Mathematics Review for Economists

Mathematics Review for Economists by John E. Floy University of Toronto May 9, 2013 This ocument presents a review of very basic mathematics for use by stuents who plan to stuy economics in grauate school or who have long-ago complete their grauate stuy an nee a quick review of what they have learne. The first section covers variables an equations, the secon eals with functions, the thir reviews some elementary principles of calculus an the fourth section reviews basic matrix algebra. Reaers can work through whatever parts they think necessary as well as o the exercises provie at the en of each section. Finally, at the very en there is an important exercise in matrix calculations using the statistical program XLispStat. 1

3. Differentiation an Integration 1 We now review some basics of calculus in particular, the ifferentiation an integration of functions. A total revenue function was constructe in the previous section from the eman function Q = α βp (1) which was rearrange to move the price P, which equals average revenue, to the left sie an the quantity Q to the right sie as follows P = A(Q) = α β 1 β Q. (2) To get total revenue, we multiply the above function by Q to obtain T (Q) = P Q = A(Q) Q = α β Q 1 β Q 2. (3) where Q is the quantity supplie by a potential monopolist who, of course, is intereste in equating the marginal revenue with the marginal cost. The marginal revenue is the increase in total revenue that results from selling one aitional unit which, as can easily be seen from Figure 1, must equal the slope of the total revenue curve that is, the change in total revenue ivie by a one-unit change in the quantity. The marginal revenue from selling the first unit equals both the price an the total revenue from selling that unit. The sale of an aitional unit requires that the price be lower (given that the eman curve slopes ownwar) an, hence, the marginal revenue is also lower. This is clear from the fact that the slope of the total revenue curve gets smaller as the quantity sol increases. In the special case plotte, where α = 100 an β = 2, total revenue is maximum at an output of 50 units an, since its slope at that point is zero, marginal revenue is also zero. Accoringly, the marginal revenue curve plotte as the otte line in the Figure crosses zero at an output of 50 which turns out to be onehalf of the 100 units of output that consumers woul purchase at a price of 1 An appropriate backgroun for the material covere in this section can be obtaine by reaing Alpha C.Chiang, Funamental Methos of Mathematcial Economics, McGraw Hill, Thir Eition, 1984, Chapter 6, Chapter 7 except for the part on Jacobian eterminants, the sections of Chapter 8 entitle Differentials, Total Differentials, Rules of Differentials, an Total Derivatives, an all but the growth moel section of Chapter 13. Equivalent chapters an sections, sometimes with ifferent chapter an section numbers, are available both in earlier eitions of this book an in the Fourth Eition, joint with Kevin Wainwright, publishe in 2005. 2

Figure 1: Total, average an marginal revenue when α = 100, β = 2. zero, at which point total revenue woul also be zero. We o not bother to plot the marginal revenue curve where marginal revenue is negative because no firm woul prouce an sell output uner those conitions, given that marginal cost is always positive. The average revenue, or eman, curve is plotte in the Figure as the ashe line. To make the relationship between the eman curve an marginal revenue curve clearer, these two curves are plotte separately from the total revenue curve in Figure 2. The total revenue associate with any quantity is the area uner the marginal revenue curve that is the sum of the marginal revenues to the left of that quantity. We now state the fact that the erivative of a function y = F (x) is equal to the slope of a plot of that function with the variable x represente by the 3

Figure 2: Deman an marginal revenue curves when α = 100, β = 2. horizontal axis. This erivative can be enote in the four alternative ways y x F (x) x x F (x) F (x) where the presence of the integer in front of a variable enotes a infinitesimally small change in its quantity. We normally enote the slope of y with respect to x for a one-unit change in x by the expression y/ x. The expression y/x represents the limiting value of that slope as the change in x approaches zero that is, y lim x 0 x = y x The first rule to keep in min when calculating the erivatives of a function is that the erivative of the sum of two terms is the sum of the 4 (4)

erivatives of those terms. A secon rule is that in the case of polynomial functions like y = a (b x) n the erivative takes the form y x = x a (b x)n = a n (b x) n 1 x (b x) = b a n (b x)n 1 (5) where, you will notice, a multiplicative constant term remains unaffecte. Accoringly, the erivative of equation (3) is the function M(Q) = [ α Q β Q 1 ] β Q 2 = α Q β Q 1 Q β Q 2 = α β 2 β Q (6) which is the marginal revenue function in Figure 2. The slope of that marginal revenue function is its erivative with respect to Q, namely, M(Q) Q = α Q β Q = 0 2 β 2 β Q = 2 β. (7) which verifies that, when the eman curve is linear, the marginal revenue curve lies half the istance between the eman curve an the vertical axis. Notice also from the above that the erivative of an aitive constant term is zero. Since marginal revenue is the increase in total revenue from aing another unit, it follows that the total revenue associate with any quantity is the sum of the marginal revenues from aing all units up to an incluing that last one that is Q 0 MR Q = 0 Q 0 TR Q Q = TR Q 0. (8) When we represent marginal revenue as the erivative of the total revenue curve that is, as TR Q = lim TR 0 Q (9) 5

