<ol>
<li><p>" But then, there is clearly no steady state". How do you define steady state ? In this context I think it refers to balanced growth equilibrium. See <a href="https://economics.stackexchange.com/questions/10115/solow-model-steady-state-v-balanced-growth-path?rq=1">here</a>.</p></li>
<li><p>Yes there is a contradiction. Maybe $Y = K^\alpha (AL)^{1 - \alpha}$ ? This would mean $A$ is some measure of the efficiency of labor.</p></li>
</ol>

CC BY-SA 3.0
2017-02-08T14:09:42.703Z
I think you did not realise how your answer was helpful. If you use the formula you have, you can achieve a balanced growth path where the three variables grow at the same rate, as in the official answer. Can you write that down in your answer?
luchonacho
CC BY-SA 3.0
2017-02-08T12:17:15.287Z
2017-02-08T12:17:15.287Z
2017-04-13T12:51:38.253Z
user11629
<p>To simplistically answer your question, use the following:
$Y = K^\alpha (AL)^{1 - \alpha}$</p>

<p><strong><em>In order to prove all three can be equal:</em></strong>
We will assume that technological progress is labor augmenting (Harrod Neutral) or "labor saving", while holding the following true:</p>

<ul>
<li>Returns to capital are roughly constant over time.</li>
<li>Capital share of income is roughly constant over time.</li>
</ul>

<p>The Harrod Neutral is associated with a capital-output ratio that is constant (K/Y) in our "steady state".</p>

<p><em>Capital Accumulation</em>
In the basic Solow model, (1).
$$ \frac{\dot{k}}{k} \equiv s\frac{y}{k}-(d+n)$$</p>

<p>For growth rate per capita capital to be constant, y/k = Y/K must also be constant. (Characteristic of the Harrod Neutral)</p>

<p><em>We Need To Define The Balanced Growth Path With Technological Progress</em>
Taking the original production equation:
$Y = K^\alpha (AL)^{1 - \alpha}$</p>

<p>and dividing by AL (number of effective labor units) we are given our production function representing effective labor.
$$ \tilde{y} \equiv \tilde{k}^\alpha$$</p>

<p><em>Doing The Same For Capital Accumulation</em>
Follow steps below:
$$\frac{\tilde{\dot{k}}}{k} = \frac{dln(k/AL)}{dt}$$</p>

<p>$$ \frac{dlnK}{dt} - \frac{dlnA}{dt} - \frac{dlnL}{dt} $$</p>

<p>$$\frac{\dot{K}}{K} - \frac{\dot{A}}{A} - \frac{\dot{L}}{L} = \frac{\dot{K}}{K} - g - n$$</p>

<p>then using this to finding our Capital Accumulation equation below:</p>

<p>$$\frac{\dot{K}}{K} = S \frac{{Y}}{K} - d$$
$$\frac{\tilde{\dot{k}}}{\tilde{k}} = s\frac{Y}{K} - d - g - n$$
And, $$\frac{Y}{K} = \frac{Y}{AL} \frac{AL}{K} = \frac{\tilde{y}}{\tilde{k}}$$</p>

<p>Then, 
$$\frac{\tilde{\dot{k}}}{\tilde{k}} = s\frac{\tilde{y}}{\tilde{k}} - d - g - n$$</p>

<p>$$\tilde{\dot{k}} = s \tilde{y} - (d+g+n)\tilde{k}$$</p>

<p>Finally, combining our production function with capital accumulation we get an equation (6) that represents the Solow model with tech changes.
$$\tilde{\dot{k}} = s \tilde{k^\alpha} - (d+g+n)\tilde{k}$$</p>

<p>So our steady state is when (7) is true.
$$\tilde{\dot{k}^*} = 0$$</p>

<p>If we set this to be true for our steady state (denoted by *) levels of capital and output per effective labor. (8 and 9)
8: $$\tilde{k^*} = [\frac{s}{d+n+g}]^\frac{1}{1-\alpha}$$
9: $$\tilde{y^*} = [\frac{s}{d+n+g}]^\frac{\alpha}{1-\alpha}$$</p>

<p>In the steady state: (10)
Capital (and income) per capita grow at the rate of exogenous technological growth.</p>

<p>$$\tilde{k} = K/AL$$
$$k = K/L$$ 
$$k = A\tilde{k}$$</p>

<p>$$\tilde{k} = \tilde{k}^* = [\frac{s}{d+n+g}]^\frac{1}{1-\alpha}$$
$$k(t) = A(t)\tilde{k}^*$$
$$lnk(t) = lnA(t) + ln\tilde{k}^*$$
$$\frac{\dot{k}}{k} = \frac{\dot{A}}{A}$$</p>

<p>If you graph out
$$\tilde{\dot{k}} = s \tilde{k^\alpha} - (d+g+n)\tilde{k}$$</p>

<ul>
<li>Any savings rate changes, s, change the savings curve and therefore change $$\tilde{k^*}$$</li>
<li>Population or depreciation changes, n and d respectively, rotates the straight line, (d+g+n)k.
<strong><em>Changes in these affect the growth rate of per capita capital and per capita income only outside the steady-state.</em></strong></li>
<li>exo-tech growth changes the graph sort of like n and d by changing the level of $$\tilde{k^*}$$</li>
</ul>

