<p>We know that <span class="math-container">$g_A = \dot{A}/A$</span> and thus <span class="math-container">$g_A = \delta L_A A^{\phi-1}$</span>.</p>

<p>Take logs of both sides of <span class="math-container">$g_A = \delta L_A A^{\phi-1}$</span> and differentiating with respect to time gives us</p>

<p><span class="math-container">$$ \frac{\dot{g_A}}{g_A} = n + (\phi-1)g_A.$$</span></p>

<p>Multiply both sides of the equation by <span class="math-container">$g_A$</span>:</p>

<p><span class="math-container">$$ \dot{g_A} = n g_A + (\phi-1) g_A^2$$</span></p>

<p>Given that <span class="math-container">$\phi&lt;1$</span>, and in the steady-state BGP the growth of the growth rate <span class="math-container">$\dot{g_A} = 0$</span>, we divide both sides by <span class="math-container">$g_A$</span> to get:</p>

<p><span class="math-container">$$n+ (\phi -1) g_A = 0 $$</span> </p>

<p>Rearrange to get the desired expression:</p>

<p><span class="math-container">$$g_A^* = \frac{n}{1-\phi}$$</span></p>

<p>The second equation is just the result above multiplied by <span class="math-container">$\sigma$</span>:</p>

<p><span class="math-container">$$g_Y = g_A \sigma =  \frac{\sigma n}{1-\phi}$$</span></p>

CC BY-SA 4.0
2020-05-18T19:39:12.450Z
2020-05-19T23:51:54.197Z
2020-05-19T23:51:54.197Z
1muflon1
<p>He is just saying that there exists some exponent, reflecting  either diminishing or increasing returns to scale, theta. Research  can have nonlinear marginal returns. </p>

<p>Given that, as theta increases the growth rate must increase linearly since the exponential returns on knowledge translate to a linear increase in exponential growth rate.</p>

<p>So it's just how exponential  functions work, you could write a lengthy proof but it's really just arithmetic.</p>

CC BY-SA 4.0
2020-05-18T19:46:36.577Z
2020-05-18T19:46:36.577Z
user28027
<p>There is more than one way to derive the formula for <span class="math-container">$g_A$</span> with constant growth: the following is a way I find conceptually simple.  Starting from your <span class="math-container">$g_A=\delta L_A A^{\phi-1}$</span>, differentiate with respect to time (using the product and chain rules) and (for constant growth) set the result to zero:</p>

<p><span class="math-container">$$\dot{g_A}=0=\delta[A^{\phi-1}\dot{L_A}+L_A(\phi-1)A^{\phi-2}\dot{A}]$$</span></p>

<p>Dividing by <span class="math-container">$\delta L_AA^{\phi-1}$</span>:</p>

<p><span class="math-container">$$0=\frac{\dot{L_A}}{L_A}+(\phi-1)\frac{\dot{A}}{A}=n+(\phi-1)g_A$$</span></p>

<p><span class="math-container">$$g_A=\frac{n}{1-\phi}$$</span></p>

<p>For growth in output per worker, and assuming that <span class="math-container">$L_Y$</span> like <span class="math-container">$L$</span> grows at rate <span class="math-container">$n$</span> so that growth in <span class="math-container">$Y/L$</span> equals growth in <span class="math-container">$Y/L_Y$</span>, we have:</p>

<p><span class="math-container">$$g_y=g_{A^{\sigma}}=\frac{1}{A^{\sigma}}\frac{dA^{\sigma}}{dt}= \frac{1}{A^{\sigma}}\sigma A^{\sigma-1}\dot{A}=\sigma g_A=\frac{\sigma n}{1-\phi}$$</span></p>

CC BY-SA 4.0
2020-05-19T15:54:32.403Z
(+1) I actually like your way even better than mine
1muflon1
CC BY-SA 4.0
2020-05-19T15:44:37.770Z
2020-05-19T15:44:37.770Z
Adam Bailey
<p>Jones (1999) builds his semi-endogenous growth model where output is produced with only one input, labour, in the following "research" function:</p>

<p><span class="math-container">$$ Y = A^\alpha L_Y $$</span></p>

<p>Labour is augmented with technology <span class="math-container">$A$</span> which is also researched by
workers, <span class="math-container">$L_A$</span>, in the following production function:</p>

<p><span class="math-container">$$ \dot{A} = \delta L_A A^\phi. $$</span></p>

<p>We have that <span class="math-container">$L_A + L_Y = L$</span> and <span class="math-container">$0 &lt; \phi &lt; 1$</span>.</p>

<p>He then writes:</p>

<blockquote>
  <p>Assuming that the labour force L grows at some constant exogenous rate
  <span class="math-container">$n$</span>, it is easy to show that there exists a stable balanced growth
  path for the model where</p>
  
  <p><span class="math-container">$$ g_A = \frac{n}{1-\phi}$$</span></p>
  
  <p>and</p>
  
  <p><span class="math-container">$$ g_Y = \sigma g_A = \frac{\sigma n}{1-\phi} $$</span></p>
</blockquote>

<p>He says it's "easy to show", but I can't for the life of me show it! We know that <span class="math-container">$g_A = \dot{A}/A$</span>, and on the balanced growth path <span class="math-container">$g_A = g_Y = g_L = n$</span>. So I try to plug <span class="math-container">$g_A = \delta L_A A^{\phi-1} = n$</span> but that gets me nowhere.</p>

