July 27, 2009

Mathematical Economics: 2009-8-17(Mon)–21(Fri): TF: Kagawa

1 What to Study

Mathematical Economics. . .A study of economics, using advanced mathematical models.Cf. Economics. . .A study of a capitalist economy, or a group of economies: how equilibrium

states are determined, and how the government can shift an equilibrium position.Cf. General Equilibrium Analysis: A method in the study of a national economy, deriving

macro-relations from basic behaviors of individual agents. Initiated by L.Walras.Cf. Marxian Economics: A study on how a capitalist economy functions, and how it deteriorates

and breaks down in the long-run.

2 Warnings at the Beginning

Please note that we study economics, not mathematics. Math is a tool, and grasp it intuitively,or geometrically. Draw diagrams, and do real calculations on your notebooks by yourself withno help of computers.

3 Vector Spaces

Real numbers: this is a real vector space of one dimension, and written as R.

We consider an n-tuple of real numbers: x ≡

x1x2...xn

, y ≡y1y2...yn

, and so on. It is sometimesconvenient to write x (or y) as

x0 ≡ (x1, x2, . . . , xn)with ()0 meaning transposition. (NB. Vectors are represented in bold face.)We define the sum, difference, scalar multiplication between two n-tuples.

Sum and difference: x0 ± y0 ≡ (x1 ± y1, x2 ± y2, . . . , xn ± yn), in short, elementwise.scalar multiplication: kx0 ≡ (kx1, kx2, . . . , kxn), in short, all the elements. k ∈ R.We also define the inner-product of two n-tuples:

p · x ≡nXi=1

pi · xi = p1 · x1 + p2 · x2 + · · ·+ pn · xn,

where p = (p1, p2, . . . , pn) and x0 ≡ (x1, x2, . . . , xn). Often, p · x is written also as px.(NB. p is a row vector, while x a column vector. Hereinafter, p and q show row vectors: normallyvectors are column ones.)We then define the Euclidean distance between two n-tuples, x and y:

kx− yk ≡p(x− y)0 · (x− y) =

vuut nXi=1

(xi − yi)2 .

Def: With these operations, sum, difference, and scalar multiplication, the set of all real n-tuplesforms a vector space of n dimensions. With the inner-product and Euclidean distance defined,the space is the Euclidean space of n dimensions.

July 27, 2009

4 Convergence

A sequence of vectors is simply a map from N to Rn, and is denoted as {xn}n∈N , or simply {xn},where N is the set of natural numbers. (What you have learned is a sequence of (real) numbers,which are in R.)Def. A vector sequence {xn} is said to converge to x∗ when for any small positive number ²,there exists an m ∈ N such that kxn − x∗k < ² for all n > m.

5 Mappings

A map (or a mapping), T , between two spaces, X and Y , is a rule concerning how a particularelement in one space, X, is moved (or mapped) to an element in the other space, Y , and writtenas: T : X → Y , or T : x ∈ X → T (x) ∈ Y . A function is a map from Rn to R. (What youlearned as a function is from R to R.)When there is only one x ∈ X such that T (x) = y for y ∈ T (X) ≡ {y | y = T (x) for somex ∈ X}, T is said to be one-to-one between X and T (X).Suppose that T is one-to-one between X and T (X), then we can define the inverse map T−1 :T (X)→ X .Def. A function, f : Rn → R, is continuous, when {xn}→ x∗ implies {f(xn)}→ f(x∗).Def. A map T from a vector space X to a vector space Y is called linear, if it satisfies twoconditions:(i) [additivity] T (x1 + x2) = T (x1) + T (x2) for x1, x2 ∈ X and(ii)[homogeneity] T (kx) = kT (x) for k ∈ R and x ∈ X .

6 Differentiation

To find out a maximum or a minimum, we differentiate functions, if they are differentiable. (Mostfunctions you know are differentiable, no worry.)Rules of differentiation:(i) (f (x) + g(x))0 = f 0(x) + g0(x); (ii) (k · f (x))0 = k · f 0(x); (iii) (f (x) · g(x))0 = f 0(x) · g(x) +f(x) · g0(x);(iv) (f(g(x)))0 = f 0(u) · g0(x), [u = g(x)].NB. The property (iv) is called “chain rule”.

(v) (xn)0 = n · xn−1; (vi) (ex)0 = ex; (vii) (loge x)0 = 1/x for x > 0; (viii) (

R xcf(t, x)dt)0 =

f(x, x) +R xc(∂f/∂x)dt.

Exercises:

Q01. What are the limits for the following sequences of numbers.(i) (1 + 1

n)n; (ii) { sin(1/n)

1/n}; (iii) (A ·K−1/n + (1− A) · L−1/n)−k··n .

NB. L’Hopital’s Rule: Suppose f(x)→ 0, g(x)→ 0, f0(x)g0(x) → C ( and g0(x) 6= 0) as x→ a, then

f (x)g(x)→ C as x→ a.

Q02. Differentiate the following functions wrt. x.(i) loga x; (ii) (x

3 + 2x− 7)11; (iii) (loge x)3.

July 27, 2009

7 Optimization with a constraint

Maximization Problem with a constraint: maximize f(x) subject to g(x) = 0, wheref, g : Rn → R.Lagrange Multiplier Method: Construct the Lagrangian function L ≡ f(x)− λ · g(x), andsolve the following system of simultaneous equations:

∂ L∂xi= 0, i = 1, 2, . . . n

∂ L∂λ= 0.

There are n+ 1 equations for n + 1 variables, x1, x2, . . . , xn, and λ.NB. The last equation is the constraint itself.

NB. When conducting partial differentiation, regard the other variables as constants.

8 Theory of Consumers’ Demand

Consumers are supposed to maximize their “utility”, u(), subject to their respective budgetconstraint:maximize u(x) subject to p · x = E. Here E is a given positive amount of expenditures, andp a given price vector.NB. In this problem, our utility is simply a function of consumption basket x. Cf. Veblen effect.

NB. Never fail to transform the constraint into the form g(x) = 0.

