Chapter 4. Maximum Theorem, Implicit Function Theorem and Envelope Theorem

Chapter 4. Maximum Theorem, Implicit Function Theorem and Envelope Theorem This chapter will cover three key theorems: the maximum theorem (or the theorem of maximum), the implicit function theorem, and the envelope theorem. The maximum theorem studies the continuity of the optimizer and optimum, the implicit function theorem studies the di erentiablity of the optimizer, and the envelope theorem studies the di erentiablity of the optimum, all with respect to a group of parameters. The maximum theorem was rst stated and proven by Claude Berge (959; 963, p. 6). Augustin-Louis Cauchy is credited with the rst rigorous form of the implicit function theorem. As to the envelope theorem, the clues of inventors are not very clear; see Löfgren (20) for a summary of its history. In the previous two chapters, we are "moving along a curve", that is, the objective function f and the feasible set G are given (or the parameter value is given) and we are searching over x along f to nd the optimal x. On the contrary, we are "shifting a curve" in this chapter, i.e., we change the parameter values to shift f and G, and check how the optimizer and optimum respond to such shifting. Related materials of this chapter can be found in Chapter 9., 9.2 and 9.4 of Simon and Blume (994) and Chapter 9 of Sundaram (996). The Maximum Theorem Often in economics we are not so much interested in what the solution to a particular maximization problem is but rather wish to know how the solution to a parameterized problem depends on the parameters. Thus in our example of utility maximization in Section. of Chapter 2 we might be interested not so much in what the solution to the maximization problem is when p = 2, p 2 = 7, and y = 25, but rather in how the solution depends on p, p 2, and y. (That is, we might be interested in the demand function.) Sometimes we shall also be interested in how the maximized function depends on the parameters - in the example how the maximized utility depends on p, p 2, and y. This raises a number of questions. In order for us to speak meaningfully of a demand function it should be the case that the maximization problem has a unique solution. Further, we would like to know if the demand function is continuous - or even if it is di erentiable. Consider again the Email: pingyu@hku.hk

equality-constrained maximization problem, but this time let us explicitly add some parameters. max f(x ; ; x n ; a ; ; a k ) x ; ;x n s.t. g (x ; ; x n ; a ; ; a k ) = c ;. g m (x ; ; x n ; a ; ; a k ) = c m : () The uniqueness problem is studied in the last chapter. Now let v(a ; ; a k ) be the maximized value of f when the parameters are (a ; ; a k ). Let us suppose that the problem is such that the solution is unique and that (x (a ; ; a k ); ; x n(a ; ; a k )) are the values that maximize the function f when the parameters are a (a ; ; a k ) 0 ; then v(a ; ; a k ) = f(x (a ; ; a k ); ; x n(a ; ; a k ); a ; ; a k ): (Notice however that the function v is uniquely de ned even if there is not a unique maximizer.) The Maximum Theorem gives conditions on the problem under which the function v and the functions x ; ; x n are continuous. The constraints in the problem () de ne a set of feasible vectors x over which the function f is to be maximized. Let us call this set G(a ; ; a k ), i.e., G(a ; ; a k ) = f(x ; ; x n ) j g j (x ; ; x n ; a ; ; a k ) = c j, 8 jg: Now we can restate the problem as max f(x ; ; x n ; a ; ; a k ) x ; ;x n s.t. (x ; ; x n ) 2 G(a ; ; a k ): (2) Notice that both the function f and the feasible set G depend on the parameters a, i.e., both may change as a changes. The Maximum Theorem requires both that the function f be continuous as a function of x and a and that the feasible set G(a ; ; a k ) change continuously as a changes. We already know what it means for f to be continuous but the notion of what it means for a set to change continuously is less elementary. We call G a set valued function or a correspondence. G associates with any vector (a ; ; a k ) a subset of the vectors (x ; ; x n ). The following two de nitions de ne what we mean by a correspondence being continuous. First we de ne what it means for two sets to be close. De nition Two sets of vectors A and B are within of each other if for any vector x in one set there is a vector x 0 in the other set such that x 0 2 B (x). We can now de ne the continuity of the correspondence G in essentially the same way that we de ne the continuity of a single valued function. De nition 2 The correspondence G is continuous at (a ; ; a k ) if 8 > 0, 9 > 0 such that if (a 0 ; ; a0 k ) is within of (a ; ; a k ) then G(a 0 ; ; a0 k ) is within of G(a ; ; a k ). 2

