A NOTE ON CES FUNCTIONS Drago Bergholt, BI Norwegian Business School 2011

Transcription

1 A NOTE ON CES FUNCTIONS Drago Bergholt, BI Norwegan Busness School 2011 Functons featurng constant elastcty of substtuton CES are wdely used n appled economcs and fnance. In ths note, I do two thngs. Frst, I derve a number of condtons such as the optmal demand schedule when aggregaton technology s CES. Second, I show how the CES functon nests some partcular functonal forms as specal cases. Consder a general CES functon: 1 PRELIMINARIES Y = A J α 1 1 Y =1 1, 1 where A, > 0, α, Y > 0, and =1 α = 1. In models wth monopolstc competton between frms, Y s nterpreted as aggregate output and {Y } J =1 as output by ntermedate frms. In turn, the producton technology of frms can also have a CES structure, wth each Y beng referred to as an nput factor such as labor or captal or land. Y can also be aggregate demand or consumpton, a composte of ndvdual consumer classes or goods. In open economy macroeconomcs, one uses the CES structure to dstngush between domestcally produced goods and mports. The scalng parameter A s often normalzed to 1, but n the context of producton theory, t s typcally nterpreted as a measure of total factor productvty whch can be stochastc n a tme seres context. 2 IMPLICATIONS OF CES TECHNOLOGY What are the mplcatons of assumng a CES technology for aggregaton n economc models? To shed lght on ths queston, we start wth a maxmzaton problem. The problem s to choose the vector {Y } J =1 that maxmzes Y subect to some budget constrant P Y Z, 2 =1 where Z s total money spent. Let us state the Lagrangan where Λ s the multpler for the constrant: J 1 J L = A Λ P Y Z α 1 1 Y =1 =1 The necessary frst order condton s A 1 1 Y J, so we can wrte 1 α Y 1 = ΛP. Ths must hold for all Y = α α P P 1 Y, 3

2 2.1 A NATURAL PRICE INDEX Next, we derve an exact prce ndex for a unt ncrease n Y. Plug 3 nto the budget constrant and solve for Y : Z = Y = =1 P Y = Y P α Zα P =1 α P 1 Insert ths result nto Y now wth ndexng : =1 α P 1 Y = A α Y =1 J = A = AZ = AZ =1 J =1 J =1 α 1 Zα P =1 α P 1 1 α P 1 α P =1 1 α P Fnally, defne P as the expendture needed to buy one unt of Y,.e. P Z Y =1. From the expresson above, t naturally follows that P = 1 A Thus, P s the natural prce ndex for Y. J =1 α P THE BUDGET CONSTRAINT REVISITED Next, note that nsertng 3 nto 2 gves Z = = P Y =1 =1 P α α P P Y = P Y Y = α P P A 1 P 1 α Z A 1 P. 5 2

3 Substtute nto 1 and solve for Z. The result s Y = A α 1 P α P =1 Z A 1 P 1 Thus, Z = P Y, and we can wrte the budget constrant as P Y = P Y. =1 1 = ZP THE OPTIMAL DEMAND SCHEDULE Fnally, substtuton of Z = P Y nto 5 whch holds, J gves the optmal demand for Y : P Y Y = α 6 P A 1 One can also get rd of A and end up wth Y = α P J =1 α P α 1 1 Y =1 1. Thus, optmal demand for Y n 6 s not determned by A, but rather by the prce P relatve to all other prces, as well as by total factor demand. Equaton 6 s wdely used as demand functon n theoretcal models of optmzng economc behavor. 2.4 ELASTICITIES To understand why the functonal form n 1 exhbts CES between any par Y, Y, we defne the elastcty of substtuton as the percentage rse n Y Y followng a one percent rse n the relatve prce P P. It s clear from 3 that Y Y = α P, α P so we can wrte By dvdng by Y Y / P Y Y P P = α P α P 1. P, a constant elastcty of substtuton emerges: Elastcty of substtuton Y Y Y / = P P P / P Y Elastcty of ncome Y /Y Z/Z = α α P 1 α P α P 1 α P α P What about the ncome elastcty? Earler we defned Z as the ncome spent on Y. The ncome elastcty follows from 5: P 1 P A 1 P 3 P 1 A 1 P = 1 =

