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FUNCTIONS COST 1. INTRODUCTION TO THE COST FUNCTION 1.1. Understanding and representing technology. Economists are interested in the technology used by the rm. This is usually represented by a set or a function. For example we might describe the technology n in terms of the input correspondence which maps output vectors in Rm into subsets of 2R+ . These subsets + are vectors of inputs that will produce the given output vector. This correspondence is given by V : Rm 2R+ (1) + Alternatively we could represent the technology by a function such as the production function y = f(x1 , x2 , . . . , xn ) We typically postulate or assume certain properties for various representations of technology. For example we typically make the following assumptions on the input correspondence. 1.1.1. V.1 No Free Lunch. a: V(0) = Rn + b: 0 V(y), y > 0. 1.1.2. V.2 Weak Input Disposability. y Rm , x V (y) and 1 x V (y) + 1.1.3. V.2.S Strong Input Disposability. y Rm x V (y) and x x x + 1.1.4. V.3 Weak Output Disposability. y Rm + V (y) V ( y), 0 1. V (y) n 1.1.5. V.3.S Strong Output Disposability. y, y Rm , y y V (y ) V (y) + 1.1.6. V.4 Boundedness for vector y. If y + as + , + V (y ) = =1 If y is a scalar, V (y) = y (0,+ ) 1.1.7. V.5 T(x) is a closed set. V: Rm 2Rn is a closed correspondence. + + 1.1.8. V.6 Attainability. If x V(y), y 0 and x 0, the ray { x: 0} intersects all V( y), 0. 1.1.9. V.7 Quasi-concavity. V is quasi-concave on Rm which means y, y Rm , 0 1, V(y) V(y ) + + V( y + (1- )y ) 1.1.10. V.8 Convexity of V(y). V(y) is a convex set for all y Rm + 1.1.11. V.9 Convexity of T(x). V is convex on Rm which means y, y Rm , 0 1, V(y)+(1- )V(y ) + + V( y+ (1- )y ) Date: October 4, 2005. 1 2 COST FUNCTIONS 1.2. Understanding and representing behavior. Economists are also very interested in the behavioral relationships that arise from rms optimizing decisions. These decisions result in reduced form expressions such as supply functions y = x(p, w), input demands x = x(p, w), or Hicksian (cost minimizing) demand functions x = x(w, y). One of the most basic of these decisions is the cost minimizing decision. 2. THE COST FUNCTION AND ITS PROPERTIES 2.1. De nition of cost function. The cost function is de ned for all possible output vectors and all positive input price vectors w = (w1, w2 , . . . , wn). An output vector, y, is producible if y belongs to the effective domain of V(y), i.e, Dom V = {y Rm : V (y) = } + The cost function does not exist it there is no technical way to produce the output in question. The cost function is de ned by C(y, w) = min {wx : x V (y)}, x y Dom V, w > 0, (2) or in the case of a single output C(w, y) = min{wx : f(x) y} x (3) The cost function exists because a continuous function on a nonempty closed bounded set achieves a minimum in the set (Debreu [6, p. 16]). In gure 1,the set V(y) is closed and nonempty for y in the producible output set. The function wx is continuous. Because V(y) is non-empty if contains at least one input bundle x . We can thus consider cost minimizing points that satisfy wx wx But this set is closed and bounded given that w is strictly positive. Thus the function wx will attain a minimum in the set at x . 2.2. Solution to the cost minimization problem. The solution to the cost minimization problem 2 is a vector x which depends on output vector y and the input vector w. We denote this solution by x(y,w). This demand for inputs at for a xed level of output and input prices is often called a Hicksian demand curve. 2.3. Properties of the cost function. 2.3.1. C.1.1. Non-negativity: C(y, w) 0 for w > 0. 2.3.2. C.1.2. Nondecreasing in w: If w w then C(y, w) C(y, w ) 2.3.3. C.2. Positively linearly homogenous in w C(y, w) = C(y, w), 2.3.4. C.3. C is concave and continuous in w 2.3.5. C.4.1. No xed costs: C(0, w) = 0, w > 0. We sometimes assume we have a long run problem. 2.3.6. C.4.2. No free lunch: C(y, w) > 0, w > 0, y > 0. 2.3.7. C.5. Nondecreasing in y (proportional): C( y, w) C(y, w), 0 1, w > 0. 2.3.8. C.5.S. Nondecreasing in y: C(y , w) C(y, w), y y, w > 0. 2.3.9. C.6. For any sequence y such that || y || as and w > 0, C(y , w) as . 2.3.10. C.7. C(y,w) is lower semi-continuous in y, given w > 0. w > 0. COST FUNCTIONS 3 FIGURE 1. Existence of the Cost Function 2.3.11. C.8. If the graph of the technology (GR) or T, is convex, C(y,w) is convex in y, w > 0. 2.4. Discussion of properties of the cost function. 