Figure : Continuous and Discontinuous Correspondence Remark If G is a function, then the continuity of G as a correspondence is equivalent to the continuity as a function. Figure shows some cases where G is continuous and discontinuous. It is, unfortunately, not the case that the continuity of the functions g j necessarily implies the continuity of the feasible set. (Exercise asks you to construct a counterexample.) Remark 2 (*) In Appendix B of Chapter 3, we de ne two weaker notions of continuity, which we call upper hemicontinuity and lower hemicontinuity. A correspondence is in fact continuous in the way we have de ned it if it is both upper hemicontinuous and lower hemicontinuous. We are now in a position to state the Maximum Theorem. We assume that f is a continuous function, that G is a continuous correspondence, and that for any (a ; ; a k ) the set G(a ; ; a k ) is compact. The Weierstrass Theorem thus guarantees that there is a solution to the maximization problem (2) for any (a ; ; a k ). Theorem (The Maximum Theorem) Suppose that f(x ; ; x n ; a ; ; a k ) is continuous (in (x ; ; x n ; a ; ; a k )), that G(a ; ; a k ) is a continuous correspondence, and that for any (a ; ; a k ) the set G(a ; ; a k ) is compact. Then (i) v(a ; ; a k ) is continuous, and (ii) if (x (a ; ; a k ); ; x n(a ; ; a k )) are (single valued) functions then they are also continuous. 3

Remark 3 (*) Our maximum theorem assumes x (a ; ; a k ) to be single valued, which can be achieved by our uniqueness theorem from Chapter 3. If x (a ; ; a k ) is set valued, then we can show x (a ; ; a k ) is a upper hemicontinuous correspondence, where the concept of upper hemicontinuity is de ned in Appendix B of Chapter 3. A single-valued upper hemicontinuous correspondence must be a continuous function; this is the second result of the maximum theorem. Exercise Show by example that even if the functions g j are continuous the correspondence G may not be continuous. [Hint: Use the case n = m = k =.] 2 The Implicit Function Theorem In Chapter 2 we said things like: Now we have three equations in x ; x 2, and the new arti cial or auxiliary variable. Again we can, perhaps, solve these equations for x ; x 2, and. In this section we examine the question of when we can solve a system of n equations to give n of the variable in terms of the others. Let us suppose that we have n endogenous variables x ; : : : ; x n, m exogenous variables or parameters, b ; : : : ; b m, and n equations or equilibrium conditions f (x ; ; x n ; b ; ; b m ) = 0; f 2 (x ; ; x n ; b ; ; b m ) = 0;. f n (x ; ; x n ; b ; ; b m ) = 0; (3) or, using vector notation, f(x; b) = 0; where f : R n+m! R n, x 2 R n, b 2 R m, and 0 2 R n. When can we solve this system to obtain functions giving each x i as a function of b ; : : : ; b m? As we ll see below we only give an incomplete answer to this question, but rst let s look at the case that the function f is a linear function. Suppose that our equations are a x + + a n x n + c b + c m b m = 0; a 2 x + + a 2n x n + c 2 b + c 2m b m = 0;. a n x + + a nn x n + c n b + c nm b m = 0: We can write this, in matrix notation, as [A j C] x b! = 0; where A is an n n matrix, C is an n m matrix, x is an n (column) vector, and b is an m 4

