Mathematical Economics Part 2

Transcription

1 An Introduction to Mathematical Economics Part Loglinear Publishing Michael Sampson

2 Copyright 00 Michael Sampson. Loglinear Publications: Terms of Use This document is distributed "AS IS" and with no warranties of any kind, whether express or implied. Until November, 00 you are hereby given permission to print one () and only one hardcopy version free of charge from the electronic version of this document (i.e., the pdf file) provided that:. The printed version is for your personal use only.. You make no further copies from the hardcopy version. In particular no photocopies, electronic copies or any other form of reproduction. 3. You agree not to ever sell the hardcopy version to anyone else. 4. You agree that if you ever give the hardcopy version to anyone else that this page, in particular the Copyright Notice and the Terms of Use are included and the person to whom the copy is given accepts these Terms of Use. Until November, 00 you are hereby given permission to make (and if you wish sell) an unlimited number of copies on paper only from the electronic version (i.e., the pdf file) of this document or from a printed copy of the electronic version of this document provided that:. You agree to pay a royalty of either $3.00 Canadian or $.00 US per copy to the author within 60 days of making the copies or to destroy any copies after 60 days for which you have not paid the royalty of $3.00 Canadian or $.00 US per copy. Payment can be made either by cheque or money order and should be sent to the author at: Professor Michael Sampson Department of Economics Concordia University 455 de Maisonneuve Blvd W. Montreal, Quebec Canada, H3G M8. If you intend to make five or more copies, or if you can reasonably expect that five or more copies of the text will be made then you agree to notify the author before making any copies by at: sampson@loglinear.com or by fax at You agree to include on each paper copy of this document and at the same page number as this page on the electronic version of the document: ) the above Copyright Notice, ) the URL: and the address sampson@loglinear.com. You may then if you wish remove this Terms of Use from the paper copies you make.

3 Contents 0. Preface... vi Total Di erentials. EndogenousandExogenousVariables.... StructuralandReducedFormsofaModel..... TheStructuralFormofaModel..... TheReducedFormofaModel ImplicitandExplicitFunctions CalculatingDerivatives TotalDi erentialsforbeginners A Recipe for Calculating dy dx TotalDi erentialsforintermediates A Recipe j Example:ASimpleLinearFunction Example Example 3: The Bivariate Normal Density Example4:LabourDemand TotalDi erentialsforexperts SystemsofImplicitFunctions A Recipe j Example Example : Supply and Demand I Example 3: Supply and Demand II Example 4: The IS=LM Model TotalDi erentialsandoptimization AGeneralDiscussion A Quick Derivation of the Main Principle Pro tmaximization ConstrainedOptimization Utility Maximization CostMinimization i

4 CONTENTS ii The Envelope Theorem 38. UnconstrainedOptimization Pro t Maximization in the Short-Run I EnvelopeTheoremI Pro t Maximization in the Short-Run II Pro tmaximizationinthelong-run EnvelopeTheoremII ARecipeforUnconstrainedOptimization ThePro tfunction HomogeneousFunctions ConcavityandConvexity Concavity, Convexity and Homogeneity of f (x) PropertiesofthePro tfunction SomeSymmetryResults ConstrainedOptimization Envelope Theorem III: Constrained Optimization A Recipe for Constrained Optimization Concavity, Convexity and Homogeneity of f (x) CostMinimization PropertiesoftheCostFunction Utility Maximization ExpenditureMinimization Integration and Random Variables Introduction DiscreteSummation De nitions Example : P n i= ax i Example : P n i= (ax i + by i ) Example 3: P n i= (ax i + b) Example 4: P n i= (ax i + b) Example 5: P n i= Xi ¹X = ManipulatingtheSigmaNotation Example : P n i= (ax i + by i ) Example : P n i= (ax i + b) Xi X ¹ 3..0 Example 3: P n i= Integration The Fundamental Theorem of Integral Calculus Example : R 0 x dx Example : R 5 0 e x dx IntegrationbyParts Example : R 4 0 xe x dx Example : (n) IntegrationasSummation R b a dx asalinearoperator... 89

5 CONTENTS iii Example : R b a (cf (x)+dg (x)) dx Example : R b a (f (x)+d) dx RandomVariables Introduction DiscreteRandomVariables TheBernoulliDistribution TheBinomialDistribution ThePoissonDistribution ContinuousRandomVariables TheUniformDistribution TheStandardNormalDistribution TheNormalDistribution TheChi-SquaredDistribution The Student s t Distribution: The F Distribution: ExpectedValues E [X] fordiscreterandomvariables Example:DieRoll Example:BinomialDistribution E [X] for Continuous Random Variables Example : The Uniform Distribution TheStandardNormalDistribution E []asalinearoperator The Rules for E h [] Example : E (3X +4Y ) i... h Example : E (3X +4) i Example 3: X» N ¹; ¾ Variances De nition SomeResultsforVariances Example:DieRoll Example:UniformDistribution Example 3: The Normal Distribution Covariances De nitions Independence ResultsforCovariances LinearCombinationsofRandomVariables Linear Combinations of Two Random Variables TheCapitalAssetPricingModel(CAPM) AnExample Sums of Uncorrelated Random Variables AnExample...7

6 CONTENTS iv TheSampleMean The Mean and Variance of the Binomial Distribution The Linear Regression Model for Beginners Linear Combinations of Correlated Random Variables TheLinearRegressionModelforExperts Dynamics Introduction Trigonometry DegreesandRadians The Functions cos (x) and sin (x) The Functions tan (x) and arctan (x) ComplexNumbers Introduction ArithmeticwithComplexNumbers Complexconjugates: TheAbsoluteValueofaComplexNumber The Argument of a Complex Number The Polar Form of a Complex Number DeMoivre stheorem Euler sformula When does (a + bi) n! 0 as n!? First-OrderLinearDi erenceequations De nition Example Stability AMacroeconomicExample TheCob-WebModel Second-OrderLinearDi erenceequations The Behavior of Y t withcomplexroots ExampleRealRootsandStable ExampleRealRootsandUnstable Example3ComplexRootsandStable Example 4 Complex Roots and Unstable AMacroeconomicExample First-Order Di erential Equations De nitions The Case where b = The Case where b 6= Example: Example Second-OrderDi erentialequations Solution Stability with Real Roots

