Neutral freedom and freedom as control

Abstract

I present a model of freedom as control. Control is measured by the preferences of a decision-maker, or judge, who values flexibility and is neutral towards outcomes ex ante. Formally, I explore the consequences of adding a neutrality axiom to the Dekel et al. (Econometrica 69(4):891–934, 2001) axioms for preference for flexibility. I characterize the consensus of all neutral judges about which choice situations embody more freedom. The theory extends the freedom ranking literature to situations where agents have imperfect control, as modeled by choices among lotteries. In a voting context, the consensus of neutral judges coincides with Banzhaf power.

Appendix: Proofs and technical comments

1.1 Proof of Proposition 2

Let $\Delta ^*_{\text{ sym }}\left( U\right)$ be the set of symmetric probability measures in $\Delta ^*\left( U\right)$.

Lemma A.1

Suppose that $\preccurlyeq$ has a representation of the form (5) with some $p \in \Delta ^*\left( U\right)$. Then $\preccurlyeq$ has a representation of the form (5) with some $p \in \Delta ^*_{\text{ sym }}\left( U\right)$ if and only if $\preccurlyeq$ satisfies neutrality.

Proof

For any $p \in \Delta ^*\left( U\right)$ and $\pi \in \Pi$, define $p^{\pi } \in \Delta ^*(U)$ by:

$$\begin{aligned} p^\pi \left( S\right) := p\left( S^\pi \right) , \;\;\; \forall \text{ measurable } S. \end{aligned}$$

Choose any $M \in {\mathscr {M}}$ and $\pi \in \Pi$. The neutrality axiom holds if and only if

$$\begin{aligned} \int \max _{\ell \in M} \left( u \cdot \ell \right) p\left( du\right)&= \int \max _{\ell \in M^\pi } \left( u \cdot \ell \right) p\left( du\right) , \;\;\; \forall \pi \in \Pi , \;\forall M \in {\mathscr {M}}. \end{aligned}$$

(A.1)

Next observe that

$$\begin{aligned} \begin{aligned} \int \max _{\ell \in M^\pi } \left( u \cdot \ell \right) p\left( du\right)&= \int \max _{\ell \in M} \left( u \cdot \ell ^\pi \right) p\left( du\right) = \int \max _{\ell \in M} \left( u^{\pi ^{-1}} \cdot \ell \right) p\left( du\right) \\&= \int \max _{\ell \in M} \left( u \cdot \ell \right) p^\pi \left( du\right) . \end{aligned} \end{aligned}$$

(A.2)

In other words, if we consider an agent who values flexibility and faces menu M, permuting the menu M according to $\pi$ leads to the same ex ante expected utility as permuting the agent’s current uncertainty p over her fututre utility at the time of choice according to $\pi$.

It follows that (A.1) is equivalent to

$$\begin{aligned} \int \max _{\ell \in M} \left( u \cdot \ell \right) p\left( du\right)&= \int \max _{\ell \in M} \left( u \cdot \ell \right) p^\pi \left( du\right) , \;\;\; \forall \pi \in \Pi , \;\forall M \in {\mathscr {M}}. \end{aligned}$$

(A.3)

Define ${\bar{p}} = \frac{1}{n!} \sum _{\pi \in \Pi } p^\pi$. (A.3) implies that

$$\begin{aligned} \int \max _{\ell \in M} \left( u \cdot \ell \right) p\left( du\right) = \frac{1}{n!} \sum _{\pi \in \Pi }\int \max _{\ell \in M} \left( u \cdot \ell \right) p^\pi \left( du\right) = \int \max _{\ell \in M} \left( u \cdot \ell \right) {\bar{p}}\left( du\right) . \end{aligned}$$

(A.4)

Observe that if $p \in \Delta ^*\left( U\right)$, then ${\bar{p}} \in \Delta ^*_{\text{ sym }}\left( U\right)$. So we have established that (A.3) implies that

$$\begin{aligned} \exists p' \in \Delta^* _{\text{ sym }}\left( U\right) , \forall M \in {\mathscr {M}}, \;\;\; \int \max _{\ell \in M} \left( u \cdot \ell \right) p\left( du\right) = \int \max _{\ell \in M} \left( u \cdot \ell \right) p'\left( du\right) . \end{aligned}$$

