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\begin{document}
\title{Romer's Problem Set Ch. 1}
\author[1]{Clara Jace}%
\affil[1]{George Mason University}%
\vspace{-1em}
\date{\today}
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\sloppy
\subsection*{1.1~}
{\label{564993}}
\textbf{a)} If~\(Z(t) = X(t) Y(t)\), then~\(\frac{\dot Z(t)}{Z(t)} = [\frac{\dot X(t)}{X(t)}] + [\frac{\dot Y(t)}{Y(t)}]\)
\(\ln Z = \ln (XY)\)
Given the multiplicative property of logs:~\(\ln Z = \ln X + \ln Y\)
Remember that the growth rate of a variable equals the time derivative
of its log:
\(\frac{d \ln Z}{dt} = \frac{d \ln X}{dt} + \frac{d \ln Y}{dt} = \frac{\dot Z(t)}{Z(t)} = \frac{\dot X(t)}{X(t)} + \frac{\dot Y(t)}{Y(t)}\)
\textbf{b)~}\(\) If~\(Z(t) = \frac{X(t)}{Y(t)}\),
then~\(\frac{\dot Z(t)}{Z(t)} = [\frac{\dot X(t)}{X(t)}] - [\frac{\dot Y(t)}{Y(t)}]\)
\(\ln Z = \ln (\frac{X}{Y})\)
Given the division property of logs:~\(\ln Z = \ln X - \ln Y\)
Remember that the growth rate of a variable equals the time derivative
of its log:
\(\frac{d \ln Z}{dt} = \frac{d \ln X}{dt} - \frac{d \ln Y}{dt} = \frac{\dot Z(t)}{Z(t)} = \frac{\dot X(t)}{X(t)} - \frac{\dot Y(t)}{Y(t)}\)
\textbf{c)~\(Z(t) = X(t)^ \alpha\)}
\textbf{\(\ln Z(t) = \ln X(t)^ \alpha\)}
Given the exponent property of logs:~\(\ln Z= \alpha \ln X\)
Remember that the growth rate of a variable equals the time derivative
of its log:
\(\frac{d \ln Z}{dt} = \alpha \frac{d \ln X}{dt} = \frac{\dot Z(t)}{Z(t)} = \alpha \frac{\dot X(t)}{X(t)} \)
\subsection*{1.2}
{\label{484940}}
\textbf{a)~}
\par\null\par\null\par\null\par\null\par\null\par\null\par\null
\textbf{b)}
\par\null\par\null\par\null\par\null\par\null\par\null
\subsection*{1.3}
{\label{926651}}
\textbf{a)} The break-even investment line,~\((n+g+ \delta)k\), will
decrease in slope as~\(\delta\) falls. The actual investment
line, \(sf(k)\), is unaffected.
\textbf{b)}~ The break-even line,~\((n + g+ \delta)k\), will increase in
slope as~\(g\) increases. The actual investment line,
\(sf(k)\), is unaffected.
\textbf{c)}~ The break-even investment line will be unaffected. However,
the impact on the actual investment line can be determined by
assessing~\(\frac{\partial sk^\alpha}{\partial \alpha} = sk^\alpha \ln k\). Since~\(0< \alpha < 1\),
if~\(\ln k > 0\), then the new actual investment curve will lie
above the old one. If~\(\ln k < 0\), then the new curve will lie
below the old.~
\textbf{d)} The break-even investment line is unaffected.
However,~\(f(k)\) will increase by a factor of the increased
output per unit of effective labor, shifting the curve up.
Mathematically,~\(s C_{old} f(k) \rightarrow C_{new} = (C_{old} + 1) \therefore sC_{new} f(k) > sC_{old}f(k)\)
\subsection*{1.4}
{\label{114356}}
\textbf{a)}~Since the increase in workers is greater than the increase
in technological progress (\(n > g\)), then output per unit of
effective labor will fall at the time of the jump. This will reduce the
amount of capital per unit of effective labor,
causing~\(k^*\) to fall to~\(k_{new}\). Likewise,
the fall in capital causes a fall in output per unit of effective
labor:~\(y^* \rightarrow y_{new}\).
\textbf{b)~}After the initial change, the lower~\(k_{new}\)
means that actual investment per unit of effective labor will be higher
than the break-even line. Thus, the economy is saving/investing more
than enough to offset depreciation and technological progress, meaning
\(k_{new}\) will slowly grow back toward \(k^*\).
