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SPATIAL (EUCLIDEAN) MODEL

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Presentation on theme: "SPATIAL (EUCLIDEAN) MODEL"— Presentation transcript:

1 SPATIAL (EUCLIDEAN) MODEL
PART II SPATIAL (EUCLIDEAN) MODEL

2 Definition of Spatial Model
Voter i has ideal (bliss) point xi 2 <k Each alternative is represented by a point in <k A1 ¸i A2 iff ||xi-A1|| · || xi – A2|| Can use norms other than Euclidean e.g. ellipsoidal indifference curves

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4 Spatial Model Largely descriptive role rather than normative
The workhorse of empirical studies in political science k=1,2 are the most popular # of dimensions In U.S. k=2 gives high accuracy (~90%) , k=1 also very accurate since 1980s, and 1850s to early 20th century.

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6 What do the dimensions mean?
Different schools of thought Use expert domain knowledge or contextual information to define dimensions and/or place alternatives Fit data (e.g. roll call) to achieve best fit Maximize data fit in 1st dimension, then 2nd Impute meaning to fitted model

7 2D is qualitatively richer than 1D
x1 A1 A2 x3 x2 A3 A1 >A2 > A3 > A1

8 Condorcet’s voting paradox in Euclidean model
Hyperplane normal to and bisecting line segment A1A2

9 Even if all points in <2 are permitted alternatives, no Condorcet winner exists

10 Are there better conditions?
Major Question: Conditions for Existence of Stable Point (Undominated, Condorcet Winner) Plott (67) For case all xi distinct Slutsky(79) General case, not finite Davis, DeGroot, Hinich (72) Every hyperplane through x is median, i.e. each closed halfspace contains at least half the voter ideal points. McKelvey, Schofield (87) More general, finite, but exponential. Are there better conditions?

11 Recognizing a Stable (Undominated) Point is co-NP-complete
Theorem: Given x1…xn and x0 in <k, determining whether x0 is dominated is NP-complete. Proof: Johnson & Preparata 1978. Algorithm: In O(kn) given x_1…x_n can find x_0 which is undominated if any point is. Corollary: Majority-rule stability is co-NP-complete.

12 Implications Puts to rest efforts to find simpler necessary and sufficient conditions Computing the radius of the yolk is NP-hard Computing any other solution concept that coincides with Condorcet winner when it exists, is NP-hard


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