Comparative-Static Analysis of General Function Models

Transcription

1 Comparative-Static Analysis of General Function Models The study of partial derivatives has enabled us to handle the simpler type of comparative-static problems, in which the equilibrium solution of the model can be explicitly stated in the reduced form. This means that parameters and/or exogenous variables which appear in the reduced-form solution must be mutually independent.

2 Since these are indeed defined as predetermined data for purposes of the model, the possibility of their mutually affecting one another is inherently ruled out. The procedure of partial differentiation adopted is therefore fully justifiable. However, no such expediency should be expected when, owing to the inclusion of general functions in a model, no explicit reduced-form solution can be obtained. In such cases, we will have to find the comparative-static derivatives directly from the originally given equations in the model.

3 Take, for instance, a simple national-income model: Y = C + I 0 + G 0 ; I 0,G 0 exogenous C = C(Y,T 0 ) ; T 0 exogenous which is reducible to a single equation (an equilibrium condition): Y = C(Y,T 0 ) + I 0 + G 0

4 Under certain rather general equilibrium, the solution is Y 0 = Y 0 (I 0, G 0, T 0 ); however, we are unable to determine explicitly the form which this function takes. How would you yield Y 0 / T 0, which measures the effect on equilibrium income as a result of a change in tax, if it is not explicitly determined?

5 Differentials and Derivatives Given a function y = f(x); Δy /Δx represents the rate of change of y with respect to x. Since it is true that Δy = (Δy/Δx) Δx; the magnitude of Δy can be found, once the rate of change (Δy /Δx) and the variation in x are known.

6 Given Δx 0 then Δy /Δx dy/dx and Δy dy therefore dy = (dy/dx) dx or dy = f ' (x) dx The symbol dy and dx are called the differentials of y and x, respectively. Thus, (dy) / (dx) = (dy/dx) or (dy) / (dx) = f'(x). The result shows that the derivative f'(x) may be interpreted as the quotient of two separate differentials dy and dx.

7 Example: y = 3x 2 + 7x 5 ; dy? dy = (dy/dx) dx = (6x + 7) dx. It should be remembered, however, that the differentials dy and dx refer to infinitesimal changes only; hence, if we put an x change of substansial magnitude (Δx), the resulting dy can only serve as an approximation to the exact value of the corresponding y change (Δy).

8 As an illustration: given changes from then the value of x =5 and the value of dx = 0.01 dy =(6(5) + 7) (0.01) = 0.37 However, if we refer to the equation y = 3x 2 + 7x -5 ; at x = 5 y = 105 x = 5.01 y = which means that Δy = Therefore, the true change in y is Δy = , for which our answer dy = 0.37 constitutes an approximation with an error of

9 Geometrically, the source of error in the approximation of Δy and dy can be illustrated with the following figure: y y 1 B y 0 A C D x 0 x 1 x

10 Δx = x 1 x 0 Δy = y 1 y 0 = BC dy = (dy / dx ) Δx = (CD/AC) AC = CD From this figure, Δy = BC > dy = DC This error can be expected to become smaller, the smaller is the Δx.

11 Differential and point elasticity Given demand function Q = f(p), elasticity is defined as ratio between relative change of Q and relative change of P or (ΔQ/Q) / (ΔP/ P). If the change of ΔP is very small, then ΔQ dq and ΔP dp so that elastisity can be defined as : ε d = (dq/q) / (dp/p) = (dq/dp) / (Q / P) is often called as point elastisity

12 In general when y = f(x); point elasticity from y to x is expressed as ε yx = (dy/dx) / (y/x) = (marginal function) / (mean function) Term: If ε d > 1 elastic demand ε d < 1 non elastic demand ε d = 1 demand has a unit elasticity

13 Example: Q = 100 2P dq / dp = -2 ; Q/P = (100-2P)/P ε d = (dq/dp ) / (Q/P) = (-2P) / (100 2P) for P = 25 ε d = 1 P = 30 ε d = 1.5 > 1; elastic demand at this price It turns out that: ε d > 1 if 25 < P < 50 ε d < 1 if 0 < P < 25 What happened if P > 50?

