Positive Political Theory II

Positive Political Theory II (University of Michigan Press, 2005) regrettably contains a variety of obscurities and errors, both typographical and substantive. Most of these are apparent and the appropriate corrections evident. Unfortunately, a few of the mistakes to surface are egregious (and thus correspondingly embarrassing ...). So, with apologies to Je¤ and to those using this book, the mistakes within this category identied so far are located and corrected below. (The University of Michigan Press has kindly agreed to publish a corrected version of the ms in 2015.)

To avoid the occasional ambiguity in some subsequent examples, a sentence has been added that reads, "Hereafter, unless explicitly stated otherwise, assume an aggregation rule f satis…es unrestricted domain, D = R n ."Corresponding changes are thereafter made where appropriate.
p.5, remarks on simple rules and the Nakamura number.
The examples using majority and plurality rule on this page (from the top to the …nal paragraph) presume unrestricted domain.
As written, this is insu¢ ciently tight.Replace the current de…nition and subsequent paragraph with: De…nition 2.4 Fix a collective choice function ' : R n !X with range ' .Say that ' respects unanimity if and only if, for all 2 R n and all sets X ' such that (x; y) 2 X ' nX implies P (x; y; ) = N , '( ) 2 X .
In words, ' respects unanimity if, at any pro…le for which it is possible to partition the range of ' into two subsets such that every alternative in one subset is strictly preferred by all individuals to every alternative in the other subset, then ' surely selects an alternative from the universally more-preferred-to set.It is easy to see that if there are multiple such sets for any pro…le, then they must be ordered by set-inclusion.Although similar in spirit to the Pareto criterion, respecting unanimity and weak Pareto are not equivalent (where a choice function ' is weakly Paretian if, for all x; y 2 X and all 2 R n , P (x; y; ) = N implies '( ) 6 = y).
Should have k and n k + 1 below the equality, not i k and n i k + 1, respectively.
The constructive argument used in the text to prove Lemma 2.5 turns out to have paid insu¢ cient attention to di¢ culties at the boundaries of the feasible set of alternatives.In particular, the particular choice of pro…les for the argument in one case need not be feasible.Below is a complete restatement of the required (slightly extended) Lemma and proof.
Lemma 2.5 Let ' : S Q !X be strategy-proof and satisfy citizen sovereignty.Then ' is weakly Paretian and satis…es peak only.
(Peak only) Now …x 2 S Q , arbitrary i 2 N and any single-peaked preference ordering R 0 i with x i = x 0 i .To prove the lemma, it su¢ ces to show '( ) = '(R 0 i ; i ).If '( ) = x 0 i , then the result is trivial.So, without loss of generality, assume x 0 i = x i > '( ).There are two cases (in what follows, the alternative 0 i (y) is de…ned in exactly the same way as the alternative i (y), above, but with respect to the preference ordering R 0 i rather than to the ordering R i ). ( Then, depending on the relative position of the alternative '(R 0 i ; i ) in Figure 2.3, there are three possibilities, each of which (given single-peakedness) contradicts strategyproofness. (1a): , the second case.
(2) i ('( )) 0 i ('( )).In this case, we need to take a more indirect approach than in case (1) to proving the claim.To this end, let L = fj 2 N : 1), this contradicts ' weakly Paretian.So assume L 6 = ;.By relabeling if necessary, let L = f1; :::; `g and x 1 x 2 : : : x `.For all j 2 L, let R 0 j be such that x 0 j = x 0 i and x j 0 j ('( )) j ('( )): see Figure 2.4 for the situation in which all of the inequalities are strict for i and some j 2 L. `) 6 = '( ).Arguing similarly to Case (1), single-peaked preferences implies that ' is manipulable by `unless '(R 0 And continuing in this way iteratively for ` 2; ` 3; : : . There are …ve possibilities for the relative location of '( 0 ).In four of these, '( 0 ) 6 = '( ) and we show that i has opportunity to manipulate the outcome pro…tably; thus we must have '( 0 ) = '( ) and, in this case we show '( 0 ) = '(R 0 i ; i ) as claimed.
Exercises 2.1 and 2.9 of the original text have been eliminated.
The vote pro…les m 2 M should be understood as rationalizable preference relations (see PPT I, ch.1, on rationalizable preferences).
p.139, from the end of line 5 to the middle of line 12.