7 CONTENTS v Stability with Complex Roots Example:StablewithRealRoots Example : Unstable with Real Roots Example 3: Stable with Complex Roots Example4:UnstablewithComplexRoots...79

8 CONTENTS vi 0. Preface Thanks go to my Economics 36 students from the Winter 999 and Winter 000 semesters for working so hard on the rst and second versions of these lecture notes. I would in particular like to thank Tamarha Louis-Charles, Dana Al-Aghbar, Christopher Martin and Samnang Om for pointing out many typos and errors.

9 Chapter Total Di erentials. Endogenous and Exogenous Variables In economics we mostly work with mathematical models. By their very nature these models contain variables which can be divided into two classes: ) endogenous variables and ) exogenous variables. Endogenous variables (from Greek, endo: within and genous: born hence born or generated from within the model) are those variables which the model attempts to explain. In a supply and demand model, for example, both the equilibrium price P and quantity Q are determined by the model, that is at the intersection of the supply and demand curves. They are therefore endogenous variables. Exogenous variables (from Greek, exo: from outside, hence generated from outside the model) are those variables whose determination the model says nothing about, but which nevertheless in uence the endogenous variables. An example of an exogenous variable in a supply and demand model would be the weather. Bad weather shifts the supply curve of say oranges to the left causing P to go up and Q to go down. Weather therefore in uences P and Q but P and Q do not (presumably) in uence the weather. A supply and demand model says nothing about how the weather is determined. Often policy variables such as taxes, government expenditure or the money supply are exogenous variables. Mathematically you can think of the exogenous variables as the x 0 s or the independent variables, while endogenous variables are the dependent variables or the y 0 s: Typically in economics we do not use x 0 s and y 0 s in our notation and so you willoftenneedtothink carefully about what are the exogenous variables and what are the endogenous variables in any particular model.

10 CHAPTER. TOTAL DIFFERENTIALS. Structural and Reduced Forms of a Model.. The Structural Form of a Model When we rst write down an economic model, that is before doing any complicated derivations, it is often the case that the endogenous variables y =[y i ] (say an n vector) and exogenous variables x =[x i ] (say an m vector) are all jumbled and appear on both sides of the = sign as: M (y; x) =N (y; x) : This form of the model is often referred to as the structural form of the model since it is in this form that the basic assumptions (or structure) of the model are most apparent... The Reduced Form of a Model In economics we are often interested in how changes the x 0 s; the exogenous variables a ect the y 0 s; the endogenous variables. For example we might ask how increasing the tax rate t (an exogenous variable) a ects the price P or quantity Q (endogenous variables). It is often hard with the structural form of the model to perform such experiments. The problem is that the y 0 s and x 0 s are all jumbled together. It would instead be more useful to have the reduced form of the model which is: y = h (x) so that y is by itself on the left-hand side and some function of x is on the right-hand side. Given knowledge of the form of h (x) we can consider any number of changes in the x 0 s and then see how the y 0 s change. An Example Suppose y is the number of workers that a rm hires, x is the price of the good that the rm sells, and x is the nominal wage. From a model we nd that the marginal revenue of hiring an extra worker is: while the marginal cost is: M (y; x ;x )=x y N (y; x ;x )=x y : Equating marginal revenue M and marginal cost N we arrive at a structural model: M (y; x ;x )=N (y; x ;x ) or x y = x y :

11 CHAPTER. TOTAL DIFFERENTIALS 3 To get the reduced form we need to get y alone on the left-hand side. In this case this is not hard to do. By a little algebra we nd that: y = h (x ;x )= x : x Suppose x = x =: We then know that y =: Consider three experiments: ) doubling x while keeping x xed, ) doubling x while keeping x xed, and 3) doubling both x and x : We see from the reduced form that the rst experiment results in doubling y to ; the second experiment results in halving y to ; while in the third experiment nothing happens to y:..3 Implicit and Explicit Functions Closely related to the notion of structural and reduced forms are implicit and explicit functions. De nition An implicit function is any function written as: g (y; x) =0: so that both the y 0 s and x 0 s appear jumbled together on the left-hand side and zero is on the right-hand side. Given the structural form of a model it is very easy to rewrite it as an implicit function. Thus from the structural form M (y; x) =N (y;x) we can nd the implicit function g (y;x) as: g (y; x) =M (y; x) N (y; x) =0: Using the example of: M (y; x ;x )=x y and N (y; x ;x )=x y we have: g (y; x ;x )=x y x y =0: De nition An explicit function unscrambles the y 0 s and x 0 s so that y is alone on the left-hand side as: y = h (x) : Thus an explicit function and the reduced form are essentially the same thing. Given an explicit function or reduced form it is always possible to write it as an implicit function as: g (y; x) =y h (x) =0: However, it is often impossible to nd an explicit function or the reduced form from an implicit function or structural model.