(A.5)

On the other hand assume (A.5) holds; then (A.3) holds with $p'$ taking the role of p. It follows that if $\int \max _{\ell \in M} \left( u \cdot \ell \right) p\left( du\right)$ represents $\preccurlyeq$, then neutrality is equivalent to (A.5), which establishes the lemma. $\square$

Suppose that $\preccurlyeq$ satisfies the weak order, continuity, monotonicity, independence, and neutrality axioms. Then Proposition 1 and Lemma A.1 imply that $\preccurlyeq$ has the desired representation. Conversely, assume that $\preccurlyeq$ has a representation of the form (5) with some $p^* \in \Delta ^*_{\text{ sym }}\left( U\right)$. Then Proposition 1 and Lemma A.1 imply that $\preccurlyeq$ satisfies all of the axioms. $\square$

1.2 Proof of Theorem 1

Lemma A.2

For all $M, M' \in {\mathscr {M}}$, $M \preccurlyeq ^* M'$ if and only if:

$$\begin{aligned} \frac{1}{n!} \sum _{\pi \in \Pi } \max _{\ell \in M} (u^\pi \cdot \ell ) \le \frac{1}{n!} \sum _{\pi \in \Pi } \max _{\ell \in M'} (u^\pi \cdot \ell ),\;\;\forall u \in {\mathbb {R}}^n. \end{aligned}$$

(A.6)

Proof

First observe that for all $M \in {\mathscr {M}}$,

$$\begin{aligned} \begin{aligned} \max _{\ell \in S\left( M\right) } u \cdot \ell&= \max _{\left( \ell \left( \pi \right) : \pi \in \Pi \right) \in \times _{\pi \in \Pi } \text{ co }\left( M^\pi \right) } u \cdot \left[ \frac{1}{n!}\sum _{\pi \in \Pi } \ell \left( \pi \right) \right] \\&= \max _{\left( \ell \left( \pi \right) : \pi \in \Pi \right) \in \times _{\pi \in \Pi } \text{ co }\left( M^\pi \right) } \left[ \frac{1}{n!}\sum _{\pi \in \Pi } u \cdot \ell (\pi )\right] = \frac{1}{n!} \sum _{\pi \in \Pi } \max _{\ell \in \text{ co }\left( M\right) } \left( u \cdot \ell ^\pi \right) \\&= \frac{1}{n!} \sum _{\pi \in \Pi } \max _{\ell \in \text{ co }\left( M\right) } \left( u^{\pi ^{-1}} \cdot \ell \right) = \frac{1}{n!} \sum _{\pi \in \Pi } \max _{\ell \in \text{ co }(M)} \left( u^\pi \cdot \ell \right) . \end{aligned} \end{aligned}$$

(A.7)

So

$$\begin{aligned} M \preccurlyeq ^* M' \Leftrightarrow S\left( M\right) \subseteq S\left( M'\right) \Leftrightarrow \left( \max _{\ell \in S\left( M\right) } u \cdot \ell \le \max _{\ell \in S\left( M'\right) } u \cdot \ell , \;\; \forall u \in {\mathbb {R}}^n \right) \Leftrightarrow {(A.6)}, \end{aligned}$$

where the second equivalence follows from the fact that $S\left( M'\right)$ is closed and convex and the strong separating hyperplane theorem,³⁶ and the last equivalence follows from (A.7). $\square$

Recall that $\Delta ^*_{\text{ sym }}\left( U\right)$ is the set of all symmetric probability measures in $\Delta ^*\left( U\right)$. Proposition 2 implies that to show that that $\preccurlyeq ^*$ coincides with consensus of neutral judges, it is sufficient to show that:

$$\begin{aligned}&M \preccurlyeq ^* M' \Leftrightarrow \left[ \int \max _{\ell \in M} \left( u \cdot \ell \right) p \left( du\right) \le \int \max _{\ell \in M'} \left( u \cdot \ell \right) p \left( du\right) , \forall p \in \Delta ^*_{\text{ sym }}\left( U\right) \right] , \;\;\;\\&\quad \forall M, M' \in {\mathscr {M}}. \end{aligned}$$