Then, output per unit of effective labor will rise again toward the
balanced growth path (\(y_{new} \rightarrow y^*\)).
\textbf{c)}~At a balanced growth path, the level of output per unit of
effective labor will be as high as it needs to offset the technological
progress and depreciation, so~\(k\) returned to its
original level. This is because although the new workers joined the
workforce, capital has been accumulated toward the steady state again so
output per unit of effective labor is the same.
\subsection*{1.5}
{\label{508869}}
\textbf{a)} Cobb-Douglas:~\(f(k) = k^ \alpha\)
Substituting this into the equation describing the evolution of the
capital stock per unit of effective labor:~\(\dot k = sk^ \alpha - (n + g+ \delta)k\)
On balanced growth path,~\(\dot k = 0\) and \(sk^{*\alpha} = (n + g+ \delta)k^*\)
Solving this first for \(k^*\): \(k^* = [\frac{s}{(n+g+ \delta}]^{\frac{1}{(1 - \alpha)}}\)
Solving this for \(y^* = k^ \alpha\): ~\(y^* = [\frac{s}{(n+g+ \delta}]^{\frac{\alpha}{( 1- \alpha)}}\)
Consumption per unit of effective labor is~\(c^* = (1 - s)y*\)
Solving for~\(c^*\):~\(c^* = (1-s) [\frac{s}{(n+g+ \delta)}]^{\frac{ \alpha}{( 1- \alpha)}}\)
\textbf{b)~}Golden-rule value is the level at which consumption per unit
of effective labor is maximized. By rearranging the k* equation, we
have:
\(s = (n + g+ \delta) k^{* \alpha - 1}\)
Now substituting into the c* equation:
\(c^* = [1 - (n + g+ \delta)k^{* 1- \alpha}][\frac{(n+g+ \delta)k^{* 1 - \alpha}}{(n+g+ \delta)}] ^{\frac{\alpha}{1 - \alpha}}\)
Simplifying this we have:~
\(c^* = k^{* \alpha} - (n + g+ \delta)k^*\)
Recall that on the balanced growth path, investment is equal to
break-even investment. Now maximize c* with respect to k*:
\(\frac{\partial c^*}{\partial k^*} = \alpha k^{*\alpha - 1} - (n + g+ \delta) = 0\)
Solving equation for golden-rule level of k yields:
\(k*_{GR} = [\frac{\alpha}{(n+ g+ \delta)}]^{\frac{1}{1- \alpha}}\)
\textbf{c)} To find the savings rate, we substitute to find:~
\(s_{GR} = (n + g+ \delta)[\frac{\alpha}{(n + g + \delta)}]^{\frac{1 - \alpha}{1- \alpha}}\)
This simplifies to:~\(s_{GR} = \alpha\)
Thus, with Cobb-Douglas, the savings rate required to reach the golden
rule equals the elasticity of output with respect to capital
\subsection*{1.6}
{\label{482377}}
\textbf{a)}~Capital per worker, output per worker, and consumption per
worker would increase as a result of the fall in rate of population
growth.