14 Total Differentials Consider a saving function: S = S(Y,i) ; S: savings, Y: income, i: interest rate Remember that the partial derivative S / Y or S Y measures the rate of change of S with respect to an infinitesimal change in Y, or, in short, that it signifies the Marginal Propensity to Save (MPS).

15 As a result, the change in S due to that change in Y may be represented by the expression S / Y dy. By the same token, the change in S resulting from an infinitesimal change in i can be denoted as S / i di. The total change in S will then equal to: ds = S / Y dy + S / i di or ds = S Y dy + S i di ; The expression ds, being the sum of changes from both sources, is called the total differential of the saving function. And the process of finding such a total differential is called total differentiation.

16 Observations: 1. ds = ( S / Y) dy + ( S / i) di ds/dy = ( S / Y) + ( S / i) di/dy for a constant value of i, (ds/dy) = ( S / Y) 2. ds/di = ( S / Y) dy/di + ( S / i) 3. for a constant value of Y, (ds/di) = ( S / i) For the more general case of a function of n independent variables, U = U (x 1,x 2,..,x n ), the total differential of this function can be written as: du = ( U / x 1 ) dx 1 + ( U / x 2 ) dx ( U / x n ) dx n or du = U 1 dx 1 + U 2 dx U n dx n

17 Partial Elasticities For the saving function S = S(Y,i), the partial elasticities may be written as: ε SY = ( S / Y) / (S/Y) = ( S / Y). (Y/S) ε Si = ( S / i) / (S/i) = ( S / i). (i/s) For the utility function U = U (x 1, x 2,..,x n ), the n partial elasticities can be concisely denoted as follows: ε uxi = = ( U / x i ) / (x i /U) ; i = 1,2.,n.

18 Rules of Differentials Certain rules of differentials bear striking resemblance to rules of derivatives. Let k be a constant, u = u(x 1,x 2 ); v =v(x 1,x 2 ) ; w=(x 1,x 2 ) I. dk = 0 II. III. IV. d ( cu n ) = cn u n-1 du; c: a constant d (u ± v) = du ± dv d (u. v) = v du + u dv V. d ( u / v) = (v du u dv) / v 2 VI. d ( u ± v ± w ) = du ± dv ± dw VII. d (u. v. w) = v w du + u w dv + u v dw

19 Examples: 1. a. y = 5x x 2 dy = f 1 dx 1 + f 2 dx 2 = 10x 1 dx 1 +3 dx 2 or: b. u = 5x 1 2 ; v = 3x 2 dy = du + dv = d(5x 1 2 ) + d (3x 2 ) = 10x 1 dx 1 +3 dx 2 2. y = 3x x 1 x U = 3x 1 ; v = x 1 x 2 dy = du + dv = 6x 1 dx 1 + x 2 2 dx 1 + x 1 (2x 2 2 dx 2 ) = (6x 1 + x 2 2 )dx x 1 x 2 dx

20 3. y = (x 1 +x 2 ) / 2 x 1 2 u = x 1 +x 2 ; v = 2 x 1 2 d( u/v ) = ( v du u dv) / v 2 = {2x 1 2 d(x 1 + x 2 ) (x 1 + x 2 ) d(2 x 1 2 ) } / 4 x 1 4 = {2x 1 2 (dx 1 + dx 2 ) 4 x 1 (x 1 + x 2 )dx 1 } / 4 x 1 4 = {-2x 1 (x 1 + 2x 2 ) dx 1 +2x 1 2 dx 1 } / 4 x 1 4 = - { (x 1 + 2x 2 ) /2x 1 3 dx 1 + (1/2 x 1 2 ) dx 2

21 Total Derivatives Let y = f(x,w) ; x= g(w) The total differential: dy = f x dx + f w dw The total derivative: dy/dw = f x dx/dw + f w dw/dw = y/ x dx/dw + y/ w dy/dw: may be regarded as some measure of the rate of change of y with respect to w and interpreted as the total derivative. y/ w : measures the direct effect of w on y. y/ x 9/27/2011 dx/dw : measures the indirect effect of w on y (through x). Prepared by Nachrowi

