ECON0702: Mathematical Methods in Economics

ECON0702: Mathematical Methods in Economics

Yulei Luo

SEF of HKU

January 18, 2008

Luo, Y. (SEF of HKU) MME January 18, 2008 1 / 44 Comparative Statics and The Concept of Derivative

Comparative Statics is concerned with the comparison of di¤erent equilibrium states that are associated with di¤erent sets of values of parameters and exogenous variables. When the value of some parameter or exogenous variable that is associated with an initial equilibrium changes, we can get a new equilibrium. The question posted in the Comparative Statics analysis is: How would the new equilibrium compare with the old one? Note that in the CS analysis, we don’tconcern with the process of adjustment of the variables; we merely compare the initial equilibrium state with the …nal equilibrium.

Luo, Y. (SEF of HKU) MME January 18, 2008 2 / 44 (Conti.) The problem under consideration is essentially one of …nding a rate of change: the rate of change of the equilibrium value of an endogenous variable with respect to the change in a particular parameter or exogenous variable. Hence, the concept of derivative is the key factor in comparative statics analysis. We will study the rate of change of any variable y in response to a change in another variable x : y = f (x). (1) Note that in the CS analysis context, y represents the equilibrium value of an endogenous variable, and x represents some parameter or exogenous variable. The di¤erence quotient. We use the symbol ∆ to denote the change from one point, say x0, to another point, say x1. Thus ∆x = x1 x0. When x changes from x0 to x0 + ∆x, the value of the function y = f (x) changes from f (x0) to f (x0 + ∆x). The change in y per unit of change in x can be expressed by the di¤erence quotient:

∆y f (x0 + ∆x) f (x0) = (2) ∆x ∆x Luo, Y. (SEF of HKU) MME January 18, 2008 3 / 44 Quick Review of Derivative, Di¤erentiation, and Partial Di¤erentiation

The derivative of the function y = f (x) is the limit of the di¤erence quotient ∆y exists as ∆x 0. The derivative is denoted by ∆x ! dy ∆y = y 0 = f 0(x) = lim (3) dx ∆x 0 ∆x ! Note that (1) a derivative is also a function; (2) it is also a measure of some rate of change since it is merely a limit of the di¤erence quotient; since ∆x 0, the rate measured by the derivative is an instantaneous rate! of change; and (3) the concept of the slope of a curve is merely the geometric counterpart of the concept of derivative. Example: If y = 3x2 4, dy = y = 6x. (4) dx 0

Luo, Y. (SEF of HKU) MME January 18, 2008 4 / 44 The concept of limit. For a given function q = g(v), if, as v N (it can be any number) from the left side (from values less than !N), q approaches a …nite number L, we call L the left-side limit of q. Similarly, we call L the right-side limit of q. The left-side limit and right-side limit of q are denoted by lim q and lim + q, v N v N respectively. The limit of q at N is said! to exist if !

lim q = lim q = L (5) v N v N + ! !

and is denoted by limv N q = L. Note that L must be a …nite number. ! The concept of continuity. A function q = g(v) is said to be continuous at N if limv N q exists and limv N g(v) = g(N). Thus continuity involves the following! requirements:! (1) the point N must be in the domain of the function; (2) limv N g(v) exists; and (3) ! limv N g(v) = g(N). !

Luo, Y. (SEF of HKU) MME January 18, 2008 5 / 44 The concept of di¤erentiability. By the de…nition of the derivative of a function y = f (x), at x0, we know that f 0(x0) exists if and only if ∆y the limit of the di¤erence quotient exists at x = x0 as ∆x 0, ∆x ! that is,   ∆y f 0 (∆x) = lim ∆x 0 ∆x ! f (x0 + ∆x) f (x0) = lim (Di¤erentiation condition).(6) ∆x 0 ∆x ! dy Di¤erentiation is the process of obtaining the derivative dx . Note that the function y = f (x) is continuous at x0 if and only if

lim f (x) = f (x0) (Continuity condition). (7) x x0 !

Luo, Y. (SEF of HKU) MME January 18, 2008 6 / 44 Continuity and di¤erentiability are very closely related to each other. Continuity of a function is a necessary condition for its di¤erentiability, but this condition is not su¢ cient. Consider example: f (x) = x , which is clearly continuous at x = 0, but is not di¤erentiablej j at x = 0. Continuity rules out the presence of a gap, whereas di¤erentiability rules out sharpness as well. Therefore, di¤erentiability calls for smoothness of the function as well as its continuity. Most of the speci…c functions used in economics are di¤erentiable everywhere.

