<p>When the (exogenous) rate of technological change/efficiency is smaller, the corresponding steady-state levels of consumption and capital per unit of effective labor, <em>increase</em>.  </p>

<p>For capital, we have</p>

<p>$$f'(\hat k^*) = \rho + \theta g_x \implies \frac {\partial}{\partial g_x}f'(\hat k^*) = \theta &gt;0$$</p>

<p>So if $g_x \downarrow \implies f'(\hat k^*) \downarrow \implies \hat k^* \uparrow$ due to decreasing marginal product of capital.</p>

<p>For steady-state consumption per unit of effective labor we have</p>

<p>$$\hat c^* = f(\hat k^*) - (n+g_x)\hat k^*$$</p>

<p>$$\implies \frac {\partial \hat c^*}{\partial g_x}= f'(\hat k^*) \frac {\partial \hat k^*}{\partial g_x} - \hat k^* - (n+g_x) \frac {\partial \hat k^*}{\partial g_x}$$</p>

<p>$$=[f'(\hat k^*) - n - g_x]\cdot \frac {\partial \hat k^*}{\partial g_x} - \hat k^* $$</p>

<p>The term in brackets is assumed positive, i.e. we have already assumed</p>

<p>$$f'(\hat k^*) = \rho + \theta g_x &gt; n + g_x \implies \rho &gt; n + (1-\theta)g_x$$</p>

<p>in order to exclude infinite utility. </p>

<p>Moreover, evidently $\frac {\partial \hat k^*}{\partial g_x} &lt;0$ so in all</p>

<p>$$\frac {\partial \hat c^*}{\partial g_x} &lt;0$$</p>

<p>Therefore, if $g_x \downarrow \implies \hat c^* \uparrow $.</p>

<p>See <a href="http://down.cenet.org.cn/upfile/8/200751171644184.pdf" rel="nofollow">Barro &amp; Martin</a>, ch. 2 page 102, where they discuss the  $g_x \uparrow$ case (in p. 101 they discuss the constraint on the parameters).</p>

<p><strong>Comment:</strong> This result may appear counter-intuitive, but a deeper examination of the model shows that if $g_x$ is lower, <em>utility per capita</em> is lower. Using "consumption per unit of effective labor" is a modeling tactic, what interests us is what happens per individual. So no, the model does <em>not</em> argue in favor of lower productivity/technology.</p>

CC BY-SA 3.0
2015-03-09T02:07:42.093Z
Perfect! One more thing, the result of this in the phase diagram is that the $k^$ and $C^$ curves both move in the phase diagram (C/K coordinates). What is the intuition behind only one of the curves moving vs. both of them doing so?
Dole
CC BY-SA 3.0
2015-03-09T02:32:07.973Z
@Dole I would suggest to go after Barro's book, it is legally freely downloadable and it presents the whole model in many details and extensions, as well as intuitive explanations.
Alecos Papadopoulos
CC BY-SA 3.0
2015-03-08T21:53:39.133Z
2015-03-09T02:36:23.883Z
2015-03-09T02:36:23.883Z
Alecos Papadopoulos
<p>Is the conclusion that both decline correct? Both long and short run.</p>

<p>Let $g_x$ denote growth rate of $x$, $g_c$ the growth rate of $\hat c$. In the model there is no depreciation so,</p>

<p>$$ g_c = \frac 1{\theta}[r - \rho - \theta g_x] $$</p>

<p>$$\dot {\hat k} = f(\hat k) - \hat c - (n+g_x)\hat k$$ </p>

<p>At steady state growth of $\hat k$ and $\hat c$ are both $0$. When that is the case:</p>

<p>$$\hat c^* = f(\hat k^*) - (n+g_x)\hat k^*$$</p>

<p>Also since $r = f'(k)$</p>

<p>$$f'(\hat k^*) = \rho + \theta g_x$$</p>

<p>Somehow based on these equations I reach the opposite conclusion, i.e. that $\hat c^*$ and $\hat k^*$ will increase rather than decrease if $g_x$ is reduced.</p>

<macroeconomics><economic-growth>
What happens to effective consumption and capital in the Ramsey/Cass-Koopmans(RCK) model when technological growth decreases?
Dole
# What happens to effective consumption and capital in the Ramsey/Cass-Koopmans(RCK) model when technological growth decreases?

