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Microeconomics Problem Set 1 NCKU Economics π€ 1. Suppose π₯π (π, π€) = πΏπ for π = 1, β― , πΏ. π (a.) Show that π₯(π, π€) is homogeneous of degree zero (hd0) in (π, π€) πΌπ€ π€ π₯π (πΌπ, πΌπ€) = = = π₯π (π, π€) πΏπΌππ πΏππ (b.) Show that π₯(π, π€) satisfies Walrasβ law (WL). πΏ π β π₯(π, π€) = β πΏ ππ π₯π (π, π€) = β π=1 π=1 ππ πΏ π€ π€ =β =π€ πΏππ π=1 πΏ (c.) Let πΏ = 2. Derive its Slutsky substitution matrix. βπ€ 4π2 π(π, π€) = [ π€1 4π1 π2 π€ 4π1 π2 βπ€ ] 4π22 2. For computing the change in consumption quantities between two periods, Laspeyres quantity index ( πΏπ = π0 β π₯1 /π0 β π₯0 ) uses period 0 prices as weights while Paasche quantity index (ππ = π1 β π₯1 /π1 β π₯0 PQ=p1x1/p1x0) uses period 1 prices as weights. Show that (a.) If πΏπ < 1, then the consumer has a revealed preference for π₯0 over π₯1 . If πΏπ < 1, π0 β π₯1 < π0 β π₯0 . When one can afford both π₯0 and π₯1 , and π₯0 is chosen, π₯0 is preferred to π₯1 . (b.) If ππ > 1, then the consumer has a revealed preference for π₯1 over π₯0 . If ππ > 1, π1 β π₯1 > π1 β π₯0 . When one can afford both π₯0 and π₯1 , and π₯1 is chosen, π₯1 is preferred to π₯0 . 3. Verify if one can derive π·π π₯(π, π€)π + π·π€ π₯(π, π€)π€ = 0 simply from ππ π·π π₯(π, π€) + π₯(π, π€)π = 0 and ππ π·π€ π₯(π, π€) = 1 . If not, discuss the necessary condition for the property to hold. ππ π·π π₯(π, π€)π + π₯(π, π€)π π = 0π π; ππ π·π π₯(π, π€)π + π€ = 0; π·π€ π₯(π, π€)ππ π·π π₯(π, π€)π + π·π€ π₯(π, π€)π€ = 0 is different from π·π π₯(π, π€)π + π·π€ π₯(π, π€)π€ = 0. In order to derive π·π π₯(π, π€)π + π·π€ π₯(π, π€)π€ = 0. We need π₯(π, π€) to be hd0, i.e. π₯(πΌπ¦) = π₯(π¦). π Let π¦ = ( ). Differentiate both sides of π₯(πΌπ¦) β‘ π₯(π¦) with respect to πΌ. π€ By chain rule, π·πΌπ¦ π₯(πΌπ¦)π¦ = 0 and π·π¦ π₯(π¦)π¦ = 0 when evaluated at Ξ± = 1. π·π¦ π₯(π¦)π¦ = (π·π π₯(π, π€) π π·π€ π₯(π, π€)) ( ) = π·π π₯(π, π€)π + π·π€ π₯(π, π€)π€. π€ Microeconomics Problem Set 1 NCKU Economics 4. Suppose π₯(π, π€) satisfies WL and hd0 in (π, π€). Show that WA holds if and only if it holds for any compensated price change. (Refer to p.31 of the textbook) a. WA holds β WA for compensated price change holds. Since WA holds for any change in price and income pair, certainly it will