1. the case of two commodities and u(x) = A · xα1 · xβ2 . (parameters: A,α, β > 0.)First, form the Lagrangian L = A · xα1 · xβ2 − λ · (p1 · x1 + p2 · x2 − E). Then we get

∂ L∂x1

= A · α · xα−11 · xβ2 − λ · p1 (1)∂ L∂x2

= A · β · xα1 · xβ−12 − λ · p2 (2)

−(p1 · x1 + p2 · x2 − E) = 0 (3)

(1)× x1 ⇒ A · α · xα1 · xβ2 = λ · p1 · x1 (4)(2)× x2 ⇒ A · β · xα1 · xβ2 = λ · p2 · x2 (5)Dividing (4) by (5), we get

α

β=p1 · x1p2 · x2 .

Substitute p2 · x2 = βα· p1 · x1 into (3), and we have x1 = ( α

α+β· E) / p1. In a similar way, we

obtain x2 = (β

α+β· E) / p2.

In sum, divide the budget into two parts of the ratios, α to β. Spend these parts for commodities1 and 2 respectively.Q00. When p1 goes up, what will happen to the consumption basket?

Exercises:

Q01. Consider the case of two commodities and u(x) = αx1 + βx2, and derive the utilitymaximizing consumption basket.Q02. Consider the case of three commodities and u(x) = A · xα1 · xβ2 · xγ3 , and derive the utilitymaximizing consumption basket.

July 27, 2009

9 Theory of Consumers’ Demand (continued)

Def. An indifference curve is a curve each consumption basket on which gives the same utilitylevel.

(1) the case of Cobb=Douglas utility function (2 commodities): No Substitutionu(x, y) = A · xα · yβ ; prices px, py; expenditures E. See “mathecon 02.pdf”

x =α

a+ β· Epx, y =

β

a+ β· Epy

Exercises:Q01. When our utility function is of Cobb=Douglas type, what is the effects of the increase inpx on x and y? (No substitution at all!).Q02. Let u(x, y) = x·y, px = 2, py = 3, and E = 12. Draw indifference curves (E = 6, 10, 12, 16)and the area where the budget constraint is satisfied.

(2) the case of a linear utility function (2 commodities): u(x) = αx + βy; prices px, py;expenditures E. Sudden Perfect SubstitutionSuppose you buy only x, then you can get E

px, securing the utility level α · E

px. On the other

hand, if you buy only y, then you can get Epy, securing the utility level β · E

py. Thus, when

α · Epx> β · E

py, you buy only x as much as E

px. When α · E

px< β · E

py, you buy only y as much

as Epyyou can get E

py, securing the utility level β · E

py. When α · E

px= β · E

py, you may buy either

commodity, or mix in any ratio. So, as px changes from a very low price to a higher one, acomplete substitution takes place all of a sudden.The area where the budget constraint, px · x+ py · y ≤ E, is satisfied forms a triangle. However,only the hypotenuse is relevant.Exercises:

Q03. Consider the case of two commodities, and u(x) = 2x + 3y, px = 20, py = 100, andE = 10000. Derive the utility maximizing consumption basket. Find out the price of x abovewhich y substitutes x.Q04. Consider the case of three commodities, and u(x) = 2x+ 3y + 4z, px = 20, py = 60, pz =100 and E = 12000. Derive the utility maximizing consumption basket. Now suppose the priceof x starts to rise, and you wish to find out the price of x , above which x is substituted for byanother commodity. By which commodity, is x substituted?Answer. If we buy x only, we can get E/px = 12000/20 = 600, thus giving us the utility levelu = 2× 600 = 1200. Similarly, concentrating on y, we secure the utility level u = 3× 200 = 600,and by z, u = 480. Therefore, we buy only x in the amount of 600, with y = z = 0. As the priceof x, px, goes up, the quantity of x we can buy decreases, hence the utility level also decreasesby sticking to x. When px = 40, we can buy x as much as 300(= 12000/40), giving the utilitylevel 600(= 2× 300), which is the same level if we concentrate on y. Hence, substituting y for xtakes place when px goes over 40.

(3) the case of a min(,) utility function (2 commodities): L-shaped Indifference Curves andComplementsNow we consider the utility function u(x, y) = min(x/a, y/b), which attains the smaller valuebetween two arguments. In this case, indifference curves become L-shaped, and the highestutility level is attained at the south-west corner of an L-shaped indifference curve, which cornershould be on the hypotenuse of the budget constraint triangle. (Draw a diagram.) Thus, as the

July 27, 2009

price of x goes up, both commodities x and y decreases, and vice versa: they are complements.These are like a desk & a chair, coffee & a coffee cup etc.Exercises:Q05. Let u(x, y) = min(x/3, y/4), px = 20, py = 60, and E = 12000. What is the consumptionbasket which maximizes utility.Answer. Suppose we buy 3k of x and 4k of y, then the expense is 20× 3k + 60× 4k, whichshould be equal to E = 12000. Thus, 300k = 12000, getting k = 40. Hence, one should buyx = 120 and y = 160, yielding utility 40.

Q06. Now take up the case of three commodities, and let u = min(x/3, y/4, z/5), px = 20,py = 30, pz = 40, and E = 12000. Find out the consumption basket with the maximum utility.

10 Order

Def. An order is a binary relation ¹ on a given set which satisfies the following three properties.(i) reflexivity: a ¹ a;(ii) symmetry: If a ¹ b and b ¹ a, then a = b;(iii) transitivity: If a ¹ b and b ¹ c, then a ¹ c.

Examples. (a) The set of natural numbers, N . (b) The set of integers, Z.(c) The set of rational numbers, Q. (d) The set of real numbers, R.(e) In the Euclidean space of Rn, we consider an order ¹ which is defined as x ¹ y iff xi ≤ yifor all i. (that is, when a vector y is to the north-east of a vector x , we say y is greater thanor equal to x.(e) In the commodity space of R2+, the non-negative orthant of R

2, we consider an order ¹which is defined as in the case of Rn.Def. A set is called a partially ordered set when it has an order.Def. A map T form an partially ordered set to another partially ordered set is called “isotone”,if x ¹ y implies T (x) ¹ T (y).