4 3 SPECIAL CASES OF THE CES FUNCTION Next, we show how the CES functon nests as specal cases some partcular functonal forms. Before we start, t s useful to note that equaton 1 s equvalent to Y = A J =1 α Ỹ γ 1 γ, 7 where Ỹ Y α and γ 1. Our task s to fnd an expresson for Y when or γ 1 lnear, 1 or γ 0 Cobb-Douglas, and 0 or γ Leontef. For reasons that wll be clear below, t s convenent to work wth a log transformaton of equaton 7: ln Y = ln A + 1 γ ln J =1 α Ỹ γ = ln A + F γ G γ, 8 J where F γ ln =1 α Ỹ γ and G γ γ. Fnally, n order to derve the Cobb- Douglas and Leontef functons we need L Hôptal s rule: The L Hôptal s rule: Consder two functons F γ and G γ, whch are dfferentable on an nterval a, b, except possbly at a pont c a, b. Suppose lm γ c F γ = lm γ c G γ = 0 or ±, lm γ c F γ G γ exsts, and G γ 0 γ c a, b. Then F γ lm γ c G γ = lm F γ γ c G γ. In our case the frst dervatves are F γ = F γ G γ = =1 α Ỹ γ lnỹ =1 α Ỹ γ and G γ = 1, mplyng J =1 α Ỹ γ Ỹ ln =1 α. 9 Ỹ γ Next, we consder each of the three specal cases of the CES functon. 3.1 THE LINEAR CASE γ 1 Ths one s trval as t follows mmedately from equaton 7 that lm Y = A Y. 10 γ 1 =1 4

5 3.2 THE COBB-DOUGLAS CASE γ 0 Here we want to evaluate equaton 8 when γ 0. However, the result s J ln =1 α Ỹ γ 0 lm ln Y = ln A + lm = ln A +. γ 0 γ 0 γ 0 The last term holds as long as =1 α = 1. 1 To proceed, we apply L Hôptal s rule. Evaluatng equaton 9 n the lmt as γ 0, we can wrte 8 as or lm ln Y = ln A + lm γ 0 γ 0 = ln A + = ln A J =1 =1 Ỹ α =1 α Ỹ γ ln Ỹ =1 α Ỹ γ α ln Ỹ lm Y = A 1 γ 0 J =1 αα, J =1 Y α. 11 Ths s the wdely used Cobb-Douglas functon. The scalng parameter A s often normalzed so that A J 1 =1 αα = THE LEONTIEF CASE γ When γ, then equaton 8 becomes J ln =1 α Ỹ γ lm ln Y = ln A + lm γ γ γ Agan, usng L Hôptal s rule, we can wrte Ỹ m mn {Ỹ1,..., ỸJ lm ln Y = ln A + lm γ γ = ln A + =1 α Ỹ γ ln Ỹ =1 α Ỹ γ lnỹ =1 α Ỹ γ.. =1 However, a new problem arses because lm α Ỹ γ γ = 0 0, whch cannot be solved by multple } uses of L Hôptal s rule. 2 In order to proceed, we defne. Dvdng the nomnator and denomnator by Ỹ m, γ we get lm ln Y = ln A + lm γ γ =1 α Ỹ Ỹ m γ ln Ỹ =1 α Ỹ 1 If nstead =1 α > 1 < 1, then the last term goes to as γ 0. wll always show up both n the nomnator and n the denomnator. 2 Ỹ γ 5 Ỹ m γ

6 = ln A + ln = ln mn {Ỹ1,..., ỸJ }, A mn {Ỹ1,..., ỸJ } or { lm Y = A mn Y1,..., Y } J. 12 γ α 1 α J Ths s the Leontef functon. 3.4 FROM ISOELASTIC UTILITY TO LOG UTILITY Consder the soelastc utlty functon U = C1 σ 1 1 σ, wth C > 0. Apparently a 0 0 -ssue arses when one evaluates U as σ 1. In that case one can defne F σ C 1 σ 1 and G σ 1 σ. Because F σ = C 1 σ ln C and G σ = 1, we get lm U = lm F σ σ 1 σ 1 G σ = ln C. 13 6