<p>In deterministic growth models of closed economies with exogenous constant growth drivers (here, population and labor efficiency), the steady-state of the economy is in "growth rates", or under the newer terminology, it is a "balanced growth path".  </p>

<p>The growth rate of the economy can only be the sum of the growth rates of these exogenous growth drivers. So in the balanced growth path, aggregate magnitudes grow at $n+g$, per capita magnitudes grow at $g$ and magnitudes per efficiency unit of labor are constant.  </p>

<p>One can see this directly through the production function by considering</p>

<p>$$Y_t = K_t^\alpha (A_t L_t)^{1-\alpha} \implies \ln Y = \alpha \ln K_t + (1-\alpha)\cdot [\ln A_t + \ln L_t]$$</p>

<p>$$\implies \frac {\text{d}\ln Y}{\text{d}t} = \frac {\dot Y}{Y} = \alpha \frac {\dot K}{K} + (1-\alpha)\cdot \left[\frac {\dot A}{A} + \frac {\dot L}{L}\right]$$</p>

<p>$$\implies \frac {\text{d}\ln Y}{\text{d}t} = \frac {\dot Y}{Y} = \alpha \frac {\dot K}{K} + (1-\alpha)\cdot (g+n)$$</p>

<p>To have a balanced growth path, we must have equal and constant growth rates (so ratios like $Y/K$ etc remain constant), therefore</p>

<p>$$\frac {\dot Y}{Y} = \frac {\dot K}{K} = \gamma$$</p>

<p>$$\implies \gamma = \alpha \gamma + (1-\alpha)\cdot (g+n)$$</p>

<p>$$\implies \gamma = g+n$$</p>

<p>Note the importance of constant returns to scale, in order to obtain this result.</p>

CC BY-SA 3.0
2016-01-26T14:14:16.563Z
2016-01-26T14:14:16.563Z
Alecos Papadopoulos
<p>The last two equations are valid expressions. This version of the model supposes that all the productivity comes from labor (because A is attached to L in the first equation). So L is the number of workers, but AL is the effective work done by them. Dividing K by L just give you the capital divided by labour productivity.</p>

<p>Last equation is also correct if the economy is growing at the steady-state But they are both unnecessary tautologies. What matter in the Solow-Swan model is the output per worker and the capital per effective labor. That is just another way to look to the aggregate capital trajectory.</p>

CC BY-SA 3.0
2016-01-26T03:07:54.273Z
2016-01-26T03:07:54.273Z
Joao Luiz Pacheco
<p>The Solow growth model defines </p>

<p>$$k_t \equiv \frac{K_t}{A_t L_t}$$</p>

<p>Let our production function be Cobb Douglas </p>

<p>$$Y_t = K_t^\alpha (A_t L_t)^{1-\alpha}$$ </p>

<p>We have </p>

<p>$$A_t = A_0 e^{gt}$$
$$L_t = L_0 e^{nt}$$</p>

<p>We find the steady state is </p>

<p>$$k^* = \left(\frac{s}{n+ g + \delta}\right)^{\frac{1}{1-\alpha}}$$</p>

<p>At the steady state, we notice </p>

<p>$$\frac{K_t}{L_t} = A_t k_t = A_0 e^{gt}  \left(\frac{s}{n+ g + \delta}\right)^{\frac{1}{1-\alpha}}$$</p>

<p><strong>My Question</strong></p>

<p>Does $$\frac{K_t}{A_t} = L_t k_t = L_0 e^{nt}  \left(\frac{s}{n+ g + \delta}\right)^{\frac{1}{1-\alpha}}$$</p>

<p>also grow at the steady state? In several lecture notes I found online, no one discusses this so I wanted to clarify.</p>

<p><strong>Follow Up</strong> </p>

<p>But if $\frac{K_t}{L_t}$ and $\frac{K_t}{A_t}$ both grow this way, then </p>

<p>$$K_t = A_0 e^{gt} L_0 e^{nt}  \left(\frac{s}{n+ g + \delta}\right)^{\frac{1}{1-\alpha}}$$</p>

<p>So does this mean $K_t$ is a function of these parameters (e.g. $A_0$) or is this not a valid economic expression? </p>

<macroeconomics>
In the Solow growth model, does labor grow like technology at the steady state?
Stan Shunpike
# In the Solow growth model, does labor grow like technology at the steady state?