\par\null\par\null\par\null\par\null\par\null\par\null
\textbf{b)~}The fall in population growth would ultimately cause the
path of output to grow at a permanently lower rate since Y will
gradually shift down until K, AL, and Y are all growing at the new lower
n.~
\subsection*{1.7}
{\label{553234}}
\textbf{a)}~\(\frac{\partial y*}{\partial n} = f'(k*)[\frac{\partial k*}{\partial n}]\)
Using the evolution of capital stock per unit of effective
labor,~\(\dot k = sf(k) - (n + g + \delta)k\), and the fact that on a balanced growth
path,~\(\dot k = 0; k = k*; sf(k*) = (n + g+ \delta)k*\)
Taking the derivative with respect to n gives:~
\(\frac{\partial k*}{\partial n} = \frac{k*}{sf'(k*) - ( n + g + \delta)}\)
Then by substitution we find:
\(\frac{\partial y*}{\partial n} = f'(k*)[\frac{k*}{sf'(k*) - (n + g + \delta)}]\)
By rearranging the condition that defines k* and solving for s we find:
\(s = \frac{ (n + g + \delta)k*}{f(k*)}\)
Next, substituting this equation into the elasticity:
\(\frac{\partial y*}{\partial n} = \frac{f'(k*)k*}{\frac{[(n+ g+ \delta) f'(k*)k*}{f(k*)} - (n + g+ \delta)}\)
Multiplying both sides by~\(\frac{n}{y*}\):
\(\frac{n}{y*} \frac{\partial y*}{\partial n} = \frac{n}{(n + g+ \delta)} \frac{\frac{f'(k*)k*}{f(k*)}}{\frac{[f'(k*)k*}{f(k*)]} - 1}\)
Using the definition that~\(\alpha _K(k*) \equiv \frac{f'(k*)k*}{f(k*)}\) gives:
\(= - \frac{n}{(n + g+ \delta)} [\frac{\alpha_K (k*)}{1 - \alpha_K(k*)}]\)
By plugging in the numbers given:~\(\alpha _K = \frac{1}{3}; g = .02; \delta = .03\) we calculate the
effect on y* of a fall in n from 2\% to 1\%, using n = 0.015:
\(= \frac{0.015}{(0.015 + 0.02 + 0.03)} (\frac{1/3}{1 - 1/3}) \approx -0.12\)
\par\null
Thus, the 50\% drop in the population growth rate leads to about a 6\%
increase in the level of output per unit of effective labor
(\(-0.5 \cdot -0.12 = 0.06\)).
\subsection*{1.8}
{\label{130778}}
\textbf{a)}~We know that an increase in the fraction of output that is
devoted to investment from 0.15 to 0.18 is a 20\% increase in the saving
rate. The elasticity of output with respect to the saving rate is:
\(\frac{s}{y*} \frac{\partial y*}{\partial s} = \frac{\alpha_ K (k*)}{1 - \alpha_K (k*)}\)
Substituting the given that~\(\alpha_k(k*) = 1/3\) we have:
\(= \frac{1/3}{1 - 1/3} = \frac{1}{2}\)
So we know that the elasticity of output with respect to the saving rate
is 1/2, meaning that the 20\% increase in saving would have risen output
by about 10\%.
\textbf{b)}~Consumption rises less than output, since increasing savings
means that consumption accounts for a smaller part of output. Since
consumption on the balanced growth path is:~\(c* = (1-s)y*\)
By taking the derivative with respect to s and multiplying both sides
by~\(\frac{s}{c*}\), we have the elasticity:
\(\frac{\partial c*}{\partial s} \frac{s}{c*} = \frac{-y*s}{(1-s)y*} + (1 - s) \frac{\partial y*}{\partial s} \frac{s}{(1 - s)y*)}\)
Simplified this is:
\(\frac{\partial c*}{\partial s} \frac{s}{c*} = \frac{-s}{(1 - s)} + \frac {\partial y*}{\partial s} \frac {s}{(1 - s)y*}\)
\(\frac{-0.165}{(1 - 0.165)} + 0.5 \approx 0.030\)
Therefore the elasticity of consumption with respect to the saving rate
is approx. 0.3, meaning that consumption will be about 6\% above where
it would have been.~
\textbf{c)~}The immediate effect of the rise in investment is that
consumption falls simultaneously. Though y* doesn't rise at once, it
begins to move toward the new, higher balanced-growth path level. The
text on the speed of convergence helps to determine the time it takes
for consumption to return to what it would have been without the saving
rate increase. Since~\(c = (1-s)y\), consumption will grow at the
same rate as y on the way to the new balanced growth path. The rate of
convergence of k and y is:~\(\lambda = (1 - \alpha_k)(n + g+ \delta) \)
We know that~\((n + g+ \delta)\) equals 6\% per year
and~\(\alpha _ k = 1/3\), yielding~\(\lambda \) = 4\%. Thus, k and
y move 4\% of the remaining distance toward their balanced growth path
values each year. Since consumption falls initially by 3.5\% and will
eventually be 6\% higher than the original level, it must be 36.8\% up
the growth path (3.5/9.5). Thus, we can determine the length of time
this will take by using this formula:
\(e^{- \lambda t*} = 0.632\)
Taking the natural log:~\(- \lambda t* = ln(0.632)\)
\(t* = 11.5\) years for consumption to return to what it would
have been.