Figure 3: Total revenue visualize as the sum of sucessive marginal revenues. equation (8) can be rewritten as Q0 0 MR Q = Q0 0 TR Q Q = TR Q 0. (10) Marginal revenue is the length of an infinitesimally narrow vertical slice extening from the marginal revenue curve own to the quantity axis in the reprouction of Figure 2 above an Q is the with of that slice. We horizontally sum successive slices by calculating the integral of the marginal revenue function. In the previous section, the effect of the stock of liquiity on the fraction of output that is useable that is, not lost in the process of making exchange is a thir-egree polynomial term that multiplies the level of the 6

capital stock as follows where Y = mkω (11) Ω = 1 1 ( ϕ λ L ) 3. (12) 3λ K The erivative of income Y with respect to the stock of liquiity L takes the form Y L = mk Ω L where, taking into account the fact that ϕ an K are constants, so that Ω L = 3 1 3λ = 1 λ = 1 λ = = = λ K ( ϕ λ L ) 2 K L ) 2 ( ϕ ( ϕ λ L K ( ϕ λ L K ( ϕ λ L ) 2 ( λ K L ( ϕ λ L ) 2 λ K K ) 2 ( 0 λ L ( ϕ λ L ) K ) L L λ L K ) L K ( L L ) L K ) (13) ( ϕ λ L K ) 2 (14) Y (ϕ L = mλ λ L ) 2 (15) K which is a secon-egree polynomial function. Since Y/L is equal to the increase in the final output flow resulting from an increase in the stock of liquiity (or, in a cruer moel, money) the above equation can be interprete as a eman function for liquiity or money when we set Y L equal to the increase in the income flow to the iniviual money holer that will result from holing another unit of money. This income-flow comes at a cost of holing an aitional unit of money, normally set equal to the nominal interest rate. Also, we nee to incorporate the fact that the quantity of money 7

emane normally also epens on the level of income. It turns out that, uner conitions of full-employment, the level of income can be roughly approximate by m K so that, treating some measure of the money stock as an inicator of the level of liquiity, we can write the eman function for money as ( i = mλ ϕ λ ) M 2 (16) m Y where i is the nominal interest rate. A further moification woul have to be mae to allow for income changes resulting from changes in the utilization of the capital stock in booms an recessions. In any event, you can see that an increase in i will reuce the level of M emane at any given level of Y an an increase in Y holing i constant will also increase the quantity of M emane. Suppose now that we are presente with a fourth-egree polynomial of the form y = F (x) = α + β x + γ x 2 + δ x 3 + ϵ x 4. (17) We can ifferentiate this following the rule given by equation (5) together with the facts that the erivative of a constant is zero an the erivative of a sum of terms equals the sum of the respective erivatives to obtain F (x) = y x = β + 2 γ x + 3 δ x2 + 4 ϵ x 3. (18) Suppose, alternatively, that we are given the function F (x) in equation (18) without seeing equation (17) an are aske to integrate it. We simply follow for each term the reverse of the ifferentiation rule by aing the integer 1 to the exponent of that term an then iviing the term by this moifie exponent. Thus, for each term we obtain y x = x a xn = a n x n 1 (19) y x x = n n 1 + 1 a xn 1+1 = a x n. (20) Application of this proceure successively to the terms in equation (18) yiels F (x) x = 1 1 βx0+1 + 2 2 γ x1+1 + 3 3 δ x2+1 + 4 4 ϵ x3+1 = β x + γ x 2 + δ x 3 + ϵ x 4 (21) 8