Luo, Y. (SEF of HKU) MME January 18, 2008 7 / 44 Rules of Di¤erentiation

The central problem of comparative-static analysis, that of …nding a rate of change, can be identi…ed with the problem of …nding the derivative of some function f (x), provided only a small change in x is being considered. Constant function rule: for y = f (x) = c, where c is a constant, then

dy = f (x) = 0. (8) dx 0 Power function rule: if y = f (x) = x α where α ( ∞, ∞) is any real number, then 2 dy = αx α 1. (9) dx Power function rule generalized: if y = f (x) = cx α, then

dy = cαx α 1. (10) dx

Luo, Y. (SEF of HKU) MME January 18, 2008 8 / 44 (continue. For two or more functions of the same variable) Sum-di¤erence rule: d d d [f (x) g (x)] = f (x) g (x) = f 0 (x) + g 0 (x) , (11) dx  dx  dx which can easily extend to more functions

d n n d n ∑ fi (x) = ∑ fi (x) = ∑ fi0 (x) . (12) dx "i=1 # i=1 dx i=1 Product rule: d d d [f (x) g (x)] = f (x) g (x) + g (x) f (x) (13) dx dx dx = f (x) g 0 (x) + g (x) f 0 (x) (14)

Quotient rule:

d f (x) f (x) g (x) g (x) f (x) = 0 0 . (15) dx g (x) g 2 (x)  

Luo, Y. (SEF of HKU) MME January 18, 2008 9 / 44 Marginal-revenue Function and Average-revenue Function

Suppose that the average-revenue function (AR) is speci…ed by AR = 15 Q,then the total revenue function (TR) is TR = AR Q = 15Q Q2, (16)
which means that the marginal revenue (MR) function is given by d(TR) MR = = 15 2Q. (17) dQ In general, if AR = f (Q) , then

TR = f (Q) Q and MR = f (Q) + f 0 (Q) Q, (18)

which means that MR AR = f 0 (Q) Q. TR PQ Note that since AR = Q = Q = P, we can view AR as the inverse demand function for the product of the …rm. If the market is perfect competition, that is, the …rm takes the price as given, then P = f (Q) =constant, which means that f 0 (Q) = 0 and thus MR = AR. Luo, Y. (SEF of HKU) MME January18,2008 10/44 Marginal-cost Function and Average-cost Function

Suppose that a total cost function is

C = C (Q) , (19)

the average cost (AC) function and the marginal cost (MC) function are given by C (Q) AC = and MC = C (Q) . (20) Q 0 The rate of change of AC with respect to Q is

> 0 MC > AC d C (Q) 1 C (Q) 2 3 = 2C 0 (Q) 3 = = 0 i¤ MC = AC dQ Q Q Q 8 8 < 0 MC < AC 6 7 6 MC 7 < < 6 AC 7 6 AC 7 4 5 4| {z } 5 : : (21) | {z } | {z }

Luo, Y. (SEF of HKU) MME January18,2008 11/44 Rules of Di¤erentiation Involving Functions of Di¤erent Variables

Consider cases where there are two or more di¤erentiable functions, each of which has a distinct independent variables, Chain rule: If we have a function z = f (y), where y is in turn a function of another variable x, say, y = g(x), then the derivative of z with respect to x gives by dz dz dy = = f (y) g (x) . (22) dx dy dx 0 0 Intuition: Given a ∆x, there must result a corresponding ∆y via the function y = g(x), but this ∆y will in turn being about a ∆z via the function z = f (y). Example: Suppose that total revenue TR = f (Q) , where output Q is a function of labor input L, Q = g (L) . By the chain rule, the marginal revenue of labor is dTR dTR dQ MRL = = = f (Q) g (L) . (23) dL dQ dL 0 0

Luo, Y. (SEF of HKU) MME January18,2008 12/44 Inverse function rule: If the function y = f (x) represents a one-to-one mapping, i.e., if the function is such that a di¤erent value of x will always yield a di¤erent value of y, then function f will have an inverse function 1 x = f (y) , (24) 1 note that here the symbol f doesn’tmean the reciprocal of the function f (x). For monotonic functions, the corresponding inverse functions exist. Generally speaking, if an inverse function exists, the original and the inverse functions must be both monotonic. For inverse functions, the rule of di¤erentiation is dx 1 = . (25) dy dy dx 5 4 Examples: Suppose that y = x + x, y 0 = 5x + 1 and dx 1 = . (26) dy 5x4 + 1

Luo, Y. (SEF of HKU) MME January18,2008 13/44 Partial Di¤erentiation

In comparative static analysis, several parameters appear in a model, so that the equilibrium value of each endogenous variable may be a function of more than one parameter.