(Asked by: username_0 on 2015-03-07)

Is the conclusion that both decline correct? Both long and short run.

Let $g\_x$ denote growth rate of $x$, $g\_c$ the growth rate of $\hat c$. In the model there is no depreciation so,

$$ g\_c = \frac 1{\theta}[r - \rho - \theta g\_x] $$

$$\dot {\hat k} = f(\hat k) - \hat c - (n+g\_x)\hat k$$

At steady state growth of $\hat k$ and $\hat c$ are both $0$. When that is the case:

$$\hat c^\* = f(\hat k^\*) - (n+g\_x)\hat k^\*$$

Also since $r = f'(k)$

$$f'(\hat k^\*) = \rho + \theta g\_x$$

Somehow based on these equations I reach the opposite conclusion, i.e. that $\hat c^\*$ and $\hat k^\*$ will increase rather than decrease if $g\_x$ is reduced.

**Tags:** macroeconomics, economic-growth

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## 1 answers

### Answer by username_1 on 2015-03-08. Score: 1 | Accepted answer

When the (exogenous) rate of technological change/efficiency is smaller, the corresponding steady-state levels of consumption and capital per unit of effective labor, *increase*.

For capital, we have

$$f'(\hat k^\*) = \rho + \theta g\_x \implies \frac {\partial}{\partial g\_x}f'(\hat k^\*) = \theta >0$$

So if $g\_x \downarrow \implies f'(\hat k^\*) \downarrow \implies \hat k^\* \uparrow$ due to decreasing marginal product of capital.

For steady-state consumption per unit of effective labor we have

$$\hat c^\* = f(\hat k^\*) - (n+g\_x)\hat k^\*$$

$$\implies \frac {\partial \hat c^\*}{\partial g\_x}= f'(\hat k^\*) \frac {\partial \hat k^\*}{\partial g\_x} - \hat k^\* - (n+g\_x) \frac {\partial \hat k^\*}{\partial g\_x}$$

$$=[f'(\hat k^\*) - n - g\_x]\cdot \frac {\partial \hat k^\*}{\partial g\_x} - \hat k^\* $$

The term in brackets is assumed positive, i.e. we have already assumed

$$f'(\hat k^\*) = \rho + \theta g\_x > n + g\_x \implies \rho > n + (1-\theta)g\_x$$

in order to exclude infinite utility.

Moreover, evidently $\frac {\partial \hat k^\*}{\partial g\_x} <0$ so in all

$$\frac {\partial \hat c^\*}{\partial g\_x} <0$$

Therefore, if $g\_x \downarrow \implies \hat c^\* \uparrow $.

See [Barro & Martin](http://down.cenet.org.cn/upfile/8/200751171644184.pdf), ch. 2 page 102, where they discuss the $g\_x \uparrow$ case (in p. 101 they discuss the constraint on the parameters).

**Comment:** This result may appear counter-intuitive, but a deeper examination of the model shows that if $g\_x$ is lower, *utility per capita* is lower. Using "consumption per unit of effective labor" is a modeling tactic, what interests us is what happens per individual. So no, the model does *not* argue in favor of lower productivity/technology.

> **Comments:**
> **username_0**: Perfect! One more thing, the result of this in the phase diagram is that the $k^$ and $C^$ curves both move in the phase diagram (C/K coordinates). What is the intuition behind only one of the curves moving vs. both of them doing so?  
> **username_1**: @Dole I would suggest to go after Barro's book, it is legally freely downloadable and it presents the whole model in many details and extensions, as well as intuitive explanations.  

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