i\cdot y_i < 0$.

The truth of (ii) follows immediate from the definition of matrix multiplication. To see that (i) is true, note that the columns of A correspond with the integer allocations of $\textbf{MaxSat}({\mathcal {B}})$, where ${\mathcal {B}} = (E, N, \textbf{1}, s)$ satisfies the conditions of the Lottery Theorem. As such, x is a column of A, and hence $x^\top$ a row of $A^\top$, if and only if x is a vector of length $\mid N \mid$ with E entries equal to 1 and $\mid N \mid - E$ entries equal to 0. So, with any $X \subseteq N$ with $|X| = E$ there corresponds a column x of A, and hence a row $x^\top$ of $A^\top$, which is defined by letting $x_i = 1$ if $i \in X$ and $x_i = 0$ if $i \not \in X$. Now, from the definition of x in terms of X, it readily follows that:

$$\begin{aligned} x^\top y = \sum _{i \in N} x_i y_i = \sum _{i \in X} y_i \end{aligned}$$

(6)

As $x^\top$ is a row of $A^\top$, it follows from the supposition that $A^\top y \ge 0$ and (6) that $\sum _{i \in X} y_i \ge 0$, which establishes (i).

With k the number of negative entries of y, observe that:

$$\begin{aligned} 1 \le k \le E-1 \end{aligned}$$

(7)

The fact that $k \ge 1$ follows immediately from condition (ii) and the the fact that all entries of s are strictly positive. To see that $k \le E-1$, suppose that $k \ge E$. Then, there is a subset $X \subseteq N$ with $\mid X\mid = E$ such that $y_i < 0$ for all $i \in X$. Let $x^\top$ be the row of $A^\top$ that corresponds with this X and note that, per definition, $x^\top y < 0$. This contradicts with the supposition that $A^\top y \ge 0$.

Now rearrange the entries of y in ascending order so that, after the rearrangement, we have $y_ i \le y_j$ whenever $i \le j$. Let $Y^+$ and $Y^-$ be the sets consisting of the non-negative and negative entries of y respectively. With $k = |Y^-|$, it follows from (ii), as $y_ i < 0$ for $i \le k$, and $0< s_i < 1$ that

$$\begin{aligned} \sum \nolimits _{i = 1}^k y_i + \sum \nolimits _{i = {k+1}}^n s_i y_i< 0 \end{aligned}$$

(8)

As $y_ i \le y_j$ whenever $i \le j$, it follows from (8) that:

$$\begin{aligned} \sum \nolimits _{i = 1}^k y_i \ + \ \sum \nolimits _{i = {k+1}}^{E} s_i y_i \ + \ y_{E+1} \sum \nolimits _{i = {E +1}}^n s_i \ < 0 \end{aligned}$$

(9)

Observe that the summations in (9) are well-defined in virtue of (7) and the fact that, per definition of a lottery equation, $E < n$.

We argue that $y_{E+1} > 0$. To do so, suppose that $y_{E+1} \le 0$. Then, as$E+1 > k$ so that $y_{E+1}$ is non-negative, we have $y_{E+1} =0$. But then, when $y_{E+1} = 0$, per definition of k and the arrangement of the entries of y in ascending order, we have:

$$\begin{aligned} {\left\{ \begin{array}{ll} y_i < 0 &{} \text{ for } 1 \le i \le k \\ y_i = 0 &{} \text{ for } k+1 \le i \le E \end{array}\right. } \end{aligned}$$

(10)

It then follows from (10) that $\sum _{i = 1}^E y_i < 0$, which contradicts condition (i). Hence $y_{E+1} > 0$.

Now let $\bar{s_i} = 1 -s_i$, let $m = \sum _{i = k + 1}^n s_i$ and rewrite $\sum \nolimits _{i = {k+1}}^{E} s_i y_i \ + \ y_{E+1} \sum \nolimits _{i = {E+1}}^n s_i$, i.e. the last two terms of (9), as:

$$\begin{aligned} \sum \nolimits _{i = {k+1}}^{E} (1 - {\bar{s}}_i) y_i + y_{E+1} (m - \sum \nolimits _{i = {k+1}}^{E} (1 - {\bar{s}}_i)) \end{aligned}$$

(11)

Observe that (11), and so the last two terms of (9), may be rewritten as:

$$\begin{aligned} \sum \nolimits _{i = {k+1}}^{E} y_i \ + \ ( m y_{E+1} \ - \sum \nolimits _{i = {k+1}}^{E} y_{E+1}) \ + \ \sum \nolimits _{i = {k+1}}^{E} {\bar{s}}_i (y_{E+1} - y_i) \end{aligned}$$

(12)

The middle term of (12) equals $(m - (E- k ) ) y_{E+ 1}$. As $\sum _{i = 1}^n s_i = E$ we have $m = E - \sum \nolimits _{i = {1}}^{k} s_i$. Hence, as each $s_i < 1$, we have $m > E-k$ so that, as $y_{E+1} > 0$ the middle term of (12) is positive. Moreover, as ${\bar{s}}_i > 0$ and as, per arrangement of the entries of y in ascending order, $y_{E+1} \ge y_i$ whenever $k+1 \le i \le E$, the rightmost term of (12) is non-negative. Hence, the sum of the middle and rightmost term of (12) is positive so that (the sum of the three terms of) (12) is strictly larger than $\sum \nolimits _{i = {k+1}}^{E} y_i$. And so, as (12) equals the sum of the last two expressions of (9), we have:

$$\begin{aligned} \sum \nolimits _{i = {k+1}}^{E} s_i y_i + \ y_{E+1} \sum \nolimits _{i = {E+1}}^n s_i > \sum \nolimits _{i = {k+1}}^{E} y_i \end{aligned}$$

(13)

By adding $\sum _{i = k+ 1}^{E} y_i$ to both sides of the inequality sign of (9), we become:

$$\begin{aligned} \sum \nolimits _{i = 1}^E y_i \ + \ \sum \nolimits _{i = {k+1}}^{E} s_i y_i \ + \ y_{E+1} \sum \nolimits _{i = {E +1}}^n s_i \ < \sum _{i = k+ 1}^E y_i \end{aligned}$$

(14)

So that it follows from (13) and (14) that $\sum _{i = 1}^E y_i < 0$, which contradicts with condition (i) and which completes our reductio argument. $\square$

The Lottery Theorem readily follows from Lemma 1, Lemma 2 and Farkas’ lemma:

Lottery Theorem (proof): It follows from Farka’s lemma and Lemma 2 that any lottery equation has a non-negative solution $p \ge 0$ whose entries must, according to Lemma 1, sum to 1. That is, any lottery equation has a solution which is a probability vector, which establishes the Lottery Theorem. $\square$