Partial derivatives: Consider a function y = f (x1, x2, , xn),where


the variables xi are all independent of one another, so that each can vary by itself without a¤ecting the others. If the variable xi changes ∆xi while the other variables remain …xed, there will be a corresponding change in y, ∆y :

∆y f (x1, , xi + ∆xi , , xn) f (x1, , xi , , xn) =







. (27) ∆xi ∆x The partial derivative of y with respect to xi is de…ned as ∂y ∆y = fi = lim (28) ∂xi ∆xi 0 ∆xi ! Techniques of partial di¤erentiation: PD di¤ers from the previously discussed di¤erentiation primarily in that we must hold the other independent variables constant while allowing one variable to vary. Luo, Y. (SEF of HKU) MME 2 January18,20082 14/44 Example: suppose that y = f (x1, x2) = 3x1 + x1x2 + 4x2 , ∂y ∂y = 6x1 + x2, = x1 + 8x2. (29) ∂x1 ∂x2 Applications to Comparative-Static Analysis

How the equilibrium value of an endogenous variable will change when there is a change in any exogenous variables or parameter? The Single Market Model. The model setup is

Qd = a bP and Qs = c + dP (30)

where a, b, c, and d are all positive. Qd = Qs implies that a + c ad bc P = and Q = , (31) b + d b + d which means ∂P 1 ∂P (a + c)  = > 0,  = < 0, (32) ∂a b + d ∂b (b + d)2 ∂P 1 ∂P (a + c)  = > 0,  = < 0. (33) ∂c b + d ∂d (b + d)2

Luo, Y. (SEF of HKU) MME January18,2008 15/44 The National Income Model. The model setup is

Y = C + I0 + G0 (34) C = α + β (Y T ) where α > 0, 0 < β < 1. (35) T = γ + δY where γ > 0, 0 < δ < 1, (36)

where the …rst equation in this system gives the equilibrium condition for national income, while the second and third equations show how C and T are determined in the model. α > 0 means that consumption is positive even if disposable income is 0; β is a positive fraction because it represents the marginal propensity to consume; γ is positive because even if Y is zero the government will still have a positive revenue; δ is an income tax rate that cannot exceed 100%. Combining the three equations gives

α βγ + I0 + G0 Y  = . (37) 1 β + βδ

Luo, Y. (SEF of HKU) MME January18,2008 16/44 From the equilibrium income (37), we can obtain the following comparative-static derivatives that imply the e¤ects of government policy:

∂Y 1  = > 0 (38) ∂G0 1 β + βδ ∂Y β  = < 0 (39) ∂γ 1 β + βδ ∂Y β (α βγ + I0 + G0) βY  = =  < 0 (40) ∂δ (1 β + βδ)2 (1 β + βδ)2 Implications: (38) gives us the government-expenditure multiplier, which is positive since β < 1 and βδ > 1. (39) is the nonincome-tax multiplier because it measures the e¤ect of a change in γ on the equilibrium income. (40) measures the e¤ect of a change in the income tax rate on equilibrium income.

Luo, Y. (SEF of HKU) MME January18,2008 17/44 Note on Jacobian Determinants

Partial derivative is useful in CS analysis, but it also tests whether functional (linear or nonlinear) dependence among a set of n functions in n variables. This is related to the notion of Jacobian determinants. Consider the two functions 2 2 y1 = 2x1 + 3x2 and y2 = 4x1 + 12x1x2 + 9x2 , (41) all the four partial derivatives are