<p>This exo-tech growth changes the growth rate of per capita capital and per capita income in the steady state (along balanced growth paths).</p>

CC BY-SA 3.0
2017-02-06T17:37:13.910Z
2017-02-15T20:40:24.260Z
2017-02-15T20:40:24.260Z
Ian Brigmann
<p>Consider the following question:</p>

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

<p>So, assume the standard function for production:</p>

<p>$$Y_t = A_t K^\alpha_tL^{1-\alpha}$$</p>

<p>where $L$ is fixed.</p>

<p>Then, the growth rate of output is: </p>

<p>$$g_Y \approx g_A + \alpha g_K $$</p>

<p><strong>Case without technological change</strong></p>

<p>In the standard Solow model with constant $A$, $g_A=0$. In the steady state, $sY=dK$ (investment equal depreciation), and $g_Y = g_K = 0$, consistent with the above equation.</p>

<p><strong>Case with exogenous technological change</strong></p>

<p>Here, $g_A=2\%$ (as per the question). But then, there is clearly no steady state, as the marginal product of capital is permanently increasing, thereby expanding capital. In the standard diagram, this means a shift of the function $sY_t$ upwards every period, meaning an ever increasing "steady state", which is never achieved. <strong>Hence, to me the question does not make sense.</strong></p>

<p>Finally, here is the official answer:</p>

<p><a href="https://i.sstatic.net/z3U0q.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/z3U0q.png" alt="enter image description here"></a> </p>

<p>Now, to me this answer is wrong. How can $Y$, $A$, and $K$ growth at the same rate? The equation of growth written above denies this! </p>

<p><strong>Is the question and/or answer wrong? Am I missing something? Am I wrong?</strong></p>

CC BY-SA 3.0
2017-02-05T17:20:48.607Z
" But then, there is clearly no steady state". How do you define steady state ? In this context I think it refers to balanced growth equilibrium.
user11629
CC BY-SA 3.0
2017-02-05T17:46:30.443Z
In any case, there is no such equilibrium where $A, K, Y$ grow at 2%. According to the growth formula, and intuitively, that is impossible. What is possible instead is, for instance, $g_Y=g_A=2\%$ and $g_K=4\%$ with $\alpha = 0.5$. But then the answer is wrong.
luchonacho
CC BY-SA 3.0
2017-02-05T19:23:50.580Z
It would be useful to see you type out the exact contradiction, because I can't seem to find any. More precisely, how do you derive the growth rate of output?
Giskard
CC BY-SA 3.0
2017-02-06T11:03:51.683Z
@denesp How can $Y, K, A$ all grow at the same rate? If you could prove that, then most of my cues would be gone.
luchonacho
CC BY-SA 3.0
2017-02-08T11:07:17.267Z
The question might be improved by noting that the original Solow paper (http://piketty.pse.ens.fr/fichiers/Solow1956.pdf) did consider (p 85) a production function with neutral (rather than labour-augmenting) technical progress.
Adam Bailey
<macroeconomics><economic-growth><solow>
Steady state Solow model with exogenous technological change
luchonacho
# Steady state Solow model with exogenous technological change

(Asked by: username_0 on 2017-02-05)

Consider the following question:

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

So, assume the standard function for production:

$$Y\_t = A\_t K^\alpha\_tL^{1-\alpha}$$

where $L$ is fixed.

Then, the growth rate of output is:

$$g\_Y \approx g\_A + \alpha g\_K $$

**Case without technological change**

In the standard Solow model with constant $A$, $g\_A=0$. In the steady state, $sY=dK$ (investment equal depreciation), and $g\_Y = g\_K = 0$, consistent with the above equation.

**Case with exogenous technological change**

Here, $g\_A=2\%$ (as per the question). But then, there is clearly no steady state, as the marginal product of capital is permanently increasing, thereby expanding capital. In the standard diagram, this means a shift of the function $sY\_t$ upwards every period, meaning an ever increasing "steady state", which is never achieved. **Hence, to me the question does not make sense.**

Finally, here is the official answer:

[![enter image description here](https://i.sstatic.net/z3U0q.png)](https://i.sstatic.net/z3U0q.png)

Now, to me this answer is wrong. How can $Y$, $A$, and $K$ growth at the same rate? The equation of growth written above denies this!

**Is the question and/or answer wrong? Am I missing something? Am I wrong?**

**Tags:** macroeconomics, economic-growth, solow
> **Comments:**
> **username_1**: " But then, there is clearly no steady state". How do you define steady state ? In this context I think it refers to balanced growth equilibrium.  
> **username_0**: In any case, there is no such equilibrium where $A, K, Y$ grow at 2%. According to the growth formula, and intuitively, that is impossible. What is possible instead is, for instance, $g_Y=g_A=2\%$ and $g_K=4\%$ with $\alpha = 0.5$. But then the answer is wrong.  
> **username_2**: It would be useful to see you type out the exact contradiction, because I can't seem to find any. More precisely, how do you derive the growth rate of output?  
> **username_0**: @denesp How can $Y, K, A$ all grow at the same rate? If you could prove that, then most of my cues would be gone.  
> **username_3**: The question might be improved by noting that the original Solow paper (http://piketty.pse.ens.fr/fichiers/Solow1956.pdf) did consider (p 85) a production function with neutral (rather than labour-augmenting) technical progress.  

