<p>Following the conversation with user @denesp at the comments of my previous answer, I have to clarify the following: the usual graphical device we use related to the basic Solow growth model (see for example <a href="http://cruel.org/econthought/essays/growth/neoclass/solowgr.html" rel="nofollow noreferrer">here</a>, figure 2) is not a phase diagram, since reasonably we call "phase diagrams" those that contain zero-change loci, identify the crossing points of them as fixed points of a dynamical system, and examine their stability properties. And this is not what we do for the Solow model. So it was careless use of terminology from my part.</p>

<p>Nevertheless we can draw a "semi-Phase diagram" for the Solow growth model, in $(y,k)$ space. Understanding the symbols as "per efficiency unit of labor" we have the system of differential equations (while $y=f(k)$)</p>

<p>$$\dot k = sy - (n+\delta+g)k$$</p>

<p>$$\dot y = f'_k(k)\cdot \dot k$$
Writing the zero-change equation as a weak inequality to show also the dynamical tendencies, we have</p>

<p>$$\dot k \geq 0 \implies y \geq \frac {n+\delta+g}{s} k$$</p>

<p>$$\dot y \geq 0 \implies \dot k \geq 0$$</p>

<p>So this system gives a <em>single</em> zero change locus, a straight line. No crossing points to identify a fixed point What can we do? Draw also the production function in the diagram, since, in reality, the $(y,k)$ space is unidimensional, not an area, but a line. Then we get</p>

<p><a href="https://i.sstatic.net/i88fC.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/i88fC.png" alt="enter image description here"></a> </p>

<p>The vertical/horizontal arrows indicating the dynamical tendencies come properly from the weak inequalities above (both $y$ and $k$ tend to grow when above the zero-change locus). Then, since $y$ and $k$ are constrained to move on the dotted line (which is the production function), it follows that they move towards their fixed point, no matter where we start. Here the production function graph represents essentially the path towards long-run equilibrium, since convergence is monotonic.</p>

CC BY-SA 3.0
2016-01-13T14:11:47.877Z
2016-01-13T14:24:33.007Z
2016-01-13T14:24:33.007Z
Alecos Papadopoulos
<p>This is when the attempt at accuracy creates confusion and misunderstanding.  </p>

<p>Back in the day, growth models were not incorporating technological progress, and led to a long-run equilibrium characterized by <em>constant</em> per capita magnitudes. Verbally, the term "steady-state" seemed appropriate to describe such a situation. </p>

<p>Then Romer and endogenous growth models came along, which also pushed the older models to start including as a routine feature exogenous growth factors (apart from population). And "suddenly", per capita terms were not constant in the long-run equilibrium, but <em>growing at a constant rate</em>. Initially the literature described such a situation as "steady state in growth rates".  </p>

<p>Then it appears the profession thought something like "it is inaccurate to use the word "steady" here because per capita magnitudes are growing. What happens is that all magnitudes <em>grow</em> at a <em>balanced</em> rate (i.e at the same rate, and so their ratios remain constant). And since they grow, they follow a <em>path</em>..." Eureka!: the term "balanced growth path" was born.  </p>

<p>...To the frustration of students (at least), which have now to remember that for example, the "saddle path" is indeed a <em>path</em> in the Phase diagram, but the "balanced growth path" is only a point! (because in order to actually draw a Phase diagram and obtain a good old long-run equilibrium, we express magnitudes per effective worker, and these magnitudes do have a traditional steady-state. But we continue to call it "balanced growth path", because per capita magnitudes, which is what we are interested in, in our individualistic approach), continue to grow).</p>

<p>So "balanced growth path" = "steady state of magnitudes per efficiency unit of labor", and I guess you can figure out the rest for your phase diagram.</p>

CC BY-SA 3.0
2016-01-07T16:52:41.200Z
2016-01-07T16:52:41.200Z
Alecos Papadopoulos
<p>Okay, so I'm having real problems distinguishing between the Steady State concept and the balanced growth path in this model:</p>

<p>$$ Y = K^\beta (AL)^{1-\beta} $$</p>

<p>I have been asked to derive the steady state values for capital per effective worker:</p>

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

<p>As well as the steady state ratio of capital to output (K/Y):</p>

<p>$$ \frac{K^{SS}}{Y^{SS}} = \frac{s}{n+g+\delta } $$</p>

<p>I found both of these fine, but I have been also asked to find the "steady-state value of the marginal product of capital, dY/dK". Here is what I did:</p>

<p>$$ Y = K^\beta (AL)^{1-\beta} $$
$$ MPK = \frac{dY}{dK} = \beta K^{\beta -1}(AL)^{1-\beta } $$</p>

<p>Substituting in for K in the steady state (calculated when working out steady state for K/Y ratio above):</p>

<p>$$ K^{SS} = AL\left(\frac{s}{n+g+\delta }\right)^{\frac{1}{1-\beta }} $$</p>

<p>$$ MPK^{SS} = \beta (AL)^{1-\beta }\left[AL\left(\frac{s}{n+g+\delta }\right)^{\frac{1}{1-\beta }}\right]^{\beta -1} $$</p>

<p>$$ MPK^{SS} = \beta \left(\frac{s}{n+g+\delta }\right)^{\frac{\beta -1}{1-\beta }} $$</p>

<p>Firstly I need to know whether this calculation for the steady state value of MPK is correct?</p>

<p>Secondly, I have been asked to Sketch the time paths of the capital-output ratio and the marginal product of capital, for an economy that converges to its balanced growth path "from below".</p>