∂y1 ∂y1 ∂y2 ∂y2 = 2, = 3, = 8x1 + 12x2, = 12x1 + 18x2, (42) ∂x1 ∂x2 ∂x1 ∂x2 which can be arranged into a square matrix which is called a Jacobian matrix J ∂y1 ∂y1 2 3 ∂x1 ∂x2 J = ∂y ∂y = (43) 2 2 8x1 + 12x2 12x1 + 18x2 " ∂x1 ∂x2 #   and its determinant is known as a Jacobian determinant, J . j j Luo, Y. (SEF of HKU) MME January18,2008 18/44 More generally, if we have n di¤erentiable functions in n variables, 1 y1 = f (x1, xn) (44)

n


yn = f (x1, xn) ,


we can then derive a total of n2 partial derivatives. The Jacobian determinant is

∂y1 ∂y1 ∂x ∂x ∂ (y1 yn) 1


n J =


= (45) j j ∂ (x1 xn)
∂y
n




∂y
n



∂x1


∂xn

Theorem: J will be identically zero for all values of x1 xn if and only if the jn jfunctions are functionally (linearly or nonlinearly)


dependent. For example, for the above two equation system, J = 2 (12x1 + 18x2) 3 (8x1 + 12x2) = 0 for any values of x1 and x2, which then means that the two functions are dependent (y2 is just y1 squared, thus they are functionally nonlinearly dependent). Note that test of linear dependence of a linear equation system is a special application of the Jacobian criterion of functional dependence Luo, Y. (SEF of HKU) MME January18,2008 19/44 Comparative Static Analysis of General-functions

The study of partial derivatives allows us to handle the simple type of CS problem, in which the equilibrium solution of the model can be explicitly stated in the reduced-form. Note that the de…nition of PD requires that the absence of any functional relationship among the independent variables. As applied to CS, this means that parameters or exogenous variables must be mutually independent. However, in some cases, there is no explicit reduced-form solution due to the inclusion of general functions in a model. In such a case, we have to …nd the CS derivatives directly from the originally given equations. For example,

Y = C + I0 + G0 (Equilibrium condition) (46)

C = C (Y , T0) (T0 : Exogenous taxes), (47)

which can be reduced to a single equation Y = C (Y , T0) + I0 + G0 to be solved for Y . No explicit solution is available due to the general function C.Must …nd the CS derivatives directly from this equation.

Luo, Y. (SEF of HKU) MME January18,2008 20/44 Suppose that an equilibrium solution Y  does exist. Then under certain general conditions (will discuss later), we may take Y  to be a di¤erentiable function of the exogenous variables:

Y  = Y  (I0, G0, T0) (48) even though we are unable to determine the explicit form of this function. Furthermore, in some neighborhood of the equilibrium value, the following identical equality will hold

Y  = C (Y , T0) + I0 + G0, (49) which is called equilibrium condition because Y is replaced by its equilibrium value Y . Since Y  is a function of T0, the two arguments in the C function are not independent. Speci…cally, in this case T0 a¤ects C not only directly, but also indirectly via Y . Consequently, partial di¤erentiation is no longer appropriate for our purposes. We must resort to total di¤erentiation and total derivatives, which can be used to measure the rate of change in which the arguments are not

Luo,independent. Y. (SEF of HKU) MME January18,2008 21/44 dy The symbol, dx , for the derivative of the function y = f (x) , has been regarded as a single entity. It can also be interpret as a ratio of two quantities, dy and dx, which are called the di¤erentials of x and y, respectively. The process of …nding the di¤erential dy from a given function y = f (x) is called di¤erentiation:

dy = f 0 (x) dx. (50) Economic application (Point elasticity of the demand function). Given a demand function Q = f (P) , the point elasticity of demand is de…ned as dQ/Q dQ/dP marginal function εd = = = , (51) dP/P Q/P average function > 1 : the demand is elastic εd = 1 : the demand is of unit elastic . (52) j j 8 < < 1 : the demand is inelastic Example: Given the demand function Q = 100 2P, : P 1 when P = 25 εd = = . (53) 50 P 1.5 when P = 30  Luo, Y. (SEF of HKU) MME January18,2008 22/44 Total Di¤erentials

