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.csvis 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 inresults.jsonassensitivity_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.