Lemma A.2 now implies that to complete the proof, it is sufficient to show that (A.6) is equivalent to

$$\begin{aligned} \int \max _{\ell \in M} \left( u \cdot \ell \right) p (du) \le \int \max _{\ell \in M'} \left( u \cdot \ell \right) p \left( du\right) , \;\;\;\forall p \in \Delta ^*_{\text{ sym }}\left( U\right) . \end{aligned}$$

For any $p \in \Delta ^*\left( U\right)$ and $\pi \in \Pi$, define $p^\pi \in \Delta ^*\left( U\right)$ by $p^\pi \left( E\right) := p\left( E^\pi \right)$ for all measurable $E \subseteq U$.³⁷ Let $p\in \Delta ^*_{\text{ sym }}\left( U\right)$. Then for all measurable $E \subseteq U$, $p\left( E\right) = p\left( E^\pi \right) = p^\pi \left( E\right)$ for all $\pi \in \Pi$. So $p = p^\pi$ for all $\pi \in \Pi$. So:

$$\begin{aligned}&\forall p \in \Delta ^*_{\text{ sym }}\left( U\right) , \int \max _{\ell \in M} \left( u \cdot \ell \right) p\left( du\right) \\&\quad = \frac{1}{n!} \times n! \int \max _{\ell \in M} \left( u \cdot \ell \right) p\left( du\right) =\frac{1}{n!} \sum _{\pi \in \Pi } \int \max _{\ell \in M} \left( u \cdot \ell \right) p^{\pi ^{-1}} \left( du\right) \\&\quad = \frac{1}{n!} \sum _{\pi \in \Pi } \int \max _{\ell \in M} \left( u^\pi \cdot \ell \right) p\left( du\right) = \int \frac{1}{n!} \sum _{\pi \in \Pi } \max _{\ell \in M} \left( u^\pi \cdot \ell \right) p\left( du\right) . \end{aligned}$$

It is now easy to see that the desired equivalence holds. $\square$

1.3 Proof of Proposition 3

Kochov (2007) and Galaabaatar (2010) establish the following proposition.

Proposition A.1

Let $\preccurlyeq$ be a relation on ${\mathscr {M}}$. $\preccurlyeq$ is a preorder satisfying continuity, monotonicity, and independence if and only if there exists $P \subseteq \Delta ^*\left( U\right)$ such that for all $M, N \in {\mathscr {M}},$

$$\begin{aligned} M \preccurlyeq N \Leftrightarrow \left[ \int _U \max _{\ell \in M} (u \cdot \ell ) p (du)\le \int _U \max _{\ell \in N} (u \cdot \ell ) p (du), \;\;\forall p \in P \right] . \end{aligned}$$

(A.8)

Theorem 1, Proposition 2 and Proposition A.1 imply that the symmetry order is a preorder that satisfies continuity, independence, monotonicity, and neutrality. It is also straightforward to establish that the symmetry order satisfies these axioms via a direct proof. It is easy to see that the symmetry order is not complete.

Next consider any preorder satisfying continuity, independence, monotonicity and neutrality. Proposition A.1 implies that there exists $P \subseteq \Delta ^*\left( U\right)$ such that $\preccurlyeq$ satisfies (A.9). The neutrality axiom then implies that

$$\begin{aligned} \int _U \max _{\ell \in M} (u \cdot \ell ) p (du)= \int _U \max _{\ell \in M^\pi } (u \cdot \ell ) p (du),\;\;\; \forall p \in P, \; \forall M \in {\mathscr {M}}, \forall \pi \in \Pi . \end{aligned}$$