The concept of di¤erentials can be extended to a function of two or more independent variables. Consider a saving function S = S (Y , i) ,where S is savings, Y is national income, and i is the interest rate. The total change in S is then approximated by the di¤erential ∂S ∂S dS = dY + di = SY dY + Si di. (54) ∂Y ∂i dS is called the total di¤erential of the saving function (the process of …nding such a total di¤erential is called total di¤erentiation) and is the sum of the approximate changes from both sources. SY and Si play the role of converters that serve to convert the changes dY and di, respectively, into a corresponding change dS. Note that ∂S dS = , (55) ∂Y dY  i constant which means that the partial derivative can also be interpreted as the ratio of two di¤erentials dS and dY given that i is held constant. Luo, Y. (SEF of HKU) MME January18,2008 23/44 For a general function with n independent variables, y = f (x1, xn) , the total di¤erential of this function is


∂f ∂f n dy = dx1 + + dxn = ∑ fi dxi (56) ∂x1


∂xn i=1 in which each term on the right side indicates the amount of change in y resulting from an in…nitesimal change in one of the independent variable. As in the case of one variable, the n partial derivatives can be written as ∂f xi εyxi = where i = 1, , n. (57) ∂xi f

Luo, Y. (SEF of HKU) MME January18,2008 24/44 Rules of Di¤erentials

Rule 1: for any constant c,

dc = 0 (58)

Rule 2: α α 1 d (cu ) = cαu du (59) Rule 3: d (u v) = du dv (60)   Rule 4: d (uv) = udv + vdu Rule 5: u vdu udv d = (61) v v 2  

Luo, Y. (SEF of HKU) MME January18,2008 25/44 Total Derivatives

Consider any function y = f (x, w) where x = g (w) ,unlike partial derivative, a total derivative doesn’trequire that the argument x remains constant as w varies. w can a¤ect y via two channels: (1) indirectly, via g and then f , and (2) directly, via f . Whereas the partial derivative fw is adequate for expressing the direct e¤ect alone, a total derivative is needed to express both e¤ects jointly. To get the total derivative, we …rst get the total di¤erentials dy = fx dx + fw dw and then divide both sides by dw : dy dx ∂y dx ∂y = fx + fw = + . (62) dw dw ∂x dw ∂w

For a more general function y = f (x1, x2, w) with x1 = g (w) and x2 = h (w) , the total derivative of y is dy ∂f dx ∂f dx ∂f = 1 + 2 + . (63) dw ∂x1 dw ∂x2 dw ∂w

Luo, Y. (SEF of HKU) MME January18,2008 26/44 Application to an economic growth model: Let the production function is Q = Q (K, L, t) (64) where K is the capital input, L is the labor input, and t is the time which indicates that the production function can shift over time due to technological changes. Since capital and labor can also change over time, we have

K = K (t) and L = L(t). (65)

Thus the rate of output with respect to time t can be written as

dQ ∂Q dK ∂Q dL ∂Q = + + . (66) dt ∂K dt ∂L dt ∂t

Luo, Y. (SEF of HKU) MME January18,2008 27/44 Derivatives of Implicit Functions

A function y = f (x1, xn) is called an explicit function because the


variable y is explicitly expressed as a function of x1, xn. But in


many cases, the relationship between y and x1, xn is given by

F (y, x1, xn) = 0, (67)

which may also be de…ned as implicit function y = f (x1, xn) .Note that an explicit function can always be transformed into an

equation
F (y, x1, xn) = y f (x1, xn) = 0. The reverse transformation is not always


possible.


Hence, we have to impose a certain condition under which a given equation F (y, x1, xn) = 0 does indeed de…ne an implicit function


y = f (x1, xn). Such a result is called implicit-function theorem.

Luo, Y. (SEF of HKU) MME January18,2008 28/44 Theorem

(implicit-function theorem) Given F (y, x1, , xn) = 0, if (a) the function


F has continuous partial derivatives Fy , Fx , , Fx , and if (b) at a point 1


n (y0, x1,0, , xn,0) satisfying F (y0, x1,0, , xn,0) = 0 and Fy is nonzero,





then there exists an n-dimensional neighborhood of (x1,0, , xn,0), N, in


which y is an implicitly de…ned function of x1, , xn, in the form of for all points in N. Moreover, the implicit function f
is

continuous, and has continuous partial derivatives f1, , fn.