(Asked by: username_0 on 2016-01-26)

The Solow growth model defines

$$k\_t \equiv \frac{K\_t}{A\_t L\_t}$$

Let our production function be Cobb Douglas

$$Y\_t = K\_t^\alpha (A\_t L\_t)^{1-\alpha}$$

We have

$$A\_t = A\_0 e^{gt}$$
$$L\_t = L\_0 e^{nt}$$

We find the steady state is

$$k^\* = \left(\frac{s}{n+ g + \delta}\right)^{\frac{1}{1-\alpha}}$$

At the steady state, we notice

$$\frac{K\_t}{L\_t} = A\_t k\_t = A\_0 e^{gt} \left(\frac{s}{n+ g + \delta}\right)^{\frac{1}{1-\alpha}}$$

**My Question**

Does $$\frac{K\_t}{A\_t} = L\_t k\_t = L\_0 e^{nt} \left(\frac{s}{n+ g + \delta}\right)^{\frac{1}{1-\alpha}}$$

also grow at the steady state? In several lecture notes I found online, no one discusses this so I wanted to clarify.

**Follow Up**

But if $\frac{K\_t}{L\_t}$ and $\frac{K\_t}{A\_t}$ both grow this way, then

$$K\_t = A\_0 e^{gt} L\_0 e^{nt} \left(\frac{s}{n+ g + \delta}\right)^{\frac{1}{1-\alpha}}$$

So does this mean $K\_t$ is a function of these parameters (e.g. $A\_0$) or is this not a valid economic expression?

**Tags:** macroeconomics

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## 2 answers

### Answer by username_1 on 2016-01-26. Score: 4

In deterministic growth models of closed economies with exogenous constant growth drivers (here, population and labor efficiency), the steady-state of the economy is in "growth rates", or under the newer terminology, it is a "balanced growth path".

The growth rate of the economy can only be the sum of the growth rates of these exogenous growth drivers. So in the balanced growth path, aggregate magnitudes grow at $n+g$, per capita magnitudes grow at $g$ and magnitudes per efficiency unit of labor are constant.

One can see this directly through the production function by considering

$$Y\_t = K\_t^\alpha (A\_t L\_t)^{1-\alpha} \implies \ln Y = \alpha \ln K\_t + (1-\alpha)\cdot [\ln A\_t + \ln L\_t]$$

$$\implies \frac {\text{d}\ln Y}{\text{d}t} = \frac {\dot Y}{Y} = \alpha \frac {\dot K}{K} + (1-\alpha)\cdot \left[\frac {\dot A}{A} + \frac {\dot L}{L}\right]$$

$$\implies \frac {\text{d}\ln Y}{\text{d}t} = \frac {\dot Y}{Y} = \alpha \frac {\dot K}{K} + (1-\alpha)\cdot (g+n)$$

To have a balanced growth path, we must have equal and constant growth rates (so ratios like $Y/K$ etc remain constant), therefore

$$\frac {\dot Y}{Y} = \frac {\dot K}{K} = \gamma$$

$$\implies \gamma = \alpha \gamma + (1-\alpha)\cdot (g+n)$$

$$\implies \gamma = g+n$$

Note the importance of constant returns to scale, in order to obtain this result.


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### Answer by username_2 on 2016-01-26. Score: 1

The last two equations are valid expressions. This version of the model supposes that all the productivity comes from labor (because A is attached to L in the first equation). So L is the number of workers, but AL is the effective work done by them. Dividing K by L just give you the capital divided by labour productivity.

Last equation is also correct if the economy is growing at the steady-state But they are both unnecessary tautologies. What matter in the Solow-Swan model is the output per worker and the capital per effective labor. That is just another way to look to the aggregate capital trajectory.


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