CC BY-SA 4.0
2020-05-19T11:50:46.433Z
I managed to find Jones' paper [here](https://web.stanford.edu/~chadj/scaleffAERPP1999.pdf) (a link in your question would have been helpful).  Looking at pp 140-1 I see that: 1) the exponent of $A$ in the output function is $\sigma$, not $\alpha$; 2) the formula $\frac{\sigma n}{1-\phi}$ is for $g_y$ (growth of output per worker), not $g_Y$ (growth of output).
Adam Bailey
<economic-growth><balanced-growth>
Proving a result In Jones (1999) "Growth: With or Without Scale Effects"
Lieu Zheng Hong
# Proving a result In Jones (1999) "Growth: With or Without Scale Effects"

(Asked by: username_0 on 2020-05-18)

Jones (1999) builds his semi-endogenous growth model where output is produced with only one input, labour, in the following "research" function:

$$ Y = A^\alpha L\_Y $$

Labour is augmented with technology $A$ which is also researched by
workers, $L\_A$, in the following production function:

$$ \dot{A} = \delta L\_A A^\phi. $$

We have that $L\_A + L\_Y = L$ and $0 < \phi < 1$.

He then writes:

> Assuming that the labour force L grows at some constant exogenous rate
> $n$, it is easy to show that there exists a stable balanced growth
> path for the model where
>
> $$ g\_A = \frac{n}{1-\phi}$$
>
> and
>
> $$ g\_Y = \sigma g\_A = \frac{\sigma n}{1-\phi} $$

He says it's "easy to show", but I can't for the life of me show it! We know that $g\_A = \dot{A}/A$, and on the balanced growth path $g\_A = g\_Y = g\_L = n$. So I try to plug $g\_A = \delta L\_A A^{\phi-1} = n$ but that gets me nowhere.

**Tags:** economic-growth, balanced-growth
> **Comments:**
> **username_1**: I managed to find Jones' paper [here](https://web.stanford.edu/~chadj/scaleffAERPP1999.pdf) (a link in your question would have been helpful).  Looking at pp 140-1 I see that: 1) the exponent of $A$ in the output function is $\sigma$, not $\alpha$; 2) the formula $\frac{\sigma n}{1-\phi}$ is for $g_y$ (growth of output per worker), not $g_Y$ (growth of output).  

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## 3 answers

### Answer by username_1 on 2020-05-19. Score: 3 | Accepted answer

There is more than one way to derive the formula for $g\_A$ with constant growth: the following is a way I find conceptually simple. Starting from your $g\_A=\delta L\_A A^{\phi-1}$, differentiate with respect to time (using the product and chain rules) and (for constant growth) set the result to zero:

$$\dot{g\_A}=0=\delta[A^{\phi-1}\dot{L\_A}+L\_A(\phi-1)A^{\phi-2}\dot{A}]$$

Dividing by $\delta L\_AA^{\phi-1}$:

$$0=\frac{\dot{L\_A}}{L\_A}+(\phi-1)\frac{\dot{A}}{A}=n+(\phi-1)g\_A$$

$$g\_A=\frac{n}{1-\phi}$$

For growth in output per worker, and assuming that $L\_Y$ like $L$ grows at rate $n$ so that growth in $Y/L$ equals growth in $Y/L\_Y$, we have:

$$g\_y=g\_{A^{\sigma}}=\frac{1}{A^{\sigma}}\frac{dA^{\sigma}}{dt}= \frac{1}{A^{\sigma}}\sigma A^{\sigma-1}\dot{A}=\sigma g\_A=\frac{\sigma n}{1-\phi}$$

> **Comments:**
> **username_2**: (+1) I actually like your way even better than mine  

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### Answer by username_2 on 2020-05-18. Score: 3

We know that $g\_A = \dot{A}/A$ and thus $g\_A = \delta L\_A A^{\phi-1}$.

Take logs of both sides of $g\_A = \delta L\_A A^{\phi-1}$ and differentiating with respect to time gives us

$$ \frac{\dot{g\_A}}{g\_A} = n + (\phi-1)g\_A.$$

Multiply both sides of the equation by $g\_A$:

$$ \dot{g\_A} = n g\_A + (\phi-1) g\_A^2$$

Given that $\phi<1$, and in the steady-state BGP the growth of the growth rate $\dot{g\_A} = 0$, we divide both sides by $g\_A$ to get:

$$n+ (\phi -1) g\_A = 0 $$

Rearrange to get the desired expression:

$$g\_A^\* = \frac{n}{1-\phi}$$

The second equation is just the result above multiplied by $\sigma$:

$$g\_Y = g\_A \sigma = \frac{\sigma n}{1-\phi}$$


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### Answer by username_3 on 2020-05-18. Score: 1

He is just saying that there exists some exponent, reflecting either diminishing or increasing returns to scale, theta. Research can have nonlinear marginal returns.

Given that, as theta increases the growth rate must increase linearly since the exponential returns on knowledge translate to a linear increase in exponential growth rate.

So it's just how exponential functions work, you could write a lengthy proof but it's really just arithmetic.


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