<p>The direction of <span class="math-container">$\frac{\partial c^{*}}{\partial n}$</span> is not ambiguous.</p>
<p>An easy way to show this is taking derivative of <span class="math-container">$c^*=(1-s)f(k^*)$</span> so that <span class="math-container">$\frac{\partial c^{*}}{\partial n}=(1-s)f'\frac{\partial k^{*}}{\partial n}$</span> and because <span class="math-container">$f'&gt;0$</span> and we can prove <span class="math-container">$\frac{\partial k^{*}}{\partial n}&lt;0$</span> we thus have <span class="math-container">$\frac{\partial c^{*}}{\partial n}&lt;0$</span>.</p>
<p>Through this you can also easily see that <span class="math-container">$c^{*}$</span> will be monotone in <span class="math-container">$n$</span> and <span class="math-container">$\delta$</span> but it will not be monotone in <span class="math-container">$s$</span>.</p>
<p>It is somehow harder to prove this from <span class="math-container">$\frac{\partial c^{*}}{\partial n}=\left[f^{\prime}\left(k^{*}\right)-(n+\delta)\right] \frac{\partial k^{*}}{\partial n}-k^{*}$</span></p>

CC BY-SA 4.0
2021-09-29T11:36:09.693Z
Thank you, this makes perfect sense. For record: The sign in the square term of the bracket is actually not ambiguous. However, I notice that the response function $\frac{\partial c^*}{\partial s} = [f'(k^*) - (n + \delta)]\frac{\partial k^*}{\partial s}$. If we were to reason the same way, wouldn't steady state consumption be ever-increasing in savings, and so no such thing as the golden rule maximum?
Panhaboth K
CC BY-SA 4.0
2021-09-29T13:40:54.800Z
@PanhabothK Sorry, you are correct. I made a mistake in my initial answer. I find it is harder than I thought to prove the sign from the equation you are dealing with. But I show that you can easily prove this in an alternative way. And it helps you see the difference between $n$ and $s$.
Alalalalaki
CC BY-SA 4.0
2021-09-29T11:17:39.477Z
2021-09-29T13:47:06.330Z
2021-09-29T13:47:06.330Z
Alalalalaki
<p>Consider a Solow model without technological progress so that the steady state <span class="math-container">$k^*$</span> occurs at <span class="math-container">$sf(k^*) = (n+\delta)k^*$</span> where <span class="math-container">$n$</span> is population growth rate, <span class="math-container">$\delta$</span> is capital depreciation rate and <span class="math-container">$s$</span> is the proportion of output that is saved and invested. Given that consumption per worker is <span class="math-container">$c = f(k) - sf(k)$</span>, how does steady state consumption change if <span class="math-container">$n$</span> decreases?</p>
<p>Intuitively, output per worker rises so <span class="math-container">$c = (1-s)f(k)$</span> should also rise. However, if I try to solve for the change numerically I would do:</p>
<p><span class="math-container">$c^* = f(k^*(s, n, \delta)) - (n+\delta)k^*(s, n, \delta)$</span> at the steady state.</p>
<p>So, <span class="math-container">$\frac{\partial c^*}{\partial n} = f'(k^*)\frac{\partial k^*}{\partial n} - k^* -(n + \delta)\frac{\partial k^*}{\partial n}$</span>.</p>
<p>Re-arranging:</p>
<p><span class="math-container">$\frac{\partial c^*}{\partial n} = [f'(k^*) - (n+\delta)]\frac{\partial k^*}{\partial n} - k^*$</span>.</p>
<p>We know <span class="math-container">$\frac{\partial k^*}{\partial n}$</span> &lt; 0 since a higher slope of the breakeven investment causes it to intersect with <span class="math-container">$sf(k^*)$</span> at a lower <span class="math-container">$k^*$</span>. The sign of the term in the square bracket is ambiguous so <span class="math-container">$\frac{\partial c^*}{\partial n}$</span> itself is ambiguous.</p>
<p>There seems to be a contradiction between my intuition and the numerical analysis, so where have I gone wrong?</p>

<macroeconomics><economic-growth><solow><balanced-growth>
Solow model response function
Panhaboth K
# Solow model response function

(Asked by: username_0 on 2021-09-29)

Consider a Solow model without technological progress so that the steady state $k^\*$ occurs at $sf(k^\*) = (n+\delta)k^\*$ where $n$ is population growth rate, $\delta$ is capital depreciation rate and $s$ is the proportion of output that is saved and invested. Given that consumption per worker is $c = f(k) - sf(k)$, how does steady state consumption change if $n$ decreases?

Intuitively, output per worker rises so $c = (1-s)f(k)$ should also rise. However, if I try to solve for the change numerically I would do:

$c^\* = f(k^\*(s, n, \delta)) - (n+\delta)k^\*(s, n, \delta)$ at the steady state.

So, $\frac{\partial c^\*}{\partial n} = f'(k^\*)\frac{\partial k^\*}{\partial n} - k^\* -(n + \delta)\frac{\partial k^\*}{\partial n}$.

Re-arranging:

$\frac{\partial c^\*}{\partial n} = [f'(k^\*) - (n+\delta)]\frac{\partial k^\*}{\partial n} - k^\*$.

We know $\frac{\partial k^\*}{\partial n}$ < 0 since a higher slope of the breakeven investment causes it to intersect with $sf(k^\*)$ at a lower $k^\*$. The sign of the term in the square bracket is ambiguous so $\frac{\partial c^\*}{\partial n}$ itself is ambiguous.

There seems to be a contradiction between my intuition and the numerical analysis, so where have I gone wrong?

**Tags:** macroeconomics, economic-growth, solow, balanced-growth


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## 1 answers

### Answer by username_1 on 2021-09-29. Score: 1

The direction of $\frac{\partial c^{\*}}{\partial n}$ is not ambiguous.

An easy way to show this is taking derivative of $c^\*=(1-s)f(k^\*)$ so that $\frac{\partial c^{\*}}{\partial n}=(1-s)f'\frac{\partial k^{\*}}{\partial n}$ and because $f'>0$ and we can prove $\frac{\partial k^{\*}}{\partial n}<0$ we thus have $\frac{\partial c^{\*}}{\partial n}<0$.

Through this you can also easily see that $c^{\*}$ will be monotone in $n$ and $\delta$ but it will not be monotone in $s$.

It is somehow harder to prove this from $\frac{\partial c^{\*}}{\partial n}=\left[f^{\prime}\left(k^{\*}\right)-(n+\delta)\right] \frac{\partial k^{\*}}{\partial n}-k^{\*}$

> **Comments:**
> **username_0**: Thank you, this makes perfect sense. For record: The sign in the square term of the bracket is actually not ambiguous. However, I notice that the response function $\frac{\partial c^*}{\partial s} = [f'(k^*) - (n + \delta)]\frac{\partial k^*}{\partial s}$. If we were to reason the same way, wouldn't steady state consumption be ever-increasing in savings, and so no such thing as the golden rule maximum?  
> **username_1**: @username_0 Sorry, you are correct. I made a mistake in my initial answer. I find it is harder than I thought to prove the sign from the equation you are dealing with. But I show that you can easily prove this in an alternative way. And it helps you see the difference between $n$ and $s$.  

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