ECON 4010: Advanced Microeconomic Theory
Final Exam answers.
1. Not necessarily. Exception: corner solutions.
2. False. Indirect utility function is homogenous of degree zero. If prices and income
double, the budget constraint and utility remains the same.
3. False. For inferior goods it is the other way round.
4. False. As wealth increases, marginal utility of potential loss from the gamble falls
but so does the marginal utility of winning the gamble. So the net result is
5. True if non-credible threat or sub-game-imperfect equilibrium is ruled out.
Example battle of sexes. If the wife moves first, boxing/boxing is not a sub-game
6. False. Marginal products for linearly homogenous functions are inversely related
to the relative amount of the factor used.
7. True. The Shepard’s lemma is used to derive the conditional factor demand
functions, which allow1/2s to find the relationship between inputs and output.
Example: C = 2(wv) q. Taking the partial derivatives wrt w & v gives the
conditional factor demands: L = (v/w) q, K = (v/w) q. Thus, L/q = q/K or q =
8. True. Follows from the envelope theorem (Hotelling lemma): the partial
derivative of the profit function with respect to product price gives the product
9. True only for constant cost industry (firm’s cost curves are independent of the
number of firms).
10. True. Follows from the fact that MC = MR = P( 1 + 1/e ).
11. Not necessarily. Exception: first degree price discrimination.
12. True. P = MC under Bertrand duopoly when firm’s product and cost are identical
but P > MC under Cournot duopoly.
1. a) 2,2 and 1,1
b) Let q (1-q) be the probability column plays left (right).
EU (row) for top = q (2) + (1-q) (-1) = 3q – 1 (i)
EU (row) for bottom = q (-1) + (1-q) (1) = 1 – 2q (ii)
Equating (i) and (ii): q* = 2/5
Similraly, let p (1-p) be the probability row plays top (bottom)
EU (column) for left = p (2) + (1-p) (-1) = 3p – 1 (iii)
EU (column) for right = p (-1) + (1-p) (1) = 1- 2p (iv)
Equating (iii) and (iv), p* = 2/5
c) Top, left 2. a) Profit Π = P ( L + K ) – wL – vK
FOC for maximizing Π wrt L and K:
∂ Π/∂ L = P ½L -1/– w = 0 or L = (p/(2w)) 2
∂ Π/∂ K = P ½K – v = 0 or K = (p/(2v)