which, it turns out, iffers from equation(17) in that it oes not contain the constant term α. Without seeing (17), we ha no way of knowing the magnitue of any constant term it containe an in the integration process in (21) inappropriately gave that constant term a value of zero. Accoringly, when integrating functions we have to automatically a to the integral an unknown constant term the resulting integral is calle an inefinite integral. To make it a efinite integral, we have to have some initial information about the level of the resulting function an thereby be able to assign the correct value to the constant term. This problem oes not arise when we are integrating from some initial x-value x o an know the value of the function at that value of x. The erivative of the logarithmic function y = α + β ln(x) (22) is y x = α x + ln(x) β ln(x) = 0 + β = β 1 x x x = β x. (23) As you can see, the erivative of ln(x) is the reciprocal of x x ln(x) = 1 (24) x which is consistent with the fact that the logarithm of a variable is the cumulation of its past relative changes. The inefinite integral of the function F (x) = β x is thus simply β ln(x). Then there is the exponential function with base e, y = e x, which has the istinguishe characteristic of being its own erivative that is, y x = x e x = e x. (25) This property can be verifie using the fact, shown in the previous section, that y = e x x = ln(y). (26) Given the existance of a smooth relationship between a variable an its logarithm, the reciprocal of the erivative of that function will also exist so that, x y = y ln(y) = 1 y y x = y y x = ex. (27) 9

Of course, we will often have to eal with more complicate exponential functions such as, for example, y = α e β x + c (28) for which the erivative can be calculate accoring to the stanar rules outline above as y x (β x + c) 1 (β x + c) 1 = α (β x + c) e (β x + c) = α β (β x + c) e x = α (β 2 x + β c) e β x + c 1. (29) At this point it is useful to collect together the rules for ifferentiating functions that have been presente thus far an to a some important aitional ones. Rules for Differentiating Functions 1) The erivative of a constant term is zero. 2) The erivative of a sum of terms equals the sum of the erivatives of the iniviual terms. 3) The erivative of a polynomial where a variable x is to the power n is x xn = n x n 1. In the case of a function that is to the power n, the erivative is x α F (x)n = n α F (x) n 1 F (x). 4) The erivative of the exponential function y = b x is x bx = b x ln(b) or, in the case of y = e x, since ln(e) = 1, x ex = e x. In the case where y = e F (x) the erivative is x ef (x) = e F (x) F (x). 10

5) Where y is the logarithm of x to the base e, the erivative with respect to x is y x = x ln(x) = 1 x. 6) The chain rule. If then y = F y (z) an z = F z (x) x F y(f z (x)) = F y (F z (x)). 7) The erivative of the prouct of two terms equals the first term multiplie by the erivative of the secon term plus the secon term multiplie by the erivative of the first term. x [F 1(x) F 2 (x)] = F 1 (x) F 2 (x) + F 2 (x) F 1 (x) 8) The erivative of the ratio of two terms equals the enominator times the erivative of the numerator minus the numerator times the erivative of the enominator all ivie by the square of the enominator. x [ ] F1 (x) = F 2(x) F F 2 (x) 1 (x) F 1(x) F 2 (x) [F 2 (x)] 2 9) The total ifferential of a function of more than a single variable such as, for example, y = F (x 1, x 2, x 3 ) is y = y x 1 x 1 + y x 2 x 2 + y x 3 x 3 where the term y/ x i is the partial erivative of the function with respect to the variable x i that is, the change in y that occurs as a result of a change in x i holing all other x variables constant. We en this section with some applications of the total ifferential rule immeiately above. An interesting application is with respect to the stanar Cobb-Douglas prouction function below which states that the level of output of a firm, enote by X, epens in the following way upon the inputs of labour L an capital K Y = A L α K 1 α (30) 11

where α is a parameter an A is a constant representing the level of technology. The total ifferential of this function is Y = A [K 1 α αl α 1 L + L α (1 α) K 1 α 1 K] = (A αl α 1 K 1 α ) L + (A (1 α) L α K α ) K (31) where you shoul note that (A αl α 1 K 1 α ) an (A (1 α) L α K α ) are the respective marginal proucts of labour an capital. The increase in output associate with changes in the labour an capital inputs thus equals the marginal prouct of labour times the change in the input of labour plus the marginal prouct of capital times the change in the input of capital. Note also that, using a bit of manipulation, these marginal proucts can also be expresse as an Y L = L A αl α 1 K 1 α L = α A L α K 1 α L = α Y L (32) Y K = K A (1 α) L α K α K = (1 α) A L α K 1 α K Substituting the above two equations back into (31), we obtain = (1 α) Y K. (33) Y = α Y L L + (1 α) Y K (34) K which, upon ivision of both sies by Y becomes Y Y = α L L + (1 α)k K. (35) Actually, a simpler way to obtain this equation is to take the logarithm of equation (30) to obtain ln(y ) = ln(a) + α ln(l) + (1 α) ln(k) (36) an then take the total ifferential to yiel ln(y ) = α ln(l) L ln(k) L + (1 α) K K = α 1 L L + (1 α) 1 K K = α L L + (1 α) K K (37) 12

