Algebraic Approximations (@cite{luce-1959}, §1.G, pp. 34–37) @cite{luce-1959} #
@cite{luce-1959} develops the connection between choice probabilities and ordinal preference structures via just noticeable differences (jnds).
When stimuli are "close" in value, subjects cannot reliably discriminate
between them — the choice probability P(x, {x,y}) is near 1/2. A jnd
threshold π ∈ (1/2, 1) defines two relations on the alternative set:
- L(π):
xis discriminably preferred toy(Definition 3) - I(π):
xandyare indistinguishable (Definition 3)
Key results #
Semiorder (Theorem 5): Under Axiom 1 with imperfect discrimination,
(L(π), I(π))satisfies the semiorder axioms — trichotomy, I-reflexivity, L-transitivity, and the interval conditionxLy ∧ yIz ∧ zLw → xLw.Trace (Definition 4):
x ≥_T yiffP(x, z) ≥ P(y, z)for allz. The trace extracts the "underlying" preference by requiring dominance in all pairwise comparisons against any third alternative.Weak order (Theorem 6): Under Axiom 1, the trace is a weak order (total preorder), and
x ≥_T yiffv(x) ≥ v(y)iffP(x, y) ≥ 1/2.
Connection to the choice axiom #
The semiorder captures the observable preference structure (what a subject
can discriminate), while the trace recovers the latent ratio scale ordering.
Theorem 6 shows that Axiom 1 forces these to align: the trace is exactly the
ordering induced by the scale values v.
The pairwise choice probability P(x, {x,y}) under a ratio scale v:
P(x, y) = v(x) / (v(x) + v(y)).
This is the Luce model prediction for binary forced choice.
The function is symmetric in the sense that P(x,y) + P(y,x) = 1.
Equations
- Core.pairwiseProb v x y = v x / (v x + v y)
Instances For
Pairwise probabilities are non-negative for positive scales.
Pairwise probabilities are at most 1 for positive scales.
Complementarity: P(x, y) + P(y, x) = 1 for positive scales.
P(x, x) = 1/2 for positive scales (indifference with self).
P(x, y) > 1/2 iff v(x) > v(y): the higher-scale alternative is
chosen more than half the time.
P(x, y) ≥ 1/2 iff v(x) ≥ v(y).
Monotonicity: P(x, z) ≥ P(y, z) iff v(x) ≥ v(y).
The function t ↦ t/(t+c) is monotone increasing for c > 0,
so the ordering of pairwise probabilities against any fixed z
mirrors the ordering of scale values.
The L(π) relation (Definition 3, @cite{luce-1959}, p. 34):
x L(π) y iff P(x, {x,y}) > π.
This means x is discriminably preferred to y at threshold π:
the observer can reliably tell that x is "better" than y. The
threshold π must satisfy 1/2 < π < 1; it represents the minimum
probability that constitutes a "noticeable difference."
Equations
- Core.jndL v thr x y = (thr < Core.pairwiseProb v x y)
Instances For
The I(π) relation (Definition 3, @cite{luce-1959}, p. 34):
x I(π) y iff 1 - π ≤ P(x, {x,y}) ≤ π.
This means x and y are indistinguishable at threshold π:
neither is reliably discriminated from the other. By complementarity,
x I(π) y iff 1 - π ≤ P(x,y) ≤ π iff 1 - π ≤ P(y,x) ≤ π,
so I is symmetric.
Equations
- Core.jndI v thr x y = (1 - thr ≤ Core.pairwiseProb v x y ∧ Core.pairwiseProb v x y ≤ thr)
Instances For
Semiorder axioms #
@cite{luce-1959} defines a semiordering of a set U as a pair
(L, I) of relations satisfying, for all x, y, z, w ∈ U:
(i) Trichotomy: exactly one of xLy, yLx, or xIy holds
(ii) I-reflexivity: xIx
(iii) Interval condition: xLy ∧ yIz ∧ zLw → xLw
(iv) No sandwiching: xLy ∧ yLz → ¬(xIw ∧ wIz)
Theorem 5 proves these hold for (L(π), I(π)) under Axiom 1.