12 CHAPTER. TOTAL DIFFERENTIALS 4 For example given the implicit function: g (y; x) =y ln (y) ln (x) =0;x> there is no way of getting y alone on the left-hand side. The best we could do is: or y ln (y) =ln(x) y y = x: Nevertheless, g (y; x) = y ln (y) ln (x) = 0 is a perfectly respectable function. We can for example show that it is increasing and concave. We could even plot it as shown below..8.6 y x g (y; x) =y ln (y) ln (x) =0for x>..4 Calculating Derivatives In economics we are often interested in calculating the partial j which tells us the relationship between the j th exogenous variable and the i th endogenous variable when all other exogenous variables are held constant. This corresponds to the experiment of keeping all other exogenous variables except x j constant, changing x j by a small amount and seeing how much y j changes. De nition j > 0 then a positive relationship exists between x i and y j ; that is, an increase (decrease) in x j will result in an increase (decrease) in y i : De nition j < 0 then a negative relationship exists between x i and y j ; that is, an increase (decrease) in x j will result in an decrease (increase ) in y i :

13 CHAPTER. TOTAL DIFFERENTIALS 5 In the example above where we found that y = h (x ;x )= x x we > 0; = (x ) < 0: Thus there is a positive relationship between x and y and a negative relationship between x and y: Since it is often not possible to nd the reduced form of the model, that is, we often cannot unscramble the model enough to get it where y appears alone on the left-hand side and h (x) appears on the right-hand side, we cannot always j in the usual manner. Fortunately, no matter how jumbled up the structural form of the model or the implicit function, it is still possible j indirectly using total di erentials. In what follows we will learn this technique starting with the simplest cases and building up to the more general. We begin with beginners total di erentials where there is one exogenous variable x and one endogenous variable y: We then move on to intermediate total di erentials where there are many exogenous variables but still only one endogenous variable. Finally we move on to advanced total di erentials where there are both many endogenous and exogenous variables..3 Total Di erentials for Beginners Suppose we are given an implicit function: g (y; x) =0 where both y and x are scalars so that there is one endogenous variable y and one exogenous variable x: Under one condition this de nes an explicit function: y = h (x) : The reason we know this is the Implicit Function Theorem that: which states Theorem 5 Implicit Function Theorem I: The explicit function h (x) is guaranteed to exist as long (y; x) 6= The problem then is to calculate: dy dx = h0 (x) : We do this by taking the total di erential of g (y; x) which is:

14 CHAPTER. TOTAL DIFFERENTIALS 6 De nition 6 The total di erential is de ned (y; x) (y; x) dx The total di erential can be used to calculate dy dx using the following recipe:.3. A Recipe for Calculating dy dx Step. If it is not already clear to you, identify the endogenous variable y and the exogenous variable x in your model. Step. If the structural model is not written as an implicit function of the form g (y; x) =0; rewriteitsothatitisinthisform. Step 3. Take the partial derivative of g (y; x) with respect to y and call this a: Thus: : Verify that a 6= 0which insures that the explicit function exists by the Implicit Function Theorem. Multiply a by dy to get: a dy: Step 4. Take the partial derivative of g (y; x) with respect to x and call this b: Thus: : Multiply b by dx to get: b dx: Step 5. Add the results of steps 3 and 4 together and set them equal to zero to get the total di erential: a dy + b dx =0: Step 6. Solvefortheratio dy dx to get: dy dx = b a or alternatively using the de nition of a and b dy dx Note that if a =0this ratio would be unde ned. Example : A Simple Linear Function Let us begin with an easy structural model written 3Q 4R =9: where Q is say the quantity of apples produced in an orchard and R is the amount of rainfall.

15 CHAPTER. TOTAL DIFFERENTIALS 7 Applying Step in the recipe Q is the endogenous variable and R is the exogenous variable. This follows from the common sense notion that while rainfall might a ect apple production, apple production is unlikely to a ect rainfall. Applying Step we can write this as an implicit function by putting the 9 on the other side of the equal sign to get: since: g (Q; R) =3Q 4R 9=0: From steps 3; 4 and 5; the total di erential for g (Q; R) is: a (Q; 3dQ 4dR =0 (Q; = 4: Note that a =36= 0and so the explicit function exists by the Implicit Function Theorem. Solving for dq dr from the total di erential it then follows that: dq dr = 4 3 : This is the same answer we would get from working with the explicit form of the function or the reduced form. We obtain this by solving g (Q; R) = 3Q 4R 9=0for Q to obtain: from which it easily follows that: Example Consider the implicit function: Q = 4 3 R +3 dq dr = 4 3 : g (y; x) =ln(y)+ x + ln (¼) =0: This can be written explicitly or as a reduced form as: y = h (x) = p e x ¼ which is the standard normal density. From h (x) we can directly calculate h 0 (x) as: dy dx = x p e x = xy: ¼ {z } =y

16 CHAPTER. TOTAL DIFFERENTIALS 8 Let us obtain now attempt to obtain the derivative via the total di erential. Using Steps 3, 4, and 5 we obtain y and b = x so that the total di erential is: dy + xdx =0: y Solving for dy dx we then get: dy dx = xy: Example 3 Suppose we were given a structural equation of the form: y y = x: Here from the notation y is the endogenous variable and x is the exogenous variable. From Step ; by taking the ln ( ) of both sides we can write this as an implicit function: g (y; x) =y ln (y) ln (x) =0 which we have already seen cannot be written as an explicit function y = h (x) : From Steps 3; 4 and 5 the total di erential of g (y; x) is: since: Solving for dy dx : a (y; ( + ln (y)) dy dx =0: x =+ln(y) ;b= dy dx = x ( + ln (y)) (y; = x : We can use this result to show that 0 < dy dx < for x>and hence a positive relationship exists between x and y: This follows since: x>=) ln (y) y =ln(x) > 0 implies that y>since if y<then ln (y) < 0: Therefore from x> and ( + ln (y)) > we have x>=) 0 < dy dx = x ( + ln (y)) < : Note that we could have just as correctly used g (y; x) =y y x =0; that is there are many di erent correct ways of writing down an implicit function.

