SOLUTIONS MANUAL TO ROMER'S ADVANCED
MACROECONOMICS 4TH EDITION. COMPLETE
SOLUTION MANUAL DAVID ROMER
. Here is the complete transcription of every question, its options (where applicable), the correct
answer, and the rationale from the provided solutions manual.
---
**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
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d\ln Z(t)}{dt} = \frac{d\ln[X(t)Y(t)]}{dt}.
\]
Since the log of the product of two variables equals the sum of their logs, we have
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d[\ln X(t) + \ln Y(t)]}{dt} = \frac{d\ln X(t)}{dt} + \frac{d\ln Y(t)}{dt},
\]
or simply
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{\dot{X}(t)}{X(t)} + \frac{\dot{Y}(t)}{Y(t)}.
\]
**(b)** Again, since the growth rate of a variable equals the time derivative of its log, we can write
\[
,\frac{\dot{Z}(t)}{Z(t)} = \frac{d\ln Z(t)}{dt} = \frac{d\ln[X(t)/Y(t)]}{dt}.
\]
Since the log of the ratio of two variables equals the difference in their logs, we have
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d[\ln X(t) - \ln Y(t)]}{dt} = \frac{d\ln X(t)}{dt} - \frac{d\ln Y(t)}{dt},
\]
or simply
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{\dot{X}(t)}{X(t)} - \frac{\dot{Y}(t)}{Y(t)}.
\]
**(c)** We have
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d\ln Z(t)}{dt} = \frac{d\ln[X(t)^\alpha]}{dt}.
\]
Using the fact that \(\ln[X(t)^\alpha] = \alpha\ln X(t)\), we have
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d[\alpha\ln X(t)]}{dt} = \alpha\frac{d\ln X(t)}{dt} =
\alpha\frac{\dot{X}(t)}{X(t)},
\]
where we have used the fact that \(\alpha\) is a constant.
**Rationale:** The solution uses the mathematical property that the growth rate of a variable is the
time derivative of its natural logarithm. It then applies the properties of logarithms: the log of a product
is the sum of logs, the log of a ratio is the difference of logs, and the log of a variable raised to a constant
power is the constant times the log of the variable.
---
**Problem 1.2**
,**(a)** Using the information provided in the question, the path of the growth rate of X,
\(\dot{X}(t)/X(t)\), is depicted in the figure at right.
From time 0 to time \(t_1\), the growth rate of X is constant and equal to a > 0. At time \(t_1\), the
growth rate of X drops to 0. From time \(t_1\) to time \(t_2\), the growth rate of X rises gradually from 0
to a. Note that we have made the assumption that \(\dot{X}(t)/X(t)\) rises at a constant rate from \(t_1\)
to \(t_2\). Finally, after time \(t_2\), the growth rate of X is constant and equal to a again.
**(b)** Note that the slope of \(\ln X(t)\) plotted against time is equal to the growth rate of \(X(t)\).
That is, we know
\[
\frac{d\ln X(t)}{dt} = \frac{\dot{X}(t)}{X(t)}
\]
(See equation (1.10) in the text.)
From time 0 to time \(t_1\) the slope of \(\ln X(t)\) equals a > 0. The \(\ln X(t)\) locus has an inflection
point at \(t_1\) when the growth rate of \(X(t)\) changes discontinuously from a to 0. Between \(t_1\)
and \(t_2\), the slope of \(\ln X(t)\) rises gradually from 0 to a. After time \(t_2\) the slope of \(\ln X(t)\)
is constant and equal to a > 0 again.
**Rationale:** The problem asks to graph the growth rate of a variable X over time. The solution
describes the shape of the graph based on the given information: a constant positive growth rate, a drop
to zero, a gradual rise back to the original rate, and then a constant rate again. The second part asks to
graph the log of X, noting that the slope of this graph is the growth rate, so the shape of the log graph
reflects the changes in the growth rate over time.
