p E p E d ( ) , we have: [ ] [ ] [ ] Using the law of iterated expectations, we have:

Similar documents
1 Constant Real Rate C 1

Graduate Macroeconomics 2 Problem set 5. - Solutions

) from i = 0, instead of i = 1, we have =

FIRMS IN THE TWO-PERIOD FRAMEWORK (CONTINUED)

University of California, Davis Date: June xx, PRELIMINARY EXAMINATION FOR THE Ph.D. DEGREE ANSWER KEY

5-1. We apply Newton s second law (specifically, Eq. 5-2). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( )

Name of the Student:

The balanced budget multiplier and labour intensity in home production

Answers to Tutorial Questions

Monetary policy and models

CHAPTER 10: LINEAR DISCRIMINATION

Chapter 3: Vectors and Two-Dimensional Motion

( ) ( )) ' j, k. These restrictions in turn imply a corresponding set of sample moment conditions:

Chapter Finite Difference Method for Ordinary Differential Equations

UNIVERSITAT AUTÒNOMA DE BARCELONA MARCH 2017 EXAMINATION

If there are k binding constraints at x then re-label these constraints so that they are the first k constraints.

s = rθ Chapter 10: Rotation 10.1: What is physics?

Sections 3.1 and 3.4 Exponential Functions (Growth and Decay)

ANSWERS TO ODD NUMBERED EXERCISES IN CHAPTER 2

I-POLYA PROCESS AND APPLICATIONS Leda D. Minkova

Modern Energy Functional for Nuclei and Nuclear Matter. By: Alberto Hinojosa, Texas A&M University REU Cyclotron 2008 Mentor: Dr.

to Assess Climate Change Mitigation International Energy Workshop, Paris, June 2013

( ) () we define the interaction representation by the unitary transformation () = ()

Fall 2009 Social Sciences 7418 University of Wisconsin-Madison. Problem Set 2 Answers (4) (6) di = D (10)

Solution in semi infinite diffusion couples (error function analysis)

Handling Fuzzy Constraints in Flow Shop Problem

Lecture 5. Plane Wave Reflection and Transmission

In the complete model, these slopes are ANALYSIS OF VARIANCE FOR THE COMPLETE TWO-WAY MODEL. (! i+1 -! i ) + [(!") i+1,q - [(!

Numerical Study of Large-area Anti-Resonant Reflecting Optical Waveguide (ARROW) Vertical-Cavity Semiconductor Optical Amplifiers (VCSOAs)

. The geometric multiplicity is dim[ker( λi. number of linearly independent eigenvectors associated with this eigenvalue.

L4:4. motion from the accelerometer. to recover the simple flutter. Later, we will work out how. readings L4:3

. The geometric multiplicity is dim[ker( λi. A )], i.e. the number of linearly independent eigenvectors associated with this eigenvalue.

Political Economy of Institutions and Development: Problem Set 2 Due Date: Thursday, March 15, 2019.

Let s treat the problem of the response of a system to an applied external force. Again,

Outline. GW approximation. Electrons in solids. The Green Function. Total energy---well solved Single particle excitation---under developing

Mechanics Physics 151

Lecture-V Stochastic Processes and the Basic Term-Structure Equation 1 Stochastic Processes Any variable whose value changes over time in an uncertain

Physics 201 Lecture 15

Online Appendix for. Strategic safety stocks in supply chains with evolving forecasts

The Unique Solution of Stochastic Differential Equations. Dietrich Ryter. Midartweg 3 CH-4500 Solothurn Switzerland

Linear Response Theory: The connection between QFT and experiments

Lecture 18: Kinetics of Phase Growth in a Two-component System: general kinetics analysis based on the dilute-solution approximation

Suppose we have observed values t 1, t 2, t n of a random variable T.