Derivatives of implicit functions. Di¤erentiating F , we have dF = 0 :

Fy dy + F1dx1 + + Fndxn = 0. (68)

Luo, Y. (SEF of HKU) MME January18,2008 29/44 Suppose that only y and xi are allowed to vary, then we have

Fy dy + Fi dxi = 0, (69)

which means that

dy ∂y Fi other variables constant = = for any i. (70) dxi j ∂xi Fy For a simple case F (y, x) = 0, the rule gives dy F = x . (71) dx Fy Assume that the equation F (Q, K, L) = 0 implicitly de…nes a production function Q = Q (K, L) . We can then get the marginal product of capital and labor, MPK and MPL, as follows:

∂Q F ∂Q F MPK = = K and MPL = = L . (72) ∂K FQ ∂L FQ

Luo, Y. (SEF of HKU) MME January18,2008 30/44 Extension to the Simultaneous-equation Case

Consider a set of simultaneous equations, 1 F (y1, , yn; x1, , xm ) = 0 (73)

n




F (y1, , yn; x1, , xm ) = 0. (74)





Suppose that F 1, , F n are di¤erentiable, we have


∂F 1 ∂F 1 ∂F 1 ∂F 1 dy1 + + dyn = dx1 + + dxm (75) ∂y1


∂yn ∂x1


∂xm   ∂F n ∂F n



∂
F n ∂F n dy1 + + dyn = dx1 + + dxm ,(76) ∂y1


∂yn ∂x1


∂xm   or in matrix form ∂F 1 ∂F 1 ∂F 1 ∂F 1 dy1 dx1 ∂y1


∂yn ∂x1


∂xm 2 3 = 2 3
∂F
n




∂F
n
2


3
∂F
n




∂F
n
2


3 dyn dxm ∂y1


∂yn ∂x1


∂xm 6 7 4 5 6 7 4 (77)5 Luo,4 Y. (SEF of HKU) 5 MME 4 January18,20085 31/44 If we want to obtain partial derivatives with respect to xi , let dxj = 0 for any j = i and we have the following equation: 6 ∂F 1 ∂F 1 ∂F 1 dy1 ∂y1


∂yn ∂xi 2 3 = 2 3 dxi =
∂F
n




∂F
n
2


3
∂F
n
) dyn 6 ∂y1


∂yn 7 6 ∂xi 7 1 1 4 5 1 4 ∂F ∂F 5 ∂y1 4 ∂F 5 ∂y1


∂yn ∂xi ∂xi = . (78) 2 n n 3 2 3 2 n 3
∂F





∂F


∂y
n

∂F

6 ∂y1


∂yn 7 6 ∂xi 7 6 ∂xi 7 4 J 5 4 5 4 5

Suppose| that the{z Jacobian} determinant is nonzero: J = 0. Then by the Cramer rule, we have j j 6

∂y Ji j = j j j where i = 1, , m; j = 1. , n. (79) ∂xi J





j j

Luo, Y. (SEF of HKU) MME January18,2008 32/44 The National Income Model. The model setup is 1 F = Y (C + I0 + G0 ) = 0 (80) F 2 = C [α + β (Y T )] = 0 (81) F 3 = T (γ + δY ) = 0. (82) ∂F 1 ∂F 1 ∂F 1 ∂Y ∂C ∂T 1 1 0 J = ∂F 2 ∂F 2 ∂F 2 = 1 = 1 + . (83) ∂Y ∂C ∂T β β β βδ j j 3 3 3 ∂F ∂F ∂F δ 0 1 ∂Y ∂C ∂T Suppose all exogenous variables and parameters are …xed except G , 0 then we have (Note that here we bypassed the step of solving endogenous variables explicitly):

1 1 0 ∂Y  1 1 1 0 ∂G0 Y ∂C  ∂  β 1 β ∂G = 0 = = 0 1 β /J 2 3 2 0 3 2 3 ) ∂G0 δ 0 1 ∂T  0 0 0 1 ∂G0 4 5 6 7 4 5 4 5 1 = . 1 β + βδ Luo, Y. (SEF of HKU) MME January18,2008 33/44 The Market Model (with general functions). The model setup is ∂D ∂D Qd = D (P, Y0) ( < 0, > 0) (84) ∂P ∂Y0 dS Q = S (P) ( > 0) (85) s dP where Qd is a function not only of price P but also of an exogenous determined income Y0, and both demand and supply functions have continuous derivatives. The equilibrium condition Qd = Qs implies that F (P, Y0) = D (P, Y0) S (P) = 0. (86) We assume that there does exist a static equilibrium (for otherwise there would be no point in raising the equation of CS) and expect that