11 Fixed Point Theorems

References:Tom M. Apostol, Mathematical Analysis, 2nd ed., Addison Wesley (Menlo Park), 1974. *detailedAlpha C. Chiang, Fundamental Methods of Mathematical Economics, 2nd ed.,McGraw-Hill, 1967, 1974. (A translation from McGraw-Hill Shuppan, Tokyo.) * a starter

Robert Dorfman, Paul A. Samuelson, Robert M. Solow, Linear Programming and EconomicAnalysis, McGraw-Hill (New York), 1958.

Kelvin Lancaster, Mathematical Economics, New York: Dover (New York), 1968. * a starterWilliam Novshek, Mathematics for Economists, Academic Press (San Diego), 1993. *not easyEugene Silberberg, The Structure of Economics, McGraw Hill (New York), 1990.Akira Takayama, Mathematical Economics, Cambridge Univ Press,1985.

Def. A segment [x, z] in Rn is defined as the set {y | y ∈ Rn such that x ¹ y ¹ z}

July 27, 2009

Def. In a partially ordered set Y , an upper bound u of a subset S ⊂ Y is an element such thatx ¹ u for ∀x ∈ S .Def. The least upper bound of a subset S, supS, in a partially ordered set Y , is the minimumelement, if any, in the set of all upper bounds of S.Tarski’s Fixed Point Theorem. If a map T from a segment, S, of Rn into itself is isotone,it has at least one fixed point.Proof. Consider the set B ≡ {x |x ¹ Tx, x ∈ S}, and let the least upper bound of S be x∗.From x ¹ x∗ for ∀x ∈ S, we have x ¹ Tx ¹ Tx∗ by isotoneness and the fact x ∈ S. Thus, itfollows

x∗ ¹ Tx∗.

Again, by isotoneness, Tx∗ ¹ T (Tx∗), which shows that Tx∗ ∈ S, therefore we get

Tx∗ ¹ x∗.

From the above two inequalities, we have the desired result, x∗ = Tx∗. 2

Def. A map T from a subset of Rn into Rn is said to be continuous , if a sequence {xn}converges x∗, then {T (xn)} converges to T (x∗).Brouwer’s Fixed Point Theorem. If a map T from a segment, S, ofRn into itself is contin-uous, it has at least one fixed point.Proof. (Omitted, because we need many pages.)2

Red Herring: A great many people regard the above theorem as “spine-tingling”. This says:(1) When you stir coffee or tea in a cup quietly, making no splash, there is at least one moleculeof water which does not move. (Continuous, because you stir quietly with no drop of waterjumping out of the surface.) (The surface of coffee in a cup can be continuously transformed toa segment in R2.)(2) You spread a sheet of plastic wrap on the table, and then put it in your palm and crease itwithout tearing. When you return that rucked piece where it was on the table in the beginning,there is at least one molecule which is at the same place where it was.Reading: Robert Louis Stevenson, The Bottle Imp, 1893.(You can read at: http://gaslight.mtroyal.ab.ca/bottlimp.htm and many other sites.By the way, Stevenson preferred gas light to electric one.)

12 Fixed Point Theorem (continued): Brouwer and Kaku-

tani

Def. A set which can be included in a ball of a sufficiently large radius is bounded.Def. A set is closed iff any converging sequence in the set has a limit point also in it.Def. A bounded and closed set is compact. (This is valid only in the space of a finite dimension.)Brouwer’s Fixed Point Theorem can be generalized to a continuous map T from a compactconvex set into itself.

Def. A correspondence from X to Y is a map which assigns a subset in Y for a point in X.Def. A correspondence f from X to Y is upper semi-continuous iff given any sequence in X,{xn}→ x∗, yn ∈ f(xn), and {yn}→ y∗ imply y∗ ∈ f (x∗).

July 27, 2009

Kakutani’s Fixed Point Theorem: An upper semi-continuous correspondence from a com-pact convex set into itself has at least one fixed point.

Application to Game Theory : n player non-cooperative gamesEach player has a finite number of pure strategies, which she/he can mix with certain probabili-ties. Then each player has a compact convex set of strategies. The direct product S of strategysets of all players also forms a compact convex set.Now given a point in this S, players modify their strategy so as to maximize their pay-off. Thiscorrespondence can be regarded as upper semi-continuous. Hence, there is a fixed point.Existence of a Nash Equilibrium: In a finite game with mixed strategies, there is a Nashequilibrium.NB. John Nash proved the existence in 1950, using Kakutani’s fixed point theorem. He receiveda Nobel Prize in 1994.

13 Production

Production : Supply = Demand : Consumption and InvestmentSupply of commodities ⇐ profit maximizationDef. profit = sales - costs

Production function: F (x1, x2, . . . , xn, k1, k2, . . . , km, `1, `2, . . . , `s);xi: material inputs, kj: durable machines, `t: labor powerpi: price of material inputs, rj: rental of durable machines, wt: wage rate of labor power of

type t.p = the price of product

Profit = p·F (x1, x2, . . . , xn, k1, k2, . . . , km, `1, `2, . . . , `s)−(Pn

i=1 pi ·xi+Pm

j=1 rj ·kj+Ps

t=1wt ·`t)

14 Framework of General Equilibrium Analysis

Economic Agents: Consumers, Workers, and Firms, all mixed: the government behindthe scene.

Behavioral Assumptions: Every agent tries to maximize her/his objective function in aselfish way. Utility functions for consumers and workers, profit functions for firms.Each agent decides her/his demand/supply independently others, taking prices as given.Can there be an equilibrium, where every market sees its supply and demand in balance?Can every market reach and stay at an equilibrium?

Three Classical Problems:(1) Existence, (2) Uniqueness, and (3) Stability of Equilibrium.

15 Production Function

A typical production function for a firm is S-shaped: first exhibiting increasing returns toscale(irs) and then when the scale of production reaches a certain level, showing decreasingreturns to scale(drs).

Def. A function f from Rn+ to R is positively homogeneous of degree r iff f (kx) = kr ·f(x)for x ∈Rn+, k ∈ R+.

July 27, 2009

Euler’s Theorem: If f is positively homogeneous of degree r and differentiable at x, then

nXi=1

∂f (x)

∂xi· xi = r · f(x).

