3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I managed to find Jones' paper here (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). $\endgroup$ – Adam Bailey May 19 at 11:50
3
$\begingroup$

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:

$$g_y=g_{A^{\sigma}}=\frac{1}{A^{\sigma}}\frac{dA^{\sigma}}{dt}= \frac{1}{A^{\sigma}}\sigma A^{\sigma-1}\dot{A}=\sigma g_A=\frac{\sigma n}{1-\phi}$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ (+1) I actually like your way even better than mine $\endgroup$ – 1muflon1 May 19 at 15:54
2
$\begingroup$

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 $\sigma$:

$$g_Y = g_A \sigma = \frac{\sigma n}{1-\phi}$$

| improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.