\subsection*{1.9}
{\label{548201}}
\textbf{a)} First we define the marginal product of labor to
be~\(w \equiv \frac{\partial F(K, AL)}{\partial L}\)
The partial derivative of output with respect to L yields:
\(w \equiv \frac{\partial Y}{\partial L} = ALf'(k)[\frac{K}{AL^2}] + Af(k) = A[f(k) - kf'(k)]\)
\textbf{b)~}The marginal product of capital is~\(r \equiv [\frac{\partial F(K, AL)}{\partial K}] - \delta\)
\(wL + rK = A[f(k) - kf'(k) L + [f'(k) - \delta]K = ALf(k) - f'(k)[\frac{K}{AL}] AL + f'(k)K- \delta K\)
Simplifying gives:~\(= ALF(\frac{K}{AL}, 1) - \delta K\)
And knowing F is constant returns to scale~\(= F(K, AL) - \delta K\)
\textbf{c)} On balanced growth,~\(\frac{\dot r}{r} = 0\), and the share of
output gong to capital is~\(\frac{rK}{Y}\), meaning:
\(\frac{\frac{rK}{Y}}{rK}{Y} = \frac{\dot r}{r} + \frac{\dot K}{K} - \frac{\dot Y}{Y} = 0 + (n + g) - (n + g) = 0\)
So we know the share of output going to capital is constant. Then, we
can take the time derivative of the log of the marginal product of labor
to find the growth rate:~
\(\frac{\dot w}{w} = \frac{\dot A}{A} + \frac{\dot[f(k) - kf'(k)]}{[f(k) - kf'(k)]} = g + \frac{-kf"(k)\dot k}{f(k) - kf'(k)}\)
So, on a balanced growth path,~\(\dot k = 0 ; \frac{\dot w}{w} = g\). This means that
marginal product of labor rises at the growth rate of the effectiveness
of labor.~
\textbf{d)~}As we showed, the growth rate of the marginal product of
labor is:~~\(= g + \frac{-kf"(k)\dot k}{f(k) - kf'(k)}\). Thus, if~\(k > k*; \frac{\dot w}{w} > g\). As we move
from k to k*, the amount of capital per unit of effective labor also
rises meaning labor is more productive and the marginal product of labor
is increased even more. The growth rate of the marginal product of
capital is:~\(\frac{\dot r}{r} = \frac{f"(k) \dot k}{f'(k)}\). Therefore, as k rises toward k*, the
growth rate is negative and the marginal product of capital falls.~
\subsection*{1.10}
{\label{417274}}
\textbf{a)~}Balanced growth occurs when all the variables are growing at
constant rates. Taking the time derivative of k gives us:
\(\dot k = (\frac{\dot K}{AL}) = \frac{\dot K(AL) - K[\dot LA - \dot AL]}{(AL^2)} = \frac{\dot K}{AL} -k(n + g)\)
Then we can substitute to find:~\(\dot k = [f'(k) - (n + g + \delta)]k\). This means that the
balanced growth path level of capital stock per unit of effective labor
is implicitly in~\(f'(k*) = (n + g+ \delta)\). Therefore, all variables of the
model grow at constant rates (n+ g).~
~To show that the economy actually converges to this balanced growth
path, we first assume that if~\(f"(k) < 0 \) then f'(k) will fall
as k rises (and vice versa). Thus, if~\(k > k*; f'(k) < (n + g + \delta)\) and vice
versa. Therefore, regardless of the original value of k the economy will
converge to a balanced growth path at k* .