2.4.1. C.1.1. Non-negativity: C(y, w) 0 for w > 0. Because it requires inputs to produce an output and w is positive then C(y, w) = wx(y,w) > 0 (y > 0) where x(y,w) is the optimal level of x for a given y. 2.4.2. C.1.2. Nondecreasing in w: If w w then C(y, w) C(y, w ). Let w1 w2. Let x1 be the cost minimizing input bundle with w1 and x2 be the cost minimizing input bundle with w2. Then w2x2 w2x1 because x1 is not cost minimizing with prices w2. Now w1x1 w2x1 because w1 w2 by assumption so that C(w1, y) = w1 x1 w2 x1 w2 x2 = C(w2, y) 4 COST FUNCTIONS 2.4.3. C.2. Positively linearly homogenous in w C(y, w) = C(y, w), w > 0. Let the cost minimization problem with prices w be given by C(y, w) = min {wx : x V (y)}, x y Dom V, w > 0, (4) The x vector that solves this problem will be a function of y and w, and is usually denoted x(y,w). The cost function is then given by C(y, w) = w x(y, w) Now consider the problem with prices tw (t >0) C(y, tw) = min {twx : x V (y)}, x (5) y Dom V, w > 0 (6) y Dom V, w > 0 = t min {wx : x V (y)}, x The x vector that solves this problem will be the same as the vector which solves the problem in equation 4, i.e., x(y,w). The cost function for the revised problem is then given by C(y, tw) = tw x(y, w) = tC(y, w) 2.4.4. C.3. C is concave and continuous in w To demonstrate concavity let (w, x) and (w , x ) be two cost-minimizing price-factor combinations and let w = tw + (1-t)w for any 0 t 1. Concavity implies that C(w y) tC(w, y) + (1-t) C(w , y). We can prove this as follows. We have that C(w y) = w x = tw x + (1-t)w x where x is the optimal choice of x at prices w Because x is not necessarily the cheapest way to produce y at prices w or w,we have w x C(w, y) and w x C(w y) so that by substitution C(w y) tC(w, y) + (1-t) C(w , y). The point is that if w x and w x are each larger than the corresponding term in the linear combination then C(w ,y) is larger than the linear combination. Because C(y, w)is concave it will be continuous by the property of concavity. Rockafellar [14, p. 82] shows that a concave function de ned on an open set (w > 0) is continuous. Consider gure 2. Let x be the cost minimizing bundle at prices w . Let the price of x change. At input i prices (w ), costs are at the level C(w ). If we hold input levels xed at (x ) and change wi, we move along the tangent line denoted by C(wi, w, y), where w and x represent all the prices and inputs other than the ith. Costs are higher along this line than along the cost function because we are not adjusting inputs. Along the cost function, as the price of input i increases, we probably use less of input xi and more of other inputs. 2.4.5. C.4.1. No xed costs: C(0, w) = 0, w > 0. We assume this axiom if the problem is long run, in the short run xed costs may be positive with zero output. Speci cally V.1a implies that to produce zero output, any input vector in Rn will do, including the + zero vector with zero costs. 2.4.6. C.4.2. No free lunch: C(y, w) > 0, w > 0, y > 0. Because we cannot produce outputs without inputs (V.1b: no free lunch with the technology), costs for any positive output are positive for strictly positive input prices. (7) COST FUNCTIONS 5 FIGURE 2. The Cost Function in Concave Cost C wi,w ,y wixi w x C w,y C w ,y wi wi 2.4.7. C.5. Nondecreasing in y (proportional): C( y, w) C(y, w), 0 1, w > 0. If outputs go down proportionately, costs cannot rise. This is clear because V(y1 ) is a subset of V(y2) if y1 y2 from V.3, then C(y1 , w) = min wx|x V(y1) min wx | x V(y2) = C(y2 , w). The point is that if we have a smaller set of possible x s to choose from then cost must increase. 2.4.8. C.5.S. Nondecreasing in y: C(y , w) C(y, w), y y, w > 0. If any output goes down, costs cannot increase. 2.4.9. C.6. For any sequence y such that || y || as and w > 0, C(y , w) as . This axiom implies that if outputs increase without bound, so will costs. 2.4.10. C.7. C(y,w) is lower semi-continuous in y, given w > 0. The cost function may not be continuous in output as it is in input prices, but if there are any jump points, it will take the lower value at these points. 2.4.11. C.8. If the graph of the technology (GR) or T, is convex, C(y,w) is convex in y, w > 0. If the technology is convex (more or less decreasing returns to scale as long as V.1 holds), costs will rise at an increasing rate as y increases. 2.5. Shephard s Lemma. 6 COST FUNCTIONS 2.5.1. De nition of Shephard s lemma. In the case where V is strictly quasi-concave and V(y) is strictly convex the cost minimizing point is unique. Rockafellar [14, p. 242] shows that the cost function is differentiable in w, w > 0 at (y,w) if and only if the cost minimizing set of inputs at (y,w) is a singleton, i.e., the cost minimizing point is unique. In such cases. C(y, w) = xi(y, w) (8) wi which is the Hicksian Demand Curve. As with Hotelling s lemma in the case of the pro t function, this lemma allows us to obtain the input demand functions as derivatives of the cost function. 2.5.2. Constructive proof of Shephard s lemma in the case of a single output. For a single output, the cost minimization problem is given by C(y, w) = min wx : f(x) y = 0 x (9) The associated Lagrangian is given by L = wx (f(x) y) The rst order conditions are as follows L f = wi = 0, xi xi L = (f(x) y) = 0 Solving for the optimal x s yields xi (y, w) with C(y,w) given by C(w, y) = wx(y, w) If we now differentiate 13 with respect to w we obtain (13) i = 1, . . . , n (10) (11a) (11b) (12) C xj (y, w) = n wj + xi (y, w) (14) j=1 wi wi From the rst order conditions in equation 11a (assuming that the constraint is satis ed as an equality) we have f xj Substitute the expression for wj from equation 15 into equation 14 to obtain wj = (15) C f(x) xj (y, w) = n + xi (y, w) (16) j=1 wi xj wi If > 0 then equation 11b implies(f(x)-y) = 0. Now differentiate equation 11b with respect to wi to obtain f(x(y, w)) xj (y, w) =0 xj wi which implies that the rst term in equation 16 is equal to zero and that n j=1 (17) COST FUNCTIONS 7 C(y, w) = xi (y, w) wi 2.5.3. A Silberberg [17] type proof of Shephard s lemma. Set up a function L as follows L(y, w, x) = w x C(w, y) (18) (19) where x is the cost minimizing choice of inputs for prices w. Because C(w, y) is the cheapest way to produce y at prices w, L 0. If w = w, then L will be equal to zero. Because this is the minimum value of L, the derivative of L at this point is zero so L(y, w, x) C(w, y) = xi =0 wi wi xi = C(w, y) wi (20) The second order necessary conditions for minimizing L imply that 2 C wi wj is positive semi-de nite so that 2 C wi wj is negative semi-de nite which implies that C is concave. 2.5.4. Graphical representation of Shephard s lemma. In gure 3 we hold all input prices except the jth xed at w. We assume that when the jth price is wj , the optimal input vector is ( 1, x2, . . . , xj , . . . , n ). The cost x x function lies below the tangent line in gure 3 but conincides with the tangent line at wj By differentiability the slope of the cost function at this point is the slope of its tangent i.e., C (tangent) = = xj wj wj 2.6. Sensitivity analysis. 2.6.1. Demand slopes. If the Hessian of the cost function is negative semi-de nite, the diagonal elements all must be non-positive(Hadley [11, p. 260-262]) so we have 2 C(y, w) xi (y, w) = 0 i 2 wi wi (24) (23) (22) (21) This implies then that Hicksian demand curves slope down because the diagonal elements of the Hessian of the cost function are just the derivatives of input demands with respect to their own prices. 8 COST FUNCTIONS FIGURE 3. Shephard s Lemma Cost Tangent wjxj w i xi ij C y,wj,wi C y,w j wj wj 2.6.2. Cross price effects. By homogeneity of degree zero of input demands and Euler s theorem we have n j=1 And we know that xi 0 wi by the concavity of C. Therefore j=i . xi wj 0 wj (27) (26) xi xi xi wj = j=i wj + wi = 0 wj wj wi (25) This implies that at least one cross price effect is positive. 2.6.3. Symmetry of input demand response to input prices. By Young s theorem 2 C 2C = wi wj wj wi xi xj = wj wi So cross derivatives in input prices are symmetric. (28) COST FUNCTIONS 9 2.6.4. Marginal cost. Consider the rst order conditions for cost minimization. If we differentiate 13 with respect to y we obtain C xj (y, w) = n wj (29) j=1 y y From the rst order conditions in equation 11a (assuming that the constraint is satis ed as an equality) we have f xj Substitute the expression for wj from equation 30 into equation 29 to obtain wj = C f(x) xj (y, w) = n j=1 y xj y n j=1 (30) (31) f(x) xj (y, w) = xj y If > 0 then equation 11b implies(f(x)-y) = 0. Now differentiate equation 11b (ignoring ) with respect to y to obtain n j=1 f(x(y, w)) xj (y, w) 1 =0 xj y f(x(y, w)) n j=1 xj xj (y, w) =1 y (32) This then implies that the rst term in equation 16 that C(y, w) = (y, w) (33) y The Lagrangian multiplier from the cost minimization problem is equal to marginal cost or the increase in cost from increasing the targeted level of y in the constraint. 2.6.5. Symmetry of input demand response to changes in output and changes in marginal cost with respect to input prices. Remember that 2 C 2C = y wi wi y This then implies that xi (35) = y wi The change in xi (y, w) from an increase in y equals the change in marginal cost due to an increase in wi . 2.6.6. Marginal cost is homogeneous of degree one in input prices. Marginal cost is homogeneous of degree one in input prices. C (ty, w) C(y, w) =t , t > 0. y y We can show using the Euler equation. If a function is homogeneous of degree one, then (36) (34) 10 COST FUNCTIONS n i=1 Applying this to marginal cost we obtain f xi = f(x) xi (37) M C(y, w) (y, w) 2 C(y, w) xi(y, w) 2 C(y, w) = = = = wi wi wi y y wi y i (y, w) 2 C(y, w) 2 C(y, w) wi = i wi wi = i wi wi y y wi = = y i C(y, w) wi wi by homogeneity of C(y,w) in w (38) C(y, w ), y = M C(y, w) = (y, w) COST FUNCTIONS 11 3. TRADITIONAL APPROACH TO THE COST FUNCTION 3.1. First order conditions for cost minimization. In the case of a single output, the cost function can be obtained by carrying out the maximization problem C(y, w) = min wx : f(x) y = 0 x (39) with associated Lagrangian function L = wx (f(x) y) The rst order conditions are as follows L = w1 x1 L = w2 x2 . . . L = wn xn f =0 x1 f =0 x2 . . . f =0 xn (41b) (40) (41a) = (f(x) y) = 0 If we take the ratio of any of the rst order conditions we obtain f xj f xi = wj wi (42) This implies that the RTS between inputs i and j is equal to the negative inverse factor price ratio because RT S = Substituting we obtain xi wj = (44) xj wi Graphically this implies that the slope of an isocost line is equal to the slope of the lower boundary of V(y). Note that an isocost line is given by RT S = cost = w1 x1 + w2 x2 + . . . + wn xn wi xi = cost w1 x1 w2 x2 . . . wi 1 xi 1 wi+1 xi+1 . . . wj xj . . . wn xn cost w1 w2 wi 1 wi+1 wj wn x1 x2 . . . xi 1 xi+1 . . . xj . . . xn wi wi wi wi wi wi wi wj Slope of isocost = wi (45) xj = f xj xi = f xj x i (43) 12 COST FUNCTIONS In gure 4, we can see that slope of the isocost line is tangent to the the slope of the lower boundary of V(y) at the cost minmimizing point. FIGURE 4. Cost Minimizing Input Combination xj 40 35 30 25 Cost i wi xi Vy 20 15 10 5 xi 10 20 30 40 50 In gure 5, we can see that slope of the isocost line is tangent to the the slope of the lower boundary of V(y) at different levels of output. COST FUNCTIONS 13 FIGURE 5. Cost Minimizing Input Combinations at Alternative Output Levels xj 3 i 35 C wi xi 30 25 C 2 i wi xi V y3 20 C 1 i wi xi 15 V y2 V y1 10 5 xi 10 3.2. Notes on quasiconcavity. De nition 1. A real valued function f, de ned on a convex set X n 20 30 40 50 60 70 , a said to be quasiconcave if (46) f( x1 + +(1 ) x2) min[f(x1 ), f(x2 ) ] A function f is said to be quasiconvex if - f is quasiconcave. 14 COST FUNCTIONS Theorem 1. Let f be a real valued function de ned on a convex set X Rn. The upper contour sets {(x, y): x S, f(x)} of f are convex for every R if and only if f is a quasiconcave function. Proof. Suppose that S(f, ) is a convex set for every R and let x1 X, x2 X, = min[f(x1 ), f(x2 )]. Then x1 S(f, ) and x2 S(f, ), and because S(f, ) is convex, ( x1 + (1- )x2) S(f, ) for arbitrary . Hence f( x1 + (1 ) x2 ) = min[f(x1 ), f(x2 ) ] Conversely, let S(f, ) be any level set of f. Let x1 S(f, ) and x2 S(f, ). Then f(x1 ) , and because f is quasiconcave, we have f(x2 ) (47) (48) f( x1 + (1 ) x2 ) and ( x1 + (1- ) x2 ) S(f, ). (49) Theorem 2. Let f be differentiable on an open convex set X Rn. Then f is quasiconcave if and only if for any x1 X, x2 X such that f(x1 ) f(x2 ) we have (x1 x2) f(x2 ) 0 (51) (50) De nition 2. The kth-ordered bordered determinant Dk (f,x) of a twice differentiable function f at at point x Rn is de ned as 2 f 2f 2f f x1 xk x1 x1 x2 x2 21 2f 2f f f x2 xk x2 2 x2 x1 x2 . . . . k = 1, 2, . . . , n . . . Dk (f, x) = det . (52) . . . . 2f 2 2 f f f xk x1 xk x2 xk x2 k f f f 0 x1 x2 xk De nition 3. Some authors de ne the kth-ordered bordered determinant Dk (f,x) of a twice differentiable function f at at point x Rn in a different fashion where the rst derivatives of the function f border the Hessian of the function on the top and left as compared to in the bottom and right as in equation 52. f x 1 f Dk (f, x) = det x2 . . . 0 f x1 2f x2 1 2f x2 x1 f x2 2f x1 x2 2f x2 2 . . . . . . 2f xk x1 2f xk x2 2f x1 xk 2 f x2 xk f xk . . . f xk 2f x2 k k = 1, 2, . . . , n (53) The determinant in equation 52 and the determinant in equation 53 will be the same. If we interchange any two rows or any two columns of a determinant, the determinant will change sign but keep its absolute COST FUNCTIONS 15 value. A certain number of row exchanges will be necessary to move the bottom row to the top. A like number of column exchanges will be necessary to move the rightmost column to the left. Given that equations 52 and 53 are the same except for this even number of row and column exchanges, the determinants will be the same. You can illustrate this to yourself using the following three variable example. 2f f f f 2f 2f f 0 x1 x2 x3 x1 x2 x1 x3 x1 x2 21 2f 2f 2f f 2f 2f f f x1 x1 x2 x1 x3 x2 2 x2 x1 1 x2 x3 x2 x2 HB = 2 HB = f 2f 2f 2f 2f 2f f f 2 x2 x2 x1 x2 x3 x2 x3 x1 x3 x2 x3 x2 3 f 2f 2f 2f f f f 0 x3 x3 x1 x3 x2 x2 x1 x2 x3 3 3.2.1. Quasi-concavity and bordered hessians. a.: If f(x) is quasi-concave on a solid (non-empty interior) convex set X Rn, then ( 1)k Dk (f, x) 0, for every x X. b.: If ( 1)k Dk (f, x) > 0, k = 1, 2, . . . , n (55) for every x X,then f(x) is quasi-concave on X (Avriel [3, p.149], Arrow and Enthoven [2, p. 781-782] If f is quasiconcave, then when k is odd, Dk (f,x) will be negative and when k is even, Dk (f,x) will be positive. Thus Dk (f,x)will alternate in sign beginning with positive in the case of two variables. 3.2.2. Relationship of quasi-concavity to signs of minors (cofactors) of a matrix. Let 0 f x1 f x2 f x1 2f x2 1 2f x2 x1 f x2 2f x1 x2 2f x2 2 k = 1, 2, . . . , n (54) . . . F= f xn 2f x1 xn 2f x2 xn = det HB (56) . . . f xn 2f xi xj . . . 2f xn x1 2f xn x2 . . . 2f x2 n where det HB is the determinant of the bordered Hessian of the function f. Now let Fij be the cofactor of in the matrix HB . It is clear that Fnn and F have opposite signs because F includes the last row and column of HB and Fnn does not. If the (-1)n in front of the cofactors is positive then Fnn must be positive with F negative and vice versa. Since the ordering of rows since is arbitrary it is also clear that Fii and F have opposite signs. Thus when a function is quasi-concave Fii will have a negative sign. F 3.3. Second order conditions for cost minimization. 