---

## 2 answers

### Answer by username_4 on 2017-02-06. Score: 3 | Accepted answer

To simplistically answer your question, use the following:
$Y = K^\alpha (AL)^{1 - \alpha}$

***In order to prove all three can be equal:***
We will assume that technological progress is labor augmenting (Harrod Neutral) or "labor saving", while holding the following true:

* Returns to capital are roughly constant over time.
* Capital share of income is roughly constant over time.

The Harrod Neutral is associated with a capital-output ratio that is constant (K/Y) in our "steady state".

*Capital Accumulation*
In the basic Solow model, (1).
$$ \frac{\dot{k}}{k} \equiv s\frac{y}{k}-(d+n)$$

For growth rate per capita capital to be constant, y/k = Y/K must also be constant. (Characteristic of the Harrod Neutral)

*We Need To Define The Balanced Growth Path With Technological Progress*
Taking the original production equation:
$Y = K^\alpha (AL)^{1 - \alpha}$

and dividing by AL (number of effective labor units) we are given our production function representing effective labor.
$$ \tilde{y} \equiv \tilde{k}^\alpha$$

*Doing The Same For Capital Accumulation*
Follow steps below:
$$\frac{\tilde{\dot{k}}}{k} = \frac{dln(k/AL)}{dt}$$

$$ \frac{dlnK}{dt} - \frac{dlnA}{dt} - \frac{dlnL}{dt} $$

$$\frac{\dot{K}}{K} - \frac{\dot{A}}{A} - \frac{\dot{L}}{L} = \frac{\dot{K}}{K} - g - n$$

then using this to finding our Capital Accumulation equation below:

$$\frac{\dot{K}}{K} = S \frac{{Y}}{K} - d$$
$$\frac{\tilde{\dot{k}}}{\tilde{k}} = s\frac{Y}{K} - d - g - n$$
And, $$\frac{Y}{K} = \frac{Y}{AL} \frac{AL}{K} = \frac{\tilde{y}}{\tilde{k}}$$

Then,
$$\frac{\tilde{\dot{k}}}{\tilde{k}} = s\frac{\tilde{y}}{\tilde{k}} - d - g - n$$

$$\tilde{\dot{k}} = s \tilde{y} - (d+g+n)\tilde{k}$$

Finally, combining our production function with capital accumulation we get an equation (6) that represents the Solow model with tech changes.
$$\tilde{\dot{k}} = s \tilde{k^\alpha} - (d+g+n)\tilde{k}$$

So our steady state is when (7) is true.
$$\tilde{\dot{k}^\*} = 0$$

If we set this to be true for our steady state (denoted by \*) levels of capital and output per effective labor. (8 and 9)
8: $$\tilde{k^\*} = [\frac{s}{d+n+g}]^\frac{1}{1-\alpha}$$
9: $$\tilde{y^\*} = [\frac{s}{d+n+g}]^\frac{\alpha}{1-\alpha}$$

In the steady state: (10)
Capital (and income) per capita grow at the rate of exogenous technological growth.

$$\tilde{k} = K/AL$$
$$k = K/L$$
$$k = A\tilde{k}$$

$$\tilde{k} = \tilde{k}^\* = [\frac{s}{d+n+g}]^\frac{1}{1-\alpha}$$
$$k(t) = A(t)\tilde{k}^\*$$
$$lnk(t) = lnA(t) + ln\tilde{k}^\*$$
$$\frac{\dot{k}}{k} = \frac{\dot{A}}{A}$$

If you graph out
$$\tilde{\dot{k}} = s \tilde{k^\alpha} - (d+g+n)\tilde{k}$$

* Any savings rate changes, s, change the savings curve and therefore change $$\tilde{k^\*}$$
* Population or depreciation changes, n and d respectively, rotates the straight line, (d+g+n)k.
  ***Changes in these affect the growth rate of per capita capital and per capita income only outside the steady-state.***
* exo-tech growth changes the graph sort of like n and d by changing the level of $$\tilde{k^\*}$$

This exo-tech growth changes the growth rate of per capita capital and per capita income in the steady state (along balanced growth paths).


---

### Answer by username_1 on 2017-02-08. Score: 0

1. " But then, there is clearly no steady state". How do you define steady state ? In this context I think it refers to balanced growth equilibrium. See [here](https://economics.stackexchange.com/questions/10115/solow-model-steady-state-v-balanced-growth-path?rq=1).
2. Yes there is a contradiction. Maybe $Y = K^\alpha (AL)^{1 - \alpha}$ ? This would mean $A$ is some measure of the efficiency of labor.

> **Comments:**
> **username_0**: I think you did not realise how your answer was helpful. If you use the formula you have, you can achieve a balanced growth path where the three variables grow at the same rate, as in the official answer. Can you write that down in your answer?  

---