22 Examples: 1. a y = f(x,w) = 3x-w 2 ; x=g(w) =2w 2 +w +4 The total derivative: dy/dw = y/ x dx/dw + y/ w = 3(4w +1) + (-2w) = 10w + 3 b. We may substitute the function g into the function f, to get: y = 3(2w 2 +w +4) - w 2 dy/dw = 3 (4w + 1) 2w = 10w U = U (c,s); c: coffee consumption; s: sugar consumption. s = g (c) du/dc = U/ c + U/ s. ds/dc 9/27/2011 = U/ c + U/ s. g' Prepared (c) by Nachrowi

23 A Variation on The Theme y = f (x 1, x 2, w) ; x 1 = g(w); x 2 = h(w) the total derivative: dy/dw = f 1 dx 1 /dw + f 2 dx 2 /dw + f w dw/dw = ( y/ x 1 ) (dx 1 /dw) + ( y/ x 2 ) (dx 2 /dw) + ( y/ w)

24 Example: Q = Q(K, L, t); K = K(t); L = L(t) t: time; K: capital; L: labor dq/dt = Q/ k. dk/dt + Q/ L. dl/dt + Q/ t or dq/dt = Q K K'(t) + Q L L'(t) +Q t

25 Another Variation on the Theme y = f(x 1, x 2, u, v); x 1 = g(u, v); x 2 = h(u, v) dy/du = y/ x 1. dx 1 /du + y/ x 2. dx 2 /du + y/ u. du/du + y/ v. dv/du = y/ x 1. dx 1 /du + y/ x 2. dx 2 /du + y/ u because du/du = 1 and dv/du = 0

26 In this case, dy/du should also be interpreted as a partial derivative because y is a function of x 1, x 2, u, and v. Also, the derivatives dx 1 /du and dx 2 /du should also be interpreted as the partial derivatives as well because x 1 and x 2 are both functions of u and v. Therefore, dy/du should also be referred to as the partial total derivative and denoted by y / u = y/ x 1. x 1 / u + y/ x 2. x 2 / u + y/ u.

27 Examples: 1. Find the total derivative dz/dt, given: a. z = x 2 8xy y 3 ; x= 3t; y = 1-t dz/dt = z/ x. dx/dt + z/ y. dy/dt =(2x-8y). (3) + (-8x-3y 2 ) (-1) = 6x 24y + 8x + 3y 2 = 14x + 3y 2 24y

28 b. z = 3u + vt; u =2t 2 ; v = t+1 (i). dz/dt = z/ u. du/dt + { z/ v. v/ t + z/ t) = (3) (4t) + t. (1) + v = 14t + 1 (ii). substitute: z = 3(2t 2 ) + (t+1) t = 6 t 2 + t 2 + t = 7 t 2 + t dz/dt = 14t +1

29 c. z = f(x, y, t); x= a+bt; y= c+dt dz/dt = z/ x. x/ t + z/ y. y/ t + z/ t = f x (b) + f y (d) + f t = b f x + d f y + f t

30 2. Q = A(t) K α L β Where A (t) is a monotonically increasing function of t K = K 0 + at L = L 0 + bt Find dq/dt; the rate of change of output with respect to t. dq/dt = Q/ A. da/dt + Q/ K. dk/dt + Q/ L. dl/dt = K α L β ( A' (t) ) + A(t) α K α-1 L β (a) + A (t) K α L β-1 (b) 9/27/2011 = K α L β { A' (t) + a α A(t) Prepared / K by Nachrowi + b β A(t) / L}

31 3. Total Partial Derivatives a. w = ax 2 + bxy + cu; x= αu + βv; y=γu w/ u = w/ x. x/ u + w/ y. y/ u + w/ u. u/ u = (2ax + by) (α) + (bx) (γ) + c w/ v = (2ax + by) (β) + (bx)(0) b. w = f(x 1,x 2 ); x 1 = 5u 2 + 3v dan x 2 = u- 4v 3 w/ u = w/ x 1. x 1 / u + w/ x 2. x 2 / u = 10u f 1 + f 2 w/ v = w/ x 1. x 1 / v + w/ x 2. x 2 / v = 3 f 1 12v 2 f 2

32 The end of the lesson Created by Nachrowi D. Nachrowi.