vector. This we can rewrite as Ax + Cb = 0; and solve this to give x = A Cb: And we can do this as long as the matrix A can be inverted, that is, as long as the matrix A is of full rank. By Cramer s rule, we can obtain each component of x as the ratio of two determinants: x i = ja ij jaj ; where A i is the matrix obtained by replacing the ith column of A by the vector Cb. As a result, @x @b 0 = A C: Our answer to the general question in which the function f may not be linear is that if there are some values (x; b) for which f(x; b) = 0 then if, when we take a linear approximation to f we can solve the approximate linear system as we did above, then we can solve the true nonlinear system, at least in a neighborhood of (x; b). By this last phrase we mean that if b is not close to b we may not be able to solve the system, and that for a particular value of b there may be many values of x that solve the system, but there is only one close to x. To see why we can t, in general, do better than this consider the equation f : R 2! R given by f(x; b) = g(x) b, where the function g is graphed in Figure 2. Notice that the values (x; b) satisfy the equation f(x; b) = 0. For all values of b close to b we can nd a unique value of x close to x such that f(x; b) = 0. However, () for each value of b there are other values of x far away from x that also satisfy f(x; b) = 0, and (2) there are values of b, such as ~ b for which there are no values of x that satisfy f(x; b) = 0. Let us consider again the system of equations (3). Theorem 2 (Implicit Function Theorem) Suppose that f : R n+m! R n is a C function on an open set A R n+m and that (x; b) in A is such that f(x; b) = 0. Suppose also that is of full rank. 0 @f(x; b) @x 0 = B @ @f (x;b) @x. @f n(x;b) @x @f (x;b) @x n.... @f n(x;b) @x n Then there are open sets A R n and A 2 R m with x 2 A, b 2 A 2 and A A 2 A such that for each b in A 2 there is exactly one g(b) in A such that f(g(b); b) = 0. Moreover, g : A 2! A is a C function and @g(b) @b 0 nm = @f(g(b); b) @x 0 nn C A @f(g(b); b) @b 0 ; nm 5

Figure 2: Intuition for the Implicit Function Theorem or more compactly, D b g(b) = [D x f(g(b); b)] D b f(g(b); b): How to understand this theorem? Suppose g(b) satis es f(g(b); b) = 0. By di erentiating the system of equations with respect to b, we have so @f(g(b); b) @g(b) @f(g(b); b) @x 0 @b 0 + @b 0 = 0 nm ; @g(b) @b 0 = @f(g(b); b) @f(g(b); b) @x 0 @b 0 : The implicit function theorem (IFT) is very useful in the so-called comparative statics. Suppose the equality-constrained problem we are considering is max f(x; c) x s.t. h(x; d) = 0; where x 2 R n, c 2 R L, d 2 R M, and h 2 R K. We can also assume there is some overlap between 6

c and d. Form the Lagrangian L(x; ; c; d) = f(x; c) + h(x; d); then the FOCs are Casting in the notation of the IFT, @L @x (x ; ; c; d) = @f @x (x ; c) + @h(x ; d) 0 = 0; @x @L @ (x ; ; c; d) = h(x ; d) = 0: f(x; b) = @f @x @h(x;d)0 (x; c) + @x h(x; d)! = D x; L(x ; ; c; d) 0 ; where x (x; ), b (c; d), n n + K, and m (L + M), i.e., there are (n + K) unknowns (x; ) and (L + M) parameters. Suppose f and h are C 2 functions; then f(x; b) is a C function as required by the IFT. Suppose that (x ; ) satis es @f @x (x ; c) + @h(x ;d) 0 @x h(x ; d)! = 0; and D 2 xf(x ; c) + D 2 x (h(x ; d) 0 ) (D x h(x ; d)) 0 D x h(x ; d) 0 KK! is of full rank, where D 2 x (h(x ; d) 0 ) = P K k= k D2 xh k (x ; d). Then the assumptions of the IFT are satis ed, and we can apply the theorem to have @ x 0 ; 0 0 @ (c 0 ; d 0 ) = D 2 xf(x ; c) + D 2 x (h(x ; d) 0 ) (D x h(x ; d)) 0 D x h(x ; d) 0 KK! @ 2 f @h(x ;d) 0 @x@d 0 @h(x 0 ;d) KL @d 0 @x@c 0 (x ; c) where (x ; ) are treated as functions of (c; d). It is interesting to observe that D 2 xf(x ; c) + D 2 x (h(x ; d) 0 ) is exactly D 2 xl(x ; ; c; d) in Section 3 of Chapter 3, the rst matrix on the right hand side is D 2 x; L(x ; ; c; d), and the second matrix is @2 L(x ; ;c;d) @(x 0 ; 0 ) 0 @(c 0 ;d 0 ). Exercise 2 Consider the general utility maximization problem max u(x ; ; x n ) x ; ;x n s.t. p x + + p n x n = w: Suppose that for some price vector p the maximization problem has a utility maximizing bundle x. Find conditions on the utility function such that in a neighborhood of (x; p) we can solve for the demand functions x(p). Find the derivatives of the demand functions, @x=@p 0.! ; 7

