Expenditure Minimization and Duality

This scene finds the cheapest way to reach a target utility.

Now the problem runs in reverse. Fix the satisfaction level first, then shrink the cost. The cheapest bundle sits where an isocost line touches the curve. Turn on the mirror and it lands exactly on the earlier optimum. Two problems mirroring each other is duality.

For example, you want the satisfaction of 🍞 bread, two slices, and 🥛 milk, one glass. You shop for that satisfaction at the lowest cost. To save 💰 money, you shift toward the cheaper good.

min⁡xp⋅xs.t.u(x)≥u\min_x p \cdot x \quad \text{s.t.} \quad u(x) \ge u
h(p,u)=x(p,e(p,u)),x(p,w)=h(p,v(p,w))h(p, u) = x(p, e(p, u)), \qquad x(p, w) = h(p, v(p, w))

Part of these learning paths

Examples & Exercises

Differentiate the Cobb-Douglas expenditure function e(p,u)=u(p1a)a(p21−a)1−ae(p, u) = u\left(\frac{p_1}{a}\right)^a\left(\frac{p_2}{1-a}\right)^{1-a} in p1p_1 to verify Shephard's lemma, the formula below, directly.

e(p,u)=u(p1a)a(p21−a)1−ae(p, u) = u \left(\frac{p_1}{a}\right)^a \left(\frac{p_2}{1-a}\right)^{1-a}
h1=∂e/∂p1h_1 = \partial e/\partial p_1
a=1/2a = 1/2, p1=2p_1 = 2, target utility u=6u = 6.
Show proof

  1. Differentiating in p1p_1 brings down the exponent aa, which cancels: ∂e/∂p1=u(p1a)a−1(p21−a)1−a\partial e/\partial p_1 = u \left(\frac{p_1}{a}\right)^{a-1} \left(\frac{p_2}{1-a}\right)^{1-a}.

  2. Hicksian demand is h1=ea/p1h_1 = ea/p_1. Substituting ee gives h1=u(p1a)a−1(p21−a)1−ah_1 = u\left(\frac{p_1}{a}\right)^{a-1}\left(\frac{p_2}{1-a}\right)^{1-a} — matching the derivative.

Verify to full precision that the identity below holds in the preset numbers, and explain why this is the content of duality.

e=6⋅41/2⋅21/2=68≈16.97e = 6 \cdot 4^{1/2} \cdot 2^{1/2} = 6\sqrt{8} \approx 16.97
h(p,u)=(eap1,e(1−a)p2)≈(4.24,8.49)h(p, u) = \left(\frac{ea}{p_1}, \frac{e(1-a)}{p_2}\right) \approx (4.24, 8.49)
h(p,u)=x(p,e(p,u))h(p, u) = x(p, e(p, u))
UMP mirror on — the two points coincide.
Show proof

  1. The expenditure-minimizing bundle is h≈(4.24,8.49)h \approx (4.24, 8.49). Solving the UMP with wealth w=e≈16.97w = e \approx 16.97 gives Walrasian demand x(p,w)=(aw/p1,(1−a)w/p2)≈(4.24,8.49)x(p, w) = (aw/p_1, (1-a)w/p_2) \approx (4.24, 8.49) — the same point.

  2. That is duality: achieving target utility uu at minimum cost has the same solution as maximizing utility with that minimum cost as wealth. In the figure the dashed circle (UMP mirror) sits exactly on the solid dot h∗h^*.

4. Classical Demand Theory