Now substitute this value for P 0 into the general solution (2) above, so that P 0 = 55 = 45 + A(−1) 0 55 = 45 + A 10 = A The specific solution to the difference equation for this example is therefore P t = 45 + 10(−1) t Using this formula to calculate the first few values of P t we see that P 0 = 45 + 10(−1) 0 = 45 + 10 = 55 P 1 = 45 + 10(−1) 1 = 45 − 10 = 35 P 2 = 45 + 10(−1) 2 = 45 + 10 = 55 P 3 = 45 + 10(−1) 3 = 45 − 10 = 35 P 4 = 45 + 10(−1) 4 = 45 + 10 = 55 etc. Price therefore continually fluctuates between 35 and 55. This is the third possibility in the stability conditions examined earlier. In this example d b = 2 −2 =|−1|=1 Therefore, as t →∞, P t neither converges on its equilibrium level nor explodes until the market collapses. This fluctuation between two price levels from year to year is sometimes observed in certain agricultural markets. Test Yourself, Exercise 13.2 (Assume that the usual cobweb assumptions apply in these questions.) 1. In a market where Q d t = 160 −20P t and Q s t =−80 +40P t−1 quantity unexpectedly drops from its equilibrium value to 75. Derive the difference equation which will calculate price in the time periods following this event. 2. If Q d t = 180 − 0.9P t and Q s t =−24 + 0.8P t−1 say whether or not the long-run equilibrium price is stable and then use the difference equation method to calculate price in the thirtieth time period after a sudden one-off increase in quantity to 117. © 1993, 2003 Mike Rosser 3. Given the demand and supply schedules Q d t = 3450 −6P t and Q s t =−729 + 4.5P t−1 use difference equations to predict what price will be in the tenth time period after an unexpected drop in quantity to 354, assuming that the market was previously in long-run equilibrium. 13.4 The lagged Keynesian macroeconomic model In the basic Keynesian model of the determination of national income, if foreign trade and government taxation and expenditure are excluded, the model reduces to the accounting identity, Y = C + I (1) and the consumption function C = a + bY (2) To determine the equilibrium level of national income Y ∗ we substitute (2) into (1), giving Y ∗ = a + bY ∗ + I Y ∗ (1 − b) = a + I Y ∗ = a + I 1 − b This can be evaluated for given values of parameters a and b and exogenously determined investment I . If there is a disturbance from this equilibrium, e.g. exogenous investment I alters, then the adjustment to a new equilibrium will not be instantaneous. This is the basis of the well-known multiplier effect. An initial injection of expenditure will become income for another sector of the economy. A proportion of this will be passed on as a further round of expenditure, and so on until the ‘ripple effect’ dies away. Because consumer expenditure may not adjust instantaneously to new levels of income, a lagged effect may be introduced. If it is assumed that consumers’ expenditure in one time period depends on the income that they received in the previous time period, then the consumption function becomes C t = a + bY t−1 (3) where the subscripts denote the time period. National income, however, will still be determined by the sum of all expenditure within the current time period. Therefore the accounting identity (1), when time subscripts are introduced, can be written as Y t = C t + I t (4) © 1993, 2003 Mike Rosser From (3) and (4) we can derive a difference equation that explains how Y t depends on Y t−1 . Substituting (3) into (4) we get Y t = (a + bY t−1 ) + I t Y t = bY t−1 + a + I t (5) This difference equation (5) can be solved using the method explained in Section 13.3 above. However, let us first illustrate how this lagged effect works using a numerical example. Example 13.7 In a basic Keynesian macroeconomic model it is assumed that initially Y t = C t + I t where I t = 134 is exogenously determined, and C t = 40 +0.6Y t−1 The level of investment I t then falls to 110 and remains at this level each time period. Trace out the pattern of adjustment to the new equilibrium value of Y , assuming that the model was initially in equilibrium. Solution Although this pattern of adjustment can best be viewed using a spreadsheet, let us first work out the first