<p>I am having problems understanding exactly what the balanced growth path is, as opposed to the steady state, and how to use my calculations to figure out what these graphs should look like.</p>

<p>Sorry for the mammoth post, any help is greatly appreciated! Thanks in advance.</p>

<economic-growth><production-function><capital-returns><steady-state>
Solow Model: Steady State v Balanced Growth Path
James Baker
# Solow Model: Steady State v Balanced Growth Path

(Asked by: username_0 on 2016-01-07)

Okay, so I'm having real problems distinguishing between the Steady State concept and the balanced growth path in this model:

$$ Y = K^\beta (AL)^{1-\beta} $$

I have been asked to derive the steady state values for capital per effective worker:

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

As well as the steady state ratio of capital to output (K/Y):

$$ \frac{K^{SS}}{Y^{SS}} = \frac{s}{n+g+\delta } $$

I found both of these fine, but I have been also asked to find the "steady-state value of the marginal product of capital, dY/dK". Here is what I did:

$$ Y = K^\beta (AL)^{1-\beta} $$
$$ MPK = \frac{dY}{dK} = \beta K^{\beta -1}(AL)^{1-\beta } $$

Substituting in for K in the steady state (calculated when working out steady state for K/Y ratio above):

$$ K^{SS} = AL\left(\frac{s}{n+g+\delta }\right)^{\frac{1}{1-\beta }} $$

$$ MPK^{SS} = \beta (AL)^{1-\beta }\left[AL\left(\frac{s}{n+g+\delta }\right)^{\frac{1}{1-\beta }}\right]^{\beta -1} $$

$$ MPK^{SS} = \beta \left(\frac{s}{n+g+\delta }\right)^{\frac{\beta -1}{1-\beta }} $$

Firstly I need to know whether this calculation for the steady state value of MPK is correct?

Secondly, I have been asked to Sketch the time paths of the capital-output ratio and the marginal product of capital, for an economy that converges to its balanced growth path "from below".

I am having problems understanding exactly what the balanced growth path is, as opposed to the steady state, and how to use my calculations to figure out what these graphs should look like.

Sorry for the mammoth post, any help is greatly appreciated! Thanks in advance.

**Tags:** economic-growth, production-function, capital-returns, steady-state

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## 2 answers

### Answer by username_1 on 2016-01-07. Score: 17 | Accepted answer

This is when the attempt at accuracy creates confusion and misunderstanding.

Back in the day, growth models were not incorporating technological progress, and led to a long-run equilibrium characterized by *constant* per capita magnitudes. Verbally, the term "steady-state" seemed appropriate to describe such a situation.

Then Romer and endogenous growth models came along, which also pushed the older models to start including as a routine feature exogenous growth factors (apart from population). And "suddenly", per capita terms were not constant in the long-run equilibrium, but *growing at a constant rate*. Initially the literature described such a situation as "steady state in growth rates".

Then it appears the profession thought something like "it is inaccurate to use the word "steady" here because per capita magnitudes are growing. What happens is that all magnitudes *grow* at a *balanced* rate (i.e at the same rate, and so their ratios remain constant). And since they grow, they follow a *path*..." Eureka!: the term "balanced growth path" was born.

...To the frustration of students (at least), which have now to remember that for example, the "saddle path" is indeed a *path* in the Phase diagram, but the "balanced growth path" is only a point! (because in order to actually draw a Phase diagram and obtain a good old long-run equilibrium, we express magnitudes per effective worker, and these magnitudes do have a traditional steady-state. But we continue to call it "balanced growth path", because per capita magnitudes, which is what we are interested in, in our individualistic approach), continue to grow).

So "balanced growth path" = "steady state of magnitudes per efficiency unit of labor", and I guess you can figure out the rest for your phase diagram.


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### Answer by username_1 on 2016-01-13. Score: 5

Following the conversation with user @denesp at the comments of my previous answer, I have to clarify the following: the usual graphical device we use related to the basic Solow growth model (see for example [here](http://cruel.org/econthought/essays/growth/neoclass/solowgr.html), figure 2) is not a phase diagram, since reasonably we call "phase diagrams" those that contain zero-change loci, identify the crossing points of them as fixed points of a dynamical system, and examine their stability properties. And this is not what we do for the Solow model. So it was careless use of terminology from my part.

Nevertheless we can draw a "semi-Phase diagram" for the Solow growth model, in $(y,k)$ space. Understanding the symbols as "per efficiency unit of labor" we have the system of differential equations (while $y=f(k)$)

$$\dot k = sy - (n+\delta+g)k$$

$$\dot y = f'\_k(k)\cdot \dot k$$
Writing the zero-change equation as a weak inequality to show also the dynamical tendencies, we have

$$\dot k \geq 0 \implies y \geq \frac {n+\delta+g}{s} k$$

$$\dot y \geq 0 \implies \dot k \geq 0$$

So this system gives a *single* zero change locus, a straight line. No crossing points to identify a fixed point What can we do? Draw also the production function in the diagram, since, in reality, the $(y,k)$ space is unidimensional, not an area, but a line. Then we get

[![enter image description here](https://i.sstatic.net/i88fC.png)](https://i.sstatic.net/i88fC.png)

The vertical/horizontal arrows indicating the dynamical tendencies come properly from the weak inequalities above (both $y$ and $k$ tend to grow when above the zero-change locus). Then, since $y$ and $k$ are constrained to move on the dotted line (which is the production function), it follows that they move towards their fixed point, no matter where we start. Here the production function graph represents essentially the path towards long-run equilibrium, since convergence is monotonic.


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