CHAPTER 2: Mathematics for Microeconomics

Transcription

1 CHAPTER : Mahemaics for Microeconomics The problems in his chaper are primarily mahemaical. They are inended o give sudens some pracice wih he conceps inroduced in Chaper, bu he problems in hemselves offer few economic insighs. Consequenly, no commenary is provided. Resuls from some of he analyical problems are used in laer chapers, however, and in hose cases he suden will be direced back o his chaper. Soluions. f y y (, ) = a. f = 8, f = 6 y. y b. Consraining f (, y ) = 6 creaes an implici funcion beween he variables. The dy f 8 slope of his funcion is given by = = for combinaions of and y d f 6y ha saisfy he consrain. c. Since f (, ) = 6, we know ha a his poin y dy d 8 = =. 6 3 d. The f (, y ) = 6 conour line is an ellipse cenered a he origin. The slope of he line a any poin is given by dy d = 8 6 y. Noice ha his slope becomes more negaive as increases and y decreases.. a. Profis are given by π = R C = q + 40q 00. The maimum value is found by seing he derivaive equal o 0: d π = 4 q + 40 = 0, dq implies q = 0 and 00. π = b. Since d = π dq 4 < 0, his is a global maimum. c. MR = dr dq = 70 q. MC = dc dq = q So, MR = MC = 50. q = 0 obeys

2 Chaper : Mahemaics for Microeconomics.3 Firs, use he subsiuion mehod. Subsiuing y = yields f % ( ) = f (, ) = ( ) =. Taking he firs-order condiion, f % ( ) = = 0, and solving yields = 0.5, y = 0.5, and f % ( ) = f (, y ) = 0.5. Since f % ( ) = < 0, his is a local and global maimum. Ne, use he Lagrange mehod. The Lagrangian is L = y + λ( y). The firs-order condiions are L = y λ = 0, L = λ = 0, y L = y = 0. λ Solving simulaneously, = y. Using he consrain gives y = 0.5. = y = 0.5, λ = 0.5, and.4 Seing up he Lagrangian, L = + y + λ(0.5 y). The firs-order condiions are L = λ y, L L y λ = λ, = 0.5 y = 0. So = y. Using he consrain ( y = = 0.5) gives = y = 0.5 and λ =. Noe ha he soluion is he same here as in Problem.3, bu here he value for he Lagrangian muliplier is he reciprocal of he value in Problem.3..5 a. The heigh of he ball is given by f ( ) = 0.5g The value of for which heigh is maimized is found by using he firs-order condiion: df d = g + 40 = 0, implying = 40 g. b. Subsiuing for, Hence, f ( ) = 0.5g + 40 =. g g g df dg ( ) = 800. g c. Differeniaion of he original funcion a is opimal value yields df ( ) = 0.5( ). dg Because he opimal value of depends on g,

3 Chaper : Mahemaics for Microeconomics 3 ( ) = = = df 0.5( ) 0.5, dg g g as was also shown in par (c). d. If g = 3, = 5 4. Maimum heigh is = 5. If g = 3., maimum heigh is = 4.9, a reducion of This could have been prediced from he envelope heorem, since df ( ) = dg (0.0) = a. This is he volume of a recangular solid made from a piece of meal, which is by 3 wih he defined corner squares removed. b. The firs-order condiion for maimum volume is given by V = = 0. Applying he quadraic formula o his epression yields 6 ± ± 0.6 = = = The second value given by he quadraic (. ) is obviously eraneous c. If = 0.5, V So volume increases wihou limi. d. This would require a soluion using he Lagrangian mehod. The opimal soluion requires solving hree nonlinear simulaneous equaions, a ask no underaken here. Bu i seems clear ha he soluion would involve a differen relaionship beween and han in pars (a c)..7 a. Se up he Lagrangian: L = + 5ln + λ( k ). The firsorder condiions are L λ = 0, L L = λ 5 = λ = 0, = k = 0. Hence, λ = = 5. Wih k = 0, he opimal soluion is b. Wih k = 4, solving he firs-order condiions yields = and = = 5. = 5.

4 4 Chaper : Mahemaics for Microeconomics c. If all variables mus be nonnegaive, i is clear ha any posiive value for reduces y. Hence, he opimal soluion is = = and y = 5ln 4. 0, 4, d. If k = 0, opimal soluion is = 5, = 5, and y = 5 + 5ln 5. Because provides a diminishing marginal incremen o y as is value increases, whereas does no, all opimal soluions require ha once reaches 5, any era amouns be devoed enirely o. In consumer heory, his funcion can be used o illusrae how diminishing marginal usefulness can be modeled in a very simple seing..8 a. Because MC is he derivaive of TC, TC is an aniderivaive of MC. By he fundamenal heorem of calculus, q 0 MC( ) d = TC( q) TC(0), where TC (0) is he fied cos, which we will denoe TC(0) = K for shor. Rearranging, TC( q) = MC( ) d + K q = ( + ) d + K 0 = + + K q 0 = q = 0 q = + q + K. b. For profi maimizaion, p = MC( q) = q +, implying q = p. Bu p = 5 implies q = 4. Profi are TR TC = pq TC( q) 4 = K = 98 K. If he firm is jus breaking even, profi equals 0, implying fied cos is K = 98. c. When p = 0 and q = 9, follow he same seps as in par (b), subsiuing fied cos K = 98. Profi are