17 CHAPTER. TOTAL DIFFERENTIALS 9 As an exercise di erentiate this one more time with respect to x and show that: d µ y dx = x ( + ln (y)) x + dy < 0 y ( + ln (y)) dx so that the function is concave as the plot above illustrates. Example 4: The Demand for Labour in the Short-Run Suppose a rm facing a short-run production function Q = f (L) where f 0 (L) > 0 and f 00 (L) < 0 and where Q is output and L is the amount of labour. If w = W P is the real wage then the rm will chose that L which satis es: f 0 (L )=w: Here w is the exogenous variable and L is the endogenous variable. We can write this structural form of the model as an implicit function as: g (L ;w)=f 0 (L ) w =0: This de nes the labour demand function or the explicit function or the reduced form as: since The total di erential is: a (L L = L (w) : f 00 (L ) dl dw =0 = f 00 (L ) (L = : Note that a<0 and so the explicit function exists. Note that the coe cient on dl is negative since the diminishing marginal product of labour implies that: f 00 (L ) < 0. Solving for the ratio dl dw we nd that: dl dw = f 00 (L ) < 0 and so the labour demand curve is downward sloping..4 Total Di erentials for Intermediates Consider now an implicit function were there are m exogenous variables x ;x ;:::x m (which we may write as an m vector x) and one endogenous variable y. The implicit function is now written as: g (y; x ;x ;:::x m )=0

18 CHAPTER. TOTAL DIFFERENTIALS 0 which determines the explicit function: y = h (x ;x ;:::x m ) as long as the Implicit Function Theorem is satis ed which states that: Theorem 7 Implicit Function Theorem II: The explicit function y = h (x ;x ;:::x m ) is guaranteed to exist as long as We wish to (y; x ;x ;:::x @x j 6= 0: in order to determine how a change in the j th exogenous variable x j will a ect the endogenous variable y. To do this from the implicit function g (y; x) =0we calculate the total di erential de ned by: De nition 8 The total di erential of g (y; x) =0is given by: a dy + b dx + b dx + + b m dx m =0 where a; b ;b ;:::b m are de ned below in the following recipe:.4. A Recipe j We proceed as follows: Step. If it is not already clear to you, identify the endogenous variable y and the exogenous variables x ;x ;:::x m in your model. Step. Write the structural form of the model as an implicit function of the form: g (y; x ;x ;:::x m )=0: This generally means just collecting all terms on the left-hand side of the structural model and setting them equal to zero. Step 3. Take the partial derivative of g (y; x) with respect to y and call this a: Make sure that a 6= 0so that the explicit function exists by the Implicit Function Theorem. Multiply a by dy to get: a dy: Step 4. Take the partial derivative of g (y; x) with respect to x and call this b : Multiply b by dx to get b dx : Repeat this for x ;x 3 ;:::x m and add them together to get: b dx + b x + + b m dx m :

19 CHAPTER. TOTAL DIFFERENTIALS Step 5. Add the results of steps 3 and 4 together and set them equal to zero to get the total di erential: a dy + b dx + b x + + b m dx m =0: Step 6. To get the j set all dx 0 s equal to zero except dx j ; and then replacing the remaining d 0 s in the total di erential 0 s: Thus we set dy dx j j and dx i =0for i 6= j in the total di erential to get: Now solve for j : a@y + b j = b j @y.4. Example : A Simple Linear Function Consider the structural model: 3Q 4R 7S =9 where Q is the production of apples, R is rainfall and S is the amount of sunshine. Following Step, Q is the endogenous variable and R and S are the exogenous variables. The structural model can be re-written as an implicit function as: g (Q; R; S) =3Q 4R 7S 9=0: Following Steps 3, 4 and 5 the total di erential is since 3dQ 4dR 7S =0 a (Q; R; (Q; R; S) =3;b = b (Q; R; S) = Suppose we wish : Following step 6 rst set dr =0and then replace the remaining d 0 s 0 s so that: dq and ds We then obtain: 3@Q 7@S =0: :

20 CHAPTER. TOTAL DIFFERENTIALS Solving for = 7 3 : In this example we could also from the reduced form. From 3Q 4R 7S 9=0you can arrive at the reduced form (or explicit function) as: Q = h (R; S) = 4 3 R S +3 from which it is trivial = 7 3 :.4.3 Example Consider the implicit function: g (y; x ;x )=x y x y =0: We have already seen that this has a reduced form: y = h (x ;x )= x x from which it is easy @x : Let us nd these partial derivatives from the implicit function. The total di erential is: ³ x y 3 + x y dy + y dx y dx =0 since: a = g (y; x ;x b = g (y; x ;x = y b = g (y; x ;x = y : = x y 3 x y set dx =0and replace the remaining d 0 s 0 s so that dy and dx : We then obtain: ³ x y 3 + x + =0:

21 CHAPTER. TOTAL DIFFERENTIALS 3 Solving for @x = y ³ x y 3 + x y As an exercise use the reduced form y = x x to prove that the above is equal to x : set dx =0and replace the remaining d 0 s 0 s to obtain: yields: : ³ x y 3 + x @x = y ³ x y 3 + x y As an exercise use the reduced form y = x is equal to x (x ) : < 0: to prove that the above expression.4.4 Example 3: The Bivariate Normal Density Consider the implicit function: g (y; x ;x )=ln(y)+ x + x +ln(¼) =0 from which we can nd the explicit function: y = h (x ;x )= ¼ e (x +x ) : This is the bivariate standard normal which is plotted below: w x y = h (x ;x )= ¼ e (x +x )