---
**Problem 1.3**
**(a)** The slope of the break-even investment line is given by \((n + g + \delta)\) and thus a fall in the
rate of depreciation, \(\delta\), decreases the slope of the break-even investment line. 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 rises from \(k^*\) to \(k^*_{NEW}\).
, **(b)** Since the slope of the break-even investment line is given by \((n + g + \delta)\), 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 \(k^*_{NEW}\).
**(c)** The break-even investment line, \((n + g + \delta)k\) is unaffected by the rise in capital's share,
\(\alpha\). The effect of a change in \(\alpha\) on the actual investment curve, \(sk^\alpha\), can be
determined by examining the derivative \(\partial(sk^\alpha)/\partial\alpha\). It is possible to show that
\[
(1) \frac{\partial sk^\alpha}{\partial\alpha} = sk^\alpha \ln k.
\]
For \(0 < \alpha < 1\), and for positive values of k, the sign of \(\partial(sk^\alpha)/\partial\alpha\) is
determined by the sign of \(\ln k\). For \(\ln k > 0\), or \(k > 1\), \(\partial sk^\alpha/\partial\alpha > 0\)
and so the new actual investment curve lies above the old one. For \(\ln k < 0\) or \(k < 1\), \(\partial
sk^\alpha/\partial\alpha < 0\) and so the new actual investment curve lies below the old one. At \(k =
1\) so that \(\ln k = 0\), the new actual investment curve intersects the old one.
In addition, the effect of a rise in \(\alpha\) on \(k^*\) is ambiguous and depends on the relative
magnitudes of s and \((n + g + \delta)\). It is possible to show that a rise in capital's share, \(\alpha\), will
cause \(k^*\) to rise if \(s > (n + g + \delta)\). 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, \((n + g + \delta)k\), 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 \(k^*_{NEW}\).
**Rationale:** The solution analyzes the Solow growth model graphically. It examines how changes in
parameters like the depreciation rate, technological progress, and capital's share affect the break-even
investment line and the actual investment curve, and consequently, the steady-state level of capital per
effective worker.
---
MACROECONOMICS 4TH EDITION. COMPLETE
SOLUTION MANUAL DAVID ROMER
. Here is the complete transcription of every question, its options (where applicable), the correct
answer, and the rationale from the provided solutions manual.
---
**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
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d\ln Z(t)}{dt} = \frac{d\ln[X(t)Y(t)]}{dt}.
\]
Since the log of the product of two variables equals the sum of their logs, we have
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d[\ln X(t) + \ln Y(t)]}{dt} = \frac{d\ln X(t)}{dt} + \frac{d\ln Y(t)}{dt},
\]
or simply
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{\dot{X}(t)}{X(t)} + \frac{\dot{Y}(t)}{Y(t)}.
\]
**(b)** Again, since the growth rate of a variable equals the time derivative of its log, we can write
\[
,\frac{\dot{Z}(t)}{Z(t)} = \frac{d\ln Z(t)}{dt} = \frac{d\ln[X(t)/Y(t)]}{dt}.
\]
Since the log of the ratio of two variables equals the difference in their logs, we have
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d[\ln X(t) - \ln Y(t)]}{dt} = \frac{d\ln X(t)}{dt} - \frac{d\ln Y(t)}{dt},
\]
or simply
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{\dot{X}(t)}{X(t)} - \frac{\dot{Y}(t)}{Y(t)}.
\]
**(c)** We have
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d\ln Z(t)}{dt} = \frac{d\ln[X(t)^\alpha]}{dt}.
\]
Using the fact that \(\ln[X(t)^\alpha] = \alpha\ln X(t)\), we have
\[
\frac{\dot{Z}(t)}{Z(t)} = \frac{d[\alpha\ln X(t)]}{dt} = \alpha\frac{d\ln X(t)}{dt} =
\alpha\frac{\dot{X}(t)}{X(t)},
\]
where we have used the fact that \(\alpha\) is a constant.