5 Chaper : Mahemaics for Microeconomics 5 TR TC = pq TC( q) 9 = K = = 8.5. d. Assuming profi maimizaion, we have π ( p) = pq TC( q) ( p ) = p( p ) + ( p ) + 98 ( p ) = 98. e. i. Using he above equaion, π ( p = 0) π ( p = 5) = = 8.5. ii. The envelope heorem saes ha dπ dp = q ( p). Tha is, he derivaive of he profi funcion yields his firm s supply funcion. Inegraing over p shows he change in profis by he fundamenal heorem of calculus: π (0) π (5) = = ( p ) dp 5 dπ dp dp p = p = = 8.5. p= 0 p= 5 Analyical Problems.9 Concave and quasi-concave funcions The proof is mos easily accomplished hrough he use of he mari algebra of quadraic forms. See, for eample, Mas Colell e al.,995, pp Inuiively, because concave funcions lie below any angen plane, heir level curves mus also be conve. Bu he converse is no rue. Quasi-concave funcions may ehibi increasing reurns o scale ; even hough heir level curves are conve, hey may rise above he angen plane when all variables are increased ogeher.

6 6 Chaper : Mahemaics for Microeconomics A couner eample would be he Cobb Douglas funcion, which is always quasiconcave, bu conve when α + β >..0 The Cobb Douglas funcion a. f = α > 0, α β = β > 0, α β f = α ( α ) < 0, f f α β α β = β ( β ) < 0, α β f = f = αβ > 0. Clearly, all he erms in Equaion.4 are negaive. b. A conour line is found by seing he funcion equal o a consan: y c α β = =, β α β implying = c. Hence, d 0. d < Furher, d < 0, d implying he counour line is conve. c. Using Equaion.98, α + β >. f f f = αβ( β α), which is negaive for α β. The power funcion a. Since y > 0 and y < 0, he funcion is concave. b. Because f, f < 0 and f = f = 0, Equaion.98 is saisfied, and he funcion is concave. Because f, f > 0, Equaion.4 is also saisfied, so he funcion is quasi-concave. c. y is quasi-concave as is y γ. However, y is no concave for γδ >. This can be γδ shown mos easily by f (, ) = f (, ).. Proof of envelope heorem

7 Chaper : Mahemaics for Microeconomics 7 a. The Lagrangian for his problem is L(,, a) = f (,, a) + λg(,, a). The firs-order condiions are L = f + λg = 0, L L = f + λg = 0, λ = g = 0. b., c. Muliplicaion of each firs-order condiion by he appropriae deriviaive yields f d + f d d d + λ g + g = 0. da da da da d. The opimal value of f is given by ( ) f ( a), ( a), a. Differeniaion of his wih respec o a shows how his opimal value changes wih a : df d d = f + f + fa. da da da e. Differeniaion of he consrain ( ) dg d d = = + + da da da 0 g g ga. g ( a), ( a), a = 0 yields f. Muliplying he resuls from par (e) by λ and using pars (b) and (c) yields df = f. a + λga = La da This proves he envelope heorem. g. In Eample.8, we showed ha λ = P 8. This shows how much an era uni of perimeer would raise he enclosed area. Direc differeniaion of he original Lagrangian shows also ha da = LP = λ. dp This shows ha he Lagrange muliplier does indeed show his incremenal gain in his problem..3 Taylor approimaions a. A funcion in one variable is concave if f ( ) < 0. Using he quadraic Taylor formula o approimae his funcion a poin a : f ( ) f ( a) + f ( a)( a) f ( a)( a) f ( a) + f ( a)( a). The inequaliy holds because f ( a) < 0. Bu he righ-hand side of his equaion is