few steps of the process manually and relate them to the familiar 45 ◦ -line income-expendituregraph(illustratedinFigure13.3)oftenusedtoshowhowYisdetermined in introductory economics texts. If the system is initially in equilibrium then income in one time period is equal to expenditure in the previous time period, and income is the same each time period. Thus Y t = Y t−1 = Y ∗ where Y ∗ is the equilibrium level of Y . Therefore, when the original value of I t of 134 is inserted into the accounting identity the model becomes Y ∗ = C t + 134 (1) C t = 40 +0.6Y ∗ (2) By substitution of (2) into (1) Y ∗ = (40 +0.6Y ∗ ) + 134 Y ∗ (1 − 0.6) = 40 + 134 0.4Y ∗ = 174 Y ∗ = 435 © 1993, 2003 Mike Rosser A B 0 Y*375 E Y* Y 0 45° EЈ (I = 110) E (I = 134) Figure 13.3 This is the initial equilibrium value of Y before the change in I . Assume time period 0 is the one in which the drop in I to 110 occurs. Consumption in time period 0 will be based on income earned the previous time period, i.e. when Y was still at the old equilibrium level of 435. Thus C 0 = 40 +0.6(435) = 40 + 261 = 301 Therefore Y 0 = C 0 + I 0 = 301 + 110 = 411 In the next time period, the lagged consumption function means that C 1 will be based on Y 0 . Thus Y 1 = C 1 + 110 = (40 +0.6Y 0 ) + 110 = 40 +0.6(411) + 110 = 40 +246.6 +110 = 396.6 The value of Y for other time periods can be calculated in a similar fashion: Y 2 = C 2 + I 2 = (40 +0.6Y 1 ) + 110 = 40 +0.6(396.6) + 110 = 387.96 © 1993, 2003 Mike Rosser Y 3 = C 3 + I 3 = (40 +0.6Y 2 ) + 110 = 40 +0.6(387.96) + 110 = 382.776 and so on. It can be seen that in each time period Y decreases by smaller and smaller amounts as it readjusts towards the new equilibrium value. This new equilibrium value can easily be calculated using the same method as that used above to work out the initial equilibrium. When I = 110 and Y t = Y t−1 = Y ∗ then the model becomes Y ∗ = C t + I = C t + 110 C t = 40 +0.6Y ∗ By substitution Y ∗ = (40 +0.6Y)+ 110 (1 − 0.6)Y ∗ = 150 Y ∗ = 150 0.4 = 375 ThispathofadjustmentisillustratedinFigure13.3bythezigzaglinewitharrowswhich joins the old equilibrium at A with the new equilibrium at B. (Note that this diagram is not to scale and just shows the direction and relative magnitude of the steps in the adjustment process.) Unlike the cobweb model described earlier, the adjustment in this Keynesian model is always in the same direction, instead of alternating on either side of the final equilibrium. Successive values of Y just approach the equilibrium by smaller and smaller increments because the ratio in the complementary function to the difference equation (explained below) is not negative as it was in the cobweb model. If the initial equilibrium had been below the new equilibrium then, of course, Y would have approached its new equilibrium from below instead of from above. Further steps in the adjustment of Y in this model are shown in the Excel spreadsheet in Table13.4,whichisconstructedasexplainedinTable13.5.ThisclearlyshowsYclosingin on its new equilibrium as time increases. Difference equation solution Letusnowreturntotheproblemofhowtosolvethedifferenceequation Y t = bY t−1 + a + I t (1) Thegeneralsolutioncanthenbeappliedtonumericalproblems,suchasExample13.7above. By ‘solving’ this difference equation we mean putting it in the format Y t = f(t) so that the value of Y t can be determined for any given value of t. © 1993, 2003 Mike Rosser Table 13.4 A B C D E 1 Ex. LAGGED KEYNESIAN MODEL 2 13.7 where Yt = Ct + I t 3 Ct = a + bYt-1 4 Parameters 5 a = 40 Old I value = 134 6 b = 0.6 New I value = 110 7 Old Equil Y = 435 8 Time New Equil Y = 375 9 t C Y 10 0 301.00 411.00 11 1 286.60 396.60 12 2 277.96 387.96 13 3 272.78 382.78 14 4 269.67 379.67 15 5 267.80 377.80 16 6 266.68 376.68 17 7 266.01 376.01 18 8 265.60 375.60 19 9 265.36 375.36 20 10 265.22 375.22 21 11 265.13 375.13 22 12 265.08 375.08 23 13 265.05 375.05 24 14 265.03 375.03 25 15 265.02 375.02 26 16 265.01 375.01 27 17 265.01 375.01 28 18 265.00 375.00 The basic method is the same as that explained earlier, i.e. the solution is split into two components: the equilibrium or particular solution and the complementary function. We