I(π) is symmetric: if x and y are indistinguishable, so are y and x.
Since P(y,x) = 1 - P(x,y), the condition 1-π ≤ P(x,y) ≤ π
is equivalent to 1-π ≤ P(y,x) ≤ π.
I-reflexivity: x I(π) x, since P(x, x) = 1/2 and 1-π < 1/2 < π
whenever 1/2 < π < 1.
Trichotomy: for any x, y, exactly one of xLy, yLx, or xIy holds.
Since P(x,y) + P(y,x) = 1, the three conditions P(x,y) > π,
P(y,x) > π (i.e., P(x,y) < 1-π), and 1-π ≤ P(x,y) ≤ π
partition the interval [0, 1].
L-transitivity: xLy ∧ yLz → xLz.
Under Axiom 1, P(x,y) > π means v(x)/(v(x)+v(y)) > π, i.e.,
v(x)/v(y) > π/(1-π). If also v(y)/v(z) > π/(1-π), then
v(x)/v(z) > (π/(1-π))² > π/(1-π) (since π/(1-π) > 1),
so P(x,z) > π.
Interval condition: xLy ∧ yIz ∧ zLw → xLw.
Under Axiom 1: xLy gives v(x)/v(y) > π/(1-π), yIz gives
v(y)/v(z) ≥ (1-π)/π (from P(y,z) ≥ 1-π), and zLw gives
v(z)/v(w) > π/(1-π). Multiplying the first and third ratios and
using the bound on v(y)/v(z):
v(x)/v(w) = (v(x)/v(y)) · (v(y)/v(z)) · (v(z)/v(w)) > π/(1-π)
since the middle factor is ≥ (1-π)/π and the outer factors are > π/(1-π),
giving a product > (π/(1-π))·((1-π)/π)·(π/(1-π)) = π/(1-π).
No sandwiching: xLy ∧ yLz → ¬(xIw ∧ wIz).
If v(x) ≫ v(y) ≫ v(z) (both with ratio > π/(1-π)), then no
w can be indistinguishable from both x and z: such a w
would need v(w) ≈ v(x) and v(w) ≈ v(z) simultaneously, but
v(x)/v(z) > (π/(1-π))² ≫ 1 prevents this.
The trace relation (Definition 4, @cite{luce-1959}, p. 37):
x ≥_T y iff P(x, z) ≥ P(y, z) for all z.
The trace extracts the "underlying" preference ordering by requiring
that x is at least as preferred as y in every pairwise
comparison against a common reference z. This is a stronger condition
than just P(x, y) ≥ 1/2.
Equations
- Core.traceGe v x y = ∀ (z : A), Core.pairwiseProb v y z ≤ Core.pairwiseProb v x z
Instances For
Theorem 6: Under Axiom 1, the trace relation
is equivalent to v(x) ≥ v(y).
Proof sketch: Under Axiom 1, P(x,z) = v(x)/(v(x)+v(z)). Since
t ↦ t/(t+c) is monotone increasing for c > 0, we have
P(x,z) ≥ P(y,z) for all z iff v(x) ≥ v(y).
(→) Take z = y: P(x,y) ≥ P(y,y) = 1/2, hence v(x) ≥ v(y).
(←) If v(x) ≥ v(y), monotonicity of t/(t+c) gives P(x,z) ≥ P(y,z).
Corollary: x ≥_T y iff P(x, y) ≥ 1/2.
The trace is reflexive: x ≥_T x.
The trace is transitive: x ≥_T y ∧ y ≥_T z → x ≥_T z.
The trace is total: for any x, y, either x ≥_T y or y ≥_T x.
This completes the proof that the trace is a weak order (total preorder). Under Axiom 1, the trace is determined by the ratio scale values, which are totally ordered reals.
The trace agrees with L: if xLy for any π, then x ≥_T y.
P(x,y) > π > 1/2 implies v(x) > v(y) implies x ≥_T y.