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# F5-conf-01 — Competitive equilibrium of a 2-agent, 2-good Cobb–Douglas exchange economy

# Question

Find the Walrasian equilibrium relative price p = p1/p2 (good 2 is the numeraire, p2 = 1) and the equilibrium allocation. Check that the good-1 market clears and that Walras' law holds. Then check the closed-form price against a separate numerical root of the excess-demand function.

# Data and exclusions

  • data/endowments.csv (2 rows): A = (8, 2), B = (4, 8).
  • data/parameters.csv (4 rows): alpha_A = 0.4, alpha_B = 0.6, and summary-sheet aggregates of 14 (good 1) and 10 (good 2).
  • Data issue: the summary sheet gives 14 as the aggregate endowment of good 1, but the individual endowments add up to 8 + 4 = 12. The good-2 aggregate matches (2 + 8 = 10). In an exchange economy, aggregate supply is by definition the sum of the individual endowments, and endowments.csv is declared the primary data. I therefore used 12 and treat the 14 as a transcription error in the summary sheet. If 14 were used as supply, market clearing would be impossible with these endowments: excess demand for good 1 at the equilibrium price would be −2 (reported in results.json as sensitivity_excess_demand_good1_vs_summary_sheet_aggregate).
  • No rows were excluded. Both agents are analysed (n_analysed = 2).

# Method

  • With Cobb–Douglas preferences, agent i's income is m_i = p·e1_i + e2_i. Demands are x1_i = α_i m_i / p and x2_i = (1−α_i) m_i.
  • Market clearing for good 1, Σ_i α_i(p e1_i + e2_i) = p Σ_i e1_i, gives the closed form p* = Σ_i α_i e2_i / Σ_i (1−α_i) e1_i = (0.4·2 + 0.6·8)/(0.6·8 + 0.4·4) = 5.6/6.4 = 0.875.
  • Assumptions behind the closed form:
    • α_i ∈ (0,1) and all endowments strictly positive (both checked in code). This means every agent has positive income at any p > 0, demands are interior, and the excess demand is continuous with a unique zero.
    • Prices are normalised with p2 = 1, so only relative prices are determined.
    • Aggregate endowments are the sums of the individual endowments.
  • Numerical validation: Brent's method (scipy.optimize.brentq) finds the root of the aggregate excess demand z1(p) on the bracket [1e-8, 1e8]. It works directly on the demand functions and does not use the closed-form formula.
  • Walras' law (p·z1 + z2 = 0) is checked at p*. It is also checked as an identity on a grid of 41 log-spaced prices in [0.01, 100], away from equilibrium.

# Results

quantity value
p1/p2 (closed form) 0.875
p1/p2 (numerical root) 0.875 (abs. difference 2.2e-16)
x1_A, x2_A 4.1143 (= 144/35), 5.4
x1_B, x2_B 7.8857 (= 276/35), 4.6
excess demand good 1 at p* 0.0
excess demand good 2 at p* 0.0
Walras' law residual at p* 0.0
max abs Walras residual on off-equilibrium grid 5.7e-14 (floating-point rounding)
aggregate endowments used 12 (good 1), 10 (good 2)

The allocation is feasible: 4.1143 + 7.8857 = 12 and 5.4 + 4.6 = 10. Agent A is endowed mostly with good 1 and prefers good 2, so A sells good 1. Agent B does the reverse.

# Uncertainty

This is a deterministic theoretical exercise, so there is no sampling uncertainty and no random seed. The only error is floating-point rounding (about 1e-16 for the price, about 1e-14 for the Walras identity on the grid). The main uncertainty about the answer is the input inconsistency described above. The result rests on treating the individual endowments as authoritative.

# Limitations and open issues

  • The good-1 aggregate inconsistency (14 vs 12) was resolved by assumption. If the individual endowments are wrong instead (for example, if one agent really holds 2 more units of good 1), the equilibrium would be different. That case cannot be identified from the data provided.
  • The numerical check covers only the single-market root of z1 on a positive-price bracket. Because of Walras' law, clearing of good 1 implies clearing of good 2. Uniqueness here follows from gross substitutability of Cobb–Douglas demands. It was not checked separately by a global search.
  • The closed form is valid only for interior Cobb–Douglas solutions with strictly positive endowments, under the p2 = 1 normalisation.
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