\textbf{b)} The golden-rule level of k is where consumption is maximized
per unit of effective labor (\(f'(k^{GR}) = (n + g + \delta)\)). This means that the
slope of the production function is equal to the slope of the break even
line. In this model, we save capital's contribution to output, and if it
exceeds break-even then k rises. We see that it will always settle to
the point on the balanced growth path where \(f'(k) = (n + g+ \delta)\).~
\subsection*{1.11}
{\label{394674}}
We can write that~\(\dot y = \dot y(y); k = k*; y = y* ; \dot y = 0\). Using a first-order
Taylor-series approximation:
\(\dot y \cong [\frac{\partial \dot y}{\partial y}|_{y=y*}] (y - y*)\)
Letting \(\lambda \equiv -\frac{\partial \dot y (y)}{\partial y}|_{y=y*} \Longrightarrow \dot y(t) \cong -\lambda [y(t) - y*)]\)~
This implies that y moves toward y* at a speed approx. proportional to
its distance from y*. Then, we can say:
\(y(t) \cong y* + e^{- \lambda t}[y(0) - y*]\)
Now to determine~\(\lambda\). First, taking the time derivative
of the production function:
\(\dot y = f'(k) \dot k\)
The equation for motion of capital is: \(\dot k = sf(k) - (n + g + \delta) k\)
Substituting yields: ~\(\dot y = f'(k) [sf(k) - (n + g+ \delta)k]\). And then writing in terms of
g(y) :
\(\frac{\partial \dot y}{\partial y}|_{y = y*} = [\frac{\partial \dot y}{\partial k}|_{y = y*}][\frac{\partial k}{\partial y}|_y = y*]\)
Taking the derivative with respect to k gives:~\(\frac{\partial \dot y}{\partial k} = f"(k) [sf(k) - (n + g+ \delta)k] + f'(k)[sf(k) - (n + g+ \delta)]\)
Balanced growth means \(= f'(k*)[sf'(k*) - (n + g + \delta)]\)
Now, since k = g(y) where~\(g(\cdot) = f^-1(\cdot)\)
\(\frac{\partial k}{\partial y}|_{y = y*} = \frac{1}{\frac{\partial y}{\partial k}|_{y=y*}} = \frac{a}{f'(k*)}\)
By substitution we find:
\(\frac{\partial \dot y}{\partial y}|_{y = y*} = sf'(k*) - (n + g+ \delta)\)
Thus,~\(\lambda \equiv -\frac{\partial \dot y}{\partial y}|_{y = y*} = (n + g + \delta) - sf'(k*)\)
Since we know s on the balanced growth path we have:~
\(\lambda \equiv -\frac{\partial \dot y}{\partial y}|_{y = y*} = (n + g + \delta) - \frac{(n + g+ \delta)k* f'(k*)}{f(k*)}\)
Now using our definition for~\(\alpha_K = \frac{ kf'(k)}{f(k)}\):
\(\lambda \equiv -\frac{\partial \dot y}{\partial y}|_{y = y*} = [1 - \alpha_K(k*)](n + g+ \delta)\)
Thus y converges to its balanced growth path value at the rate above,
the same at which k converges.~
\subsection*{1.12}
{\label{362710}}
\textbf{a)} Start with the production function~\(Y(t) = [A(t)K(t)]^\alpha L(t)^{1 - \alpha}\)
Dividing both sides by the right:~
\(\frac{Y(t)}{ A(t)^{\frac{\alpha}{1 - \alpha}} L(t)} = [\frac{A(t)K(t)}{A(t)^{\frac{\alpha}{1 - \alpha}}L(t)}] ^\alpha [\frac{L(t)}{A(t) ^{\frac{\alpha}{1 - \alpha}}L(t)}] ^{1 - \alpha}\)
This simplifies down to:
\(\frac{Y(t)}{ A(t)^{\frac{\alpha}{1 - \alpha}} L(t)} = [\frac{K(t)}{A(t)^{\frac{\alpha}{1 - \alpha}}L(t)}]^\alpha\)
Now define~\(\phi \equiv \frac{\alpha}{1 - \alpha}; k(t) \equiv \frac{K(t)}{A(t)^\phi L(t)} ; y(t) \equiv \frac{Y(t)}{A(t)^\phi L(t)}\), yielding~\(y(t) = k(t)^ \phi\)
Taking the time derivative:
\(\dot k(t) = \frac{\dot K(t)[A(t)^ \phi L(t)] - K(t)[\phi A(t) ^{\phi - 1} \dot A(t)L(t) + \dot L(t) A(t)^\phi]}{[A(t)^\phi L(t)]^2}\)
By using k(t), n and \(\mu\) for the changed
variables:~\(\dot k(t) = \frac{ \dot K(t)}{A(t)^\phi L(t) - (\phi \mu + n)k(t)}\)
Finally, by using the equation~\(y(t) = k(t) ^ \alpha\) we have:
\(\dot k (t) = sk(t) ^\alpha - (\phi \mu + n + \delta)k(t)\)
This means that when actual investment per unit of effective labor
exceeds break even, k rises toward k*. Since alpha is constant, y will
also be constant when the economy converges to k*.~
\par\null\par\null\par\null\par\null\par\null\par\null
\textbf{b)~}Now the production function is~\(Y(t) = [A(t) \bar J(t)]^ \alpha L(t) ^{1 - \alpha}\)
Just as in part a, we divide and simplify to obtain:\(\frac{Y(t)}{A(t)^{\frac{\alpha}{1 - \alpha}} L(t)} = [\frac{\bar J(t)}{A(t)^{frac{ \alpha}{1 - \alpha} L(t)}}]^ \alpha\)
Now with the similar definitions as part a:~\(y(t) = \bar j(t) ^ \alpha\)
To analyze the time dynamics of j, we take the time derivative:
\(\bar j = \frac{\bar J(t)[A(t)^phi L(t)] - \bar J(t)[\phi A(t)^{\phi - 1} \dot A(t) L(t) + \dot L(t) A(t) ^ \phi]}{[A(t) ^ \phi L(t)]^2}\)
Then, using the short-cut definitions again we
obtain:\(\dot J(t) = \frac{sA(t)Y(t)}{A(t)} - \frac{\delta J(t)}{A(t)} - \mu \bar J(t)\)
Substituting:~
\(\bar j(t) = s \bar j(t) ^ alpha - [n + \delta + \mu (1+ \phi)] \bar j(t)\)
Just like the Solow model, we can graph this case where the economy does
converge to a balanced growth path since all the variables of the model
are growing at constant rates:
\par\null\par\null\par\null\par\null\par\null\par\null
\textbf{c)} On a balanced growth path,~\(\bar j(t) = 0\) and so,
\(\bar j^{1 - \alpha} = \frac{s}{[n+ \delta + \mu(1 + \phi)]}\) and then we have,
\(\bar j* = [\frac{s}{n + \delta + \mu(1 + \phi)}]^{\frac{1}{1 - \alpha}}\)
Knowing this is equal to y, we take the derivative of y* with respect to
s:~
\(\frac{\partial y*}{\partial s} = [\frac{\alpha}{1- \alpha}] [\frac{s}{n + \delta + \mu(1 + \phi)}]^{\frac{\alpha}{1 - \alpha}} [\frac{1}{n + \delta + \mu(1 + \phi)}]\)
Turning this into an elasticity gives:
\(\frac{ \partial y*}{ \partial s} \frac{s}{y*} = [\frac{ \alpha}{1 - \alpha}] [\frac{s}{n + \delta \mu(1 + \phi)}]^{\frac{ \alpha}{(1 - \alpha) - 1}}\)
Simplifying yields:~\(\frac{\partial y}{\partial s} \frac{s}{y} = \frac{\alpha}{1 - \alpha}\)
\textbf{d)} A first-order Taylor approximation of~\(\dot y\)
around the balanced growth path value will be:~
\(\dot y \cong \frac{\partial \dot y}{\partial y}|_{y = y*} [y - y*]\)
Taking the time derivative of both sides and then substituting:
\(\dot y = \alpha \bar j ^{\alpha - 1} \bar j \Longrightarrow \dot y = s \alpha \bar j^{2 \alpha - 1} - \alpha \bar j^\alpha [n + \delta + \mu(1 + \phi)]\)
Since we can write~\(\bar y = y^{\frac{1}{\alpha}}\), evaluating we find:
\(\frac{\partial \dot y}{\partial y}|_{y = y*} = [\frac{\partial \dot y}{\partial \bar j} _ {y = y*}] [\frac{\partial \bar j}{\partial y}|_{y + y*}] = [s \alpha(2 \alpha - 1) ^{-2(\alpha - 1)} - \alpha ^2 \bar j^{\alpha - 1} (n + \delta + \mu(1 + \phi))] [\frac{1}{\alpha} y ^{\frac{1 - \alpha}{\alpha}}]\)
Finally, by remembering that~\(y^{\frac{1 - \alpha}{\alpha}} = \bar j^{1 - \alpha}\) and substituting:
\(\frac{\partial \dot y}{\partial y}|_{y = y*} = s(2 \alpha - 1) \bar j^{-2(\alpha - 1) + (1 + \alpha)} - \alpha \bar j^{\alpha -1 + (1 - \alpha)} [n + \delta + \mu(1 + \phi)] = s(2 \alpha - 1) \bar j^{\alpha - 1} - \alpha [n + \delta + \mu(1 + \phi)]\)
Now, by rearranging for s we find:
\(\frac{\partial \dot y}{\partial y} = -(1 - \alpha)[n + \delta + \mu(1 + \phi)]\)
Then, finding the first-order Taylor expansion:
\(\dot y \cong -(1 - \alpha)[n + \delta + \mu(1 + \phi)] [y - y*]\)