hold for any compensated price change. b. WA doesnβt hold β WA for compensated price change doesnβt hold. If WA doesnβt hold, for any change from (πβ² , π€β²) to (πβ²β² , π€ β²β² ) such that π₯(πβ² , π€β²) β π₯(πβ²β² , π€ β²β² ), we have πβ² β π₯(πβ²β² , π€ β²β² ) β€ π€β² and πβ²β² β π₯(πβ² , π€ β² ) β€ π€β²β². If one of the two weak inequalities holds with equality, it is indeed a violation of WA for compensated price change. Next focus on the case with πβ² β π₯(πβ²β² , π€ β²β² ) < π€β² and πβ²β² β π₯(πβ² , π€ β² ) < π€β²β². First, by above inequalities, there exists Ξ± β (0,1) such that (Ξ±πβ² + (1 β πΌ)πβ²β² ) β π₯(πβ² , π€ β² ) = (Ξ±πβ² + (1 β πΌ)πβ²β² ) β π₯(πβ²β² , π€ β²β² ) . (Note when πΌ = 0 , (Ξ±πβ² + (1 β πΌ)πβ²β² ) β π₯(πβ² , π€ β² ) < (Ξ±πβ² + (1 β πΌ)πβ²β² ) β π₯(πβ²β² , π€ β²β² ) and when πΌ = 1 , (Ξ±πβ² + (1 β πΌ)πβ²β² ) β π₯(πβ² , π€ β² ) > (Ξ±πβ² + (1 β πΌ)πβ²β² ) β π₯(πβ²β² , π€ β²β² )) Let πβ²β²β² = (Ξ±πβ² + (1 β πΌ)πβ²β² ), π€ β²β²β² = πβ²β²β² β π₯(πβ² , π€ β² ) = πβ²β²β² β π₯(πβ²β² , π€ β²β² ). Ξ±π€ β² + (1 β πΌ)πβ²β² > Ξ±πβ² β π₯(πβ² , π€ β² ) + (1 β πΌ)πβ²β² β π(πβ² , πβ² ) = πβ²β²β² β π₯(πβ² , π€ β² ) = π€ β²β²β² = πβ²β²β² β π₯(πβ²β²β² , π€ β²β²β² ) = Ξ±πβ² β π₯(πβ²β²β² , π€ β²β²β² ) + (1 β πΌ)πβ²β² β π₯(πβ²β²β² , π€ β²β²β² ) When π₯(πβ²β²β² , π€ β²β²β² ) β π₯(πβ² , π€ β² ) and π₯(πβ²β²β² , π€ β²β²β² ) β π₯(πβ²β² , π€ β²β² ) either π€ β² > πβ² β π₯(πβ²β²β² , π€ β²β²β² ) or π€ β²β² > πβ²β² β π₯(πβ²β²β² , π€ β²β²β² ) implies the violation of WA for compensated price change since π€ β²β²β² = πβ²β²β² β π₯(πβ² , π€ β² ) = πβ²β²β² β π₯(πβ²β² , π€ β²β² ). 5. Show that if π₯(π, π€) satisfies WA, it must be hd0 in (π, π€). π β π₯(π, π€) β€ π€ and πΌπ β π₯(πΌπ, πΌπ€) β€ πΌπ€ because π₯(π, π€) is affordable. Equivalently we have πΌπ β π₯(π, π€) β€ πΌπ€ and π β π₯(πΌπ, πΌπ€) β€ π€. If WA holds, then it must be that π₯(π, π€) = π₯(πΌπ, πΌπ€) so that π₯(π, π€) is hd0. 6. Show that for L=2, the Slutsky matrix, π(π, π€), is always symmetric. Since π(π, π€)π = 0 and ππ π(π, π€) = 0, we have [ π11 π21 [π1 π π +π π π12 π1 ] [ ]=[ 11 1 12 2 ]=0 π22 π2 π21 π1 + π22 π2 π2 ] [π11 π21 π12 ]=[π11 π1 + π21 π2 π22 π12 π1 + π22 π2 ]=0 Since the above two both are true, we must have S21=S12. Microeconomics Problem Set 1 NCKU Economics 7. Walrasian