Chapter 6, Problem 9E

a)

To determine

Economic interpretation for the equation.

Explanation of Solution

If the output elasticity with respect to idea is ½, it implies that the percentage point increase of idea will lead to a 0.5 percentage increase in the output level. Also, the discovery of new ideas will lead to a growth rate in idea to diminish. It can be derived as follows:

$\begin{array}{c}{\text{Y}}_{\text{t}}={\text{A}}_{\text{t}}{}^{\left(\frac{1}{2}\right)}{\text{L}}_{\text{yt}}\\ \frac{\partial {\text{Y}}_{\text{t}}}{\partial {\text{A}}_{\text{t}}}=\frac{1}{2}{\text{A}}_{\text{t}}{}^{\left(-\frac{1}{2}\right)}{\text{L}}_{\text{yt}}\\ \frac{\partial {\text{Y}}_{\text{t}}}{\partial {\text{A}}_{\text{t}}}=-\frac{1}{4}{\text{A}}_{\text{t}}{}^{\left(-\frac{3}{2}\right)}{\text{L}}_{\text{yt}}<0\end{array}$

b)

To determine

Growth rate of knowledge in the economy.

Explanation of Solution

The growth rate of knowledge implies the growth rate of idea in the economy. Then, derive the growth rate of idea as follows:

 $\begin{array}{c}{\text{A}}_{\text{t}+1}=\overline{\text{z}}{\text{A}}_{\text{t}}{\text{L}}_{\text{at}}\\ \Delta {\text{A}}_{\text{t}+1}=\overline{\text{z}}{\text{A}}_{\text{t}}\overline{\text{l}}\overline{\text{L}}\\ \frac{\Delta {\text{A}}_{\text{t}+1}}{{\text{A}}_{\text{t}}}=\overline{\text{z}}\overline{\text{l}}\overline{\text{L}}\\ \text{Growth rate of idea}\left({\text{g}}_{\text{a}}\right)=\overline{\text{z}}\overline{\text{l}}\overline{\text{L}}\end{array}$

c)

To determine

Growth rate of output per person in the economy.

Explanation of Solution

Derive the growth rate of output per person in the economy as follows:

$\begin{array}{c}{\text{L}}_{\text{yt}}={\text{L}}_{\text{at}}=\overline{\text{L}}\\ \Rightarrow {\text{L}}_{\text{yt}}=\overline{\text{l}}\overline{\text{L}}=\overline{\text{L}}\\ \Rightarrow {\text{L}}_{\text{yt}}=\left(1-\overline{\text{l}}\right)=\overline{\text{L}}\\ {\text{Y}}_{\text{t}}={\text{A}}_{\text{t}}{}^{\frac{1}{2}}{\text{L}}_{\text{yt}}\\ \Rightarrow {\text{Y}}_{\text{t}}={\text{A}}_{\text{t}}{}^{\frac{1}{2}}\left(1-\overline{\text{l}}\right)\overline{\text{L}}\\ \Rightarrow \frac{{\text{Y}}_{\text{t}}}{\overline{\text{L}}}={\text{A}}_{\text{t}}{}^{\frac{1}{2}}\left(1-\overline{\text{l}}\right)\end{array}$

Where, ${\text{Y}}_{\text{t}}=\frac{{\text{Y}}_{\text{t}}}{\overline{\text{L}}}$

$\Rightarrow {\text{Y}}_{\text{t}}={\text{A}}_{\text{t}}{}^{\frac{1}{2}}\left(1-\overline{\text{l}}\right)$

Take logs and rewrite the equation as follows:

${\text{lnY}}_{\text{t}}=\frac{1}{2}{\text{lnA}}_{\text{t}}+\text{ln}\left(1-\overline{\text{l}}\right)$

Suppose $\text{ln}\left(1-\overline{\text{l}}\right)$ is a constant term, with respect to the time, differentiate it as follows:

$\begin{array}{l}\frac{\Delta {\text{Y}}_{\text{t}}}{{\text{Y}}_{\text{t}}}=\frac{1}{2}\frac{\Delta {\text{A}}_{\text{t}}}{{\text{A}}_{\text{t}}}+0\\ \Rightarrow {\text{g}}_{\text{y}}=\frac{1}{2}{\text{g}}_{\text{a}}\\ \Rightarrow {\text{g}}_{\text{y}}=\frac{1}{2}\overline{\text{z}}\overline{\text{l}}\overline{\text{L}}\end{array}$

Therefore, the growth rate of output per person is calculated as follows:

${\text{g}}_{\text{y}}=\frac{1}{2}{\text{g}}_{\text{a}}=\frac{\overline{\text{z}}\overline{\text{l}}\overline{\text{L}}}{2}$

d)

To determine

Output per person at each point in time.

Explanation of Solution

Suppose,

${\text{g}}_{\text{a}}=\overline{\text{z}}\overline{\text{l}}\overline{\text{L}}$

The stock of knowledge can be equated as follows:

${\text{A}}_{\text{t}}={\text{A}}_{\text{0}}{\left(1+{\text{g}}_{\text{a}}\right)}^{\text{t}}$

The production function per worker can be derived as follows:

$\begin{array}{c}{\text{Y}}_{\text{t}}={\text{A}}_{\text{t}}{}^{\frac{1}{2}}\left(1-\overline{\text{l}}\right)\\ \Rightarrow {\text{Y}}_{\text{t}}={\left[{\text{A}}_{\text{0}}{\left(1+{\text{g}}_{\text{a}}\right)}^{\text{t}}\right]}^{\frac{1}{2}}\left(1-\overline{\text{l}}\right)\\ \Rightarrow {\text{Y}}_{\text{t}}={\text{A}}_{\text{0}}{}^{\frac{1}{2}}{\left(1+{\text{g}}_{\text{a}}\right)}^{\frac{\text{t}}{2}}\left(1-\overline{\text{l}}\right)\end{array}$