Recall that $\Delta ^*_{\text{ sym }}\left( U\right)$ is the set of all symmetric probability measures in $\Delta \left( U\right)$. An argument similar to the proof of Lemma A.1 now implies that for all $p \in P$, there exists $p' \in \Delta _{\text{ sym }}\left( U\right)$ such that

$$\begin{aligned} \int _U \max _{\ell \in M} (u \cdot \ell ) p (du)= \int _U \max _{\ell \in M} (u \cdot \ell ) p' (du),\quad \forall M \in {\mathscr {M}}. \end{aligned}$$

It follows that there exists $P' \subseteq \Delta _{\text{ sym }}\left( U\right)$ such that for all $M, N \in {\mathscr {M}}$,

$$\begin{aligned} M \preccurlyeq N \Leftrightarrow \left[ \int _U \max _{\ell \in M} (u \cdot \ell ) p (du)\le \int _U \max _{\ell \in N} (u \cdot \ell ) p (du), \forall p \in P' \right] . \end{aligned}$$

(A.9)

That the symmetry order is the coarsest preorder satisfying continuity, monotonicity, independence, and neutrality now follows from Theorem 1. $\square$

1.4 Proof of Proposition 4

For every lottery $\ell$, $S\left( \left\{ \ell \right\} \right) =\left\{ \left( \frac{1}{n},\frac{1}{n},\ldots ,\frac{1}{n}\right) \right\}$. On the other hand, for all M, $\left( \frac{1}{n},\frac{1}{n},\ldots ,\frac{1}{n}\right) \in S\left( M\right)$ and if M contains more than one lottery, then $S\left( M\right)$ contains more than one lottery. So $S\left( \left\{ \ell \right\} \right) \subseteq S\left( M\right)$ with a strict inclusion if M contains more than one lottery. The result now follows from (8), the definition of the symmetry order. $\square$

1.5 Interpretation of (9)

Here I explain why (9) does indeed give the probability that voter i is pivotal. In a stochastic voting mechanism $\alpha$, the outcome of the election is determined by the profile of votes and a random element. To more explicitly formalize the random element, define an extended voting mechanism to be a function $\alpha ': \{0,1\}^k \times [0,1] \rightarrow \{0,1\}$. Compared to a stochastic voting mechanism $\alpha$, an extended voting mechanism $\alpha '$ has one more argument, which is a real number $r \in [0,1]$. Interpret r as a random draw chosen with uniform probability independent of voter preferences. The output of extended voting mechanism $\alpha '$ is an integer, meaning that as a function of the inputs, one candidate definitely wins, so that the stochastic element is completely captured by the realization of r. For any stochastic voting mechanism $\alpha$, define the corresponding extended voting mechanism $\alpha '$ by: $\alpha '(z,r)= 1$ if $r \le \alpha (z)$, and $\alpha '(z,r)= 0$ otherwise. Then, in $\alpha '$, the probability that candidate 1 wins conditional on vote profile z is $\alpha (z)$. Also since $\alpha$ is monotone in the vote profile, so is $\alpha '$: $\forall , z, z' \in \{0,1\}^k, \forall r \in [0,1], z \le z' \Rightarrow \alpha '(z,r) \le \alpha '(z',r)$. As $\alpha$ and $\alpha '$ are essentially identical, and differ only in that the random element of $\alpha$ has been imported as an argument of $\alpha '$, to understand what it means to be pivotal in $\alpha$, one can look to what it means to be pivotal in $\alpha '$. In particular, i is pivotal in $\alpha '$ whenever $(z_{-i},r)$ is such that $\alpha '((0,z_{-i}),r) =0$ and $\alpha '((1,z_{-i}),r)=1$. (Since $\alpha '$ is monotone, it is not possible that when a voter changes her vote from 0 to 1, the outcome will change from 1 to 0.) In other words, i is pivotal whenever whichever candidate that i votes for wins. Define random variable $v_i$ on $\{0,1\}^{k-1} \times [0,1]$ by $v_i(z_{-i},r) = \alpha '((1,z_{-i}),r)-\alpha '((0,z_{-i}),r)$. $v_i=1$ if and only if i is pivotal, and $v_i = 0$ otherwise. So the expected value of $v_i$ is the probability that i is pivotal. The expected value of $v_i$ is $\sum _{z_{-i} \in \{0,1\}^{k-1} }\left( \int _0^1 \left[ \alpha '((1,z_{-i}),r)-\alpha '((0,z_{-i}),r)\right] dr\right) \mu (z_{-i}) = \sum _{z_{-i} \in \{0,1\}^{k-1}} \left( \int _0^1 \alpha '((1,z_{-i}),r)dr-\int _0^1 \alpha '((0,z_{-i}),r)dr\right) \mu (z_{-i})$, which is equal to the right-hand side of (9).