22 CHAPTER. TOTAL DIFFERENTIALS 4 If we start from the explicit function we can = x ¼ e (x ¼ e (x +x ) = x y +x ) = x y: If instead we calculate the total di erential we nd that: y dy + x dx + x dx =0 since a (y; x ;x ) y ; b (y; x ;x ) = x ;b (y; x ;x ) = to obtain: set dy dx and dx =0in the total di erential + =0 so that solving for we = x to obtain: set dy dx and dx =0in the total di erential + =0 so that solving for we = x Example 4: Labour Demand Suppose a rm facing a short-run production function Q = f (L) where Q is output and L is the amount of labour. With price P and nominal wage W, pro ts ¼ as a function of L are given by: ¼ (L) =Pf (L) WL:

23 CHAPTER. TOTAL DIFFERENTIALS 5 The pro t maximizing L then satis es the rst-order condition: ¼ 0 (L )=0=Pf 0 (L ) W with second-order condition: Pf 00 (L ) < 0: Here P and W are the exogenous variables and L is the endogenous variable. Note that the rst-order condition gives us the appropriate implicit function: g (L ;W;P)=Pf 0 (L ) W =0: This de nes the labour demand function or the explicit function or reduced form as: L = L (W; P ) : From the implicit function g (L ;W;P)=0the total di erential is: Pf 00 (L ) dl dw + f 0 (L ) dp =0 since a (L = Pf 00 (L ) ;b (L ;W;P) = b (L ;W;P) = f 0 (L ) Note that the coe cient on dl is negative from the second-order condition; that is a = Pf 00 (L ) < 0. set dp =0in the total di erential and replace the remaining d 0 s 0 s so that dl and dw to get: Pf 00 =0 or Pf 00 (L Dividing both sides and (W; P = Pf 00 (L (W; P )) < 0 and so the labour demand curve is downward sloping. set dw =0in the total di erential and replace the remaining d 0 s 0 s so that dl and dp to get: Pf 00 (L + f 0 (L =0: Solving we = f 0 (L ) Pf 00 (L ) > 0 since f 0 (L ) > 0 (the marginal product of labour is positive) and Pf 00 (L ) < 0 from the second-order conditions.

24 CHAPTER. TOTAL DIFFERENTIALS 6.5 Total Di erentials for Experts.5. Systems of Implicit Functions Suppose we have a model where there are n endogenous variables: y ;y ; y 3 ;:::y n and m exogenous variables x ;x ;:::x m : We can write this more compactly by letting y =[y i ] be an n vector of the endogenous variables and x =[x i ] an m vector of exogenous variables. Suppose you are given n implicit functions: g i (y; x) =0; for i =; ;:::n: This de nes n reduced form or explicit functions given by: y i = h i (x) ; for i =; ;:::n as long as the Implicit Function Theorem is satis ed which states that: Theorem 9 Implicit Function Theorem III: The explicit functions y i = h i (x ;x ;:::x m ) for i =; ;:::n are guaranteed to exist as long as where: A = 6 (y;x) (y;x) det [A] 6= (y;x) n n(y;x) n For example suppose that we have a structural model where apple production Q and honey production Q are related to the amount of rainfall R; sunshine S and temperature T as: 3Q +Q +3S +5T = R 7 Q Q +6R 3T = S +4: Here Q and Q are the endogenous variables and R; S; and T are the exogenous variables. Note that we have two equations and two endogenous variables. This can then be written as two implicit functions as: : g (Q ;Q ;R;S;T) = 3Q +Q R +3S +5T +7=0 g (Q ;Q ;R;S;T) = Q Q +6R +S 3T 4=0: Here: " @g (y;x) # = 3

25 CHAPTER. TOTAL DIFFERENTIALS 7 and 3 det = 8 6= 0 and so the explicit functions: Q = h (R; S; T) Q = h (R; S; T) are guaranteed to exist by the Implicit Function Theorem. We might therefore want ; that is how sunshine a ects honey production. In general we might want j ; in order to determine the nature of the relationship between the j th exogenous variable x j and the i th endogenous variable y i : We do this by calculating the total di erential de ned as De nition 0 The total di erential of g i (y; x) =0for i =; ;:::n is de ned as: Ady + Bdx =0 where: dy = 6 4 dy dy. dy n ;dx= 6 4 dx dx. dx m and A is the n n matrix de ned above and B is an n m matrix given (y;x) m n m The recipe for calculating the total di erential and j is as follows:.5. A Recipe for j Step. Identifythen endogenous variables y ;y ;:::y n and the m exogenous variables x ;x ;:::x m ; in your model.

26 CHAPTER. TOTAL DIFFERENTIALS 8 Step. If the structural model is not written as a system of implicit functions write them in the form: g (y ;y ;:::y n ;x ;x ;:::x m ) = 0; g (y ;y ;:::y n ;x ;x ;:::x m ) = 0;. g n (y ;y ;:::y n ;x ;x ;:::x m ) = 0; Note you should have as many implicit functions as endogenous variables! Step 3. Take the partial derivative of the rst implicit function g (y; x) =0 with respect to y. Call this a so that : Multiply a by dy to get a dy : Continue this with the remaining endogenous variables y ;y 3 ;:::y n and add the results together to obtain: a dy + a dy + + a n dy n : Note that a j is the coe cient on dy j in the rst implicit function. Step 4. Take the partial derivative of g (y; x) with respect to x and call this b : Multiply b by dx to get b dx : Repeat for x ;x ;:::x m to get: b dx + b dx + + b m dx m : Step 5. Add the results of steps 3 and 4 together and set them equal to zero. This gives you the total di erential for the rst implicit function. a dy + a dy + + a n dy n + b dx + b dx + + b m dx m =0 Now repeat this for the second implicit function to get: a dy + a dy + + a n dy n + b dx + b dx + + b m dx m =0 so that a j and b j. Keep doing this for all n implicit equations to obtain n total di erentials: a dy + a dy + + a n dy n + b dx + b dx + + b m dx m = 0 a dy + a dy + + a n dy n + b dx + b dx + + b m dx m = 0. a n dy + a n dy + + a nn dy n + b n dx + b n dx + + b nm dx m = 0 so that a j and b j are the coe cients on dy j and dx j in the i th total di erential: Thus there are n total di erentials, one for each implicit equation. Step 6. j from the total di erential set all dx 0 s equal to zero except dx j and replace all remaining d 0 s 0 s to obtain:

27 CHAPTER. TOTAL DIFFERENTIALS a n + b j = a n + b j = 0. a + a + + a n + b j = 0: Now put the terms j on the right-hand side to get: a n = b j a n = b j. a + a + + a n = b j or: 6 4 a a a n a a a n a n a n a nn @y n = 6 4 b j b j. b nj 3 7 j: Divide both sides by the j and re-write this in matrix j = b j and b j are n column vectors given j = @x j and b j = 6 4 Note that b j is the j th column of the matrix B: To solve j use Cramer s rule to i = det [A i ( b j j det [A] where A i ( b j ) is the matrix you obtain by replacing the i th column of A with the vector b j : b j b j. b nj :

28 CHAPTER. TOTAL DIFFERENTIALS Example Given the apple/honey production example above, we have two linear implicit equations: g (Q ;Q ;R;S;T) = 3Q +Q R +3S +5T +7=0 g (Q ;Q ;R;S;T) = Q Q +6R +S 3T 4=0: Suppose we wish : Using Total Di erentials Following steps 3; 4; and 5 we have: a =3;a =; = =3;b =5 so that the total di erential for g (Q ;Q ;R;S;T) is: Similarly since: 3dQ +dq dr +3dS +5dT =0: a =;a =;b = 3 the total di erential for g (Q ;Q ;R;S;T) is: dq dq +6dR +ds 3dT =0: Putting these two total di erentials together and we have: 3dQ +dq dr +3dS +5dT = 0 dq dq +6dR +ds 3dT = 0: from the total di erential set dr = dt =0and replace the remaining d 0 s 0 s so that ds dq and dq to obtain: 3@Q +@Q +3@S +@S = 0 or in matrix =

29 CHAPTER. TOTAL DIFFERENTIALS Dividing both sides we then = 3 Solving by Cramer s rule then yields: 3 = = : det Similar if we wish set ds = dt =0in the total di erential and replace the remaining d 0 s 0 s so that dr dq and dq : We therefore obtain: : 3@Q +6@R = 0 or in matrix = Dividing both sides @R = 6 Solving by Cramer s rule then = 6 3 det = : From the Reduced Form Since the structural model is linear, we can easily nd the reduced form using matrix algebra. We thus have from the implicit functions: 3 Q R 3 S =0: Q T :

30 CHAPTER. TOTAL DIFFERENTIALS so that: Q Q 3 = = R 3 S 5 + T R S T : Writing these out we get the reduced form: Q = h (R; S; T) = R 5 4 S T 3 4 so that: Q = h (R; S; T) = 5 R S 7 4 T = 3 8 :.5.4 Example : Supply and Demand I First consider a demand and supply model given by: Q d = D (P ) Q s = S (P ) Q s = Q d = Q D 0 (P ) < 0 (the demand curve) S 0 (P ) > 0 (the supply curve) (the equilibrium condition). Here there are no exogenous variables, the rst equation says that the demand curve slopes downward, the second that the supply curve slopes upward and the third that in equilibrium supply equals demand. We can re-write this as two implicit functions with two endogenous variables Q and P as: g (Q; P ) = Q D (P )=0 g (Q; P ) = Q S (P )=0: At this stage there are no exogenous variables. Let us now introduce two exogenous variables: a tax T H paid by households and a tax T F paid by rms so that the model becomes: Q d = D (P + T H ) ; Q s = S (P T F ) ; Q s = Q d = Q:

31 CHAPTER. TOTAL DIFFERENTIALS 3 or as implicit functions: g (Q; P; T H ;T F ) = Q D (P + T H )=0 g (Q; P; T H ;T F ) = Q S (P T F )=0: This determines the equilibrium quantity Q and price P as functions of T H and T F as: To simplify the notation de ne: Q = h (T H ;T F ) P = h (T H ;T F ) : D P D 0 (P + T H ) S P S 0 (P T F ) : To nd the e ect of the tax T H on the endogenous variables P and Q; we calculate the total di erential which is: since: dq D P dp D P dt H = 0 dq S P dp + S P dt F = 0 and: a = a = D P b H = D P b F =0 a ;TF = a = S P b H =0 b F = S P : H set dt F =0and replace all the remaining d 0 s in the total di erential 0 s to D D H = S = 0 or S = DP H : Dividing both sides H we obtain: H H # = DP 0

32 CHAPTER. TOTAL DIFFERENTIALS 4 so that from Cramer H = = DP D det P 0 S P DP det S P D P S P < 0 S P + D P so that the tax on households reduces the equilibrium quantity..5.5 Example 3: Supply and Demand II There are of course many more variables than taxes which shift demand and supply curves. Let a be any variable which shifts the demand curve to the right and b any variable which shifts the supply curve to the right. We can write a more general supply and demand model as: Q d = D (P; a) Q s = S (P; b) Q s = Q d = Q: < 0 > 0 > 0 > 0 For example if a = T H and b = T F and D (P; a) =D (P + T H ) and S (P; b) = S (P T F ) then we would have the previous supply and demand model. Here the exogenous variables are a and b and the endogenous variables are Q and P: Written as a system of implicit equations we have: The total di erential is then: g (Q; P; a; b) = Q D (P; a) =0 g (Q; P; a; b) = Q S (P; b) =0: dq D P dp D a da +0db = 0 dq S P dp +0da S b db = 0: Suppose we want to see what happens if the demand curve shifts to the right while the supply curve stays xed. Set db and dp so that from the total di erential: DP = Da 0