P = P (Y0) . (87) Note that with the implicit-function theorem, the satisfaction of the conditions of the IF theorem will guarantee that every Y0 will yield a unique value of P in the neighborhood around a point satisfying (86) that de…nes the initial equilibrium. In that case, we can write the implicit function P = P Y . Luo, Y. (SEF of HKU)   ( 0)MME January18,2008 34/44 (conti.) A straightforward application of the IF theorem gives

dP ∂F /∂Y ∂D/∂Y  = 0 = 0 > 0, (88) dY0 ∂F /∂P ∂D/∂P dS/dP    in which all derivatives are evaluated at the equilibrium point. Economic (qualitative) implication: an increase (decrease) in the income level will always result in an increase (decrease) in the equilibrium price P. If we know the values of the derivatives at the equilibrium, this formula gives a quantitative conclusion.

Furthermore, at the equilibrium, we have Q = S (P) and apply the chain rule gives dQ dS dP  =  > 0, (89) dY0 dP dY0 >0 which also means that the equilibrium|{z} quantity is positively related to Y0.

Luo, Y. (SEF of HKU) MME January18,2008 35/44 Simultaneous-equation Approach

The above analysis was carried out on the basis of a single equation. We …rst derived dP  and then infer dQ  . Now we shall show how both dY0 dY0 can be determined simultaneously. As there are two endogenous variables, we need accordingly set up a two-equation system. First, let Q = Qd = Qs in the market model and rearranging gives 1 F (P, Q; Y0) = D (P, Y0) Q = 0 (90) 2 F (P, Q; Y0) = S (P) Q = 0, (91) which is in the form of (73), ,(74), with n = 2 and m = 1. First, we check the conditions of the


implicit-function theorem: (1) since the demand and supply functions are both assumed to have continuous derivatives, so must F 1 and F 2; (2) the endogenous-variable Jacobian (the one involving P and Q) indeed turns out to be nonzero regardless of where is evaluated because 1 1 ∂F ∂F ∂D dS D ∂P ∂Q ∂P 1 ∂ J = ∂F 2 ∂F 2 = dS = > 0. (92) j j dP 1 dP ∂P ∂P ∂Q Luo, Y. (SEF of HKU) MME January18,2008 36/44

(conti.) Hence, if an equilibrium solution exists, the implicit-function theorem implies that

P = P (Y0) and Q = Q (Y0)

even though we cannot solve for P and Q explicitly. These implicit functions are known to have continuous derivatives. Furthermore, (90) and (91) will have the status of a pair of identities in some neighborhood of the equilibrium state, so we may also have

1 F (P, Q; Y0) = D (P, Y0) Q = 0 (93) 2 F (P, Q; Y0) = S (P) Q = 0, (94) from which, dP  and dQ  can be found simultaneously by using the dY0 dY0 implicit function rule (78) (Note that here we have two endogenous variable and one exogenous variable):

1 1 1 ∂F ∂F dP  ∂F ∂P ∂Q dY0 ∂Y0 2 2 = 1 . (95) ∂F ∂F dQ  ∂F " #" dY0 # " # ∂P  ∂Q  ∂Y0

Luo, Y. (SEF of HKU) MME January18,2008 37/44 (conti.) More speci…cally,

∂D dP  ∂D 1 dY0 Y ∂P  = ∂ 0 . (96) dS 1 dQ  0  dP   " dY0 #   By Cramer rule, we have

dP ∂D dQ dS ∂D  = ∂Y0 and  = dP  ∂Y0 (97) dY0 J dY0 J j j j j where all the derivatives of the demand and supply functions are to be evaluated at the initial equilibrium. Note that the results here are identical to those from solving the single equation. Instead of directly applying for the implicit function rule, we can also reach the same result by …rst di¤erentiating totally (93) and (94) to get a linear equation system in terms of dP and dQ : ∂D ∂D dS dP dQ = dY0 and dP dQ = 0 (98) ∂P ∂Y0 dP

Luo, Y. (SEF of HKU) MME January18,2008 38/44 Use of Total Derivatives

In both the single equation and the simultaneous equation approaches illustrated above, we have taken the total di¤erentials of both sides of an equation equation and then equated the two results to arrive at the implicit function rule. However, it is possible to take and equate the total derivatives of the two sides of the equilibrium equation with respect to a particular exogenous variable or parameter. For example, in the single equation approach,