first need to find the particular solution, which will be the new equilibrium value of Y ∗ . When this is equilibrium achieved Y t = Y t−1 = Y ∗ In equilibrium, the single lag Keynesian model C t = a + bY t−1 (2) and Y t = C t + I t (3) can therefore be written as C t = a + bY ∗ Y ∗ = C t + I t © 1993, 2003 Mike Rosser Table 13.5 CELL Enter Explanation As in Table 13.4 Enter all labels and column headings B5 40 B6 0.6 These are given parameter values for consumption function in this example. E5 134 Original given investment level. E6 110 New investment level. D6 160 This is initial “shock” quantity in time period 0. A10 to A28 Enter numbers from 0 to 18 These are the time periods used. E7 =(B5+E5)/(1-B6) Calculates initial equilibrium value of Y using formula Y = (a + I)/(1 - b ). E8 =(B5+E6)/(1-B6) Same formula calculates new equilibrium value of Y, using new value of I in cell E6. B10 =B5+B6*E7 Calculates consumption in time period 0 using formula C = a + bY t – 1 where Y t – 1 is the old equilibrium value in cell E7. C10 =B10+E$6 Calculates Y in time period 0 as sum of current consumption value in cell B10 and new investment value. Note the $ on cell E6 to anchor when copied. B11 =B$5+B$6*C10 Calculates consumption in time period 1 based on Y 0 value in cell C10. Note the $ on cells B5 and B6 to anchor when copied. B12 to B28 Copy formula from B11 down column. Calculates consumption in each time period. C11 to C28 Copy formula from C10 down column. Calculates national income Y t in each time period. By substitution Y ∗ = a + bY ∗ + I t (1 − b)Y ∗ = a + I t Y ∗ = a + I t 1 − b (4) If the given values of a,b and I t are put into (4) then the equilibrium value of Y is determined. This is the first part of the difference equation solution. Returning to the difference equation (1) which we are trying to solve Y t = bY t−1 + a + I t (1) If the two constant terms a and I t are removed then this becomes Y t = bY t−1 (5) To find the complementary function we use the standard method and assume that this solution is in the format Y t = Ak t (6) © 1993, 2003 Mike Rosser where A and k are unknown parameters. This means that Y t−1 = Ak t−1 (7) Substituting (6) and (7) into (5) gives Ak t = bAk t−1 k = b Thus the complementary function is Y t = Ab t (8) The general solution to the difference equation is the sum of the particular solution (4) and the complementary function (8). Hence Y t = a + I t 1 − b + Ab t (9) If t is increased, then the value of b t in the general solution (9) will diminish as long as |b| < 1. This condition will be met since b is the marginal propensity to consume which has been estimated to lie between 0 and 1 in empirical studies. Therefore Y t will always head towards its new equilibrium value. The value of the constant A can be determined if an initial value Y 0 is known. Substituting into (9), this gives Y 0 = a + I t 1 − b + Ab 0 Remembering that b 0 = 1, this means that A = Y 0 − a + I t 1 − b (10) Thus A is the value of the difference between the initial level of income Y 0 , immediately after the shock, and its final equilibrium value Y ∗ . Putting this result into (9) above, the general solution to our difference equation becomes Y t = a + I t 1 − b + Y 0 − a + I t 1 − b b t (11) This may seem to be a rather cumbersome formula but it is straightforward to use. If you remember that a + I t 1 − b = Y ∗ is the equilibrium value of Y t and rewrite (11) as Y t = Y ∗ + (Y 0 − Y ∗ )b t (12) you will find it easier to work with. © 1993, 2003 Mike Rosser We can now check that this solution to the lagged Keynesian model difference equation workswiththenumericalExample13.7consideredabove.Thismodelassumed Y t = C t + I t where I t was initially 134 and C t = 40 +0.6Y t−1 which corresponded to an initial equilibrium of Y t of 435. When I t was exogenously decreased to 110, the adjustment path towards the new equilib- rium value of Y of 375 was worked out by an iterative method. Now let us see what values our difference equation will give. We have to be careful in determining the initial value Y 0 , immediately after the increase in investment has taken place. This depends on I 0 , which will