L β 2 k 1 2 β k 1 1 β 8 β kl 1 2 β y 2 2 y 2 1 2

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L β 2 K +(1 - 2 β ) k 1 ) 1 - β = 8 β KL - (1 - 2 β ) y 2 2 + y 2 ((1 - 2 β ) 2 y 2 2 +8 β L (1 - β )2 K 4 β L (1 - β )2 L β 2 K +(1 - 2 β ) 8 β KL - (1 - 2 β ) y 2 2 + y 2 ((1 - 2 β ) 2 y 2 2 +8 β L (1 - β )2 K 4 β L (1 - β ) = 8 β KL - (1 - 2 β ) y 2 2 + y 2 ((1 - 2 β ) 2 y 2 2 +8 β L (1 - β )2 K 4 β L 8 β (1- β )2 L 2 (1 - β )8 β KL +(1 - 2 β ) - (1 - 2 β ) y 2 2 + y 2 ((1 - 2 β ) 2 y 2 2 +8 β L (1 - β )2 K (1 - β ) Now you are thinking: “This looks so MESSY how the [email protected]#$ am I supposed to di ff erentiate this mess?” Answer: Envelope Theorem Part (c)- Solving for the general β case: Remember, that by the envelope theorem, MRT = MP 1 k 1 MP 2 k 2 = f 1 k 1 f 2 k 2 = α ( l 1 k 1 ) 1 - β 1 2 l 2 k 2 = α ( (1 - β )2 L β 2 K +(1 - 2 β ) k 1 ) 1 - β 1 2 β 2 L (1 - β )2 K +(1 - 2 β ) k 2 where I plugged in for l 1 and l 2 using (^) and (^^^^^) = α ( (1 - β )2 L β 2 K +(1 - 2 β )(2 K - k 2 ) ) 1 - β 1 2 β 2 L (1 - β )2 K +(1 - 2 β ) k 2 where k 2 = (1 - 2 β ) y 2 2 + y 2 ((1 - 2 β ) 2 y 2 2 +16 β (1 - β ) LK 4 β L . Long story short : MRT= dy 2 dy 1 = MP 1 k 1 MP 2 k 2 = f 1 k 1 f 2 k 2 (d) FROM HERE ON OUT ASSUME β = 1 2 First, an allocation ( x A 1 ,x A 2 , x B 1 ,x B 2 , y 1 ,y 2 ), gives agent B some utility. Let’s call it ¯ u . Thus, for the allocation to be Pareto optimal ( x A 1 ,x A 2 ) must maximize persons A’s utility and ( x A 1 , x A 2 , x B 1 ,x B 2 , y 1 ,y 2 ) must satisfy: 1. u B ( x B 1 , x B 2 ) = ¯ u 2. The feasibility constraints x A 1 + x B 1 = y 1 and x A 2 , + x B 1 = y 2 3. From (a) of this homework we know y 1 = Y 1 ( y 2 ) , so T ( y 1 , y 2 ) = y 1 - Y 1 ( y 2 ) = 0 Thus, the optimization problem is: U A u ) = max x A 1 ,x A 2 ,x B 1 ,x B 2 ,y 1 ,y 2 u A ( x A 1 , x A 2 ) subject to u B ( x B 1 , x B 2 ) = ¯ u 6
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x A 1 + x B 1 = y 1 x A 2 + x B 2 = y 2 T ( y 1 , y 2 ) = y 1 - Y 1 ( y 2 ) = 0 Thus, the Lagrangian is: L ( x A 1 , x A 2 , x B 1 , x B 2 , y 1 , y 2 , λ , μ 1 , μ 2 ) = u A ( x A 1 , x A 2 ) + λ u - u B ( x B 1 , x B 2 )) + μ 1 ( y 1 - ( x A 1 + x B 1 ) ) + μ 2 ( y 2 - ( x A 2 + x B 2 ))+ μ 3 ( Y 1 ( y 2 ) - y 1 ) Now for the FOCs: L x A 1 = 0, L x A 2 = 0, so u A x A 1 = μ 1 , u A x A 2 = μ 2 and (i) -MRS A = u A x A 1 u A x A 2 = μ 1 μ 2 L x B 1 = 0, L x B 2 = 0, so - λ u B x B 1 = μ 1 , - λ u B x B 2 = μ 2 and (ii) -MRS B = u B x B 1 u B x B 2 = μ 1 μ 2 . Thus, MRS A = MRS B But there is more: L y 1 = 0, so μ 1 = μ 3 and L y 2 = 0 = so - μ 3 Y 1 ( y 2 ) = μ 2 . So using μ 1 = μ 3 we get - μ 1 Y 1 ( y 2 ) = μ 2 and (iii) μ 1 μ 2 = - 1 Y 1 ( y 2 ) = -MRT So using (i), (ii), and (iii) we get: μ 1 μ 2 = -MRT = -MRS A = -MRS B So MRT = MRS A = MRS B and we are done! (e) Remember that β = 1 2 so MRT = -1. So using part (d) we get MRT = -1 = MRS A = - x A 2 x A 1 = - x B 2 x B 1 = MRS B and Let’s get rid of the - and square everything so we get 1= x A 2 x A 1 = x B 2 x B 1 = y 2 - x A 2 y 1 - x A 1 , using the feasibility constraints form part (d). Now from part (b) we know that the PPF equation is y 1 = 2 KL - y 2 , so we have: 1= x A 2 x A 1 = y 2 - x A 2 2 KL - y 2 - x A 1 . No we basically got it. Since 1= x A 2 x A 1 , then multiplying both sides by x A 1 we get x A 1 = x A 2 . Now let’s use this in: y 2 - x A 2 2 KL - y 2 - x A 1 = 1 If we plug in x A 1 = x A 2 , we get y 2 - x A 1 2 KL - y 2 - x A 1 = 1 So y 2 - x A 1 = 2 KL - y 2 - x A 1 Solving for y 2 in terms of the x term we get: 7
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y 2 = KL - x A 1 . Thus, y 1 = 2 KL - ( KL - x A 1 ) and x A 1 = x A 2 .
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