Figure 3: Intuition for the Envelope Theorem Exercise 3 Now suppose that there are only two goods and the utility function is given by u(x ; x 2 ) = (x ) 3 (x2 ) 2 3 : Solve this utility maximization problem, as you learned to do in Chapter 2, and then di erentiate the demand functions that you nd to nd the partial derivative with respect to p, p 2, and w of each demand function. Also nd the same derivatives using the method of the previous exercise. 3 The Envelope Theorem In this section we examine a theorem that is particularly useful in the study of consumer and producer theory. There is in fact nothing mysterious about this theorem. You will see that the proof of this theorem is simply calculation and a number of substitutions. Moreover the theorem has a very clear intuition. It is this: Suppose we are at a maximum (in an unconstrained problem) and we change the data of the problem by a very small amount. Now both the solution of the problem and the value at the maximum will change. However at a maximum the function is at (the rst derivative is zero). Thus when we want to know by how much the maximized value has changed it does not matter (very much) whether or not we take account of how the maximizer changes or not. See Figure 3. The intuition for a constrained problem is similar and only a little more complicated. 8

To motivate our discussion of the Envelope Theorem we will rst consider a particular case, viz., the relation between short and long run average cost curves. Recall that, in general we assume that the average cost of producing some good is a function of the amount of the good to be produced. The short run average cost function is de ned to be the function which for any quantity, Q, gives the average cost of producing that quantity, taking as given the scale of operation, i.e., the size and number of plants and other xed capital which we assume cannot be changed in the short run (whatever that is). The long run average cost function on the other hand gives, as a function of Q, the average cost of producing Q units of the good, with the scale of operation selected to be the optimal scale for that level of production. That is, if we let the scale of operation be measured by a single variable k, say, and we let the short run average cost of producing Q units when the scale is k be given by SRAC(Q; k) and the long run average cost of producing Q units by LRAC(Q) then we have LRAC(Q) = min SRAC(Q; k): k Let us denote, for a given value Q, the optimal level of k by k(q). That is, k(q) is the value of k that minimizes the right hand side of the above equation. Graphically, for any xed level of k the short run average cost function can be represented by a curve (normally assumed to be U-shaped) drawn in two dimensions with quantity on the horizontal axis and cost on the vertical axis. Now think about drawing one short run average cost curve for each of the (in nite) possible values of k. One way of thinking about the long run average cost curve is as the bottom or envelope of these short run average cost curves. Suppose that we consider a point on this long run or envelope curve. What can be said about the slope of the long run average cost curve at this point. A little thought should convince you that it should be the same as the slope of the short run curve through the same point. (If it were not then that short run curve would come below the long run curve, a contradiction.) That is, d LRAC(Q) dq = @ SRAC(Q; k(q)) : @Q See Figure 4. The envelope theorem is a general statement of the result of which this is a special case. We will consider not only cases in which Q and k are vectors, but also cases in which the maximization or minimization problem includes some constraints. Let us consider again the maximization problem (). Recall: max f(x ; ; x n ; a ; ; a k ) x ; ;x n s.t. g (x ; ; x n ; a ; ; a k ) = c ;. g m (x ; ; x n ; a ; ; a k ) = c m : 9

Figure 4: The Relationship Between SRAC and LRAC Again let L(x ; ; x n ; ; ; m ; a ; ; a k ) be the Lagrangian function: mx L(x ; ; x n ; ; ; m ; a ; ; a k ) = f(x ; ; x n ; a ; ; a k )+ j (c j g j (x ; ; x n ; a ; ; a k )): Let (x (a ; ; a k ); ; x n(a ; ; a k )) and ( (a ; ; a k ); ; m(a ; ; a k )) be the values of x and that solve this problem. Now let j= v(a ; ; a k ) = f(x (a ; ; a k ); ; x n(a ; ; a k ); a ; ; a k ): That is, v(a ; ; a k ) is the maximized value of the function f when the parameters are (a ; ; a k ). The envelope theorem says that the derivative of v is equal to the derivative of L at the maximizing values of x and. Or, more precisely Theorem 3 (The Envelope Theorem) If all functions are de ned as above and the problem is 0