Proof. Differentiate f (kx) = kr · f (x) with respect to k, and set k = 1, then we have thedesired result. 2

Theorem: If f is (positively) homogeneous of degree r and differentiable, ∂f (x)/∂xi is homo-geneous of degree r − 1.Proof. Clear from the definition of partial derivative. 2

Def. When the production function exhibits Constant Returns to Scale (crs), it is positivelyhomogeneous of degree one (or has the first degree of homogeneity). It means that if materialinputs, machines, and labour inputs are increased by the same percentage, then the output willalso increase by the same proportion.Profit Maximization: Profit maximization is compatible with constant returns to scale onlywhen maximized profit is zero.Calculation of Profit Maximization:Production function: F (x1, x2, . . . , xn, k1, k2, . . . , km, `1, `2, . . . , `s);xi: material inputs, kj: durable machines, `t: labor powerpi: price of material inputs, rj: rental of durable machines, wt: wage rate of labor power of

type t.p = the price of product

Profit π ≡ p·F (x1, x2, . . . , xn, k1, k2, . . . , km, `1, `2, . . . , `s)−(Pn

i=1 pi·xi+Pm

j=1 rj ·kj+Ps

t=1wt·`t)Solve the following system of simultaneous equations:

∂π∂xi= 0, i = 1, . . . , n

∂π∂ki= 0, i = 1, . . . ,m

∂π∂`i= 0, i = 1, . . . , s

.

As a typical equation, we have

p · ∂F∂xi− pi = 0.

This says the value of marginal product of material input i is equal to the price of the inputi: the latter being the marginal cost of the material input.NB. marginal revenue = marginal cost, at the profit maximizing operation.Def. Fixed cost is the expense necessary, independent of production level.NB. The existence of fixed cost does not damage the homogeneity of degree zero of supplyfunctions. Why?Def. Variable cost is the expenditure which varies with production level.

Exercises:Q01. Let the production function be F (x1, x2) = A · xa1 · xb2, the price of product p, the costfunction (q1 · x1 + q2 · x2). Find out x∗1 and x∗2 which maximize profits. The prices p, q1,andq2 are given, while A, a, and b are constants.. ( a > 0, b > 0, and a+ b < 1. Thus, decreasingreturns to scale (drs) obtain.)Answer. The profits can be expressed as π ≡ p · A · xa1 · xb2− (q1 · x1 + q2 · x2). We differentiatethis wrt x1 and x2, getting

July 27, 2009

½ ∂π∂x1

= p · A · a · xa−11 · xb2 − q1 = 0 (1)∂π∂x2

= p · A · b · xa1 · xb−12 − q2 = 0 (2).

From eqs.(1) and (2), we have½p · A · a · xa1 · xb2 = q1 · x1 (3)p · A · b · xa1 · xb2 = q2 · x2 (4)

.

Thus, we have the now familiar equation in maximizing a function of Cobb-Douglas type:

a

b=q1 · x1q2 · x2 .

Multiplying eq.(3) by qb2, we obtain p · A · a · xa1 · (q2 · x2)b = q1 · qb2 · x1, form which it followsp ·A · a · xa1 · ( ba · q1 · x1)b = q1 · qb2 · x1. Thus,

x∗1 = (pA

q1−b1 · qb2· a1−b · bb) 1

1−(a+b) .

Similarly,

x∗2 = (pA

qa1 · q1−a2

· aa · b1−a) 11−(a+b) .2

Q02. Confirm the solutions x∗1 and x∗2 are homogeneous of degree zero wrt prices p, q1, and q2.

Q03. Confirm the optimum output is also homogeneous of degree zero wrt prices p, q1, and q2.

Cost Functions Approach:Often, profit is described by the following, using the cost function which depends upon thequantity produced.

π ≡ p · x− C(x), C(x) ≡ ax3 + bx2 + cx+ d, where a > 0, b < 0, c > 0, d ≥ 0, b2 − 3ac ≤ 0.Here d is the fixed cost. The output x is the sole variable with the parameters p, a, b, c, d given.Exercises:Q04. Depict a graph of the cost function above.Q05. Suppose p = 10, and C(x) = x3− 5x2 +10x+10. Compute the profit maximizing output

and the profit.Q06. Suppose p = 7, and C(x) = x3 − 5x2 + 10x+ 10. Compute the profit maximizing

output and the profit. Should the firm carry out the production?Q07. Suppose p = 5, and C(x) = x3 − 5x2 + 10x+ 10. Compute the profit maximizing

output and the profit. Should the firm carry out the production?NB. Even if the maximum profit is negative, it is better than doing nothing because of thefixed cost. This is in the short-run phenomenon, though. Consider how such a situation candisappear.

“Red” Herring:Important formula of differentiation: the Quotient Rule:

d(u(x)v(x))

dx=u0 · v − u · v0

v2, where u0 ≡ du(x)

dxand v0 ≡ dv(x)

dx.

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Convergence of sequence:

{xn}→ x∗ iff for any ² > 0, there exists a N > 0 such that kxn − x∗k < ².This can be written, using logics symbols, as

{xn}→ x∗ iff ∀² , ∃N such that kxn − x∗k < ².Exercises:Q08. Prove the sequence {2−n} , (n = 1, 2, . . .) converges to 0.Q09. Consider the sequence {(−1)n · (1− 1

n)}, (n = 1, 2, . . .). Prove that this does not converge

to 1. The subsequence {(−1)k} (k = 2, 4, . . . , 2m, . . .), however, converges to 1.

16 Cost Function

Suppose C(x) = x3 − 5x2 + 10x + 10. Find out the price of product at which the maximizedprofit is zero. Thus, the profit maximizing output is ‘break-even’. Find out also the price atwhich the firm stops production.NB. Please produce a report about this problem.Answer.Let the cost function be, in general,

C(x) = x3 + bx2 + cx+ d. ( b < 0, c > 0, d > 0, and b2 − 3c ≤ 0).