demand π₯(π, π€) is said to satisfy the uncompensated law of demand (ULD) if for any two prices, πβ² , πβ²β² and the given income, π€ β² , we have (πβ²β² β πβ² ) β (π₯(πβ²β² , π€ β² ) β π₯(πβ² , π€ β² )) β€ 0 , with strict inequality if π₯(πβ² , π€ β² ) β π₯(πβ²β² , π€ β² ). Show that if π₯(π, π€) satisfies WL, hd0, and ULD for any πβ² , πβ²β² , and π€ β² , it will also satisfy WA. (Refer to p.112 of the textbook) Take any (πβ² , π€ β² ) and (πβ²β² , π€ β²β² ) with π₯(πβ² , π€ β² ) β π₯(πβ²β² , π€ β²β² ) and πβ² β π₯(πβ²β² , π€ β²β² ) β€ π€ β² . To show WA, we have to show πβ²β² β π₯(πβ² , π€ β² ) > π€ β²β² . π€β² π€β² π€β² Let πβ²β²β² = π€β²β² πβ²β² , first note that π₯(πβ²β²β² , π€ β² ) = π₯ (π€β²β² πβ²β² , π€β²β² π€ β²β² ) = π₯(πβ²β² , π€ β²β² ) by hd0. By ULD, (πβ²β²β² β πβ² ) β (π₯(πβ²β²β² , π€ β² ) β π₯(πβ² , π€ β² )) β€ 0, or equivalently πβ²β²β² β π₯(πβ²β²β² , π€ β² ) β πβ²β²β² β π₯(πβ² , π€ β² ) β πβ² β π₯(πβ²β²β² , π€ β² ) + πβ² β π₯(πβ² , π€ β² ) β€ 0. Since πβ² β π₯(πβ²β² , π€ β²β² ) β€ π€ β² or equivalently βπβ² β π₯(πβ²β²β² , π€ β² ) + πβ² β π₯(πβ² , π€ β² ) β₯ 0, π€β² we must have πβ²β²β² β π₯(πβ²β²β² , π€ β² ) β πβ²β²β² β π₯(πβ² , π€ β² ) < 0, thus π€β² π€ β²β² πβ²β² β π₯(πβ²β² , π€ β²β² ) β πβ²β² β π₯(πβ² , π€ β² ) < 0 or πβ²β² β π₯(πβ² , π€ β² ) > π€ β²β² . 8. Let [ π€ β²β² π₯1 π₯ π₯ = [ 2] , π₯3 π1 (π₯) π₯1 + π₯2 π2 (π₯) π₯ +π₯ π(π₯) = = [ π₯2 + π₯3 ] , π3 (π₯) 1 3 π₯1 + π₯2 + π₯3 [ π4(π₯) ] π(π) = [ π1 (π) ]= π2 (π) βπ1 β π2 β π3 + 2π4 ]. Show that π·π₯ π(π(π₯)) = π·π π(π)π·π₯ π(π₯). π1 + π2 + π3 + π4 π(π₯) = [ 0 0 ]; π·π₯ π(π(π₯)) = [ 3π₯1 + 3π₯2 + 3π₯3 3 β1 β1 π·π π(π)π·π₯ π(π₯) = [ 1 1 1 β1 2 0 ][ 1 1 1 1 1 1 0 1 0 0 ]; 3 3 0 0 1 ]=[ 3 1 1 π₯1 9. Let π₯ = [π₯ ], and π(π₯) = π₯1 + π₯2 + π₯1 π₯2 . Show that 2 0 0 ];; 3 3 ππ(πΌπ₯) ππΌ = π·πΌπ₯ π(πΌπ₯)π₯ , where π·πΌπ₯ π(πΌπ₯) is the derivative of π(πΌπ₯) with respect πΌπ₯ evaluated at πΌπ₯ . Note that π(πΌπ₯) and π(π₯) are evaluating the same function at different values. π(πΌπ₯) = πΌπ₯1 + πΌπ₯2 + πΌ 2 π₯1 π₯2 ; ππ(πΌπ₯) ππΌ = π₯1 + π₯2 + 2πΌπ₯1 π₯2 ; Microeconomics Problem Set 1 NCKU Economics π¦1 Let π¦ = [π¦ ] = πΌπ₯, 2 π·πΌπ₯ π(πΌπ₯) = π·π¦ π(π¦) = [1 + π¦2 π·πΌπ₯ π(πΌπ₯)π₯=[1 + πΌπ₯2 1 + π¦1 ] = [1 + πΌπ₯2 1 + πΌπ₯1 ]; π₯ 1 + πΌπ₯1 ] [π₯1 ]=π₯1 + π₯2 + 2πΌπ₯1 π₯2 2