Section 5

Walrasian Equilibrium

What we're solving for

A Walrasian equilibrium is a price vector $(p_1, p_2)$ and an allocation $(x^A, x^B)$ such that two conditions hold simultaneously. First, each agent chooses the bundle that maximises utility on their budget set — no one can do better given prices. Second, markets clear: $x^A_1 + x^B_1 = \Omega_1$ and $x^A_2 + x^B_2 = \Omega_2$ . Only the relative price of the two goods pins down behaviour, so we normalise $p_2 = 1$ throughout — good 2 serves as the numéraire.

Wealth from endowment

Each agent's wealth is the market value of what they bring to the exchange. With $p_2 = 1$ , agent $i$ 's wealth is:

m^i = p_1 \, \omega^i_1 + \omega^i_2

Cobb-Douglas demands

Agent $i$ has preferences $u^i(x_1, x_2) = x_1^{\alpha_i} x_2^{1-\alpha_i}$ with $\alpha_i \in (0,1)$ . Solving $\max u^i$ subject to the budget constraint $p_1 x_1 + x_2 = m^i$ yields closed-form demands via the Lagrangian method.

Step 1
$\mathcal{L} = x_1^{\alpha_i} x_2^{1-\alpha_i} - \lambda \, (p_1 x_1 + x_2 - m^i)$
Lagrangian for utility maximisation subject to the budget constraint.
Step 2
$\alpha_i \, x_1^{\alpha_i - 1} x_2^{1-\alpha_i} = \lambda p_1, \quad (1-\alpha_i) \, x_1^{\alpha_i} x_2^{-\alpha_i} = \lambda$
First-order conditions. Dividing eliminates $\lambda$ and reproduces $\mathrm{MRS}^i = p_1$ .
Step 3
$x^i_1(p_1, m^i) = \frac{\alpha_i \, m^i}{p_1}, \quad x^i_2(p_1, m^i) = (1-\alpha_i) \, m^i$
Cobb-Douglas spends a constant fraction of wealth on each good — the $\alpha_i$ goes to good 1, the rest to good 2.

Market clearing

Substitute the Cobb-Douglas demands into the good-1 market-clearing condition. Walras's Law guarantees that if good 1 clears, good 2 clears automatically, so we need only solve one equation for $p_1$ .

Step 1
$x^A_1(p_1, m^A) + x^B_1(p_1, m^B) = \Omega_1$
Total demand equals total endowment.
Step 2
$\frac{\alpha_A (p_1 \omega^A_1 + \omega^A_2)}{p_1} + \frac{\alpha_B (p_1 \omega^B_1 + \omega^B_2)}{p_1} = \Omega_1$
Substitute demands and wealth.
Step 3
$\alpha_A (p_1 \omega^A_1 + \omega^A_2) + \alpha_B (p_1 \omega^B_1 + \omega^B_2) = p_1 (\omega^A_1 + \omega^B_1)$
Multiply through by $p_1$ .
Step 4
$p_1^* = \frac{\alpha_A \, \omega^A_2 + \alpha_B \, \omega^B_2}{(1-\alpha_A) \, \omega^A_1 + (1-\alpha_B) \, \omega^B_1}$
Collect $p_1$ terms and solve. Numerator is the value of good 2 endowments, weighted by tastes for good 1; denominator is the value of good 1 endowments, weighted by tastes for good 2.

The equilibrium allocation

With $p_1^*$ in hand, plug it back into the demand functions to read off where each agent ends up. Agent A's equilibrium bundle is:

x^{A*}_1 = \frac{\alpha_A (p_1^* \omega^A_1 + \omega^A_2)}{p_1^*}, \quad x^{A*}_2 = (1-\alpha_A)(p_1^* \omega^A_1 + \omega^A_2)

x^{B*}_1 = \frac{\alpha_B (p_1^* \omega^B_1 + \omega^B_2)}{p_1^*}, \quad x^{B*}_2 = (1-\alpha_B)(p_1^* \omega^B_1 + \omega^B_2)

Each agent spends exactly their $\alpha_i$ share of wealth on good 1 and the remaining $(1-\alpha_i)$ share on good 2, evaluated at the equilibrium price. The market-clearing conditions are satisfied by construction.

What happens in the box

In the Edgeworth box, the equilibrium allocation sits at the unique point where three objects coincide. The budget line through the endowment point has slope $-p_1^*$ ; both agents' indifference curves are tangent to that line at the same point; and that tangency lies on the contract curve, where the two agents' marginal rates of substitution are equal. The uniqueness here follows directly from the linear expenditure structure of Cobb-Douglas preferences — each agent's demand is a linear function of wealth, so the aggregate excess demand for good 1 is strictly decreasing in $p_1$ , guaranteeing exactly one market-clearing price. The next section animates this: drag the sliders and watch the budget line, the indifference curves, and the contract-curve tangency all move together.