Technical Appendix to The Equivalence of Wage and Price Staggering in Monetary Business Cycle Models

Midterm Exam. Thursday, April hour, 15 minutes

Chapter Fifiteen. Surfaces Revisited

Reinforcement learning

Field due to a collection of N discrete point charges: r is in the direction from

Ordinary Differential Equations in Neuroscience with Matlab examples. Aim 1- Gain understanding of how to set up and solve ODE s

Reflection and Refraction

Chapter 6: AC Circuits

Lecture 11 SVM cont

Chapters 2 Kinematics. Position, Distance, Displacement

CptS 570 Machine Learning School of EECS Washington State University. CptS Machine Learning 1

I-Hsuan Hong Hsi-Mei Hsu Yi-Mu Wu Chun-Shao Yeh

Notes on Inductance and Circuit Transients Joe Wolfe, Physics UNSW. Circuits with R and C. τ = RC = time constant

FI 3103 Quantum Physics

Knowing What Others Know: Coordination Motives in Information Acquisition Additional Notes

Notes on the stability of dynamic systems and the use of Eigen Values.

Nanoparticles. Educts. Nucleus formation. Nucleus. Growth. Primary particle. Agglomeration Deagglomeration. Agglomerate

Lecture 17: Kinetics of Phase Growth in a Two-component System:

Inventory Policy Implications of On-Line Customer Purchase Behavior

Hierarchical Production Planning in Make to Order System Based on Work Load Control Method

Measuring capital market integration

8 Baire Category Theorem and Uniform Boundedness

PHYS 1443 Section 001 Lecture #4

PHYS 705: Classical Mechanics. Derivation of Lagrange Equations from D Alembert s Principle

Department of Economics University of Toronto

Rotations.

Exponential and Logarithmic Equations and Properties of Logarithms. Properties. Properties. log. Exponential. Logarithmic.

24-2: Electric Potential Energy. 24-1: What is physics

Variability Aware Network Utility Maximization

Motion in Two Dimensions

Market inefficiency and implied cost of capital models

dm dt = 1 V The number of moles in any volume is M = CV, where C = concentration in M/L V = liters. dcv v

Comparative Study of Inventory Model for Duopolistic Market under Trade Credit Deepa H Kandpal *, Khimya S Tinani #

ajanuary't I11 F or,'.

SCIENCE CHINA Technological Sciences

Technical Appendix for Inventory Management for an Assembly System with Product or Component Returns, DeCroix and Zipkin, Management Science 2005.

Bethe-Salpeter Equation Green s Function and the Bethe-Salpeter Equation for Effective Interaction in the Ladder Approximation

General Non-Arbitrage Model. I. Partial Differential Equation for Pricing A. Traded Underlying Security

Simulation of Non-normal Autocorrelated Variables

The shortest path between two truths in the real domain passes through the complex domain. J. Hadamard

A Brief Guide to Recognizing and Coping With Failures of the Classical Regression Assumptions

Mechanics Physics 151

John Geweke a and Gianni Amisano b a Departments of Economics and Statistics, University of Iowa, USA b European Central Bank, Frankfurt, Germany

Computer Propagation Analysis Tools

Mechanics Physics 151

Maximum Likelihood Estimation

Chapter 23: Electric Potential

Example: MOSFET Amplifier Distortion

THE PREDICTION OF COMPETITIVE ENVIRONMENT IN BUSINESS

, t 1. Transitions - this one was easy, but in general the hardest part is choosing the which variables are state and control variables

(,,, ) (,,, ). In addition, there are three other consumers, -2, -1, and 0. Consumer -2 has the utility function

156 There are 9 books stacked on a shelf. The thickness of each book is either 1 inch or 2

Lecture 2 M/G/1 queues. M/G/1-queue

Advanced Macroeconomics II: Exchange economy

Demographics in Dynamic Heckscher-Ohlin Models: Overlapping Generations versus Infinitely Lived Consumers*

calculating electromagnetic

Bundling with Customer Self-Selection: A Simple Approach to Bundling Low Marginal Cost Goods On-Line Appendix

Transcription:

Poblem Se #3 Soluons Couse 4.454 Maco IV TA: Todd Gomley, gomley@m.edu sbued: Novembe 23, 2004 Ths poblem se does no need o be uned n Queson #: Sock Pces, vdends and Bubbles Assume you ae n an economy whee he sock pce, p, s gven by he sandad abage equaon () and he pocess fo dvdends a me, d, s gven by equaon (2) below: p p d = [ ] [ ] ( ) d d = ρ d d ε, ε d... and [ ε] = 0 (a) Use eaed expecaons o solve fo he pce, p, as a funcon of ONLY fuue expeced dvdends. Wha assumpon do you mplcly need o do hs? Fs, by pluggng n fo [ p ], we have: p = [ p ] [ d ] p = p d d [ ] [ ] [ ] 2 2 Usng he law of eaed expecaons, we have: 2 2 p = p d [ ] [ ] [ ] 2 d 2 Connung hs ype of pocess of pluggng n fo fuue expeced pces, we ge: T p d = [ ] [ p ] T = Takng T o and assumng ha [ p ] = T T lm T = 0, we have ou soluon: T p = [ ] d () (2)

(b) Assume ha ρ < /( ). Use eaed expecaons o fnd an expesson fo he expecaon (as of me ) fo dvdends a me, [ d ], ha s a funcon of only d, ρ, and d. Fom equaon (2), we know, ( ) d d = ρ d d ε Takng expecaons and eaangng, we have, Pluggng n fo [ d ], we fnd: Connung he eaon, ( ) [ ] ρ [ ] d = d d d ( ) ρ( ) 2 ( 2 ) [ ] ρ [ ] d = d d d ( ) [ ] ρ [ ] d = d d d d d [ ] ρ [ ] d = d d d [ ] ρ ( ) d = d d d (c) Use you answes fom (a) and (b) o fnd an expesson fo and d p, as a funcon of d, ρ,. Call hs soluon o he abage equaon he fundamenal pce, Pluggng [ ] ρ ( ) d = d d d no ou answe fom pa (a), we have: ρ p = ( d d) d = p = ρ ( d d) d = 0 ρ ρ d p = ( d d) = 0 ρ d p = ( d d) ρ (d) Now assume he pce of he sock has a bubble componen, p., whee b = ( ) b b 0 and b 0 > 0. Pove ha he pce p = p b s also a soluon o he abage condon () and ha ou assumpon fom pa (a) s no longe necessay. Takng ou nal abage equaon () and pluggng n fo p = p b, we have: 2

p b = p b [ d ] p b p b d [ ] [ ] = Bu, snce [ b ] = ( ) b, hs equaon educes o: Pluggng n fo ou p and p p d [ ] = p usng ou answe fom pa (c), we have: ρ d ρ d ρ ρ ( d d) = ( d d) [ d ] Wh a lle algeba, we can educe hs expesson n he followng way: ρ ρ ( ) d d d = d d d ρ ρ ( ) ( ) [ ] ρ ρ ρ ( ) d d d = d ρ ρ ρ ( ) [ ] [ d ] ρ ρ ( ) d d d = d ρ ρ ρ ρ ρ [ d ] = ρd d d ( ) [ ] [ ] ρ ( ) d = d d d Ths las expesson s clealy ue based ou on answe fom pa (b). Thus, we have successfully shown ha he abage condon holds even n he pesence of he bubble! Moeove, we wee able o ge hs soluon whou he need fo ou eale assumpon fom pa (a) ha: T lm [ p T] = 0 T (e) Why ae ndvduals wllng o pay a hghe pce,, fo he sock han he fundamenal pce coespondng o he pesen value of he dvdends, The abage condon can be sasfed despe he bubble, because ndvduals ae wllng o pay a hghe pce fo he sock han he fundamenal pce (whch epesens he dscouned fuue dvdends) because hey coecly ancpae ha he pce wll connue o se because of he bubble componen of he pce. The sng pce yelds capal gans ha exacly offse he lowe dvdend o pce ao. p p? 3