( b < 0 to have the positive inflection point; b2 − 3c ≤ 0 for the cost not to decrease.)At the output level xo, the tangent line to the cost curve is:

y − (x3o + bx2o + cxo + d) = (3x2o + 2bxo + c) · (x− xo). (L)In order for this tangent line to go through the origin, i.e., when x = 0, y = 0, it is necessary tohave

−(x3o + bx2o + cxo + d) = −(3x3o + 2bx2o + cxo).That is,

2x3o + bx2o − d = 0. (*)

In the special case where b = −5, c = 10, and d = 10, the equation (*) becomes

2x3o − 5x2o − 10 = 0This has only one positive solution, while other two are complex numbers. The positive solutionis: xo ∼= 3. 0408. Then the price of product is:

3(3.04)2 + 2(−5)(3.04) + 10 ∼= 7.3248.So, at the price 7.3248 (or about 7.3), the maximized profit is 0 at the output about3.04.Next, let us find out the price at which the firm stop its operation. This is when the tangent

line at the profit maximizing output go through the point (0, d). Thus, we substitute 0 for x,and d for y in the equation (L) above.

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d− (x3o + bx2o + cxo + d) = (3x2o + 2bxo + c) · (0− xo). (L)

That is,

2x3o + bx2o = 0. (**)

In our particular case,

2x3o − 5x2o = 0.

We solve this equation, and find out xo = 2.5. (Certainly, xo = 0 also, but no operation at anyprice lower than 2.5.)The price of the product at which the firm makes loss as large as the fixed cost, 10, is:

3(2.5)2 + 2(−5)(2.5) + 10 = 3.75.

So, at the price 3.75, the maximized profit is −10 at the output 2.5.2

17 Cost Function: the Case of Quadratic Equation

Let us treat an easier case. Suppose C(x) = ax2 + bx+ c, with a > 0, b ≥ 0, and c ≥ 0. Then,the profit function is:

π(x) ≡ p · x−C(x),

where p is the price of product. In this case, there is no positive price of the product, abovewhich the production is carried out even making losses in order to avoid the full expenses onfixed cost.In general, the profit is maximized at

dπ

dx= −2ax+ (p− b) = 0, i.e., x = p− b

2a.

When the cost function is given as above, the tangent line at the output level, xo,is

y − (ax2o + bxo + c) = (2axo + b)(x− xo).

This tangent line go through the origin if y = 0 at x = 0. This means

−(ax2o + bxo + c) = −2ax2o − bxo, i.e., ax2o − c = 0.

Thus,

xo =

rc

a.

The slope at this point is

2a

rc

a+ b = 2

√ac+ b.

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Therefore, given the cost function C(x) = ax2 + bx+ c, if the price satisfies

p = 2√ac+ b,

then the maximum profit is zero at the output xo =p

ca.

Exercises:Q01. Suppose the cost function is C(x) = x2 + 2x + 5. The price of product is p = 10̇. Findout the profit maximizing output and the profit there.Q02. Suppose the cost function is C(x) = x2 + 2x+ 7. Find out the price of product at whichthe maximized profit becomes zero.

18 Cubic Equation

In the calculation of zero-profit price, we have come to the following cubic equation:

2x3 + bx2 − d = 0.

A real positive solution for this is :x = 3

sµ1

4d− 1

216b3 +

1

36

p(81d2 − 3db3)

¶+1

36

b2

3

r³14d− 1

216b3 + 1

36

p(81d2 − 3db3)

´ − 16b .

In general, a real solution to

x3 + bx2 + d = 0,

is:x = 3

sµ−12d− 1

27b3 +

1

18

p(81d2 + 12db3)

¶+1

9

b2

3

r³−12d− 1

27b3 + 1

18

p(81d2 + 12db3)

´ − 13b ,

while a real solution to

x3 + cx+ d = 0,

is: x = 3

sµ−12d +

1

18

p(12c3 + 81d2)

¶− 13

c

3

r³−12d+ 1

18

p(12c3 + 81d2)

´ .

July 27, 2009

19 Newton-Raphson Method

To solve linear or nonlinear equations, we can use the Newton-Raphson method under someconditions. This is an iterative method, and we choose an initial value near the wanted solution.We draw a tangent line at (x(0), f(x(0))) , and the point where this line cut the x-axis is thenext point. This process is repeated until a point gets near the wanted solution.The method is described by:

x(i+1) = x(i) − f(x(i))

f 0(x(i)), (1)

where f 0 = dfdx.

NB. The Newton-Raphson method may not converge. You need some conditions to secure theconvergence.

The method was generalized to the n variable case by Leonid Kantrovich (Nobel Prize in1975):

x(i+1) = x(i) − (JF (x(j)))−1 · F (x(i)),

where JF (x(j)) is the Jacobian of F with respect to x evaluated at x(i).

20 Gale-Nikaido’s Proof of Existence of General Equilib-

rium

We prove the following system of equations has a solution, using the Brouwer Fixed PointTheorem.

e1(p1, p2, . . . , pn) ≤ 0,e2(p1, p2, . . . , pn) ≤ 0,

· · ·en(p1, p2, . . . , pn) ≤ 0,

(GE)

where ei(p) ≡ di(p) − si(p) is the excess demand function for commodity i, with p ≡(p1, p2, . . . , pn). We know ei’s are continuous on the non-negative orthant of R

n, and homo-geneous of degree zero. We also have the Walras Law:

nXi=1

pi · ei(p) = 0.

Therefore, a solution to the system (GE), p, satisfies the following:

(1) if pi > 0, then ei(p) = 0, and (2) if ej(p) < 0, then pj = 0.

We define the (n− 1)-simplex S to be

S ≡ {x | x ∈ Rn+,nXi=1

xi = 1}.

July 27, 2009

We construct the following map T from S into itself:

T : p ∈ S → (pi +max(0, ei(p))

1 +Pn

i=1max(0, ei(p))).

This map T is continuous, and the set S is compact convex, and so there is at least one fixedpoint p∗. This should be an equilibrium price vector. First we show at the fixed point p∗, thedenominator is 1. Suppose the denominator is k, and k > 1. (It is evident k ≥ 1.) Then weget

(k − 1) · pi = max(0, ei(p)) for all i.

This means that if pi > 0, then ei(p) > 0, which is a contradiction to the Walras Law.2

Reference:Uzawa, H. (1962): “Walras’ Existence Theorem and Brouwer’s Fixed Point Theory”, Economic

Studies Quarterly, vol. 31.Exercise:Q01. Try to devise out your own map whose fixed point can be an equilibrium.

21 Supplement: Excess Demand Functions–Examples

Reference:T. Fujimoto, “Numerical Examples of Systems of Excess Demand Functions”, Kagawa University

Economic Review, vol.60, 1988, pp.801-805.

