Exam (elaborations) TEST BANK FOR Advanced Macroeconomics By David Romer (Solution Manual)-Converted

(a) Since the growth rate of a variable equals the time derivative of its log, as shown by equation (1.10) in the text, we can write (1)  ( )   ( ) Z t ln ( ) ln ( ) ( ) Z t d Z t dt d X t Y t dt   . Since the log of the product of two variables equals the sum of their logs, we have (2)  ( )   ( ) Z t ln ( ) ln ( ) ln ( ) ln ( ) Z t d X t Y t dt d X t dt d Y t dt     , or simply (3)  ( ) ( )  ( ) ( )  ( ) ( ) Z t Z t X t X t Y t Y t   . (b) Again, since the growth rate of a variable equals the time derivative of its log, we can write (4)  ( )   ( ) Z t ln ( ) ln ( ) ( ) Z t d Z t dt d X t Y t dt   . Since the log of the ratio of two variables equals the difference in their logs, we have (5)  ( )   ( ) Z t ln ( ) ln ( ) ln ( ) ln ( ) Z t d X t Y t dt d X t dt d Y t dt     , or simply (6)  ( ) ( )  ( ) ( )  ( ) ( ) Z t Z t X t X t Y t Y t   . (c) We have (7)  ( ) ( ) Z t ln ( ) ln[ ( ) ] Z t d Z t dt d X t dt    . Using the fact that ln[X(t) ] = lnX(t), we have (8)  ( )   ( ) ln ( ) ln ( )  ( ) ( ) Z t Z t d X t dt d X t dt X t X t       , where we have used the fact that  is a constant. Problem 1.2 (a) Using the information provided in the question, the path of the growth rate of X, X (t) X(t), is depicted in the figure at right. From time 0 to time t1 , the growth rate of X is constant and equal to a 0. At time t1 , the growth rate of X drops to 0. From time t1 to time t2 , the growth rate of X rises gradually from 0 to a. Note that we have made the assumption that X (t) X(t) rises at a constant rate from t1 to t2 . Finally, after time t2 , the growth rate of X is constant and equal to a again.  ( ) ( ) X t X t a 0 t1 t2 time © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1-2 Solutions to Chapter 1 (b) Note that the slope of lnX(t) plotted against time is equal to the growth rate of X(t). That is, we know d X t dt X t X t ln ( )  ( ) ( )  (See equation (1.10) in the text.) From time 0 to time t1 the slope of lnX(t) equals a 0. The lnX(t) locus has an inflection point at t1 , when the growth rate of X(t) changes discontinuously from a to 0. Between t1 and t2 , the slope of lnX(t) rises gradually from 0 to a. After time t2 the slope of lnX(t) is constant and equal to a 0 again. Problem 1.3 (a) The slope of the break-even investment line is given by (n + g + ) and thus a fall in the rate of depreciation, , decreases the slope of the breakeven investment line. The actual investment curve, sf(k) is unaffected. From the figure at right we can see that the balancedgrowth- path level of capital per unit of effective labor rises from k* to k*NEW . (b) Since the slope of the break-even investment line is given by (n + g + ), a rise in the rate of technological progress, g, makes the break-even investment line steeper. The actual investment curve, sf(k), is unaffected. From the figure at right we can see that the balanced-growth-path level of capital per unit of effective labor falls from k* to k*NEW . lnX(t) slope = a slope = a lnX(0) 0 t1 t2 time Inv/ (n + g + )k eff lab (n + g + NEW)k sf(k) k* k*NEW k Inv/ (n + gNEW + )k eff lab (n + g + )k sf(k) k*NEW k* k © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Solutions to Chapter 1 1-3 (c) The break-even investment line, (n + g + )k, is unaffected by the rise in capital's share, . The effect of a change in  on the actual investment curve, sk, can be determined by examining the derivative (sk)/. It is possible to show that (1)    sk   sk ln k . For 0  1, and for positive values of k, the sign of (sk)/ is determined by the sign of lnk. For lnk 0, or k 1, sk   0 and so the new actual investment curve lies above the old one. For lnk 0 or k 1, sk   0 and so the new actual investment curve lies below the old one. At k = 1, so that lnk = 0, the new actual investment curve intersects the old one. In addition, the effect of a rise in  on k* is ambiguous and depends on the relative magnitudes of s and (n + g + ). It is possible to show that a rise in capital's share, , will cause k* to rise if s (n + g + ). This is the case depicted in the figure above. (d) Suppose we modify the intensive form of the production function to include a non-negative constant, B, so that the actual investment curve is given by sBf(k), B 0. Then workers exerting more effort, so that output per unit of effective labor is higher than before, can be modeled as an increase in B. This increase in B shifts the actual investment curve up. The break-even investment line, (n + g + )k, is unaffected. From the figure at right we can see that the balanced-growth-path level of capital per unit of effective labor rises from k* to k*NEW . Problem 1.4 (a) At some time, call it t0 , there is a discrete upward jump in the number of workers. This reduces the amount of capital per unit of effective labor from k* to kNEW . We can see this by simply looking at the definition, k  K/AL . An increase in L without a jump in K or A causes k to fall. Since f ' (k) 0, this fall in the amount of capital per unit of effective labor reduces the amount of output per unit of effective labor as well. In the figure below, y falls from y* to yNEW . Inv/ eff lab (n + g + )k sk NEW sk k* k*NEW k Inv/ eff lab (n + g + )k sBNEW f(k) sBf(k) k* k*NEW k © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 1-4 Solutions to Chapter 1 (b) Now at this lower kNEW , actual investment per unit of effective labor exceeds break-even investment per unit of effective labor. That is, sf(kNEW ) (g + )kNEW . The economy is now saving and investing more than enough to offset depreciation and technological progress at this lower kNEW . Thus k begins rising back toward k*. As capital per unit of effective labor begins rising, so does output per unit of effective labor. That is, y begins rising from yNEW back toward y*. (c) Capital per unit of effective labor will continue to rise until it eventually returns to the original level of k*. At k*, investment per unit of effective labor is again just enough to offset technological progress and depreciation and keep k constant. Since k returns to its original value of k* once the economy again returns to a balanced growth path, output per unit of effective labor also returns to its original value of y* = f(k*). Problem 1.5 (a) The equation describing the evolution of the capital stock per unit of effective labor is given by (1) k  sf (k)  (n  g  )k. Substituting in for the intensive form of the Cobb-Douglas, f(k) = k, yields (2) k  sk  (n  g  )k . On the balanced growth path, k is zero; investment per unit of effective labor is equal to break-even investment per unit of effective labor and so k remains constant. Denoting the balanced-growth-path value of k as k*, we have sk* = (n + g + )k*. Rearranging to solve for k* yields (3) k* s (n g ) ( )      1 1  . To get the balanced-growth-path value of output per unit of effective labor, substitute equation (3) into the intensive form of the production function, y = k: (4) y* s (n g ) ( )       1  . Consumption per unit of effective labor on the balanced growth path is given by c* = (1 - s)y*. Substituting equation (4) into this expression yields (5) c* ( s) s (n g ) ( )      1 1    . (b) By definition, the golden-rule level of the capital stock is that level at which consumption per unit of effective labor is maximized. To derive this level of k, take equation (3), which expresses the balancedgrowth- path level of k, and rearrange it to solve for s: (6) s = (n + g + )k*1-. Now substitute equation (6) into equation (5): (7) c*  (n g )k * (n g )k * (n g ) ( )            1 1 1 1        . After some straightforward algebraic manipulation, this simplifies to (8) c* = k* - (n + g + )k*. Investment