such that the functions x and are well de ned, then (a ; ; a k ) = @L (x @a h @a (a ; ; a k ); ; x n(a ; ; a k ); (a ; ; a k ); ; m(a ; ; a k ); a ; ; a k ) h = @f (x @a (a ; ; a k ); ; x n(a ; ; a k ); a ; ; a k ) h mx j(a ; ; a k ) @g h (x @a (a ; ; a k ); ; x n(a ; ; a k ); a ; ; a k ) h for all h. j= In order to show the advantages of using matrix and vector notation we shall restate the theorem in that notation before returning to give a proof of the theorem. (In proving the theorem we shall return to using mainly scalar notation.) @L (a) = @a @a (x (a); (a); a) = @f @a (x (a); a) Proof. From the de nition of the function v we have Thus Now, from the FOCs we have @g(x (a); a) 0 (a): @a v(a ; ; a k ) = f(x (a ; ; a k ); ; x n(a ; ; a k ); a ; ; a k ): (a) = @f (x (a); a) + @a h @a h nx i= @f (x (a); a) @a h (a): (4) @f (x (a); a) mx j= j (a) @g j (x (a); a) = 0: Or @f (x (a); a) = mx j= Also, since x (a) satis es the constraints we have, for each j g j (x (a); ; x n(a); a ; ; a k ) = c j : j (a) @g j (x (a); a): (5) And, since this holds as an identity, we may di erentiate both sides with respect to a h giving nx i= @g j (x (a); a) @a h (a) + @g j @a h (x (a); a) = 0:

Or nx i= Substituting (5) into (4) gives @g j (x (a); a) @a h (a) = @a h (a) = @f @a h (x (a); a) + Changing the order of summation gives (a) = @f (x (a); a) + @a h @a h And now substituting (6) into (7) gives @g j @a h (x (a); a): (6) 2 3 nx mx 4 j (a) @g j (x (a); a) 5 (a): @a h i= j= j= " mx nx # @g j j (a) (x (a); a) (a) @a h i= : (7) (a) = @f (x (a); a) @a h @a h mx j= j (a) @g j @a h (x (a); a); which is the required result. Exercise 4 Rewrite this proof using matrix notation. Go through your proof and identify the dimension of each of the vectors or matrices you use. For example, D x f is a n vector, and D x g is an m n matrix. 3. Interpretation of the Lagrange Multiplier A straightforward corollary of the envelope theorem is the meaning of the Lagrange multiplier. We have actually already interpreted the Lagrange multiplier intuitively as the penalty on violating the constraint in Section. of Chapter 2; here is a more rigorous statement. Treating c j, j = ; ; m, also as parameters, then by the envelope theorem, @c j (a ; ; a k ; c ; ; c m ) = j(a ; ; a k ; c ; ; c m ): How to understand this result? Think of f as the pro t function of a rm, the g j equation as the resource constraint, and c j as the amount of input j available to the rm. In this situation, @c j (a ; ; a k ; c ; ; c m ) represents the change in the optimal pro t resulting from availability of one more unit of input j. Alternatively, it tells the maximum amount the rm would be willing to pay to acquire another unit of input j. For this reason, j is often called the internal value or imputed value, or more frequently, the shadow price of input j. It may be a more important index to the rm than the external market price of input j. We consider only the equality-constrained problem in this chapter. All the results can be extended to the inequality-constrained problems. For example, in the envelope theorem, if some 2

inequality constraint is not binding, then the corresponding multiplier is zero, and we can consider only the binding constraints without loss of generality. 4 Applications to Microeconomic Theory (*) In this section, we will cover a few applications of the envelope theorem in microeconomic theory. 4. Utility Maximization Let us again consider the utility maximization problem in Chapter 2, max u(x ; x 2 ) x ;x 2 s.t. p x + p 2 x 2 y = 0: Let v(p ; p 2 ; y) be the maximized value of u when prices and income are p, p 2, and y. Let us consider the e ect of a change in y with p and p 2 remaining constant. By the Envelope Theorem @y = @ @y fu(x ; x 2 ) + (y p x + p 2 x 2 )g = 0 + = : This is the familiar result that is the marginal utility of income. 4.2 Expenditure Minimization Let us consider the problem of minimizing expenditure subject to attaining a given level of utility, i.e., P min n x ; ;x i= p ix i n s.t. u(x ; ; x n ) u 0 = 0: Let the minimized value of the expenditure function be denoted by e(p ; ; p n ; u 0 ). Then by the Envelope Theorem we obtain @e @p i = ( nx ) @ p i x i + (u 0 u(x ; ; x n )) = x i 0 = x i @p i i= when evaluated at the point which solves the minimization problem which we write as h i (p ; ; p n ; u 0 ) to distinguish this (compensated) value of the demand for good i as a function of prices and utility from the (uncompensated) value of the demand for good i as a function of prices and income. This result is known as Hotelling s Theorem. 4.3 The Hicks-Slutsky Equations It can be shown that the compensated demand at utility u 0, i.e., h i (p ; ; p n ; u 0 ) is equal to the uncompensated demand at income e(p ; ; p n ; u 0 ), i.e., x i (p ; ; p n ; e(p ; ; p n ; u 0 )). (This 3