3.3.1. Note on requirements for a minimum of a constrained problem. Consider a general constrained optimization problem with one constraint. maximize f(x) x subject to g(x) = 0 16 COST FUNCTIONS where g(x) = 0 denotes a constraint, We can also write this as max f(x1 , x2, . . . , xn ) x1 , x2 ,...xn subject to g (x1, x2, . . . , xn) = 0 The solution can be obtained using the Lagrangian function L(x; ) = f(x1 , x2 , . . .) g(x) Notice that the gradient of L will involve a set of derivatives, i.e. xL (57) (58) = x f(x) x g(x) There will be one equation for each x. There will also an equation involving the derivative of L with respect to . The necessary conditions for an extremum of f with the equality constraints g(x) = 0 are that L(x , ) = 0 (59) where it is implicit that the gradient in (59) is with respect to both x and . The typical suf cient conditions for a maximum or minimum of f(x1, x2 , . . . , xn ) subject to g(x1 , x2 ,. . . , xn ) = 0 require that f and g be twice continuously differentiable real-valued functions on Rn . Then if there exist vectors x Rn, and R1 such that L(x , ) = 0 and for every non-zero vector z Rn satisfying z it follows that z 2 x (60) g(x ) = 0 (61) L(x , )z > 0 (62) then f has a strict local minimum at x , subject to g(x) = 0. If the inequality in (62) is reversed, then f has strict local maximum at x . For the cost minimization problem the Lagrangian is given by L = wx (f(x) y) (63) where the objective function is wx and the constraint is f(x) - y = 0. Differentiating equation 63 twice with respect to x we obtain 2 x L(x , ) = = 2 f(x) xi xj 2 f(x) xi xj (64) And so the condition in equation 62 will imply that COST FUNCTIONS 17 z (65) 2 f(x) z z<0 xi xj for all z satisfying z f(x) = 0. This is also the condition for f to be a quasi-concave function. (Avriel [3, p.149]. Thus these suf cient conditions imply that f must be quasi-concave. 3.3.2. Checking the suf cient conditions for cost minimization. Consider the general constrained minimization problem where f and g are twice continuously differentiable real valued functions. If there exist vectors x Rn, Rm , such that L(x , ) = 0 and if 2L(x , ) 2 L(x , ) ... x1 x1 x1 xp D(p) = ( 1) det 2 L(x , ) 2 L(x , ) ... x x xp xp p 1 g(x ) g(x ) ... x1 xp for p = 2, 3, . . ., n, then f has a strict local minimum at x , such that g(x ) x1 >0 g(x ) xp 0 (66) 2 f(x) z>0 xi xj (67) g(x ) = 0 We can also write this as follows where after multiplying both sides by negative one. 2 L(x , ) 2 L(x , ) g(x ) ... x1 x1 x1 xp x1 <0 D(p) = det 2 L(x , ) 2 L(x , ) g(x ) ... x x xp xp xp p 1 g(x ) g(x ) ... 0 x1 xp for p = 2, 3, . . ., n, then f has a strict local minimum at x , such that (68) (69) g(x ) = 0 (70) We check the determinants in (69) starting with the one that has 2 elements in each row and column of the Hessian and 1 element in each row or column of the derivative of the constraint with respect to x. For the cost minimization problem, D(p) is given by 18 COST FUNCTIONS 2 f x2 1 2 f x2 x1 2 f x1 x2 2 f x2 2 2 f x1 xp 2 f x2 xp f x1 f x2 D(p) = . . . f xp x1 f x1 2 . . . f xp x2 f x2 2 . . . f x2 p f xp 2 . . . f xp (71) In 0 order to factor - out of the determinant and write the second order conditions in terms of the bordered Hessian of the production function multiply last row and last column of D(p) by - , multiply out front by 1 2 , and then compute the determinant. 2 f x2 1 2 f x2 x1 2 f x1 x2 2 f x2 2 D(p) = 1 2 2 f x1 xp 2 f x2 xp f x1 f x2 . . . f xp . . . 2 f xp x1 f x1 . . . 2 f xp x2 f x2 . . . 2 f x2 p f xp (72) 0 Now factor - out of all p+1 rows of D(p). 2f x2 1 2f x2 x1 2f x1 x2 2f x2 2 D(p) = ( )p+1 1 2 2f x1 xp 2f x2 xp f x1 f x2 . . . f xp x1 f x1 2 . . . f xp x2 f x2 2 . . . f x2 p f xp 2 . . . f xp (73) 0 Because f(x) must be greater than or equal to y in the cost minimization problem, we can assume > 0 so that the sign of the D(p) in equation 73 is equal to the sign of 2f x2 1 2f x2 x1 2f x1 x2 2f x2 2 2f x1 xp 2f x2 xp f x1 f x2 ( 1) p+1 . . . f xp x1 f x1 2 . . . f xp x2 f x2 2 . . . f x2 p f xp 2 . . . f xp (74) 0 For a minimum we want |D| < 0 for p = 2,...,n. Therefore for a minimum, a suf cient condition is that COST FUNCTIONS 19 2f x2 1 2f x2 x1 2f x1 x2 2f x2 2 2f x1 xp 2f x2 xp f x1 f x2 ( 1)p+1 . . . f xp x1 f x1 2 . . . f xp x2 f x2 2 . . . f x2 p f xp 2 . . . f xp <0 (75) 0 or 2f x2 1 2f x2 x1 2f x1 x2 2f x2 2 2f x1 xp 2f x2 xp f x1 f x2 ( 1)p . . . f xp x1 f x1 2 . . . f xp x2 f x2 2 . . . f x2 p f xp 2 . . . f xp >0 (76) 0 This condition in equation 76 is the condition for the quasi-concavity of the production function f from equation 55. (Avriel [3, p. 149]. 3.3.3. Example problem with two variable inputs. The Lagrangian function is given by L = w1 x1 + w2 x2 (f(x1 , x2) y) The rst order conditions are as follows L f = w1 =0 x1 x1 L f = w2 =0 x2 x2 (f(x) y) = 0 The bordered Hessian for the Lagrangian is given by 2 L(x , ) 2 L(x , ) x1 x1 x1 x2 2 L(x , ) 2 L(x , ) HB = x2 x1 x2 x2 g(x ) g(x ) x1 x2 2 f 2 f f x1 2 = x xf 21 f x1 2 (77) (78a) (78b) (78c) g(x ) x1 g(x ) x2 0 (79) x1 x2 2 f x2 2 f x2 x1 f x2 0 20 COST FUNCTIONS The determinant of this matrix must be negative for this solution to be a minimum. To see how this relates to the bordered Hessian of the production function multiply the last row and column by - and the 2 whole determinant by 1 as follows 2f 2 x1 2f x2 x1 f x1 2 f x1 x2 2 f x2 2 f x2 f x1 f x2 0 = 1 2 2 f x2 1 2 f x2 x1 f x1 2 f x1 x2 2 f x2 2 f x2 f x1 f x2 = ( )3 1 2 0 2f x2 1 2f x2 x1 f x1 2f x1 x2 2f x2 2 f x2 f x1 f x2 0 (80) With > 0 this gives 2f x2 1 2f x2 x1 f x1 2f x1 x2 2f x2 2 f x2 f x1 f x2 f x1 f x2 ( 1)3 so that (81) 0 0 for a minimum. This is the condition for a quasi-concave function with two variables. If there were three variables, the determinant of the next bordered Hessian would be negative. 