The claim made in this section is that condition (*) is a necessary condition for an alternative y to be agenda independent.This is false and the comments immediately following (*) are thereby nonsense.The text beginning on line 5 with the sentence "In other words ..." to the end of the sentence concluding on line 12 with "... S( ; P ) = y", should be deleted and replaced with the following: "On the other hand, a su¢ cient condition for an alternative y 2 Xnfx 0 g to be the sophisticated outcome irrespective of the agenda 2 A(x 0 ), is that y 2 P (x 0 ) and, 8z 2 P (x 0 )nfyg; y 2 P (z): That is, if y satis…es (*) then, for all 2 A(x 0 ), S( ; P ) = y.To see this, recall that every terminal node of an amendment agenda pairs the status quo x 0 against an alternative from the agenda, with every such alternative appearing on at least one terminal node.By the …rst property of (*), that y 2 P (x 0 ), and the earlier logic for solving binary voting games, y must be the sophisticated equivalent of every terminal node at which y is compared with x 0 .Now consider any alternative z 6 = y.If z is the sophisticated equivalent of some terminal node, then either z = x 0 or z 2 P (x 0 ).In either case, (*) implies that y must be the sophisticated equivalent of any pairwise comparison between y and z at the next stage; and so on back up the voting tree, thus establishing the claim.In other words ... " p.172, Figure 5.12.
The payo¤s identi…ed in this diagram for Example 5.8 are incorrect.A corrected example is as follows.
Example 5.8 Consider the extensive form game G summarized in Figure 5.12, where the outcome associated with each terminal node of the game tree is de…ned in terms of the payo¤s (u 1 ; u 2 ) to individuals 1 and 2 respectively.To check and are both Nash equilibria note that, given 2 = l, the best 1 can do is play A at his …rst decision node to obtain a payo¤ of one rather than zero and, given 1 chooses A, no subsequent decision nodes are reached, so specifying 2 plays l is as good as anything else since 2 receives payo¤ one in any case; similarly, given 2 plays l, 1 is indi¤erent between L or R at the …nal decision node.That is also a Nash equilibrium follows easily from the observation that it yields the unique best payo¤ for each individual.Moreover, we claim is the only subgame perfect Nash equilibrium to G.
To see that and are not subgame perfect, consider the two proper subgames, G and G 0 , in turn.The restriction of to G is the strategy pair (L; l) which is easily checked to be a Nash equilibrium for G .On the other hand, the restriction of to G 0 is the decision "L"for individual 1, which is clearly not a best response at this decision node.Thus is not a subgame perfect equilibrium strategy pro…le for G. Similar reasoning shows is not subgame perfect: here, the restriction of to the (trivial) subgame G 0 has individual 1 choosing a best response, but then the restriction of to the subgame G is not a Nash equilibrium for G .This is because individual 2 choosing l is not a best response in G to 1's choice of R, despite the fact that 2 = l is a best response in G to 1 = (A; R).Finally, it is easy to check that is a subgame perfect Nash equilibrium to G as claimed.
p.288, last expression in the proof of Theorem 7:10.By de…nition of (a ; y), there should be no minus sign on the terms !i ( ) under the summation.The same is true in the argument for Corollary 7.3.p.288, line 2 of Corollary 7.3.
The assumption that "p i ( ) = p( )"should be replaced by "! c i ( ) is independent of i"; the …rst line of the proof is then redundant.p.336, Example 8.2 (5).
Proportional representation, even as described here, is not a scoring rule.
The last conditional, "... if m is odd" should read "... if m = 3".Furthermore, the exercise is intended to concern only pure strategy equilibria.p.213, Example 6.4 Example 6.4 contains both a typo.and a mystery matrix (pointed out to me by Diwen Si).The corrected version is as follows.
Example 6.4 Let N = f1; 2; 3g, i = all i and q = 2.By Theorem 6.2, any individual i 2 N recognized in the …rst period proposes an allocation giving a strictly positive amount V j to exactly one other committee member j 6 = i and nothing to the remaining individual.Consequently, the following three equations must hold in any stationary equilibrium: Suppose two individuals use non-degenerate mixed strategies, say r 12 ; r 23 2 (0; 1).Then both individuals 1 and 2 must be indi¤erent over to whom to make an o¤er, implying that, for all j, V j = V in equilibrium.In particular, assuming there is a dollar to be distributed, V = 1=3.Substituting V for V j , noting r 12 = 1 r 13 etc, and collecting terms, the preceding system of three equations can be written where [v] is a 3 1 vector of terms independent of (r 12 ; r 23 ; r 31 ).It is easy to check that the 3 3 matrix is singular.Therefore, there can be a continuum of randomization choices r satisfying ( ), that is, there can be a continuum of equilibria di¤ering only in the relative likelihoods of any particular minimal winning majority forming.If = 0:8 and p = (0:45; 0:35; 0:2), for instance, any r such that r 12 = 0:19 + 0:44r 31 ; r 23 = 0:96 + 0:57r 31 ; r 31 2 [0; 0:0625] constitutes stationary equilibrium behavior.
The italicized statement is incorrect.Lemma 6.1(3) only shows that if p i = p k and i < j then i V i j V j , not that V i V j .Thus, more patient players do not necessarily receive greater equilibrium payo¤s than less patient players.See Tomohiko Kawamori (2005), "Players'Patience and Equilibrium Payo¤s in the Baron-Ferejohn Model", Economic Bulletin 3, 1-5).

Figure 2 .
Figure 2.5 illustrates the construction.

Figure 5 .
Figure 5.12: The game G