33 CHAPTER. TOTAL DIFFERENTIALS 5 so that using Cramer s rule: Da D det 0 S P D = a S P = > DP S det P + D P S P Da 0 D = a = > DP S det P + D P S P Thus anything which shifts the demand curve to the right increases Q and P: Verify this using a diagram!.5.6 Example 4: The IS=LM Model The Goods and Services Market In the IS/LM model the goods and services market is described by: c d = c (y) ; 0 <c 0 (y) < ; i d = i (r) ;i 0 (r) < 0 g d = g o y d = c d + i d + g d y = y d where c d is consumption demand (with c 0 (y) the marginal propensity to consume), i d is investment demand (which depends negatively on the rate of interest), g d is government expenditure y d is aggregate demand and y is GNP: Here government expenditure is exogenous and equal to some constant g o (the o subscript here is used to denote exogenous variables) while all other variables are endogenous, i.e., c, i; y and r: In equilibrium we require that y = y d = c (y)+i (r)+g o so that: y = c (y)+i (r)+g o which is the structural form of the IS curve. We can write this as an implicit function as: g (y; r; g o ;m s o)=y c (y) i (r) g o =0 where y and r are the endogenous variables and g o and (the not yet introduced exogenous money supply) m s o are the exogenous variables. The total di erential for the IS curve is then: ( c 0 (y)) dy i 0 (r) dr dg o +0dm s o =0:

34 CHAPTER. TOTAL DIFFERENTIALS 6 TheSlopeoftheISCurve Let us now show that the IS curve is downward sloping. The IS curve is all r; y combinations which satisfy this implicit function for xed values of g o and m s o: We can therefore think of y as the independent or exogenous variable and r as the dependent variable or endogenous variable and take the total di erential with dg o = dm s o =0to obtain: ( c 0 (y)) dy i 0 (r) dr =0: dr Solving for the slope of the IS curve given by: dy we obtain: dr dy = c0 (y) i 0 < 0 (r) since c 0 (y) > 0 and i 0 (r) < 0: Thus the IS curve is downward sloping. The Money Market The money market is described by: m d = k (y)+l(r);k 0 (y) > 0; l 0 (r) < 0 m s = m s o where k (y) is the transactions demand for money, l (r) is the speculative demand for money, which depends negatively on r; and m s o is the exogenous supply of money. In equilibrium: m s = m d so that we have: m s o = k (y)+l(r) which is the structural form of the LM curve. We can write the LM curve as an implicit function as: g (y; r; g o ;m s o )=k(y)+l(r) ms o =0: This has a total di erential given by: k 0 (y) dy + l 0 (r) dr +0dg o dm s o =0: TheSlopeoftheLMCurve Let us now show that the LM curve is downward sloping. The LM curve is all r; y combinations which satisfy this implicit function for xed values of g o and m s o: We can therefore think of y as the independent or exogenous variable and r as the dependent variable or endogenous variable and take the total di erential with dg o = dm s o =0to obtain: k 0 (y) dy + l 0 (r) dr =0 and solve for dr dy ; the slope of the LM curve as: dr dy = k0 (y) l 0 (r) > 0

35 CHAPTER. TOTAL DIFFERENTIALS 7 The IS and LM Curves as Implicit Functions Taken together IS and LM equations determine the structural form of the model with two endogenous variables y and r and two exogenous variables g o and m s o. We can write the IS and LM equations respectively as: g (y; r; g o ;m s o) = y c (y) i (r) g o =0 g (y; r; g o ;m s o ) = k (y)+l(r) ms o =0: These two equations then determine a reduced form: y = f (g o ;m s o) r = f (g o ;m s o ) : We wish to calculate such partial o ; which tells us how responds to changes in government expenditure, which tells us how s o interest rates respond to changes in the money supply. The Total Di erential The total di erential for the IS=LM model is given by: since: ( c 0 (y)) dy i 0 (r) dr dg o = 0 k 0 (y) dy + l 0 (r) dr dm s o = 0 and = c0 (y) = i0 (r) ; b = ; s =0 o = k0 (y) = l0 (r) ; b s = : o The E ect of Changes in Government Expenditure o set dm s o =0and replace the d0 s 0 s to obtain: ( c 0 i o = 0 k 0 + l 0 = 0 or writing this in matrix notation: c 0 (y) i 0 (r) k 0 (y) l = o :

36 CHAPTER. TOTAL DIFFERENTIALS 8 Dividing both sides o we obtain: c 0 (y) " i 0 (r) k 0 (y) l 0 (r) Finally, using Cramer o = = so that simplifying @g o # = 0 i det 0 (r) 0 l 0 (r) c det 0 (y) i 0 (r) k 0 (y) l 0 (r) : l 0 (r) ( c 0 (y)) l 0 (r)+i 0 (r) k 0 (y) ( c 0 (y)) + i0 (r)k 0 (y) l 0 (r) Since 0 <c 0 (y) < it follows that 0 < c 0 (y) < so the rst term in the denominator is positive. The second term: i0 (r)k 0 (y) l 0 (r) represents the crowding out e ect and is also positive since i 0 (r) < 0, l 0 (r) < 0 and k 0 (y) > 0: We therefore o > 0 so that an increase in government expenditure increases GNP: Using Cramer s rule to o we nd that: c det 0 k 0 (y) 0 o c det 0 (y) i 0 (r) k 0 (y) l 0 (r) = k 0 (y) c det 0 (y) i 0 > 0 (r) k 0 (y) l 0 (r) so that an increase in government expenditure increases interest rates. It follows from this that investment falls when government expenditure increases. Using the chain rule we (g o ;m s o )) = i 0 (r (g o ;m s (g o;m s o ) < 0 o since i 0 (r) < 0: The E ect of Changes in the Money Supply Similarly s o set dg o =0and replace the d 0 s 0 s to obtain: ( c 0 i 0 = 0 k 0 + l s o = 0: :