Assumption: We consider three commodity economies only.

The equilibrium price vector is common among the first 4 examples and it is P ∗ = (1, 1, 1)̇. Ineach case, verify the homogeneity of degree zero and the Walras Law.

(1) Gross Substitutes E1(P) = (

1/P1A− 1) + P2/P

21

A+

P3/P21

A,

E2(P) =P1/P 22A

+ (1/P2A− 1) + P3/P 22

A,

E3(P) =P1/P 23A

+P2/P 23A

+ (1/P3A− 1),

where A ≡ 1/P1 + 1/P2 + 1/P3.The sign pattern of the Jacobian matrix: − + +

+ − ++ + −

.NB. When there are three consumers, and the common utility function of individual is u(x1, x2, x3) =√x1 +

√x2 +

√x3. with the initial endowment of individual i being one unit of commodity i.

(2) Negative Dominant Diagonal

July 27, 2009

E1(P) =

P 22P 21+P1P2

+P 23

P 21+P1P3− 1,

E2(P) =P1

P2+P3− P2

P1+P2,

E3(P) =P1

P2+P3− P3

P1+P3.

The sign pattern of the Jacobian matrix: − + ++ − −+ − −

.NB. This system also satisfies the weak axiom of revealed preference.

(3) Giffen Goods (NB. Unbounded from below)E1(P) = 1− P 22

P 21+P1P2− P 23

P 21+P1P3,

E2(P) =P2

P1+P2− P1

P2+P3,

E3(P) =P3

P1+P3− P1

P2+P3.

The sign pattern of the Jacobian matrix: + − −− + +− + +

.Cf. inferior goods.

(4) Scarf’s Example (NB. Global Instability)H. Scarf, “Some Examples of Global Instability of Competitive Equilibrium”, International

Economic Review, vol.1, 1960, pp.157-172.E1(P) =

P3P3+P1

− P2P1+P2

,

E2(P) =P1

P1+P2− P3

P2+P3,

E3(P) =P2

P2+P3− P1

P3+P1.

The sign pattern of the Jacobian matrix: ± − ++ ± −− + ±

.

(5) Nikaido’s Example (Gross substitutes)H. Nikaido, Convex Structures and Economic Analysis, Academic Press, New York, 1968.

Ei(P) =nXj=1

aijPj/Pi −nXj=1

aji ., for a given n × n positive matrix A ≡ (aij).

When A is given as

July 27, 2009

A =

1 2 34 3 25 5 5

, E1(P) = (P1 + 2 · P2 + 3 · P3)/P1 − 10,E2(P) = (4 · P1 + 3 · P2 + 2 · P3)/P2 − 10,E3(P) = (5 · P1 + 5 · P2 + 5 · P3)/P3 − 10.

NB. The equilibrium price vector is P ∗ = (5, 6, 11).

Exercises:Q01. Verify the weak axiom of revealed preference for the system in (2) above.Q02. Compute the equilibrium price vector in your own way for the example in (4) Nikaido’sexample.

22 Input-Output Models: A Simple Leontief Model

References:Leontief, Wassily: The Structure of the American Economy, 1919-1939, Oxford Univ Press, 1941.Dorfman, Robert, Paul A. Samuelson and Robert M. Solow: Linear Programming and EconomicAnalysis, McGraw-Hill, 1958.

Nikaido, Hukukane, Linear Algebra for Economics (in Japanese “Keizai no tameno SenkeiSuugaku”), Baifukan, 1961.

Assumptions:(1) the absence of joint production; (2) the absence of durable capital goods;(3) the homogeneous labor; (4) constant returns to scale;(5) the absence of externalities; (6) the uniform production period called year;

Variables and Data:

x : ann-column vector of activity levels, a gross output vector;

p : an n-row vector of wage-unit prices;

A :

the given n by n input-output matrix:aij = the amount of commodity i necessary to produce

one unit of commodity j in industry j.

` : the given n-row vecor of labor input coefficients;

g : a given rate of steady balanced growth;

r : a given rate of uniform profit;

d : a given n-column vector of final demands.

Quantity Equation: the gross output of each commodity at the end of production periodequals the input at the beginning of the next period plus the final demand.

x = (1 + g) · Ax+ d.

July 27, 2009

Price Equation: the price of a commodity equals its cost, that is, the cost of material inputswith a mark-up ratio, r, plus the labor cost.

p = (1 + r) · pA+ `.

23 A Two Commodity Model

We consider a special case of two commodities, say, corn and tractor.

x = (x1, x2)0,

p = (p1, p2),

A =

µa11 a12a21 a22

¶,

` = (`1, `2),

d = (d1, d2)0, and

g = r = 0.

Then, the quantity equation, x = Ax+ d, becomes½x1 = a11 · x1 + a12 · x2 + d1,x2 = a21 · x1 + a22 · x2 + d2.

And the price equation, p = pA+ `,½p1 = a11 · p1 + a21 · p2 + `1,p2 = a12 · p1 + a22 · p2 + `2.

NB. The columns of A show the inputs of industries.

Exercises:Q01. Let

A ≡µ0.2 0.50.3 0.6

¶, ` ≡ (1, 1), and d ≡

µ10

¶.

(1) Solve the quantity and price equations with g = r = 0.(2) When d becomes d = (0, 1)0 , what are solutions to the quantity equation?

Q02. Let A, `, and d be as in Q01. Does the quantity equation have a meaningful solutionwhen g = 0.1? How about when g = 0.3? What is the maximum rate of growth below whichthe quantity equation has a semi-positive solution?Q03. Compute the inverse of the matrix (I − A).

24 Input-Output Models: How to Solve It

(1) In general:

Let us consider the system of simultaneous equation Mx = d , where

July 27, 2009

M ≡

m11 m12 · · · m1n

m21 m22 · · · m2n

· · · · · · . . . · · ·mn1 mn2 · · · mnn

, x =x1x2...xn

, and d =

d1d2...dn

.The matrix M and the vector d are given, while x is a vector of unknowns.This system can be solved by use of the inverse of M , if M is regular, i.e., |M | 6= 0. That

is,

x =M−1 · d.You are reminded of the solution x = m−1 · d for a scalar equation mx = d. Another formulato solve the equation is Cramer’s Rule:

(Cramer’s Rule) xi =

i∨¯̄̄... d

...

