Joh Riley 8 Jue 03 ANSWERS TO EXERCISES IN APPENDIX B Sectio B VECTORS AND SETS Exercise B-: Covex sets (a) Let 0 x, x X, X, hece 0 x, x X ad 0 x, x X Sice X ad X are covex, x X ad x X The x X X, which implies that the itersectio of the two sets is covex (b) Let X = { a} ad X { b} = where ab, ad a b You should covice yourself that both sets are covex The ay covex combiatio betwee a ad b, with 0 < <, does ot belog to X X { ab} =, (c) Take 0 x, x X ad 3 x, x X, so 0 0 y = x + x Y ad 3 y = x + x Y Sice X ad X are covex, 0 ( )x x X + ad +, for all [ 0,] 3 ( )x x X Hece ( ) ( ) ( ) (( ) ) y y 0 y x 0 x x x 3 Y = + = + + +, which shows that Y is covex Usig a similar argumet is easy to show that Z is also covex Sectio B FUNCTIONS OF VECTORS Exercise B-: Positive semi-defiite quadratic form If qx ( ) is everywhere o-egative, the qx ( ) is everywhere o-positive This meas that ecessary ad sufficiet coditios are a, a 0 ad ( a )( a ) ( a )( a ) 0, that is a, a 0 ad ( a )( a ) ( a )( a ) 0 Aswers to Exercises i Appedix B page
Joh Riley 8 Jue 03 Exercise B-: Positive defiite quadratic form If q( x ) > 0 for all x 0, the q( x) 0 ad ( a )( a ) ( a )( a ) 0 < for all x 0 This is the case where a > 0 >, which is the same as a < 0 ad ( a )( a ) ( a )( a ) > 0 Exercise B-3: Cocave ad Quasi-cocave fuctios Proof of Propositio B5: Let f ad g be cocave o the covex set X The 0 ( ) ( ) ( ) + ( ) ad g( x ) ( ) g( x 0 ) g( x ) f x f x f x expressios we have ( ) ( ) ( ) ( ) ( ) gives the desired result + Summig both 0 0 ( ) ( ( ) ( )) f x + g x f x + g x + f x + g x, which Proof of Propositio B6: (a) Sice f is cocave we kow f ( x ) ( ) f ( x 0 ) f ( x ) 0 ( ( )) ( ) ( ) ( ) ( ) + Sice g is icreasig g f x g f x + f x Now, usig the cocavity of g, (( ) ( ) ( )) ( ) ( ) ( ) ( ( )) g f x f x g f x g f x 0 0 + + Hece 0 ( ) ( ) ( ) ( ) h x h x + h x, which proves the statemet Aswers to Exercises i Appedix B page
Joh Riley 8 Jue 03 (b) By liearity of f, we have that 0 0 ( ) (( ) ( ) ( )) ( ) (( ) ) h x = g f x + x = g f x + f x Sice g is cocave ( ) ( ) ( ) 0 0 ( ) ( ( )) ( ) ( ) ( ) h x g f x + g f x = h x + h x Exercise B-4: Covex lower cotour sets (a) The proof is almost idetical to the proof for quasi-cocave fuctios We demostrate ecessity (oly if) Suppose the lower cotour sets of f are covex ad cosider ay 0 x ad x such that 0 f( x ) f( x ) The both vectors are i the set CL ( x ) = { x f ( x ) f ( x )} By hypothesis, this set is covex Thus all covex combiatios x lie i CL ( x ) It follows immediately that f( x ) f( x ) (b) Lq ( ) = f ( q ) Sice f is covex, 0 f ( q ) ( ) f ( q ) + f ( q ) Hece 0 L( q ) ( ) f ( q ) + f ( q ) 0 0 f q f q Lq Lq = ( ) ( ) + ( ) = ( ) ( ) + ( ) Thus Lq ( ) is covex ad so quasi-covex It follows that the lower cotour sets of L are covex (c) Let L be the fixed supply of labor ad choose q so that Lq ( ) = L The the set of feasible outputs must satisfy the costrait Lq ( ) Lq ( ) This is a lower cotour set ad is therefore covex Aswers to Exercises i Appedix B page 3
Joh Riley 8 Jue 03 Exercise B-5: Cotour sets (a) O (b) (c) f( x ) = 0 if ad oly if either x = x or x = x For x above (below) both these lies f( x ) > 0 Otherwise f( x ) < 0 Aswers to Exercises i Appedix B page 4
Joh Riley 8 Jue 03 *(d) Completig the square, f( x) = x 6xx + 7 x = ( x + 6xx + 9 x ) + 6x Hece f( x) = (4 x ) ( x 3 x ) But a b = ( a b)( a+ b) Therefore f( x) = (7 x x )( x + x ) We ca the proceed as i part (c) Begi by drawig the lies f( x ) = 0 Sectio B3 TRANSFORMATIONS OF VECTORS Exercise B3-: Liear Ecoomy (a) If gross output is ( x, x ) the iput requiremets of commodity are ax + ax Net output of commodity is therefore y = x a x a x = ( a ) x a x Similarly et output of commodity is y = a x + ( a ) x Expressig this i matrix form, a a x 0 x a a x y= = = ( I Ax ) a a x 0 x a a x Aswers to Exercises i Appedix B page 5