Queson #2: Makups va Scky Pces n he Goods Make Ths poblem s based heavly on secons, 4., and 4.3 of Roembeg and Woodfod s Handbook chape. You may fnd vey helpful o ead hese secons of he chape befoe poceedng wh hs poblem se. PART : -- Makups: Wha ae hey and why do hey mae? Consde a connuum of mpefecly compeve fms. Le he magnal cos of each fm be gven by Pc( y ), whee y s he quany suppled, P s he geneal pce level, and c'( y ) > 0. Snce c'( y ) > 0, an ncease n he quany suppled by ndusy, wll be assocaed wh an ncease n magnal cos. Because of mpefec compeon, each fm faces a downwad slopng demand fo he good, and can chage a pce geae han magnal cos. The makup, µ, of a fm s smply gven as s pce ove magnal cos. In hs example, he makup by fm s gven by µ = P /[ Pc( y)]. So, f a fm wshes o ncease s oupu and manan a consan makup, wll need o ase s elave pce ( P / P ). Howeve, n a symmec equlbum, we know ha mus be ha: cy ( ) µ = (3) whee he common level of oupu (and hence aggegae) wll be gven by (and hence aveage) makup. Y, and µ s he common (a) Analyze equaon (3). Wha s he nuon fo why aveage makups and aggegae oupu move n oppose decons n hs equaon? If we wan o popagae/amplfy he busness cycle, wha ype of movemen n makups mus ou models geneae? To see whee (3) comes fom, smply noe ha n a symmec equlbum, mus be ha y = Y, P = Pfo all. Thus, P Pc( y ) µ = mples P µ = = Pc( Y ) c( Y ) The aveage oupu and aggegae oupu move n he oppose decons fo he followng eason: If all fms y o ncease he oupu, wll no be possble fo hem all o ncease he elave pce as a means of mananng he makup. Why? Because f all fms nceased he elave pce, hen he geneal pce of all fms would also se, hus nceasng magnal coss and bngng he makups back down. So, n geneal equlbum, an ncease n oupu by each fm can only be possble f he fms allow he makup o fall. To popagae o amplfy he busness cycle, we wll wan o geneae counecyclcal movemens n he makup. Fo nsance, f makups fall dung economc booms, hen hee wll be an even geae ncease n he aggegae level of oupu as seen n equaon (3). 4

Agan, assume monopolsc compeon among a lage numbe of supples of dffeenaed goods. ach fm faces a downwad-slopng demand cuve fo s poduc of he fom: P Y = Y (4) P whee P s he pce of fm a me, P s an aggegae pce ndex, Y s an ndex of aggegae sales a me, and s a deceasng funcon. Assume a consan elascy of demand, ε = x '( x) / ( x ) >, and assume each fm faces he same level of (nomnal) magnal coss C n a gven peod. Neglecng fxed coss, pofs of fm a me ae gven by: P Π = ( P C ) Y P (5) (b) Assume compleely flexble pces: Maxmze he fm s pofs o fnd s opmal makup, µ, as a funcon of he elascy of demand. Is he makup an nceasng o deceasng funcon of he elascy? xplan he nuon of hs esul. As usual, monopolsc fms maxmze he pofs by choose he pce: The FOC wll be: max ( ) P P P C Y ( P C) ( P C) P P P Y ' Y = 0 P P P P P P = ' P P P P P P ' P P P = = ε P C P P P ε µ = C ε Thus, we can clealy see ha he opmal makup s a funcon only of he elascy of he subsuon, and s a deceasng funcon of he elascy. Ths makes sense: As cusomes become moe elasc and moe esponsve o pces changes, he monopolsc fm wll no be able o chage as hgh of a makup. If cusomes ae nfnely elasc, a monopolsc fm wll have no ably o chage a makup on s poduc. PART 2 Geneang movemen n Makups va Scky Pces, evng he Mah Now, we ae gong o look a scky pce model ha wll geneae movemens n he makups chaged by fms ove he busness cycle. Now assume ha n each peod, a facon ( α) of fms ae able o change he pces whle he es mus keep he pces consan. A fm ha changes s pce, chooses n ode o maxmze: 5