**Rationale:** The solution uses the mathematical property that the growth rate of a variable is the
time derivative of its natural logarithm. It then applies the properties of logarithms: the log of a product
is the sum of logs, the log of a ratio is the difference of logs, and the log of a variable raised to a constant
power is the constant times the log of the variable.
---
**Problem 1.2**
,**(a)** Using the information provided in the question, the path of the growth rate of X,
\(\dot{X}(t)/X(t)\), is depicted in the figure at right.
From time 0 to time \(t_1\), the growth rate of X is constant and equal to a > 0. At time \(t_1\), the
growth rate of X drops to 0. From time \(t_1\) to time \(t_2\), the growth rate of X rises gradually from 0
to a. Note that we have made the assumption that \(\dot{X}(t)/X(t)\) rises at a constant rate from \(t_1\)
to \(t_2\). Finally, after time \(t_2\), the growth rate of X is constant and equal to a again.
**(b)** Note that the slope of \(\ln X(t)\) plotted against time is equal to the growth rate of \(X(t)\).
That is, we know
\[
\frac{d\ln X(t)}{dt} = \frac{\dot{X}(t)}{X(t)}
\]
(See equation (1.10) in the text.)
From time 0 to time \(t_1\) the slope of \(\ln X(t)\) equals a > 0. The \(\ln X(t)\) locus has an inflection
point at \(t_1\) when the growth rate of \(X(t)\) changes discontinuously from a to 0. Between \(t_1\)
and \(t_2\), the slope of \(\ln X(t)\) rises gradually from 0 to a. After time \(t_2\) the slope of \(\ln X(t)\)
is constant and equal to a > 0 again.
**Rationale:** The problem asks to graph the growth rate of a variable X over time. The solution
describes the shape of the graph based on the given information: a constant positive growth rate, a drop
to zero, a gradual rise back to the original rate, and then a constant rate again. The second part asks to
graph the log of X, noting that the slope of this graph is the growth rate, so the shape of the log graph
reflects the changes in the growth rate over time.
---
**Problem 1.3**
**(a)** The slope of the break-even investment line is given by \((n + g + \delta)\) and thus a fall in the
rate of depreciation, \(\delta\), decreases the slope of the break-even investment line. 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 rises from \(k^*\) to \(k^*_{NEW}\).
, **(b)** Since the slope of the break-even investment line is given by \((n + g + \delta)\), 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 \(k^*_{NEW}\).
**(c)** The break-even investment line, \((n + g + \delta)k\) is unaffected by the rise in capital's share,
\(\alpha\). The effect of a change in \(\alpha\) on the actual investment curve, \(sk^\alpha\), can be
determined by examining the derivative \(\partial(sk^\alpha)/\partial\alpha\). It is possible to show that
\[
(1) \frac{\partial sk^\alpha}{\partial\alpha} = sk^\alpha \ln k.
\]
For \(0 < \alpha < 1\), and for positive values of k, the sign of \(\partial(sk^\alpha)/\partial\alpha\) is
determined by the sign of \(\ln k\). For \(\ln k > 0\), or \(k > 1\), \(\partial sk^\alpha/\partial\alpha > 0\)
and so the new actual investment curve lies above the old one. For \(\ln k < 0\) or \(k < 1\), \(\partial
sk^\alpha/\partial\alpha < 0\) and so the new actual investment curve lies below the old one. At \(k =
1\) so that \(\ln k = 0\), the new actual investment curve intersects the old one.
In addition, the effect of a rise in \(\alpha\) on \(k^*\) is ambiguous and depends on the relative
magnitudes of s and \((n + g + \delta)\). It is possible to show that a rise in capital's share, \(\alpha\), will
cause \(k^*\) to rise if \(s > (n + g + \delta)\). 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, \((n + g + \delta)k\), 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 \(k^*_{NEW}\).
**Rationale:** The solution analyzes the Solow growth model graphically. It examines how changes in
parameters like the depreciation rate, technological progress, and capital's share affect the break-even
investment line and the actual investment curve, and consequently, the steady-state level of capital per
effective worker.
---
