SOLUTIONS MANUAL TO ROMER'S
ADVANCED MACROECONOMICS 4TH
EDITION. COMPLETE SOLUTION
MANUAL DAVID ROMER.
SOLUTIONS TO CHAPTER 1
Problem 1.1
(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
Z(t) d ln Z(t) d ln X(t)Y(t)
(1) .
Z(t) dt dt
Since the log of the product of two variables equals the sum of their logs, we have
Z(t) d ln X(t) ln Y(t) d ln X(t) d ln Y(t)
(2) ,
Z(t) dt dt dt
or simply
Z(t) X(t) Y(t)
(3) .
Z(t) X(t) Y(t)
(b) Again, since the growth rate of a variable equals the time derivative of its log, we can write
Z(t) d ln Z(t) d lnX(t) Y(t)
(4) .
Z(t) dt dt
Since the log of the ratio of two variables equals the difference in their logs, we have
Z(t) d ln X(t) ln Y(t) d ln X(t) d ln Y(t)
(5) ,
Z(t) dt dt dt
or simply
Z(t) X(t) Y(t)
(6) .
Z(t) X(t) Y(t)
(c) We have
Z(t) d ln Z(t) d ln[X(t) ]
(7) .
Z(t) dt dt
Using the fact that ln[X(t) ] = lnX(t), we have
Z(t) d ln X(t) d ln X(t) X(t)
(8) ,
Z(t) dt dt 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 X(t)
depicted in the figure at right. X(t)
From time 0 to time t1 , the growth rate of X is
constant and equal to a > 0. At time t1 , the growth
© 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
a
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
,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.
© 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 lnX(t)
d ln X(t) X(t) slope = a
dt X(t)
(See equation (1.10) in the text.) slope = a
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 lnX(0)
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. 0 t1 t2 time
Problem 1.3
(a) The slope of the break-even investment line is
Inv/ (n + g + )k
given by (n + g + ) and thus a fall in the rate of eff lab
depreciation, , decreases the slope of the break-
even investment line. (n + g + NEW)k
The actual investment curve, sf(k) is unaffected.
sf(k)
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 .
k* k*NEW k
(b) Since the slope of the break-even investment
line is given by (n + g + ), a rise in the rate of Inv/ (n + gNEW + )k
technological progress, g, makes the break-even eff lab
investment line steeper.
(n + g + )k
The actual investment curve, sf(k), is unaffected.
sf(k)
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 .
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
Inv/
unaffected by the rise in capital's share, . eff lab
The effect of a change in on the actual investment
curve, sk, can be determined by examining the (n + g + )k
derivative
(sk )/. It is possible to show that
sk sk
NEW
(1) sk ln k .
sk
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
k* k*NEW k
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 Inv/
constant, B, so that the actual investment curve is eff lab
given by sBf(k), B > 0. (n + g + )k
sBNEW f(k)
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 sBf(k)
shifts the actual investment curve up.
The break-even investment line, (n + g + )k, is
unaffected.
k* k*NEW k
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 .
© 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.
ADVANCED MACROECONOMICS 4TH
EDITION. COMPLETE SOLUTION
MANUAL DAVID ROMER.
SOLUTIONS TO CHAPTER 1
Problem 1.1
(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
Z(t) d ln Z(t) d ln X(t)Y(t)
(1) .
Z(t) dt dt
Since the log of the product of two variables equals the sum of their logs, we have
Z(t) d ln X(t) ln Y(t) d ln X(t) d ln Y(t)
(2) ,
Z(t) dt dt dt
or simply
Z(t) X(t) Y(t)
(3) .
Z(t) X(t) Y(t)
(b) Again, since the growth rate of a variable equals the time derivative of its log, we can write
Z(t) d ln Z(t) d lnX(t) Y(t)
(4) .
Z(t) dt dt
Since the log of the ratio of two variables equals the difference in their logs, we have
Z(t) d ln X(t) ln Y(t) d ln X(t) d ln Y(t)
(5) ,
Z(t) dt dt dt
or simply
Z(t) X(t) Y(t)
(6) .
Z(t) X(t) Y(t)
(c) We have
Z(t) d ln Z(t) d ln[X(t) ]
(7) .
Z(t) dt dt
Using the fact that ln[X(t) ] = lnX(t), we have
Z(t) d ln X(t) d ln X(t) X(t)
(8) ,
Z(t) dt dt 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 X(t)
depicted in the figure at right. X(t)
From time 0 to time t1 , the growth rate of X is
constant and equal to a > 0. At time t1 , the growth
© 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
a
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
,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.
© 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 lnX(t)
d ln X(t) X(t) slope = a
dt X(t)
(See equation (1.10) in the text.) slope = a
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 lnX(0)
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. 0 t1 t2 time
Problem 1.3
(a) The slope of the break-even investment line is
Inv/ (n + g + )k
given by (n + g + ) and thus a fall in the rate of eff lab
depreciation, , decreases the slope of the break-
even investment line. (n + g + NEW)k
The actual investment curve, sf(k) is unaffected.
sf(k)
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 .
k* k*NEW k
(b) Since the slope of the break-even investment
line is given by (n + g + ), a rise in the rate of Inv/ (n + gNEW + )k
technological progress, g, makes the break-even eff lab
investment line steeper.
(n + g + )k
The actual investment curve, sf(k), is unaffected.
sf(k)
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 .
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
Inv/
unaffected by the rise in capital's share, . eff lab
The effect of a change in on the actual investment
curve, sk, can be determined by examining the (n + g + )k
derivative
(sk )/. It is possible to show that
sk sk
NEW
(1) sk ln k .
sk
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
k* k*NEW k
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 Inv/
constant, B, so that the actual investment curve is eff lab
given by sBf(k), B > 0. (n + g + )k
sBNEW f(k)
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 sBf(k)
shifts the actual investment curve up.
The break-even investment line, (n + g + )k, is
unaffected.
k* k*NEW k
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 .
© 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.