be the new level of investment of 110, and C 0 . The level of consumption in period 0 depends on the previously existing equilibrium level of Y t which was 435 in time period ‘minus one’. Therefore C 0 = a + bY t−1 = 40 +0.6(435) = 301 Y 0 = C 0 + I 0 = 301 + 110 = 411 (13) This is the same initial value Y 0 as that calculated in Example 13.7. The new equilibrium value of income is Y ∗ = a + I t 1 − b = 40 +110 1 − 0.6 = 150 0.4 = 375 (14) Substituting (13) and (14) into the formula for the general solution to the difference equation derived above Y t = Y ∗ + (Y 0 − Y ∗ )b t (15) the general solution for this numerical example becomes Y t = 375 + (411 − 375)0.6 t = 375 + 36(0.6) t The first few values of Y are thus Y 1 = 375 + 36(0.6) = 375 +21.6 = 396.6 Y 2 = 375 + 36(0.6) 2 = 375 + 12.96 = 387.96 Y 3 = 375 + 36(0.6) 3 = 375 + 7.776 = 382.776 These are exactly the same as the answers computed by the iterative method in Example 13.7 andalsothesameasthoseproducedbythespreadsheetinTable13.4,whichiswhatone would expect. This difference equation solution can now be used to calculate Y t in any given time period. For example, in time period 9 it will be Y 9 = 375 + 36(0.6) 9 = 375 + 0.3628 = 375.3628 © 1993, 2003 Mike Rosser As t increases in value, eventually the value of (0.6) t becomes so small as to make the second term negligible. In the above example we can say that for all intents and purposes Y t has effec- tively reached its equilibrium value of 375 by the ninth time period, although theoretically Y t would never actually reach 375 if infinitesimally small increments were allowed. By now, many of you may be thinking that this difference equation method of computing the different values of Y in the adjustment process in a Keynesian macroeconomic model is extremely long-winded and it would be much quicker to compute the values by the iterative method, particularly if a spreadsheet can be used. In many cases you may be right. However, you must remember that this chapter is only intended to give you an insight into the methods that can be used to trace out the time path of adjustment in dynamic economic models. The mathematical methods of solution explained here can be adapted to tackle more complex problems that cannot be illustrated on a spreadsheet. Also, economists need to set up mathematical formulations for functional relationships in order to estimate the parameters of these functions. Those of you who study econometrics after the first year of your course will discover that the algebraic solutions to difference equations can help in the setting up of models for testing certain dynamic economic relationships. Now that the general solution to the lagged Keynesian macroeconomic model has been derived, it can be applied to other numerical examples and may even allow you to compute answers more quickly than by switching on your computer and setting up a spreadsheet. Example 13.8 There is initially an equilibrium in the basic Keynesian model Y t = C t + I t C t = 650 +0.5Y t−1 with I t remaining at 300. Then I t suddenly increases to 420 and remains there. What will be the actual level of Y six time periods after this change? Solution The initial equilibrium in period ‘minus 1’ before the change is Y ∗ −1 = a + I t 1 − b = 650 +300 1 − 0.5 = 950 0.5 = 1,900 Therefore the value of C in time period 0 when the increase in I takes place is C 0 = 650 +0.5(1,900) = 650 + 950 = 1,600 and so the value of Y t immediately after this shock is Y 0 = C 0 + I 0 = 1,600 +420 = 2,020 The new equilibrium level of Y is Y ∗ = a + I t 1 − b = 650 +420 1 − 0.5 = 1,070 0.5 = 2,140 © 1993, 2003 Mike Rosser [...]