¯̄̄|M | ,

where the numerator is M with the i-th column replaced by d.Determinants:

A given matrix is 1×1, m, then its determinant is m. Then, for a n×n matrix, its determinantis defined recursively as

|M | ≡nXj=1

aij · (−1)i+ j · |Mij| ,

where Mij is the (n− 1)× (n− 1) matrix created by deleting the i-th row and the j-th columnfrom M . This is the expansion of |M |, using the i-th row. The expansion via j-th column is

|M | ≡nXi=1

aij · (−1)i+ j · |Mij| .

Normally, we use a column or a row which has many zeros as possible.

Calculation of the Inverse using the Cramer’s Rule:

When d is

d = ei =

...1......

,i.e., d has one in the i-th element with the remaining elements being all zero, the solution toMx = d is the i-th column of the inverse M−1.

(2) 2 by 2 case:

Let

July 27, 2009

M ≡µa bc d

¶,

then

M−1 =

Ãd|M |

−b|M |

−c|M |

a|M |

!.

The determinant |M | is

|M | = ad− bc.Cranes-and-Tortoises Calculation (“Tsuru-Kame Zan”):½

Cranes + Tortoises = 20,(Legs of Cranes) + (Legs of Tortoises) = 62.

Suppose there are 20 cranes, then we have 40 legs. By replacing one crane by one tortoise, wecan increase two legs, and 22(=62-40) legs should be added, and so the number of tortoises is11(=22/2). And that of cranes is 9.The above can be converted to a system of simultaneous equations:x =number of cranes, and y =number of tortoises.½

x+ y = 20,2x+ 4y = 62.

, i.e.,

½µ1, 12, 4

¶µxy

¶=

µ2062

¶.

M =

µ1, 12, 4

¶, |M | = 1× 4− 1× 2 = 2, and M−1 =

µ4/2, −1/2−2/2, 1/2

¶=

µ2, −1/2−1, 1/2

¶.

Thus, µxy

¶=M−1 ·

µ2062

¶=

µ40− 31−20 + 31

¶=

µ911

¶.

Or, using the Cramer’s Rule:

x =

¯̄̄̄20, 162, 4

¯̄̄̄|M | =

80− 622

= 9, y =

¯̄̄̄1, 202, 62

¯̄̄̄|M | =

62− 402

= 11.

(3) the case of (I-A): We know A ≥ 0. If there exists an x ∈ Rn+ such that xÀ Ax. Then,

(I − A)−1 = I +A+A2 + · · ·+Ak + · · · (Neumann series).For the quantity equation, x = Ax+ d, we have an iterative method:

x(0) = d,

x(1) = Ax(0) + d,

· · ·x(k+1) = Ax(k) + d,

· · · .

July 27, 2009

Exercises:Q01. Find out the numbers of cranes and tortoises which satisfy the following:½

Cranes + Tortoises = 30,(Legs of Cranes) + (Legs of Tortoises) = 102.

25 Answers to the Problems:

1. Consider the case of three commodities, andu(x, y, z) = 2x+3y+ 6z, px = 20, py = 60, pz =100, and E = 12000. Derive the utility maximizing consumption basket. Now suppose theprice of x starts to rise, and you wish to find out the price of x above which commodity x issubstituted for by another commodity. By which commodity, is x substituted?Hint. Lecture note (3). Draw a figure of indifference surface.Answer:If we buy only commodity x, we can buy qx =

1200020

= 600, giving the utility level, ux:only =2× 600 = 1200. Next, in a similar way, with commodity y, we can buy qy = 12000

60= 200, giving

the utility level, uy:only = 3×200 = 600. Third, with commodity z, we can buy qz = 12000100

= 120,giving the utility level, uy:only = 6× 120 = 720.So, the utility maximizing basket is: (600, 0, 0), purchasing commodity x only. The secondhighest utility is realized by buying commodity y only. Thus, when the price of commodity xrises up to a point where it gives the utility level 720, we become indifferent between x and z.This level is obtained by getting the amount 720

2= 360 of x. This amount is purchasable when

its price is 12000360

= 33. 333.The utility maximizing basket is: (600, 0, 0). At px = 34, commodity x is substitutedby z.

2. Consider the case of three commodities and u(x) = xα1 · xβ2 · xγ3 , where α > 0, β > 0, andγ > 0, and derive the utility maximizing consumption basket subject to the budget constraint ,p1 · x1 + p2 · x2 + p3 · x3 = E, where p1 > 0, p2 > 0, p3 > 0 and E > 0.Hint. Use the Lagrangian multiplier method.Answer. Rewrite the constraint into the form p1 · x1 + p2 · x2 + p3 · x3 −E = 0, and define theLagrangian functions as

L ≡ xα1 · xβ2 · xγ3 − λ · (p1 · x1 + p2 · x2 + p3 · x3 − E).Differentiate L wrt each variable xi and the Lagrangian multiplier λ, and set them to zero, thusobtaining a system of four equations with four variables.

∂L∂x1

= α · xα−11 · xβ2 · xγ3 − λ · p1 = 0, (1)∂L∂x2

= β · xα1 · xβ−12 · xγ3 − λ · p2 = 0,∂L∂x3

= γ · xα1 · xβ2 · xγ−13 − λ · p3 = 0,∂L∂ λ= −( p1 · x1 + p2 · x2 + p3 · x3 −E) = 0.

Multiply eq.(1) by x1, then we get α ·xα1 ·xβ2 ·xγ3−λ ·p1 ·x1 = 0. (NB.Revive the original utilityfunction.) Thus, it follows

p1 · x1 = α · xα1 · xβ2 · xγ3λ

.

In a similar way, we have

July 27, 2009

p2 · x2 = β · xα1 · xβ2 · xγ3λ

and p3 · x3 = γ · xα1 · xβ2 · xγ3λ

.

Hence

(p1 · x1) : (p2 · x2) : (p3 · x3) = α : β : γ .