2f x2 1 2f x2 x1 f x1 2f x1 x2 2f x2 2 f x2 >0 (82) COST FUNCTIONS 21 3.4. Input demands. If we solve the equations in 41 for xj , j = 1, 2, . . ., n, and , we obtain the optimal values of x for a given y and w. As a function of w for a xed y, this gives the vector of factor demands for x and the optimal Lagrangian multiplier (y,w). x = x(y, w) = ( 1(y, w), x2 (y, w), . . . , xn(y, w)) x = (y, w) (83) 3.5. Sensitivity analysis. We can investigate the properties of x(y,w) by substituting x(y,w) for x in equa tion 41 and then treating it as an identity. L = w1 x1 L = w2 x2 . . . L = wn xn f( 1 (y, w), x2(y, w), . . . , xn(y, w)) x =0 x1 f( 1 (y, w), x2(y, w), . . . , xn(y, w)) x =0 x2 . . . (84a) f( 1 (y, w), x2(y, w), . . . , xn(y, w)) x =0 xn (f(( 1 (y, w), x2(y, w), . . . , xn(y, w)) y) = 0 x (84b) Where it is obvious we write x(y,w) for x(y,w) and xj (y,w) for xj (y,w). If we differentiate the rst equation in 84a with respect to wj we obtain 0 2 f (x(y, w)) x1 (y, w) 2 f (x(y, w)) x2 (y, w) 2 f (x(y, w)) x3 (y, w) + + + x2 wj x2 x1 wj x3 x1 wj 1 f (x(y, w)) (y, w) 0 x1 wj x1 (y, w) wj x2 (y, w) wj x3 (y, w) wj 2 f (x(y, w)) x2 1 2 f (x(y,w)) x2 x1 2 f (x(y,w)) x3 x1 2 f (x(y,w)) xn x1 f (x(y, w)) x1 . . . xn (y, w) wj 1 (y, w) wj 0 (85) If we differentiate the second equation in 84a with respect to wj we obtain 22 COST FUNCTIONS 0 2 f (x(y, w)) x1 (p, w) 2 f (x(y, w)) x2 (y, w) 2 f (x(y, w)) x3 (y, w) + + + x1 x2 wj x2 wj x3 x2 wj 2 f (x(y, w)) (y, w) 0 x2 wj x1 (y, w) wj x2 (y, w) wj x3 (y, w) wj 2 f (x(y, w)) x1 x2 2 f (x(y,w)) x2 2 2 f (x(y,w)) x3 x2 2 f (x(y,w)) xn x2 f (x(y, w)) x2 . . . xn (y, w) wj 1 (y, w) wj 0 (86) If we differentiate the jth equation in 84a with respect to wj we obtain 1 2 f (x(y, w)) x1 (y, w) x1 xj wj + 2 f (x(y, w)) x2 (y, w) x2 xj wj + + 2 f (x(y, w)) xj (y, w) f (x(y, w)) (y, w) + 0 2 xj wj xj wj x (y, w) 1 wj x2 (y, w) wj x (y, w) 3 wj . . . x (y, w) n wj 2 f(x(y, w)) x1 xj 2 f(x(y,w)) x2 xj 2 f(x(y,w)) 2 f(x(y,w) xn xj x2 j f(x(y, w)) xj 1 (y, w) wj 1 (87) Continuing in the same fashion we obtain 2 f (x(y, w)) 2 2 x1 f (x(y, w)) x1 x2 . . . 2 f (x(y, w)) x1 xj . . . 2 f (x(y, w)) x1 xn 2 f (x(y,w) x2 x1 2 f (x(y,w) x2 2 . . . . . . 2 f (x(y,w) xj x1 2 f (x(y,w) xj x2 . . . . . . 2 f (x(y,w) xn x1 2 f (x(y,w) xn x2 f (x(y,w) x1 . . . 2 f (x(y,w) x2 xj . . . 2 f (x(y,w) x2 j . . . 2 f (x(y,w) xn xj . . . 2 f (x(y,w) x2 xn . . . 2 f (x(y,w) xj xn . . . 2 f (x(y,w) x2 n f (x(y,w) xj f (x(y,w) f (x(y,w) x2 xn x1 (y, w) wj x2 (y, w) wj . . . xj (y, w) wj . . . xn (y, w) wj 1 (y, w) wj 0 0 . . . 1 . . . 0 (88) This is a system of n equations in n+1 unknowns. Equation 84b implies that - (y,w) (f(( 1 (y, w), x2 (y, w), . . . , xn(y, w)) x If we differentiate equation 84b with respect to wj we obtain f (x(y, w)) x1 (y, w) f (x(y, w) x2 (y, w) f (x(y, w) xj (y, w) f (x(y, w) xn (y, w) + + + + + x1 wj x2 wj xj wj xj wj ( f (x(y, w) y ) (y, w) 0 wj (89) If the solution is such that (y,w) is not zero, then ( f (x(y, w) y ) is equal to zero and we can write COST FUNCTIONS 23 f (x(y, w)) x1 f (x(y,w) x2 f (x(y,w) xj f (x(y,w) xn 0 x1 (y, w) wj x2 (y, w) wj x3 (y, w) wj . . . xn (y, w) wj 1 (y, w) wj 0 (90) Combining equations 88 and 90 we obtain 2 f (x(y, w)) x1 x2 . . . 2 f (x(y, w)) x1 xj . . . 2 f (x(y, w)) x x n 1 f (x(y, w)) x1 2 f (x(y, w)) x2 1 2 f (x(y,w) x2 x1 2 f (x(y,w) x2 2 . . . . . . 2 f (x(y,w) xj x1 2 f (x(y,w) xj x2 . . . . . . 2 f (x(y,w) xn x1 2 f (x(y,w) xn x2 . . . 2 f (x(y,w) x2 xj . . . 2 f (x(y,w) x2 j . . . 2 f (x(y,w) xn xj . . . 2 f (x(y,w) x2 xn f (x(y,w) x2 . . . 2 f (x(y,w) xj xn f (x(y,w) xj . . . 2 f (x(y,w) x2 n f (x(y,w) xn x1 (y, w) 0 wj x2 (y, w) f (x(y,w) 0 x2 wj . . . . . . xj (y, w) f (x(y,w) 1 wj xj . . . . . . f (x(y,w) xn (y, w) 0 wj xn (y, w) 1 0 0 wj f (x(y,w) x1 (91) If we then consider derivatives with respect to each of the wj we obtain 2 f (x(y, w)) x x 1 2 . . . 2 f (x(y, w)) x1 xn f (x(y, w)) x1 2 f (x(y, w)) x2 1 . . . 2 f (x(y,w) xn x1 f (x(y,w) xn x2 2 . . . 2 f (x(y,w) x2 n f (x(y,w) xn f (x(y,w) x2 . . . f (x(y,w) xn f (x(y,w) x1 x1 (y, w) w1 x2 (y, w) w1 x1 (y, w) w2 x2 (y, w) w2 x1 (y, w) wn x2 (y, w) wn . . . xn (y, w) w1 1 (y, w) w1 xn (y, w) w2 1 (y, w) wn 0 xn (y, w) wn 1 (y, w) wn 1 0 . . . 0 0 0 1 . . . 0 0 . . . 0 0 . . . 1 0 (92) The matrix on the leftmost side of equation 92 is the just the bordered Hessian of the production function from equation 76 which is used in verifying that the extreme point is a minimum where p = n. We repeat it here for convenience. 2f x2 1 2f x2 x1 2f x1 x2 2f x2 2 2f x1 xp 2f x2 xp f x1 f x2 ( 1)p . . . 2f xp x1 f x1 . . . 2f xp x2 f x2 . . . 