8 8 Chaper : Mahemaics for Microeconomics he equaion for he angen o he funcion a poin a. So we have shown ha any concave funcion mus lie on or below he angen o he funcion a ha poin. b. A funcion in wo variables is concave if f f f > 0. Hence, he quadraic form ( fd + fd dy + f dy ) will also be negaive. Bu his says ha he final porion of he Taylor epansion will be negaive (by seing d = a and dy = y b ), and hence he funcion will be below is angen plane..4 More on epeced value a. The angen o g( ) a he poin E( ) will have he form c + d g( ) for all values of and c + de( ) = g( E( )). Bu, because he line c + d is above he funcion g( ), we know E( g( )) E( c + d) = c + de( ) = g( E( )). This proves Jensen s inequaliy. b. Use he same procedure as in par (a), bu reverse he inequaliies. c. Le u = F( ), du = f ( ), = v, and d = dv. = [ ( )] = [( ( )) ] [ ( )] = 0 F d F f d 0 0 = 0 + E( ) = E( ). d. Use he hin o break up he inegral defining epeced value: E( ) = f ( ) d f ( ) d + 0 = f ( ) d f ( ) d f ( ) d = P( ). e.. Show ha his funcion inegraes o :

9 Chaper : Mahemaics for Microeconomics 9 3 = ( ) = = =. = f d d. Calculae he cumulaive disribuion funcion: 3 = ( ) = = =. = F d 3. Using he resul from par (c): [ ] E( ) = F( ) d = d = =. 4. To show Markov s inequaliy use E( ) P( ) = F( ) = < =. f.. Show ha he PDF inegraes o : 3 = 8 d = = = =. Calculae he epeced value: = E( ) = d = = =. 3 4 = 3. Calculae P( 0): 0 3 = 0 d = = = 4. All we mus do is adjus he PDF so ha i now sums o over he new, smaller inerval. Since P( A ) = 8 9, = = f ( ) 3 f ( A) = = defined on The epeced value is again found hrough inegraion: 3 4 = E( A) = d = = = 0 6. Eliminaing he lowes values of increases he epeced value of he remaining values..5 More on variances a. This is jus an applicaion of he definiion of variance:

10 0 Chaper : Mahemaics for Microeconomics [ ] Var( ) = E E( ) = E E( ) + [ E( )] = E( ) [ E( )] + [ E( )] = E( ) [ E( )]. b. Here, we le y = µ and apply Markov s inequaliy o y and remember ha can only ake on posiive values. E( y ) σ P( y k) = P( y k ) =. k k c. Le i, i =,, n be n independen random variables each wih epeced value µ and variance σ. n E i = µ + L + µ = nµ. i= n Var i = σ + L + σ = nσ. i= n Now, le ( n) =. i = i nµ E( ) = = µ. n nσ σ Var( ) = =. n n d. Le X = k + ( k) and E( X ) = kµ + ( k) µ = µ. Var( X ) = k σ + ( k) σ = (k k + ) σ. dvar( X ) = (4k ) σ = 0. dk Hence, variance is minimized for k = 0.5. In his case, k = 0.7, Var( X ) = 0.58σ (no much of an increase). Var( X ) 0.5σ. = If e. Suppose ha Var( ) = σ and Var( ) rσ. = Now Var( X ) = k σ + ( k) rσ = ( + r) k kr + r σ. dvar( X ) = [ ( + r) k r] σ = 0. dk r k =. + r For eample, if r =, hen k = 3, and opimal diversificaion requires ha he lower risk asse consiue wo-hirds of he porfolio. Noe, however, ha i is sill

11 Chaper : Mahemaics for Microeconomics opimal o have some of he higher risk asse because asse reurns are independen..6 More on covariances a. This is a direc resul of he definiion of covariance: Cov(, y) = E ( E( ))( y E( y)) [ ] = E[ y E( y) ye( ) + E( ) E( y)] = E( y) E( ) E( y) E( y) E( ) + E( ) E( y) = E( y) E( ) E( y). b. Var( a ± by) = E[( a ± by) ] [ E( a ± by)] = a E( ) ± abe( y) + b E( y ) a [ E( )] ± abe( )E( y) b [E( y)] = a Var( ) + b Var( y) ± abcov(, y). The final line is a resul of Problems.5a and.6a. c. The presence of he covariance erm in he resul of Problem.6b suggess ha he resuls would differ. In he wo-variable case, however, his is no necessarily he siuaion. For eample, suppose ha and y are idenically disribued and ha Cov(, y) = rσ. Using he prior noaion, Var( X ) = k σ + ( k) σ + k( k) rσ. The firs-order condiion for a minimum is (4k + r 4 rk) σ = 0, implying r k = = r Regardless of he value of r. Wih more han wo random variables, however, covariances may indeed affec opimal weighings. d. If, = k he correlaion coefficien will be eiher + (if k is posiive) or (if k is negaive), since k will facor ou of he definiion leaving only he raio of he common variance of he wo variables. Wih less han a perfec linear relaionship y < [ y) ] 0.5 e. If y = α + β, Cov(, ) Var( )Var(. [ ] Cov(, y) = E ( E( ))( y E( y)) = E[( E( ))( α + β α β E( ))] = βvar( ).

12 Chaper : Mahemaics for Microeconomics Hence, Cov(, y) β =. Var( )