An important extension of the total ifferential analysis in equations (31) an (34) is to the process of maximization. Suppose for example that the wage pai to a unit of labour is ω an the rental rate on a unit of capital is κ. The total cost of prouction for a firm woul then be T C = ω L + κ K (38) an if the quantity of the labour input is change holing total cost constant we will have which implies that T C = 0 = ω L + κ K K = ω L (39) κ which, when substitute into the total ifferential equation (34) yiels Y = [ α Y L ω (1 α) κ ] Y L. (40) K Maximization of the level of output proucible at any given total cost requires that, starting from a very low initial level, L be increase until a level is reache for which a small change in L has no further effect on output. This will be the level of L for which α Y L ω (1 α) κ Y K = 0 α Y L = ω (1 α) κ Y K. When both sies of the above are ivie by (1 α) Y/K, the expression reuces to α (Y/L) (1 α) (Y/K) = ω κ. (41) You will recognize from equations (32) an (33) that the left sie of the above equation is simply the ratio of the marginal prouct of labour over the marginal prouct of capital otherwise known as the marginal rate of substitution of labour for capital in prouction. As note in the Figure below, this ratio is the slope of the constant output curve. Since the right sie of the equation is the ratio of the wage rate to the rental rate on capital, the conition specifies that in equilibrium the marginal rate of substitution must be equal to the ratio of factor prices, the latter being the slope of the firm s 13

buget line in the Figure below. The optimal use of factors in prouction is often state as the conition that the wage rate of each factor of prouction be equal to the value marginal prouct of that factor, efine as the marginal prouct times the price at which the prouct sells in the market. You can easily see that multiplication of the numerator an enominator of the left sie of the equation above by the price of the prouct will leave the optimality conition as state there unchange. 14

Figure 4: Cobb-Douglas prouction function Y = AL.75 K.25 at specific output level. Another concept of interest is the elasticity of substitution of labour for capital in prouction. It is efine as the elasticity of the ratio of capital to labour employe with respect to the marginal rate of substitution efine as the marginal prouct of labour ivie by the marginal prouct of capital that is, the relative change in the capital/labour ratio ivie by the relative change in the ratio of the marginal prouct of labour over the marginal prouct of capital. The elasticity of substitution in the Cobb- Douglas prouction function can be obtaine simply by ifferentiating the left sie of equation (41) above with respect to the capital/labour ratio. First, we can cancel out the variable Y an then take the logarithm of both 15

sies an ifferentiate the logarithm of the marginal rate of substitution with respect to the logarithm of the capital/labour ratio. This yiels α (Y/L) MRS = (1 α) (Y/K) = α 1 α ( ) ( ) α K ln(mrs) = ln + ln 1 α L ) ln(mrs) ln(α)/(1 α) = ln (K/L) ln(k/l) K L + ln(k/l) ln(k/l) MRS MRS / (K/L) K/L = 0 + 1 = 1/σ σ = 1 (42) where σ is the elasticity of substitution, which always equals unity when the prouction function is Cobb-Douglas. The above result makes it worthwhile to use functions having constant elasticities that are ifferent from unity. A popular function with this characteristic is the constant elasticity of substitution (CES) function which is written below as a utility function in the form [ ] U = A δ C ρ 1 + (1 δ) C ρ 1/ρ 2 (43) where U is the level of utility an C 1 an C 2 are the quantities of two goos consume. Letting Ψ enote the terms within the square brackets, the marginal utility of C 1 can be calculate as ( U = A C 1 ρ ( = A δ = A δ Ψ [ (1/ρ) 1]) ( δ ρ) C ρ 1 1 Ψ (1+ρ)/ρ) C (1+ρ) 1 [ δ C ρ 1 + (1 δ) C ρ 2 By a similar calculation, the marginal utility of C 2 is ] (1+ρ)/ρ (1+ρ) C 1. (44) U [ ] = A (1 δ) δ C ρ 2 + (1 δ) C ρ (1+ρ)/ρ (1+ρ) 2 C 2 (45) C 2 an the marginal rate of substitution is therefore MRS = U/ C 1 U/ C 2 = δ (1 δ) ( ) (1+ρ) C1 = C 2 δ (1 δ) ( C2 C 1 ) (1+ρ) (46) 16