2f x2 p f xp . . . f xp >0 0 3.5.1. Obtaining derivatives using Cramer s rule. We can obtain the various derivatives using Cramer s rule. For example to obtain x1 (y, w) we would replace the rst column of the bordered Hessian with the rst w1 column of matrix on the right of equation 91. To obtain x2 (y, w) w1 we would replace the second column of 24 COST FUNCTIONS the bordered Hessian with the rst column of matrix on the right of equation 91. To obtain x2 (y, w) , we wj would replace the second column of the bordered Hessian with the jth column of matrix on the right of equation 91 and so forth. Consider for example the case of two variable inputs. The bordered Hessian is given by 2f 2f f HB x1 2f = x x 21 f x1 2 0 The determinant of the HB in equation 93 is often referred to as F and is given by 2f x2 1 2f x2 x1 f x1 f x2 2f x1 x2 2f x2 2 x1 x2 2f x2 2 f x2 x1 f x2 (93) F = 0 ( 1) 6 f + ( 1)5 x2 2f x2 1 2f x2 x1 f x1 f x2 2f x2 1 2f x2 x1 f x1 f x2 f + ( 1)4 x1 2f x1 x2 2f x2 2 f x1 f x2 f = x1 2f x1 x2 2f x2 2 f x2 (94) f 2f 2 f f f x2 x2 x2 x1 x2 x1 1 2f x2 2 f x2 2 f f 2f f 2 f = x1 x1 x2 x2 x1 x2 2 =2 f f 2f x1 x2 x1 x2 f x1 2 2f x2 1 2 2 = 2 f1 f2 f12 f1 f22 f2 f11 We know that this expression is positive by the quasiconcavity of f or the second order conditions for 0 x cost minimization. We can determine w1 by replacing the rst column of F with the vector 1 . Doing so 2 0 we obtain 2f f 0 x1 x2 x1 2f f 1 (95) x2 x2 2 f 0 0 x2 The determinant of the matrix in equation 95 is easy to compute and is given by 0 f ( 1)5 x2 1 f x1 f x2 (96) f f f f = = 0 x2 x1 x1 x2 x x Given that F is positive, this implies that w1 is positive. To nd w2 we replace the second column of F 2 2 0 with the vector 1 . 0 COST FUNCTIONS 25 2f 2 x1 2f x2 x1 f x1 0 1 0 f x1 f x2 (97) 0 The determinant of the matrix in equation 97 is also easy to compute and is given by 0 f ( 1)4 x1 1 f x1 f x2 = x2 w2 f f x1 x1 is negative. = f x1 2 0 (98) Given that F is positive, this implies that 3.5.2. Obtaining derivatives by matrix inversion. We can also solve equation 92 directly for the response function derivatives by inverting the bordered Hessian matrix. This then implies that x (y, w) 1 w1 x2 (y, w) w1 . . . xn (y, w) w1 x1 (y, w) w2 x2 (y, w) w2 1 (y, w) w1 xn (y, w) w2 1 (y, w) w2 2 f(x(y, w)) x2 1 2 f(x(y, w)) x1 x2 1 . . . 2 f(x(y, w)) xn (y, w) wn x1 xn 1 (y, w) f(x(y, w)) x1 (y, w) wn x2 (y, w) wn 2 f(x(y,w) x2 x1 2 f(x(y, w)) x2 2 . . . 2 f(x(y,w) xn x1 2 f(x(y,w) xn x2 1 f(x(y,w) x1 1 f(x(y,w) x2 0 1 . . . 0 0 . . . 0 . . . 2 f(x(y, w)) x2 xn f(x(y, w)) x2 . . . 2 f(x(y,w) x2 n f(x(y,w) xn wn x1 f(x(y,w) xn 0 0 . . . 0 0 0 (99) 0 . . . 1 x For example w2 would be the (2,2) element of the inverse of the bordered Hessian. We can determine 2 this element using the formula for the inverse of a matrix involving the determinant of the matrix and the adjoint of the matrix. The adjoint of the matrix A denoted adj (A) or A+ is the transpose of the matrix obtained from A by replacing each element aij by its cofactor Aij . For a square nonsingular matrix A, its inverse is given by A 1 = 1 A+ |A | (100) We can compute (2,2) element of the inverse of the bordered Hessian from the cofactor of that element divided by the determinant of the bordered Hessian which we have denoted F. It is clear from section 3.2.2 that F22 (the cofactor of the (2,2) element) and F have opposite signs so that the partial derivative will be x x negative. Given that the Hessian is a symmetric matrix, it is also clear that wi = wj . j i Consider the two variable case with bordered Hessian HB = 2f 2 x1 2f x2 x1 f x1 2f x1 x2 2f x2 2 f x2 f x1 f x2 (101) 0 We found the determinant of HB in equation 94, i.e, det HB = F = 2 2f f f x1 x2 x1 x2 2 f1 f22 f x1 2 f2 f11 2 2f x2 2 f x2 2 2f x2 1 (102) = 2 f1 f2 f12 26 COST FUNCTIONS Each element of the inverse will have this expression as a denominator. To compute the numerator of the (1,1) element of the inverse we compute the cofactor of the (1,1) element of the bordered Hessian. This is given by 2f x2 2 f x2 2 f x2 ( 1) 2 0 (103) = f x2 To compute the numerator of the (1,2) element of the inverse we compute the cofactor of the (2,1) element of the bordered Hessian. This is given by 2f x1 x2 f x2 f x1 ( 1) 3 0 (104) f f = x1 x2 To compute the numerator of the (3,2) element of the inverse we compute the cofactor of the (2,3) element of the bordered Hessian. This is given by 2f x2 1 f x1 2f x1 x2 f x2 ( 1) 5 = ( 1) = 2 f f f 2f 2 x1 x2 x1 x2 x1 (105) f 2f f 2 f x1 x1 x2 x2 x2 1 To compute the numerator of the (3,3) element of the inverse we compute the cofactor of the (3,3) element of the bordered Hessian. This is given by 2f x2 1 f x2 x1 2f x1 x2 2f x2 2 ( 1) 6 = 2 f f 2f 2f 2 2 x1 x2 x1 x2 x1 x2 2 f f x2 x2 1 2 2f x1 x2 2 (106) = This expression will be positive or zero for a concave function. We can now start to ll in the inverse matrix of the bordered Hessian. COST FUNCTIONS 27 (HB ) 1 = 2 f x2 2 2 f 2f f f f f 2 x x x x x x 1 2 1 2 1 2 x2 2 f f x1 x1 2 2 2f1 f2 f12 f1 f22 f2 f11 2 2 f x2 1 f f x1 x1 2 2f1 f2 f12 f1 f22 2 f2 f11 2f1 f2 f12 ? 2 2 f1 f22 f2 f11 f 2 f x2 x2 1 2 f22 f2 f11 2 2f x1 x2 2 f22 f2 f11 2f1 f2 f12 2 f11 2 2 f1 f22 f2 f11 2 f x f 2 x2 1 2 2 f1 f22 f2 f11 2 f f x1 x1 x2 2 2f1 f2 f12 f1 2 f f x2 x2 1 2 2f1 f2 f12 2 f2 ? 