Joh Riley 8 Jue 03 (b) Usig the data, y 4 x 3 3 y = x 4 4 The if x = 0, y 4 x x y = 3 3 = 3 0 x 4 4 4 Each uit of output uses oe uit of labor thus x = 80 ad so y 40 y = 60 A similar argumet whe x = 0 establishes that i this case y 0 y = 60 (c) With half the labor goig to commodity ad the rest to commodity the et output is halved i each case so that y 40 0 0 y = + = 60 60 0 These three et output vectors are depicted below I a closed ecoomy the et output of both commodities must be o-egative Thus oly the third output vector is feasible Note that all the feasible output vectors are covex combiatios of Thus they lie alog the heavy lie 0 y ad y Aswers to Exercises i Appedix B page 6
Joh Riley 8 Jue 03 (d) We have show that y = ( I-A ) x Ivertig, x= ( I-A ) y Total labor requiremets are a 0 x a 0 x a 0 ( ) = = I-A y Thus with L uits of labor available, a 0 ( I-A ) y L Computig the iverse ad the multiplyig it by a 0,it follows that the costrait is 8y+ 4y 80 (e) As ca be see from the figure ad from the equatio the maximum feasible et output of commodity is 0 Exercise B3-: Simple Regressio 0 0 (a) Y Yˆ ( Y ax bx ) ( Y ax bx ) =, where the first matrix has dimesio ad the secod, ad each elemet is give by Y a bx The the result should be a scalar: the sum of the square distace of each observatio (b) First ote this is a positive defiite fuctio, thus the first order coditio is both ecessary ad sufficiet for a miimum This coditio is give by ( Y a bx ) = 0, that is Y a b X = 0 Hece a = Y bx (c) Replacig a i our equatio we obtai ( ) ( ) ( ( ) ) ( ) ( ) Y Yˆ = Y Y bx bx = Y Y b X X = y bx (d) Proceedig as i (b), the first order coditio is give by ( ) y bx x = 0 Solvig for b we have that b = yx x Aswers to Exercises i Appedix B page 7
Joh Riley 8 Jue 03 Exercise B3-3: Least squared error (a) X = x() x( ) b ax ( ) = is a colum vector The t-th compoet of this vector is bt = ax t+ + axt = ( xt,, xt) ( a,, a) Thus b t is the ier product of the t-th row of X ad the colum vector a It follows that b= ax ( ) = X a (b) Let e be the differece betwee y ad vectors is ee = ( y Xa)( y Xa) = y ( y Xa) ( Xa)( y X a) = y ( y Xa) ( Xa) y+ ( Xa) X a = yy y Xa ( Xa) y+ a XX a = yy ( Xa) y+ a XX a = yy a X y+ a XX a X a The the distace betwee the two (c) zc = zc i i= c Thus zc = c z z i= z It follows that a X y = X y a Also from the discussio of the quadratic form, qz ( ) = z Bz = B z Hece z z a XX a= XX a a The ee = X y + XX a a (d) ee is a cocave fuctio if it has a quadratic form that is egative defiite Appealig to Propositio B3-, ee = XX ai a The quadratic form of this matrix q = z X Xz ca be rewritte as follows q = ( Xz)( Xz) Aswers to Exercises i Appedix B page 8
Joh Riley 8 Jue 03 The product of two vectors is the sum of squares so must be positive Hece the quadratic form is egative defiite It follows that ee is a cocave fuctio of a Thus the first order coditio is both ecessary ad sufficiet for a maximum (e) Follows immediately from the first order coditio, which is ee = 0 Hece a a= ( XX) X y B4: SYSTEMS OF LINEAR DIFFERENCE EQUATIONS Exercise B4-: Polar represetatio of a complex umber iv (a) If f( z) = cos z the f ( z) = si z, f ( z) = cos z, f ( z) = si z, f ( z) = cos z Therefore iv cos z = f(0) + f (0) z+ f (0) + f (0) + f (0)! 3! 4! Similarly, z = z + z +! 4! 4 cos (b) Therefore 3! 5! 3 5 si z = z z + z + i! 3! 4! 3 4 cos z + isi z = + iz z z + z + iz iz (c) If f( z) = e the f ( z) = ie, Applyig Taylor s Expasio, 3 4 i i 3 i 4 = + iz + z + z + z +! 3! 4! ( ) =, iz f z ie =, 3 iz f ( z) ie iv 4 iz f ( z) ie = 3 4 iz i i 3 i 4 e = + iz + z + z + z +! 3! 4! Exercise B4-: Saddle poit dyamics (a) For there to be a solutio of the form xt ( + ) = xt ( ), the eigevalue must be a solutio of 3 5 5 A I = = 3 + 4 = ( )( ) = 0 Aswers to Exercises i Appedix B page 9