37 CHAPTER. TOTAL DIFFERENTIALS 9 or c 0 (y) i 0 (r) k 0 (y) l 0 s o # = 0 : Using Cramer s rule it follows that : 0 i det 0 l 0 s = o c det 0 (y) i 0 (r) k 0 (y) l 0 (r) > s o c det 0 (y) 0 k 0 (y) = c det 0 (y) i 0 (r) k 0 (y) l 0 (r) < 0 so that an increase in the money supply increases GNP and decreases interest rates. (Verify this!) It follows from this that investment increases when the money supply increases since using the chain rule we have: since i 0 (r) < (g o ;m s s o = i 0 (r (g o ;m s o (g o;m s s o > 0.6 Total Di erentials and Optimization.6. A General Discussion Economics deals with rational behavior, and rational behavior typically means maximizing or minimizing something. For example utility and pro ts get maximized while costs get minimized. Maximization and minimization imply rst-order and second-order conditions. First-order conditions say that some expression of the endogenous and exogenous variables must equal zero and so give directly implicit functions from which we can take total di erentials. Second-order conditions tell us that something about a symmetric matrix, the Hessian H: For example with an unconstrained maximization problem we know that H must be negative de nite while for an unconstrained minimization problem H must be positive de nite. It turns out that when we take the total di erential, the coe cients on the dy 0 s; the exogenous variables, always come from H; the Hessian from the secondorder conditions. This turns out to be invaluable in signing partial derivatives since if we use Cramer s rule det [H] will always appear in the denominator.

38 CHAPTER. TOTAL DIFFERENTIALS A Quick Derivation of the Main Principle Let y be an n vector and x an m vector of exogenous variables. Suppose that our structural model involves either maximizing or minimizing: f (y; x) : The optimal y; say y will then satisfy the rst-order (y ;x) =0; for i =; i Here y is the vector of endogenous variables and x is the vector of exogenous variables. The rst-order conditions then de ne a set of n implicit functions given by: g i (y (y ;x) =0; for i =; i This determines the reduced form of n equations: y i = y i (x) for i =; ;:::n: From the second-order conditions we have that the f (y ;x) H j is negative de nite for a maximum and positive de nite for a minimum. Taking the total di erential of g (y ;x)=0we nd that: Hdy + Bdx =0 where the n n matrix H and the n m matrix B are given H (y f (y ;x) ;x) = f (y ;x) B j Using Cramer s rule we then nd that as before that i = det [H i ( b j )] j det [H] From the second-order conditions we know that H is either negative de nite for a maximum or positive de nite and for a minimum hence non-singular. Furthermorewecandeterminethesignofdet [H] : For example det [H] > 0 if H is positive de nite. This can be used in j is positive or negative when we use Cramer s rule.

39 CHAPTER. TOTAL DIFFERENTIALS 3 While this derivation has skipped some details and is perhaps hard to follow in the details, the essential point is easy to grasp and is this: When taking the total di erential from an unconstrained optimization problem, the matrix of coe cients on the dy 0 s is identical to the Hessian: H from the second-order conditions. This fact can then be used in signing the partial derivatives j :.6.3 Pro t Maximization Consider the problem of pro t maximizing in the long-run for a perfectly competitive rm with technology Q = F (L; K) where ¼ is pro ts, Q is output, P is price, L is labour, K is capital, W is the nominal wage and R is the rental cost of capital. Pro t are thus given by: ¼ (L; K; P; W; R) =PF (L; K) WL RK: From the rst-order conditions we have: (L ;K ) W = (L ;K ) R = Here L and K are the endogenous variables and P; W and R are the exogenous variables. The rst-order conditions thus de ne two implicit functions: g (L ;K ;P;W;R) = (L ;K ) W g (L ;K ;P;W;R) = (L ;K ) R which determines the reduced form: L = L (P; W; R) K = K (P; W; R) : To simplify the notation let us make the following de nitions: F ;K F ;K F F (L ;K F F (L ;K F F (L ;K The second-order conditions are then that the Hessian H given by: PFLL PF H = LK PF LK PF KK is negative de nite. Therefore: PF LL (L ;K ) < 0; PF KK (L ;K ) < 0 det [H] > 0:

40 CHAPTER. TOTAL DIFFERENTIALS 3 Taking the total di erential we nd that: PF LL dl + PF LK dk + F L dp dw = 0 PF LK dl + PF KK dk + F K dp dr = 0 so that in matrix notation we have: PFLL PF LK dl FL 0 PF LK PF KK dk + F K 0 {z } =H 4 dp dw dr 3 5 = Note that the matrix of coe cients on the vector of dl and dk ; the endogenous variables, is H; the Hessian from the second-order conditions. Suppose now we wish @W : Set dp = dr =0and change the remaining d 0 s 0 s so that dl ;dk ; and dw put the terms on the right-hand side, and divide to obtain: PFLL PF LK = : KK 0 {z =H Using Cramer s rule we nd = PFLK 0 PF KK det [H] = PF KK det [H] < 0 since PF KK < 0 and det [H] > 0 from the second-order conditions. Note how we have used the second-order conditions to show that the long-run labour demand curve is downward sloping. From this we can also show using Cramer s rule = PFLL PF LK 0 = PF LK det [H] det [H] which can be either positive or negative depending on the sign of F LK : Suppose now @R : From the total di erential we can show that: PFLL 0 = PF LK PF KK @R 0 PFLK det PF KK det [H] PFLL 0 PF LK det [H] = PF LK det [H] = PF LL det [H] < 0: 0 0 :