2 2 f1 f22 f2 f11 2 f f x1 x1 x2 2f1 f2 f12 2 2f1 f2 f12 f1 f1 f2 f2 f12 f1 f22 2 f1 f1 f12 f2 f11 f1 f2 2 ? f1 f12 f2 f11 f11 f22 f12 = 2 2 2f1 f2 f12 f1 f22 f2 f11 10 If we postmultiply this matrix by 0 1 we can obtain the comparative statics matrix. 00 f f 1 x1 x1 2 2 2f1 f2 f12 f1 f22 f2 f11 f f 2 f 2 f x2 x1 x2 x1 x2 2 2 2 2f1 f2 f12 f1 f22 f2 f11 (107) f x 2 2 2 2f1 f2 f12 f1 f22 f2 f11 2 f f x1 x1 2 2 2f1 f2 f12 f1 f22 f2 f11 2 f11 2 2 2f1 f2 f12 f1 f22 f2 f11 f 2 f 2 f f x1 x1 x2 x2 x2 1 2 2 2f1 f2 f12 f1 f22 f2 f11 2f f 2 f f x 2 x2 x1 x2 1 x2 2 2 2f1 f2 f12 f1 f22 f2 f11 2f f 2 f f x 2 x1 x1 x2 2 x1 2 2 2f1 f2 f12 f1 f22 f2 f11 2 2 f 2 f 2 f x x 1 2 x2 x2 1 2 2 2 2f1 f2 f12 f1 f22 f2 f11 1 0 0 0 1 0 x1 (y, w) w1 x2 (y, w) w1 1 (y, w) w1 x1 (y, w) w2 x2 (y, w) w2 1 (y, w) w2 (108) For example, 1 x2 w1 is equal to 2 f f 2 f x f x1 x1 x2 2 x2 1 2 2 2f1 f2 f12 f1 f22 f2 f11 f f x1 x1 2 2 2f1 f2 f12 f1 f22 f2 f11 2 f11 2 2 2f1 f2 f12 f1 f22 f2 f11 f f 1 x1 x1 0 = 1 2 2 2f1 f2 f12 f1 f22 f2 f11 0 (109) or w1 is equal to 2 f f f 2 f x2 x1 x2 x1 x2 2 2 2 2f1 f2 f12 f1 f22 f2 f11 . 3.6. The impact of output on input demands. If we differentiate the rst order conditions in equation 84 with respect to y instead of with respect to wj the system in equation 91 would be as follows. 2 f (x(y, w)) x1 x2 . . . 2 f (x(y, w)) x1 xj . . . 2 f (x(y, w)) x x n 1 f (x(y, w)) x1 2 f (x(y, w)) x2 1 2 f (x(y,w) x2 x1 2 f (x(y,w) x2 2 . . . . . . 2 f (x(y,w) xj x1 2 f (x(y,w) xj x2 . . . . . . 2 f (x(y,w) xn x1 2 f (x(y,w) xn x2 . . . 2 f (x(y,w) x2 xj . . . 2 f (x(y,w) x2 j . . . 2 f (x(y,w) xn xj . . . 2 f (x(y,w) x2 xn f (x(y,w) x2 . . . 2 f (x(y,w) xj xn f (x(y,w) xj . . . 2 f (x(y,w) x2 n f (x(y,w) xn f (x(y,w) x2 f (x(y,w) xj f (x(y,w) xn f (x(y,w) x1 x1 (y, w) y x2 (y, w) y . . . xj (y, w) y . . . xn (y, w) y 1 (y, w) y 0 0 0 . . . 0 . . . 0 1 (110) 28 COST FUNCTIONS Consider the impact of an increase in output on marginal cost. To investigate this response replace the last column of the bordered Hessian with the right hand side of equation 110. This will yield 2 2 f (x(y,w) 2f f (x(y, w)) 2 f (x(y,w) xj x1 x(x(y,w) 0 x2 x1 x2 n x1 1 2 f (x(y, w)) 2 f (x(y,w) 2 2 f (x(y,w) f xj x2 x(x(y,w) 0 2 x1 x2 x2 n x2 . . . . . . . . . . . . . . . . . . 0 (111) 2 f (x(y, w)) 2 f (x(y,w) 2 f (x(y,w) 2 f (x(y,w) 0 x1 xj x2 xj xn xj x2 j 2 f (x(y, w)) 2 f (x(y,w) 2 f (x(y,w) 2 f (x(y,w) x x 0 x2 xn xj xn x2 n 1 n f (x(y, w)) f (x(y,w) f (x(y,w) f (x(y,w) 1 x1 x2 xj xn 1 We can then nd and so can be ignored. (y, w) y using Cramer s rule noting that 1 appears on both sides of the expression for (y, w) y An alternative way to investigate the impact of output on marginal cost is to totally differentiate the rst order conditions as follows. n j=1 f 2 f dxj d = 0 xi xj xi 2 f dxj f = xi xi xj d f Now substitute into the differential of the constraint by replacing xi in dy = with the above expression to obtain f dxi xi (112) (113) 2 f dxj dxi > 0 (114) xi xj because > 0 and the summation is just the quadratic form for the concave function f which is always negative. Because changes in marginal cost and output have the same signs, marginal cost is increasing as a function of output. Thus with concave production, marginal cost is always upward sloping (Shephard [16, p. 83; 88-92]). For further discussion of this traditional approach see Allen [1, p.502-504], Ferguson [10, p. 109; 137-140], or Samuelson [15, p. 64-65]. dy d = i n j=1 COST FUNCTIONS 29 4. COST FUNCTION DUALITY We can show that if we have an arbitrarily speci ed cost function that satis es the conditions in section 2.3, we can construct from it a reasonable technology (speci cally the input requirement function) that satis es the conditions we speci ed in section 1.1.1 and that would generate the speci ed cost function. 4.1. Constructing the input requirement set from the cost function. Let V (y) be de ned as V (y) {x : wx C(y, w), w > 0} w > 0 {x : w x C(y, w)} To see the intuition of why we can construct V(y) in this way consider gure 6 (115) FIGURE 6. Half-Spaces and Input Requirement Set xj 40 35 30 25 20 15 10 5 xi 10 20 30 40 50 Vy If we pick a particular set of prices then V (y) consists of all points above the line that is tangent to V(y) at the optimal input combination. The equation {x) : wx C(y, w) for a particular set of prices de nes a line in R2 or a hyperplane in Rn . Points above the line are considered to be in V (y). Now pick a different set of prices with w2 higher and construct a different line as in gure 7. The number of points in V (y) is now less than before. If we add a third set of prices with w2 lower than in the original case as in gure 8, we can reduce the size of V (y) even more. Continuing in this fashion we can recover V(y). This is an application of what is often called Minkowski s theorem. Theorem 3. Suppose V(y) satis es properties 1, 2 (strong), 5 and 7. If y 0 and is an element of the domain of V then V(y) = V (y)

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