Joh Riley 8 Jue 03 5 Thus the eigevalues are (, ) = (, ) 5 For the first eigevalue x ( t+ ) = x () t + x () t = x () t Therefore the costat growth path is x = x For the secod eigevalue x( t+ ) = x() t + x() t = x() t Therefore the costat growth path is x = x (b) To draw the phase portrait we first solve for the boudaries of the four phases 3 x x( t+ ) x() t x() t + x() t ( ) xt x = x( t ) x() t = A I = + x() t The there is o chage i x () t if x = x Note that x x () t = Thus alog this 3 boudary lie x is icreasig if ad oly if x > 0 3 Similarly, there is o chage i x if x = 0 I this case x = x Thus x is icreasig if ad oly if x > 0 The phase boudaries are depicted below ad the arrows o these boudaries depict the directio of chage o theses boudaries II I III O IV Figure B: B4-6: The The four four phases phases Aswers to Exercises i Appedix B page 0
Joh Riley 8 Jue 03 It is readily checked that iside each phase the directios of chage are as depicted Give the directio of the arrows, phases I ad III are absorbig phases That is, if the path eters oe of these phases it remais there We have already established that the costat growth paths are as give below = 5 = : x x = = : x x These are depicted as dashed lies i the figure below II I O III IV Figure B4-7: Phase portrait If x () is i phase II above the dashed lie, x iitially declies ad x icreases Evetually the path eters phases I ad both x ad x icrease I the log ru both grow at the rate give by the high eigevalue Startig i phase II below the dashed Aswers to Exercises i Appedix B page
Joh Riley 8 Jue 03 curve, the system evetually moves ito phases III ad from the o x decreases I the log ru it agai decreases at the rate of the high eigevalue A almost idetical argumet holds for paths begiig i phase IV (c) From the above argumet, the oly way the system ca approach zero is if x () lies o the slow costat growth lie x = x Exercise B4-3: Oscillatios (a) The characteristic equatio is A I = = + = Thus the 5 4 0 eigevalues are = α + iβ = i ad = α iβ = i To covert to polar coordiates cosider the figure below (b) I polar coordiates = r(cosθ + isi θ) = iad = r(cosθ isi θ) = i The (i) r cosθ = 0 ad (ii) r siθ = Coditio (i) is satisfied if θ = π Sice si π = it follows that r = Therefore = (cos π + i si π) ad = (cos π i si π) Fig B4-7: Polar coordiates The t t = (cos πt+ isi πt) ad = (cos πt isi πt) t t (c) Note that π( t+ 4) = πt+ π thus the cycles have a period of 4 Fially, sice r >, the cycles are explosive Aswers to Exercises i Appedix B page
Joh Riley 8 Jue 03 Exercise B4-4: Walrasia Dyamics xt ( + ) =A xt ( ) where A= ke (a) The eigevalues satisfy With k =, A I = ke I = 0 A I = = + + = 0 ± Solvig, = 4 4 8 3 8 4 Note that both eigevalues lie i the iterval [-,] (b) The eigevalues satisfy A I = ke I = k E µ I = 0 where µ = / k From (a), the solutio to this quadratic equatio is k = kµ = ( ± ) 4 ± µ = Hece the characteristic roots are 4 k (c) = if ( + ) =, that is 4 smaller k ad ustable for all larger k 4 k = Therefore the system is stable for all + Aswers to Exercises i Appedix B page 3