Π α β = 0 P (6) whee β s he dscoun faco pe peod of me and α epesens he pobably ha hs pces wll sll apply peods lae. The pof funcon s he same as n equaon (5). (c) enoe X as he new pce chosen a dae by any fms ha choose hen. Pove ha he fs ode condon fo he opmzaon poblem s: ( ) αβ ' = 0 (7) Y X X C P P P ε X The maxmzaon poblem of hese fms can be wen as: max X ( ) X X C Y P α β = 0 P Takng he devave wh espec o X, we have: ( ) = 0 ( P ) X X Y X C ' Y P P ( αβ ) 2 = 0 P X Y P ( X C ) ' = 0 ' P X ( αβ ) 0 P P = X P ( ) ' = 0 ' P P X Y X X P X C ( αβ ) 0 P = P P X X X ( αβ ) = 0 Y X X C ' P P P ε X = 0 NOT: Pa (d) asks you o do a log-lneazaon. If you unfamla wh hs ype of calculaon, please go o he couse web-se and download he fle Log-Lneazaon Handou found on he webpage whee you can download he poblem ses. Ths handou should help famlaze you wh log-lneazaon. (d) We now wan o ake a log-lnea appoxmaon of he fs-ode condon (7) aound a seady sae n whch all pces ae consan ove me and equal o one anohe, magnal cos s smlaly consan, and he consan ao of pce o magnal cos equals µ. Le x ˆ, ˆ ˆ π, and c denoe he pecenage devaons of he vaables X / P, P / P and C / P, especvely, fom he seady sae values. o he log lneazaon of equaon (7) o ge equaon (8) below, and nepe hs equaon. 6

( ) ˆ ˆ ˆ αβ x π k c = 0 = 0 (8) The poof of hs equaon s ahe dffcul. So, le me fs gve he nuon of he esul. Noe ha he em xˆ ˆ π k epesens he fms elave pce a me. (You wll see hs n he poof below). And, c ˆ s he elave magnal cos a me. Gven hese nepeaons, we see ha he equaon s smply ellng us ha ha devaons n he fm s opmal seady sae elave pce ae expeced o be popoonal o he devaons n magnal cos of poducon on aveage ove he me ha he pce chosen a dae s expeced o apply. Thus, f he fm ancpaes hghe aveage magnal coss o nflaon ove he peod expecs s pces o be fxed, wll choose a hghe pce. Now, le s do he log-lneazaon o show hs esul: The fs hng I wll do s ewe equaon (7) n ems of he vaables we ae gong o log-lneaze aound: Y X X C ' = 0 ( αβ ) 0 P P = P ε X X X ' Y X P P C = 0 P ( αβ ) 0 P P = X ε X Y X ε C = 0 ( αβ ) ( ε ) P 0 P = ε X Mulplyng hough by X, and eaangng, we ge: X ε X C ( αβ ) Y = 0 0 P = ε P P X ε X X C ( αβ ) Y = ( αβ ) Y = 0 P ε P = 0 P P (9) Now, noe he followng: X X P P P =... P P P P P 2 (0) Thus, f we log-lneaze equaon (0), we wll ge: xˆ π () k 7