1.6 Proof or Proposition 5

Fix a voter i. For $j,k \in \{0,1\}$, let $\ell ^k_j$ and $\ell '^k_j$ be, respectively, the probabilities that lotteries $\ell ^k_i(\alpha ,\mu )$ and $\ell ^k_i(\alpha ',\mu ')$ assign to candidate j. Let $\delta _j$ be the lottery that assigns probability 1 on candidate j.

We have:

$$\begin{aligned} \begin{aligned} M_i(\alpha ,\mu )&= \{ \ell ^0_0 \delta _0 + \ell ^0_1 \delta _1, \ell ^1_0 \delta _0 + \ell ^1_1 \delta _1\} \\&= \ell ^1_0 \{\delta _0\}+\ell ^0_1 \{ \delta _1 \} + ( \ell ^1_1-\ell ^0_1 ) \{ \delta _0,\delta _1 \} \\&\sim ^* (\ell ^1_0 + \ell ^0_1)\{\delta _0\} + ( \ell ^1_1-\ell ^0_1 ) \{ \delta _0,\delta _1 \}. \end{aligned} \end{aligned}$$

(A.10)

For the second equality, note that monotonicity of $\alpha$ implies that $\ell ^1_1 -\ell ^0_1 \ge 0$. The indifference follows from the fact that all singleton menus are indifferent according to the symmetry order (Proposition 4) and the independence axiom. Similarly,

$$\begin{aligned} M_i(\alpha ',\mu ') \sim ^* (\ell '^1_0 + \ell '^0_1) \{\delta _0\} + ( \ell '^1_1-\ell '^0_1 ) \{ \delta _0,\delta _1 \}. \end{aligned}$$

(A.11)

Definitions imply that $B_i(\alpha ,\mu ) = \ell ^1_1-\ell ^0_1$ and $B_i(\alpha ',\mu ')= \ell '^1_1-\ell '^0_1$. So

$$\begin{aligned} B_i(\alpha ,\mu ) \le B_i(\alpha ',\mu ') \Leftrightarrow \;&\ell ^1_1-\ell ^0_1 \le \ell '^1_1-\ell '^0_1\\ \Leftrightarrow \;&(\ell ^1_0 + \ell ^0_1)\{ \delta _0\} + ( \ell ^1_1-\ell ^0_1 ) \{ \delta _0,\delta _1 \} \\&\preccurlyeq ^* (\ell '^1_0 + \ell '^0_1) \{\delta _0\} + ( \ell '^1_1-\ell '^0_1 ) \{ \delta _0,\delta _1 \}.\\ \Leftrightarrow \;&M_i(\alpha ,\mu ) \preccurlyeq ^* M_i(\alpha ',\mu '), \end{aligned}$$

where the second equivalence follows from the independence axiom and the fact that singleton menus are strictly dispreferred to all menus M such that M is not a singleton (Proposition 4), and the last equivalence follows from (A.10–A.11). $\square$