result is known as the duality theorem.) Thus totally di erentiating the identity x i (p ; ; p n ; e(p ; ; p n ; u 0 )) = h i (p ; ; p n ; u 0 ) with respect to p k we obtain which by Hotelling s Theorem gives + @e = @h i ; @p k @y @p k @p k So @p k + @y h k = @h i @p k : @p k = @h i @p k h k @y for all i; k = ; ; n. These are the Hicks-Slutsky equations. 4.4 The Indirect Utility Function Again let v(p ; ; p n ; y) be the indirect utility function, that is, the maximized value of utility as described in Section 4.. Then by the Envelope Theorem since @u @p i @p i = @u @p i x i (p ; ; p n ; y) = x i (p ; ; p n ; y) = 0. Now, since we have already shown that = @y in Section 4. we have x i (p ; ; p n ; y) = =@p i =@y : This is known as Roy s Theorem. 4.5 Pro t functions Now consider the problem of a rm that maximizes pro ts subject to technology constraints. Let x = (x ; ; x n ) 0 be a vector of netputs, i.e., x i is positive if the rm is a net supplier of good i, negative if the rm is a net user of that good. Assume that we can write the technology constraints as F (x) = 0. Thus the rm s problem is P max n x ; ;x n i= p ix i s.t. F (x ; ; x n ) = 0: Let ' i (p) be the value of x i that solves this problem, i.e., the net supply of commodity i when 4

prices are p = (p ; ; p n ) 0. We call the maximized value the pro t function which is given by (p) = nx p i ' i (p): i= And so by the Envelope Theorem This result is known as Hotelling s lemma. 4.6 Cobb-Douglas Example @ @p i = ' i (p): We consider a particular Cobb-Douglas example of the utility maximization problem, max x ;x 2 p x p x2 s.t. p x + p 2 x 2 = w: (8) The Lagrangean is L(x ; x 2 ; ) = p x p x2 + (w p x p 2 x 2 ) and the FOCs are If we divide equation (9) by equation (0) we obtain @L = @x 2 x 2 x 2 2 p = 0; (9) @L = @x 2 2 x 2 x 2 2 p 2 = 0; (0) @L @ = w p x p 2 x 2 = 0: () x x 2 = p =p 2 or p x = p 2 x 2 ; and if we substitute this into equation () we obtain w p x p x = 0 or Similarly, x = w 2p : (2) x 2 = w 2p 2 : (3) 5

Substituting equations (2) and (3) into the utility function gives v(p ; p 2 ; w) = s w 2 4p p 2 = w 2 p p p 2 : (4) As a check here we can check some known properties of the indirect utility function. For example it is homogeneous of degree zero, that is, if we multiply p, p 2, and w by the same positive constant, say we do not change the value of v. You should con rm that this is the case. We now calculate the optimal value of from the FOCs by substituting equations (2) and (3) into (9), giving or or or w 2 w 2 2 2p 2p 2 r 2p w = p ; 2 w2p 2 p p p = ; 2 p2 p = 2 p p p 2 : p = 0; The application of the Envelope Theorem in Section 4. told us that this value of could be found as the derivative of the indirect utility function with respect to w. We con rm this by di erentiating the function we found above with respect to w: as we had found directly above. @w = @ @w w 2 p p p 2 = 2 p p p 2 ; Now, for the same utility function consider the expenditure minimization problem min x ;x 2 p x 2 + p 2 x 2 s.t. p x p x2 = u: The Lagrangian is and the FOCs are L(x ; x 2 ; ) = p x + p 2 x 2 + (u p x p x2 ) @L = p @x 2 x 2 x 2 2 = 0; (5) @L = p 2 @x 2 2 x 2 x 2 2 = 0; (6) @L @ = u p p x x2 = 0: (7) 6