Addonally, noce wha happens f we log-lneaze We wll ge: X Y ε xˆ π P X P k Y. (2) whee XYP,, epesen he especed seady sae values. Keepng hs n mnd, now le s log-lnea he RHS of equaon (9) ecognzng ha we can convenenly use vaaons of ou esuls shown n equaons () and (2). ong hs, we fnd: X C P P ( αβ ) Y ε ˆ ˆ ˆ x π k c (3) = 0 Now le s log-lneaze he LHS of equaon (9): ε X X ε P P ( αβ ) Y ε ˆ ˆ ˆ ˆ x π k x π k (4) = 0 Now, we can equae ou RHS n equaon (3) wh he LHS n equaon (4) o ge ou fnal esul: ε X X X C αβ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Y ε x π k x π k αβ Y ε x π k c 0 ε = = P P = 0 P P ( ) ( ) ( αβ ) ε ε cˆ = 0 ( αβ ) ( X) xˆ ˆ π k = ( αβ ) ( C) = 0 ε X ˆ ˆ ˆ = 0 x π k c = 0 ε C αβ µ xˆ ˆ π cˆ = 0 ( ) ( ) k = 0 µ x π k c = 0 ( αβ ) ˆ ˆ ˆ = 0 (e) quaon (8) can be solved fo he elave pce xˆ of fms ha have us changed he pce as a funcon of fuue nflaon and eal magnal coss. Assumng αβ <, use a quas-dffeence of hs elaon o pove he followng: xˆ = αβ ˆ π ( αβ ) cˆ αβ x (5) ˆ Rewng ou esul fom pa (d), we have: 8

αβ = αβ ˆ π k cˆ ( ) xˆ ( ) = 0 = 0 xˆ ( ) ˆ = αβ π k cˆ αβ = 0 xˆ ( ) ( ) ˆ = αβ αβ π cˆ = 0 k xˆ cˆ ˆ c = ( αβ ) ( αβ ) ( αβ ) π k ˆ = (6) Usng hs, we also now know ha: x c ˆ ( ) ( ) ˆ ˆ = αβ αβ π k = 0 αβ x αβ αβ π c ( ) ( ) ˆ ˆ ˆ = k = 0 (7) Subacng equaon (7) fom equaon (6), we have: xˆ xˆ cˆ ˆ cˆ ˆ cˆ αβ = ( αβ ) ( αβ ) ( αβ ) π k ( αβ ) ( αβ ) π k = = 0 xˆ ˆ ( ) ˆ ( ) ( ) ˆ ( ) ( ) ˆ αβ x = αβ c αβ αβ π k αβ αβ π k = = 0 xˆ ˆ αβ x = ( αβ ) cˆ ( ) ( ) ˆ ( ) ( ) ˆ αβ αβ π k αβ αβ π k = = 0 αβ = ( αβ ) ( αβ ) ( αβ ) π = 0 xˆ xˆ cˆ ˆ ( ) ( ) xˆ αβ xˆ = αβ cˆ αβ ˆ π xˆ = αβ xˆ αβ cˆ αβ ˆ π (f) Now, gven a few assumpons, we can show ha he ae of ncease of he pce ndex sasfes he followng elaon n ou log-lnea appoxmaon: α ˆ π = xˆ α (Please see page 5-6 f you wan moe deals of whee hs equaon comes fom.) Subsue (8) no (5) and use he fac ha ˆ µ ˆ = c, whee ˆ µ denoes he pecenage devaon of he aveage makup µ = P / C fom s seady sae value of µ, o show ha: whee κ ( αβ)( α) / α ˆ π β ˆ π κµ Fom equaon (8), we have: ˆ (8) = (9) ˆ π α = xˆ α Pluggng hs no ou esul fom pa (e) and usng equaon (8), we have: 9