... at an annual rate of 8.5%? Solution Given the final amount of y = 90 0, continuous growth rate r = 8.5% = 0.085, and time period t = 10, then using the initial sum formula A = ye−rt = 90 0 e−0.085(10) = 90 0 e−0.85 = 90 0 × 0.42741 49 = 384.67 Therefore, she will need to start with 385 animals, as you cannot have a fraction of an animal! © 199 3, 2003 Mike Rosser Test Yourself, Exercise 14.2 1 A statistician... equal 1.8 is 12r Therefore, ln 1.8 = 12r r= ln 1.8 0.5877867 = = 0.04 898 22 12 12 and so consumption has risen at an annual rate of 4 .9% A general formula for finding a continuous rate of growth when y, A and t are all known can be derived from the final sum formula Given y = Aert then taking natural logs y = ert A y ln = rt A giving the rate of growth formula 1 y ln =r t A © 199 3, 2003 Mike Rosser Example... value of P X ) © 199 3, 2003 Mike Rosser 3 In a duopoly where the assumptions of the Bertrand model hold, the two firms’ reaction functions are Y PtX = 95 .54 + 0.83Pt−1 X PtY = 95 .54 + 0.83Pt−1 If firm X unexpectedly changes price to 499 , derive the solution to the difference equation that determines PtX and use it to predict PtX in the twelfth time period after the initial change © 199 3, 2003 Mike Rosser... manually below for you to check against The initial equilibrium level Y ∗ would have satisfied the equations Y ∗ = Ct + 90 (1) ∗ ∗ Ct = 320 + 0.5Y + 0.3Y = 320 + 0.8Y ∗ By substitution into (1) Y ∗ = 320 + 0.8Y ∗ + 90 0.2Y ∗ = 410 Y ∗ = 2,050 = Yt−1 = Yt−2 Thus C0 = 320 + 0.5(2,050) + 0.3(2,050) = 1 ,96 0 Y0 = C0 + I0 = 1 ,96 0 + 140 = 2,100 C1 = 320 + 0.5(2,100) + 0.3(2,050) = 1 ,98 5 Y1 = C1 + I1 = 1 ,98 5 + 140... possible formulations have been suggested for the ways in which past income levels can determine current expenditure For example Ct = a + bYt−2 © 199 3, 2003 Mike Rosser or Ct = a + b1 Yt−1 + b2 Yt−2 The latter example is known as a ‘distributed lag’ model The solutions of these more complex models require more advanced mathematical methods than are explained in this basic mathematics text You should, however,... applications in mathematics These are known as ‘natural logarithms’, and the usual notation is ‘ln’ (as opposed to ‘log’ for logarithms to base 10) As with values of the exponential function, natural logarithms can be found on a mathematical calculator Using the [LN] function key on your calculator, check that you can derive the following values: ln 1 = 0 ln 2.6 = 0 .95 55114 ln 0.45 = −0. 798 5 © 199 3, 2003... can be used for multiplication Example 14.8 Multiply 5,623.76 by 441.873 using natural logarithms Solution Taking natural logarithms and performing multiplication by adding them: ln 5,623.760 = 8.6347558+ ln 441.873 = 6. 091 0225 14.725778 (to 6 dp) To transform this logarithm back to its corresponding number we find e14.725778 = 2,484, 98 7.7 This answer can be verified by carrying out a straightforward multiplication... the predicted final population is 7,057,405 © 199 3, 2003 Mike Rosser Example 14.3 An economy is forecast to grow continuously at an annual rate of 2.5% If its GNP is currently e56 billion, what will the forecast for GNP be at the end of the third quarter the year after next? Solution In this example: t = 1.75 years, r = 2.5 % = 0.025, A = 56 (e billion) Therefore, the final value of GNP will be y = Aert... answers to 2 significant decimal places Solution (i) Using the formula r = ln(1 + i) the answers are: (a) i = 0% = 0 r = ln(1 + 0) = ln 1 = 0% (b) i = 10% = 0.1 r = ln(1 + 0.1) = ln 1.1 = 0. 095 31 = 9. 53% (c) i = 50% = 0.5 r = ln(1 + 0.5) = ln 1.5 = 0.405465 = 40.55% (d) i = 100% = 1 r = ln(1 + 1) = ln 2 = 0. 693 1472 = 69. 31% (ii) Using the formula i = er − 1 the answers are (a) r = 0% = 0 i = e0 − 1 =... has the property that if y = et then dy = et dt Thus, using the chain rule for differentiation, for any constant b, if y = ebt then dy = bebt dt Therefore, if the differential equation to be solved has no constant term and has the format dy = by dt then a possible solution is y = ebt because this would give dy = bebt = by dt © 199 3, 2003 Mike Rosser . 9 t C Y 10 0 301.00 411.00 11 1 286.60 396 .60 12 2 277 .96 387 .96 13 3 272.78 382.78 14 4 2 69. 67 3 79. 67 15 5 267.80 377.80 16 6 266.68 376.68 17 7 266.01 376.01 18 8 265.60 375.60 19. 40 +246.6 +110 = 396 .6 The value of Y for other time periods can be calculated in a similar fashion: Y 2 = C 2 + I 2 = (40 +0.6Y 1 ) + 110 = 40 +0.6( 396 .6) + 110 = 387 .96 © 199 3, 2003 Mike Rosser Y 3 =. be used to calculate Y t in any given time period. For example, in time period 9 it will be Y 9 = 375 + 36(0.6) 9 = 375 + 0.3628 = 375.3628 © 199 3, 2003 Mike Rosser As t increases in value, eventually