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Exercise 9.6

Exercise 9.6. MICROECONOMICS Principles and Analysis Frank Cowell. February 2007. Ex 9.6(1): Question. purpose : to derive equilibrium prices and incomes as a function of endowment. To show the limits to redistribution within the GE model for a alternative SWFs

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Exercise 9.6

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  1. Exercise 9.6 MICROECONOMICS Principles and Analysis Frank Cowell February 2007

  2. Ex 9.6(1): Question • purpose: to derive equilibrium prices and incomes as a function of endowment. To show the limits to redistribution within the GE model for a alternative SWFs • method: find price-taking optimising demands for each of the two types, use these to compute the excess demand function and solve for r

  3. Ex 9.6(1): budget constraints • Use commodity 2 as numéraire • price of good 1 is r • price of good 2 is 1 • Evaluate incomes for the two types, given their resources: • type a has endowment (30, k) • therefore ya= 30r + k • type b has endowment (60, 210 k) • therefore yb= 60r + [210 k] • Budget constraints for the two types are therefore: • rx1a +x2a≤ 30r + k • rx1b +x2b ≤ 60r + [210 k]

  4. Ex 9.6(1): optimisation • We could jump straight to a solution • utility functions are simple… • …so we can draw on known results • Cobb-Douglas preferences imply • indifference curves do not touch the origin… • …so we need consider only interior solutions • also demand functions for the two commodities exhibit constant expenditure shares • In this case (for type a) • coefficients of Cobb-Douglas are 2 and 1 • so expenditure shares are ⅔ and ⅓ • (and for b they will be ⅓ and ⅔ ) • gives the optimal demands immediately… Jump to “equilibrium price”

  5. Ex 9.6(1): optimisation, type a • The Lagrangean is: • 2log x1a + log x2a + na[ya rx1a x2a ] • where nais the Lagrange multiplier • and ya is 30r + k • FOC for an interior solution • 2/x1a nar = 0 • 1/x2a na = 0 • ya rx1a x2a= 0 • Eliminating na from these three equations, demands are: • x1a= ⅔ ya / r • x2a = ⅓ ya

  6. Ex 9.6(1): optimisation, type b • The Lagrangean is: • log x1b + 2log x2b + nb[yb rx1b x2b ] • where nbis the Lagrange multiplier • and yb is 60r + 210 k • FOC for an interior solution • 1/x1b nbr = 0 • 2/x2b nb = 0 • yb rx1b x2b= 0 • Eliminating nb from these three equations, demands are: • x1b= ⅓ yb / r • x2b = ⅔yb

  7. Ex 9.6(1): equilibrium price • Take demand equations for the two types • substitute in the values for income • type-a demand becomes • type-b demand becomes • Excess demand for commodity 2: • [10r+⅓k]+[40r +140− ⅔k]−210 • which simplifies to 50r− ⅓k−70 • Set excess demand to 0 for equilibrium: • equilibrium price must be: • r= [210 + k] / 150

  8. Ex 9.6(2): Question and solution • Incomes for the two types are resources: • ya= 30r + k • yb= 60r + [210 k] • The equilibrium price is: • r= [210 + k] / 150 • So we can solve for incomes as: • ya= [210 + 6k] / 5 • yb= [1470 3k] / 5 • Equivalently we can write yaand ybin terms of ras • ya= 180r 210 • yb= 420  90r

  9. Ex 9.6(3): Question • purpose: to use the outcome of the GE model to plot the “income-possibility” set • method: plot incomes corresponding to extremes of allocating commodity 2, namely k = 0 and k = 210. Then fill in the gaps.

  10. Income possibility set yb • incomes for k = 0 • incomes for k = 210 • incomes for intermediate values of k • attainable set if income can be thrown away 300 • (42, 294) • yb = 315  ½ya 200 • (294, 168) 100 ya 300 0 100 200

  11. Ex 9.6(4): Question • purpose: find a welfare optimum subject to the “income-possibility” set • method: plot contours for the function W on the previous diagram.

  12. Welfare optimum: first case yb • income possibility set • Contours of W = log ya + log yb • Maximisation of W over income-possibility set 300 • W is maximised at corner • incomes are (294, 168) • here k = 210 • so optimum is where all of resource 2 is allocated to type a 200 • 100 ya 300 0 100 200

  13. Ex 9.6(5): Question • purpose: as in part 4 • method: as in part 4

  14. Welfare optimum: second case yb • income possibility set • Contours of W = ya + yb • Maximisation of W over income-possibility set 300 • again W is maximised at corner • …where k = 210 • so optimum is where all of resource 2 is allocated to type a 200 • 100 ya 300 0 100 200

  15. Ex 9.6: Points to note • Applying GE methods gives the feasible set • Limits to redistribution • natural bounds on k • asymmetric attainable set • Must take account of corners • Get the same W-maximising solution • where society is averse to inequality • where society is indifferent to inequality

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