Figure 5: CES utility function U = A[δ C ρ 1 + (1 δ) C ρ 2 ] 1/ρ at a specific utility level with δ = 0.5 an the elasticity of substitution, equal to 1/(1 + ρ), set alternatively at 0.5, 1.0, an 2.0. which, upon taking the logarithm, becomes ( ) δ ln(mrs) = ln + (1 + ρ) ln 1 δ ( ) C2 C 1 The elasticity of substitution therefore equals σ = ln(c 2/C 1 ) ln(mrs) = 1 1 + ρ = (1 + ρ) ln ( ) C2 C 1. (47) (48) an becomes equal to unity when ρ = 0, less than unity when ρ is positive an greater than unity when 1 < ρ < 0, becoming infinite as ρ 1. 17

Obviously it woul make no sense for ρ to be less than minus one. Inifference curves with elasticities of substitution ranging from zero to infinity are illustrate in Figure 5 above where δ =.5. It turns out that the Cobb-Douglas function is a special case of the CES function where ρ = 0, although equation (43) is unefine when ρ = 0 because ivision by zero is not possible. Nevertheless, we can emonstrate that as ρ 0 the CES function approaches the Cobb-Douglas function. To o this we nee to use L Hôpital s rule which hols that the ratio of two functions m(x) an n(x) approaches the ratio of their erivatives with respect to x as x 0. m(x) lim x 0 n(x) = lim m (x) x 0 n (x) (49) When we ivie both sies of equation (43) by A an take the logarithms we obtain ( ) Q ln = A ρ ln[δc1 + (1 δ)c ρ 2 ] = m(ρ) ρ n(ρ) (50) for which m (ρ) becomes, after using the chain rule an the rule for ifferentiating exponents with base b, m (ρ) = = which, in the limit as ρ 0 becomes ρ [δc1 + (1 δ)c ρ 2 ] ρ 1 [δc ρ 1 + (1 δ)c ρ 2 ] ρ [ δ C1 ln(c 1 ) (1 δ) C ρ 2 ln(c 2 )] [δc ρ 1 + (1 δ)c ρ 2 ] m (ρ) = δ ln(c 1 ) + (1 δ) ln(c 2 ). (51) Since n(ρ) = ρ, n (ρ) also equals unity, so we have ( ) Q lim ln ρ 0 A This implies that m (ρ) = lim ρ 0 n (ρ) = δ ln(c 1) + (1 δ) ln(c 2 ) 1 = δ ln(c 1 ) + (1 δ) ln(c 2 ) (52) Q = A C δ 1 C 1 δ 2 (53) 18

which is the Cobb-Douglas function. 2 Exercises 1. Explain the ifference between y/ x an y/x. 2. Differentiate the following function. y = a x 4 + b x 3 + c x 2 + x + g 3. Integrate the function that you obtaine in the previous question, assuming that you are without knowlege of the function you there ifferentiate. 4. Differentiate the function y = a + b ln(x). 5. Differentiate the following two functions an explain why the results iffer. y = a b x y = a e x 6. Given that x = ln(y), express y as a function of x. 7. Suppose that y = a + b z 2 an z = e x Use the chain rule to calculate the erivative of y with respect to x. 8. Using the two functions in the previous question, fin ( ) x (yz) an y. x z 2 Every use of L Hôpital s rule brings to min the comments of a well-known economist when aske about the quality of two university economics epartments, neither of which he like. His reply was You have to use L Hôpital s rule to compare them. 19

9. Let U(C 1 ) be the utility from consumption in perio one an U(C 2 ) be the utility from consumption in year two, where the functional form is ientical in the two years. Let the utility from consumption in both perios be the present value in year one of utilities from consumption in the current an subsequent years where the iscount rate is ω, otherwise known as the iniviual s rate of time preference. Thus, U = U(C 1 ) + 1 1 + ω U(C 2). Take the total ifferential of the present value of utility U. Let the interest rate r be the relative increase in year two consumption as a result of the sacrifice of a unit of consumption in year one, so that C 2 = (1 + r) C 1 Assuming that the iniviual maximizes utility, calculate the conition for optimal allocation of consumption between the two perios. 10. Calculate the marginal proucts of labour an capital arising from the following prouction function. Q = A[δ L ρ + (1 δ) K ρ ] 1/ρ Then calculate the elasticity of substitution between the inputs. How o the results iffer then you use the Cobb-Douglas prouction function instea of the CES prouction function above? 20