Dividing equation (5) by equation (6) gives or p p 2 = x 2 x ; And, if we substitute equation (8) into equation (5) we obtain x 2 = p x p 2 : (8) u x r p p 2 ; or Similarly, r p2 x = u : p r p x 2 = u ; p 2 and if we substitute these values back into the objective function we obtain the expenditure function r r p2 p e(p ; p 2 ; u) = p u + p 2 u = 2u p p p 2 : p p 2 Hotelling s Theorem tells us that is we di erentiate this expenditure function with respect to p i we should obtain the Hicksian demand function h i. as we had already found. And similarly for h 2. @e(p ; p 2 ; u) = @ 2u p p p 2 = 2u r r p2 p2 = u ; @p @p 2 p p Let us summarize what we have found so far. The Marshallian demand functions are x (p ; p 2 ; w) = w 2p ; x 2 (p ; p 2 ; w) = w 2p 2 : The indirect utility function is The Hicksian demand functions are v(p ; p 2 ; w) = w 2 p p p 2 : r p2 h (p ; p 2 ; u) = u ; p r p h 2 (p ; p 2 ; u) = u ; p 2 7

and the expenditure function is e(p ; p 2 ; u) = 2u p p p 2 : We now look at the application concerning the Hicks-Slutsky decomposition in Section 4.3. First let us con rm that if we substitute the expenditure function for w in the Marshallian demand function we do obtain the Hicksian demand function. x (p ; p 2 ; e(p ; p 2 ; u)) = e(p ; p 2 ; u) 2p = 2u p p p 2 2p r p2 = u ; p as required. Similarly, if we plug the indirect utility function v into the Hicksian demand function h i we obtain the Marshallian demand function x i. Con rmation of this is left as an exercise. [You should do this exercise. If you understand properly it is very easy. If you understand a bit then doing the exercise will solidify your understanding. If you can t do it then it is a message to get some further explanation.] Let us now check the Hicks-Slutsky decomposition for the e ect of a change in the price of good 2 on the demand for good. The Hicks-Slutsky decomposition tells us that Calculating these partial derivatives we have and @x = 0; @p 2 @x @w = ; 2p @h @p 2 = @x @p 2 = @h @p 2 h 2 @x @w : p u p 2 p = p2 r p2 h = u : p u 2 p p p 2 ; Substituting into the right hand side of the Hicks-Slutsky equation above gives RHS = r u 2 p p2 u = 0; p p 2 p 2p which is exactly what we had found for the left hand side of the Hicks-Slutsky equation. Finally we check Roy s Theorem, which tells us that the Marshallian demand for good can be found as x (p ; p 2 ; w) = @p : @w 8

In this case we obtain x (p ; p 2 ; w) = w 2 p p2 2 3 2 2 p q = p p 2 w 2p ; as required. Exercise 5 Consider the direct utility function u(x) = nx i log(x i i ); i= where i and i, i = ; ; n are, respectively, positive and nonpositive parameters.. Derive the indirect utility function and show that it is decreasing in its arguments. 2. Verify Roy s Theorem. 3. Derive the expenditure function and show that it is homogeneous of degree one and nondecreasing in prices. 4. Verify Hotelling s Theorem. Exercise 6 For the utility function de ned in Exercise 5,. Derive the Slutsky equation. 2. Let d i (p; y) be the demand for good i derived from the above utility function. Goods i and j are said to be gross substitutes if @d i (p; y)=@p j > 0 and gross complements if @d i (p; y)=@p j < 0. For this utility function are the various goods gross substitutes, gross complements, or can we not say? (The two previous exercises are taken from Russell and Wilkinson (979).) Exercise 7 An electric utility has two generating plants in which total costs per hour are c and c 2 respectively, where c =80 + 2x + 0:00bx 2 b >0 c 2 =90 + :5x 2 + 0:002x 2 2 with x i being the quantity generated in the i-th plant. If the utility is required to produce 2000 megawatts in a particular hour, how should it allocate this load between the plants so as to minimize costs? Use the Lagrangian method and interpret the multiplier. How do total costs vary as b changes. (That is, what is the derivative of the minimized cost with respect to b.) 9