1.7 Proof of Proposition 7

Suppose that matrix $P=(p_{st}: s \in S, t \in T)$ is more informative than $Q =(q_{st}: s \in S, t \in T)$. Let $R=(r_{vt}: v \in T, t \in T)$ be the stochastic matrix such that $Q=PR$. Choose $\gamma ^*=\gamma (Q,\eta ,\sigma ^*) \in M(Q,\eta )$, where $\sigma ^*$ is an arbitrary strategy in $\Sigma$. For each $\sigma \in \Sigma$, define $\gamma ^\sigma = \gamma (P,\eta ,\sigma )$. For each $(s,d) \in S \times D$, $\gamma ^*(s,d)$ (resp., $\gamma ^\sigma (s,d)$) is the probability that $\gamma ^*$ (resp., $\gamma ^\sigma$) puts on (s, d). It is sufficient to write $\gamma ^*$ as a convex combination of lotteries in $M(P,\eta ) =\{\gamma ^\sigma : \sigma \in \Sigma \}$. This is sufficient because it establishes that $M(Q,\eta ) \subseteq \text{ co }\left( M(P,\eta )\right)$, which implies that every agent with preference for flexibility prefers $M(P,\eta )$ to $M(Q,\eta )$, which, in turn, implies that $M(Q,\eta ) \preccurlyeq ^* M(P,\eta )$.

For $d,d' \in D$, define $\chi (d,d'):=1$ if $d =d'$ and $\chi (d,d'):=0$ otherwise. Then for all $(s,d) \in S \times D$

$$\begin{aligned} \gamma ^*(s,d) = \eta _s \sum _{t \in T} q_{st}\chi (\sigma ^*(t),d)&= \eta _s \sum _{t \in T} \sum _{v \in T} p_{sv} r_{vt}\chi (\sigma ^*(t),d). \end{aligned}$$

(A.12)

Similarly,

$$\begin{aligned} \gamma ^\sigma (s,d) = \eta _s \sum _{v \in T} p_{sv} \chi (\sigma (v),d). \end{aligned}$$

(A.13)

For any $v \in T$ and $d \in D$, define:

$$\begin{aligned} \tau (v,d) = \sum _{t \in T} r_{vt} \chi (\sigma ^*(t),d). \end{aligned}$$

(A.14)

It is easy to see that $\tau (v,d) \ge 0$ for all $v \in T$ and $d \in D$, and the for for all $v \in T$, $\sum _{d \in D} \tau (v,d) =1.$ For any $\sigma \in \Sigma$, define: $\lambda _{\sigma } = \prod _{t \in T} \tau (t,\sigma (t))$. It follows that

$$\begin{aligned} 1= \prod _{t \in T} \sum _{d \in D} \tau (t,d) = \sum _{\sigma \in \Sigma } \prod _{t \in T} \tau (t,\sigma (t)) =\sum _{\sigma \in \Sigma } \lambda _\sigma . \end{aligned}$$

Similarly:

$$\begin{aligned} \begin{aligned} \tau (v,d)&= \tau (v,d) \prod _{t \in T {\setminus } v} \sum _{d' \in D} \tau (t,d') = \sum _{\{\sigma \in \Sigma : \sigma (v) =d\}} \prod _{t \in T} \tau (t,\sigma (t)) \\&= \sum _{\{\sigma \in \Sigma : \sigma (v)=d \}} \lambda _\sigma = \sum _{\sigma \in \Sigma } \lambda _\sigma \chi (\sigma (v),d). \end{aligned} \end{aligned}$$

(A.15)

Putting it all together, we have:

$$\begin{aligned} \gamma ^*(s,d)&= \eta _s \sum _{t \in T} \sum _{v \in T} p_{sv} r_{vt}\chi (\sigma ^*(t),d) = \eta _s \sum _{v \in T} p_{sv}\sum _{t \in T} r_{vt}\chi (\sigma ^*(t),d)\\&= \eta _s \sum _{v \in T} p_{sv} \tau (v,d) = \eta _s \sum _{v \in T} p_{sv} \left[ \sum _{\sigma \in \Sigma } \lambda _\sigma \chi (\sigma (v),d) \right]\\& = \sum _{\sigma \in \Sigma } \lambda _\sigma \left[ \eta _s \sum _{v \in T} p_{sv}\chi (\sigma (v),d) \right] = \sum _{\sigma \in \Sigma } \lambda _\sigma \gamma ^\sigma (s,d), \end{aligned}$$

where the first equality follows from (A.12), the third equality follows from (A.14), the fourth equality follows from (A.15), and the last equality follows from (A.13). This shows that $\gamma ^*$ can be written as a convex combination of $\{\gamma ^\sigma : \sigma \in \Sigma \}$, as we wanted to show.