Ans so, finally, we have

x1 =1

p1· α

α+ β + γ· E, x2 = 1

p2· β

α + β + γ· E and x3 =

1

p3· γ

α+ β + γ· E .2

PS. (1) The price rise in a commodity has no effect on the consumption of other commodities.(2) When the expenditure E increases, the consumption of every commodity goes up.

2∗.Consider the case of 100 commodities and u(x) =Q100i=1 x

ii = x

11 · x22 · x33 · · · · · x100100, and derive

the utility maximizing consumption basket subject to the budget constraint , p1 · x1 + p2 · x2 +p3 · x3 + . . .+ p100 · x100 = E, wherep1 = 1, p2 = 2, p3 = 3, . . . , p100 = 100, and E = 5050.Answer. The expenditure to t-th commodity, pt ·xt = tP 100

i=1 i·E = t

5050·5050. Therefore, xt = 1

because pt = t. 2NB. A formula:

Pni=1 i =

n(n+1)2.

3. Prove that a monotone non-decreasing map from the real interval [0,1] into itself has at leastone fixed point. Hint. Lecture note (4).

Answer to 3: Omitted. (Consult your own notebook.)

4. Let the production function be F (x1, x2) = A · xα1 · xβ2 , the price of product be p, the costfunction C(x1, x2) = q1 · x1 + q2 · x2. Find out the output x = (x1, x2) which maximize profits.All the given constants are positive, i.e., A > 0, α > 0, β > 0, q1 > 0, q2 > 0, and α + β < 1.(Thus, decreasing returns to scale (drs) obtain.)Hint. Lecture note (6).

5. Suppose the cost function is C(x) = x3 − 5x2 + 10x+ 12. Find out the price of product atwhich the maximized profit is zero. Thus, the profit maximizing output is ‘break-even’.Find out also the price at which the firm stops production. Hint. Lecture note (7).

Answer to 5:Let the cost function be, in general,

C(x) = x3 + bx2 + cx+ d. ( b < 0, c > 0, d > 0, and b2 − 3c ≤ 0).At the output level xo, the tangent line to the cost curve is:

y − (x3o + bx2o + cxo + d) = (3x2o + 2bxo + c) · (x− xo). (L)In order for this tangent line to go through the origin, i.e., when x = 0, y = 0, it is necessary tohave

−(x3o + bx2o + cxo + d) = −(3x3o + 2bx2o + cxo).

July 27, 2009

That is,

2x3o + bx2o − d = 0. (*)

In the special case where b = −5, c = 10, and d = 12, the equation (*) becomes

2x3o − 5x2o − 12 = 0This has only one positive solution, while other two are complex numbers. The positive solutionis: xo ∼= 3. 1174. Then the price of product is:

3(3.12)2 + 2(−5)(3.12) + 10 ∼= 8.0032.So, at the price 8.0032 (or about 8.0), the maximized profit is 0 at the output about3.12.Next, let us find out the price at which the firm stop its operation. This is when the tangent

line at the profit maximizing output go through the point (0, d) [An error was here]. Thus, wesubstitute 0 for x, and d for y in the equation (L) above.

d− (x3o + bx2o + cxo + d) = (3x2o + 2bxo + c) · (0− xo). (L)That is,

2x3o + bx2o = 0. (**)

In our particular case,

2x3o − 5x2o = 0.We solve this equation, and find out xo = 2.5. (Certainly, xo = 0 also, but no operation at anyprice lower than 2.5.)The price of the product at which the firm makes loss as large as the fixed cost, 12, is:

3(2.5)2 + 2(−5)(2.5) + 10 = 3.75.So, at the price 3.75, the maximized profit is −12 at the output 2.5.2verification: 3.75× 2.5− ((2.5)3 − 5× (2.5)2 + 10× 2.5 + 12) = −12.0

NB. In this problem, we changed the fixed cost, i.e., d, only. Please consider the effects on theoutput, and the ‘break-even’ price of the increase in the fixed cost. Hint. Draw a diagram.

6. Explain the Newton-Raphson method to solve nonlinear equations.Hint Lecture notes (9) and (14)*.Answer: This is an iterative method to solve a nonlinear equation of a single variable, f (x).We choose an initial value, x(0), close to the desired solution. Then, we consider the tangent lineat (x(0), f(x(0))), and the cross point between this tangent line and the x-axis is made the nextiterate x(1). We continue this procedure until a satisfactory approximation is attained.NB. A generalization to many variables was made by Leonid Vitaliyevich Kantrovich (1912-1986;Nobel Prize, 1975).

Newton-Raphson Method:

2x3o − 5x2o − 12 = 0.

July 27, 2009

formula:

x(i+1) = x(i) − f(x(i))

f 0(x(i)).

x(i+1) = x(i) − 2× (x(i))3 − 5× (x(i))2 − 12

6× (x(i))2 − 10× (x(i)) .

We start from x(0) = 4. Then,

x(1) = 4− 2× (4)3 − 5× (4)2 − 12

6× (4)2 − 10× (4) ' 3.3571.

x(2) = 3.36− 2× (3.36)3 − 5× (3.36)2 − 12

6× (3.36)2 − 10× (3.36) ' 3.1427.

x(3) = 3.14− 2× (3.14)3 − 5× (3.14)2 − 12

6× (3.14)2 − 10× (3.14) ' 3.1177.

Continue the procedure until the tolerance limit you have designated.

References:Isaac Newton (1643-1727): Methodus Fluxionum et Serierum Infinitarum, (1664-1671).Joseph Raphson (1648-1715): Analysis Aequationum Universalis, London, 1690.A. S. Householder, The Numerical Treatment of a Single Nonlinear Equation, McGraw-Hill:New York, 1970.

L.V. Kantorovich and G.P. Akilov, Functional Analysis in Normed Spaces, Pergamon Press:New York, 1964.

J.M. Ortega and W.C. Rheinboldt, Iterative Solution of Non Linear Equations in SeveralVariables, Academic Press: New York, 1970.

H. Qiu, “A Robust Examination of the Newton-Raphson Method with Strong GlobalConvergence Properties”, Master’s Thesis: University of Central Florida, 1993.

Tjalling J. Ypma, “Historical Development of the Newton-Raphson Method”, SIAM Review,vol. 37(4), pp. 531-551, 1995.

¥Exam: 2009-8-21(Fri).