D (P, Y0) S (P) = 0

where P = P (Y0) . Taking total derivatives of this equation w.r.t. Y0 gives

∂D ∂D dP ∂D ∂S dP dP  +  = 0 =  = ∂Y0 . (99) ∂S ∂D ∂P dY0 ∂Y0 ∂P dY0 ) dY0 ∂P  ∂P 

Luo, Y. (SEF of HKU) MME January18,2008 39/44 Application to the IS-LM Model

Equilibrium in the IS-LM model is characterized by an income level and interest rates that simultaneously determine equilibrium in both the goods market and the money market. A goods market is described by the following set of equations: Y = C + I + G; (100) C = C (Y T ) ; I = I (r); (101) T = T (Y ); G = G0 (102) where Y is the level of GDP (gross domestic product) or total income. C, I , G, and T are consumption, investment, government spending, and taxes, respectively. Note that (1) consumption is a strictly increasing function of disposable income Y d = Y T , that is, the marginal propensity to consume is dC d = C 0 Y (0, 1) . (103) dY d 2   Luo, Y. (SEF of HKU) MME January18,2008 40/44 (conti.) (2) Investment is a strictly decreasing function of the interest rate dI = I (r) < 0. (104) dr 0 (3) The government sector is described by two variables: government spending (G) and taxes (T ). G is assumed to be set exogenously, whereas T is assumed to be an increasing function of income: the marginal tax rate dT = T (Y ) (0, 1) . dY 0 2 Slope of the IS curve. The IS equation can be written as

d Y C Y I (r) G0 = 0, (105) then take the total di¤erential with respect to Y and r gives

d dY C 0 Y 1 T 0 (Y ) dY I 0 (r) dr = 0 dr  1 C Y d
(1 T (Y )) = slope of the IS curve = 0 0 < 0(106). dY I r )
0 ( ) d where we use the fact that dY = 1 T (Y ) . dY 0 Luo, Y. (SEF of HKU) MME January18,2008 41/44 The money market can be described by the following three equations

d Money demand : M = L (Y , r) where LY (Y , r) > 0, Lr (Y , r)(107)< 0, s s s Money supply : M = M0 (M0 is determined exogenously, (108) Equilibrium condition : Ms = Md , (109)

which implies the LM equation (curve):

s L (Y , r) = M0 . (110)

Slope of the LM curve. Take the total di¤erential with respect to Y and r gives

LY dY + Lr dr = 0 dr L = slope of the LM curve = Y . (111) ) dY Lr

Luo, Y. (SEF of HKU) MME January18,2008 42/44 Comparative static analysis (examine the e¤ects of two exogenous variables on endogenous variables). The simultaneous macro equilibrium state of the goods and money markets can be characterized by the IS-LM equations:

d Y = C Y + I (r) + G0 (112) s L (Y , r) = M0.  (113) Taking the total deferential of the system gives

d dY C 0 Y 1 T 0 (Y ) dY I 0 (r) dr = dG0 (114) s  
LY dY + Lr dr = dM0 , (115) which can be written in matrix form 1 C Y d (1 T (Y )) I (r) dY dG 0 0 0 = 0 . L L dr dMs 
Y r     0  J (116) | {z } Luo, Y. (SEF of HKU) MME January18,2008 43/44 (conti.) The Jacobian determinant J is j j d J = 1 C 0 Y 1 T 0 (Y ) Lr + I 0 (r) LY < 0. (117) j j h   i Note that since J = 0, this system satis…es
the conditions of the implicit functionj theoremj 6 and the implicit functions s s Y  = Y  (G0, M0 ) and r  = r  (G0, M0 ) can be written even though we are unable to solve for them explicitly. s To do comparative static analysis, …rst we need to set dM0 = 0, and divide both sides by dG0 :

1 C Y d 1 T Y I r dY  1 0 ( 0 ( )) 0 ( ) dG0 dr = . (118) LY Lr  0 
 " dG0 #   Using Cramer rule gives

dY  1 I (r) Lr dr  LY = 0 / J = > 0 and = > 0. dG0 0 Lr j j J dG0 J

j j j j (119)

Luo, Y. (SEF of HKU) MME January18,2008 44/44