Solving this differential yields:~\(y(t) =- y* = c^{-(1 - \alpha)[n + \delta + \mu(1 + \phi)]} [y(0) - y*]\)
Now we know that the economy moves a fraction (the exponent) of the
remaining distance toward y* each year.~
\textbf{e)}~The elasticity of output with respect to s is the same in
this model as in the basic Solow model, although the speed of
convergence is faster since~\(\phi = \frac{\alpha}{1 - \alpha}\)
\subsection*{1.3}
{\label{204500}}
\textbf{a)} Growth rate of output per worker:~\(\frac{\dot Y(t)}{Y(t)} - \frac{\dot L(t)}{L(t)} = \alpha_K(t)[\frac{\dot K(t)}{K(t)} - \frac{\dot L(t)}{L(t)}] + R(t)\)
and~\(\alpha_K(t)\) is the elasticity of output with respect to
capital at time t and R(t) is the residual. If it is on the balanced
growth path, the growth rates of output and capital per worker are equal
to g, which is the growth rate of A. So, growth accounting attributes
67\% of growth in output per worker to tech. progress and 33\% to the
growth in capital per worker.
\textbf{b)} The reason that the capital-labor ratio grows at g is
because the effectiveness of labor is growing at g. Growth accounting
attributes the rise in output to the way that raising output will raise
the capital-labor ratio since it raises resources devoted to capital
accumulation. Therefore, it does not highlight the direct determinants
of this growth.~
\subsection*{1.14}
{\label{157492}}
\textbf{a)~}OLS gives a biased estimate of the slope coefficient of a
regression if there is correlation between the explanatory variable and
error term. When we substitute the equations given in section 1.7 into
each other:~
\(ln[\frac{Y}{N}_{1979}] - ln[\frac{Y}{N}_{1870}] = a + b \ln[(\frac{Y}{N})_{1870}] + [\epsilon - (1 + b)u]\)
If the value of b is -1, the error term is~\(\epsilon\). This
means that it will not be biased since the explanatory variable will not
be correlated with the error term.~
\textbf{b)~}Measuring errors for the dependent variable are not
problematic for the OLS estimate. Instead, if the measurement error is
in the year's income per capita, then there will be biases in the
results. For example, if 1870 income per capita is understated, growth
is overstated.~
\subsection*{1.15}
{\label{719368}}
When growing at a constant rate on the balanced growth
path,~\(\frac{\dot K(t)}{K(t)} = s \frac{Y(t)}{K(t)} - \delta\)
If you take a log of both sides of the production function:
\(lnY(t) = \alpha \ln K(t) + \beta \ln R(t) + \gamma \ln T(t) + (1 - \alpha - \beta - \gamma)[\ln A(t) + \ln L(t)]\)
By differentiating both sides with respect to time:~
\(g_y(t) = \alpha g_K(t) + \beta g_R(t) + \gamma g_T(t) + (1 - \alpha - \beta - \gamma)[g_A(t) + g_L(t)]\)
Substituting and simplifying that the growth rates are equal to n and A
is equal to g gives:\(y_Y(t) = \alpha g_K(t) + (1 - \alpha)n + (a - \alpha - \beta - \gamma)g\)
then, using the fact that the growth rate of output and capital are
equal on a balanced growth path means:
\((1 - \alpha)g_Y = (1 - \alpha)n + (1 - \alpha - \beta -\gamma)g\)
Thus the growth rate of output is :~\(g_Y^{\sim bpg} = \frac{(1- \alpha)n + (1 - \alpha - \beta -\gamma)g}{1 - \alpha}\)
Then, finding the growth rate of output per worker means subtracting the
growth rate of n (rate that L grows at)~
\(g_{Y/L}^{\sim bpg} = \frac{(1 - \alpha - \beta - \gamma)g}{1 - \alpha}\)
And we have arrived at the answer from the text!
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