α α ˆ π αβ ˆ π ( αβ ) ˆ αβ ˆ π α α = c ( αβ )( α ) ( ) ˆ π = αβ ˆ π cˆ α β ˆ π α ( αβ )( α ) ˆ π = β ˆ π cˆ α ˆ π = β ˆ π κµ ˆ PART 3 Inepeng he mpac of Scky Pces on Makups Okay, now he mah pa of hs poblem se s ove. Now, ha you have an dea whee hese esuls come fom, you wll need o analyze he nuon and mplcaons of hese esuls. The es of hs poblem se does no need any mah calculaons. (g) Holdng fuue expecaons of nflaon consan, consde a posve shock o aggegae demand. Usng equaon (9), how wll hs shock o aggegae demand affec he aveage makup? Ae makups po- o coune-cyclcal n hs model? Wha s he nuon fo hs esul? [Hn: You should fs consde how a posve shock o aggegae demand wll affec aveage pces and oupu. Then, holdng fuue expecaons of pces consan, you can analyze he mpac on he aveage makups usng equaon (9) ] Gong back o he sandad aggegae demand and supply famewok of Keynes (wh an upwad slopng sho-un aggegae supply), an ncease n aggegae demand should esul n an mmedae ncease n oupu and pces. Howeve, f we hold ou expecaons of fuue pces consan, hen we see fom ou equaon (9) ha mus be he case ha he aveage makup falls. Thus, we conclude ha ou scky pce model wll yeld us coune-cyclcal movemens n he makup us as we had hoped. Usng ou esul fom pa (a), we know ha hs fall n he makup of fms wll ncease aggegae oupu. Thus, he nal shock o aggegae demand wll be amplfed va he effec of he se n pces on he makup. [We could also have assumed a vecal aggegae supply cuve such ha he aggegae demand shock only affecs pces and no oupu nally. Howeve, because he aveage makup wll sll fall, we wll sll ge a posve movemen n aggegae oupu!] Wha s he nuon of hs esul? Well, fms ha can adus he pces wll ase hem upwad o mee he new demand and o manan he makup. Howeve, some fms fnd he pces fxed a he me of he aggegae demand shock and ae no be able o adus he pces upwad o manan he opmal makup. The makup wll fall because he npu pces ae nceasng n he level of poducon. Gven he fall n he aveage makup, oupu wll be pushed even hghe. If we had accouned fo changes of fuue nflaon, we wll sll ge he same esuls bu he analyss s moe complcaed. (h) How wll an ncease n α affec he magnude of movemens n he makup n esponse o a shock o he economy? (Agan, hold fuue expecaons consan o do you analyss). Please nepe hs esul. 0

An ncease n α wll decease he mulple κ n equaon (9). Ths mples ha fo any gven shock o cuen pces, he movemen n he makup wll have o be geae han befoe (agan, holdng ou expecaons of fuue nflaon consan). Thus, he coune-cyclcal movemens n he makup ae amplfed as wll be he amplfcaon effec of makups. Wha s he nuon of hs esul? Well, a hghe α mples ha ha a smalle facon of fms ae able o adus he pces each peod. Thus, moe fms wll be unable o ncease he pces n esponse o he shock and he aveage makup wll fall even fuhe han befoe! () Now suppose ha he elascy of demand sn consan. Rahe, assume ha elascy of demand s an INCRASING funcon of he fm s elave pce.. How does an aveage fm s elave pce move mmedaely followng a posve shock o aggegae demand? Agan, a posve aggegae demand shock nceases aggegae pces. Fms ha ae unable o adus he pce mmedaely wll see he elave pce fall, bu fms ha ae able o adus he pce upwads wll see an ncease n he elave pce.. Wha wll happen o an aveage fm s desed makup dung an economc boom? [Hn: Use you answe o pa (b) above.] In he economc boom, he ncease n aggegae demand nceases he demand fo npus and hus pushes up he magnal cos of fms. If a fm can adus s pce, would usually lke o ncease s pce o manan s usual desed makup. Howeve, snce ohe fms pces ae beng held down because hey ae unable o adus mmedaely, a fm ha can adus s pce upwad faces a elavely moe elasc demand when nceasng s pce elave o hose wh fxed pces. Ths elavely moe elasc demand helps dampen he desed makup fo fms able o ncease he pces, and pces wll adus upwads moe slowly.. Would allowng he elascy of demand o vay as descbed above educe o ncease amplfcaon n hs model? Clealy, he fall n he aveage desed makup dung an economc boom wll cause he aveage makup of he economy o be even moe coune-cyclcal han befoe. Ths nceases he amplfcaon of he model! (Ths s wha Roembeg-Woodfod dscuss n secon 4.3 of he chape handbook).