<ol>
<li><p><span class="math-container">$n$</span> has no effect on <span class="math-container">$K/L$</span> ratio simply because <span class="math-container">$n$</span> does not show up in the Euler. This is however, dependent on how your write the objective function. In your formulation, you have <span class="math-container">$e^{-\rho t}U(c)L$</span> under the integral.</p>
</li>
<li><p>You can (almost) avoid derivatives. Diminishing marginal product of <span class="math-container">$f(k)$</span> implies the following:</p>
</li>
</ol>
<p>When <span class="math-container">$\rho$</span> increases, <span class="math-container">$f’(k^*)$</span> rises, which then must be consistent with lower <span class="math-container">$k^*$</span>.</p>

CC BY-SA 4.0
2021-12-03T08:19:06.607Z
It might be constructive to say why you (the downvoter), think the answer is not good. :)
erik
CC BY-SA 4.0
2021-12-02T19:50:50.260Z
2021-12-02T19:50:50.260Z
erik
<p>In Acemoglu's book &quot;Introduction to Modern Economic Growth&quot; there is Proposition 8.3 which states that the impact of a change in the population growth rate on the steady state capital-labour ratio is zero. However, I was not able to prove this, which is asked in Exercise 8.17 as well. It makes more sense that this effect is negative. Can anyone offer an explanation to why this effect should be zero?</p>
<p><a href="https://i.sstatic.net/YbPmE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YbPmE.png" alt="enter image description here" /></a></p>
<p>Edit:</p>
<p>Adding my attempt:</p>
<p>We have two differential equations:</p>
<p><span class="math-container">$\frac{\dot c_t}{c_t}=\frac{1}{\epsilon_u}(A\tilde f'(k(t))-\rho-\delta)$</span></p>
<p><span class="math-container">$\dot k_t = f(k(t)) -(n+\delta)k(t) -c(t)$</span></p>
<p>For the steady state <span class="math-container">$\frac{\dot c_t}{c_t}=0$</span>, since <span class="math-container">$\epsilon_u&lt;0$</span> it has to be the case that <span class="math-container">$\tilde f'(k^*)=\frac{\rho+\delta}{A}$</span>. Then,</p>
<p><span class="math-container">$k^* = \tilde f'^{-1}(\frac{\rho+\delta}{A})$</span> .</p>
<p>Since <span class="math-container">$\tilde f(k(t))$</span> is concave in <span class="math-container">$k(t)$</span> then <span class="math-container">$\tilde f'(k(t))$</span> is strictly decreasing. Hence by the inverse function theorem, we have the result that <span class="math-container">$\frac{\partial k^{*}}{\partial A} &gt;0,\frac{\partial k^{*}}{\partial \rho} &lt;0$</span> and <span class="math-container">$\frac{\partial k^{*}}{\partial \delta} &lt; 0$</span>. Moreover <span class="math-container">$\frac{\partial k^{*}}{\partial n} = 0$</span> since <span class="math-container">$n$</span> is not an argument of <span class="math-container">$k^{*}$</span>.</p>
<p>So I am confused here because then I do not understand why Acemoglu wrote <span class="math-container">$k^* = k^*(A,\rho,n,\delta)$</span> if <span class="math-container">$n$</span> is not an argument of <span class="math-container">$k^{*}$</span>.</p>
<p>Furthermore we have to show that <span class="math-container">$\frac{\partial c^{*}}{\partial n}&lt;0$</span>. For this I use the law motion of capital and set it equal to zero and re-arrange in terms of <span class="math-container">$c^*$</span>,</p>
<p><span class="math-container">$c^* = f(k^*) -(n+\delta)k^*$</span>.</p>
<p>Differentiating w.r.t <span class="math-container">$n$</span>,</p>
<p><span class="math-container">$\frac{\partial c^*}{\partial n}= f'(k^*)\frac{\partial k^*}{\partial n} -k^* - n\frac{\partial k^*}{\partial n}$</span></p>
<p>If <span class="math-container">$\frac{\partial k^*}{\partial n}=0$</span> then the above reduces to,</p>
<p><span class="math-container">$\frac{\partial c^*}{\partial n}=  -k^* &lt;0$</span>.</p>
<p>So my question would be, am I correct in using the inverse function theorem? Are my derivations accurate?</p>

CC BY-SA 4.0
2021-12-02T10:38:48.143Z
Sure, will edit asap with what I have tried so far. This is not a homework though, it is a Proposition in the book with the proof postulated as an exercise. Fairly typical for such books but quite annoying if the proposition is not typical.
Denis Dinand
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2021-12-02T10:56:03.353Z
@DenisDinand the policy applies to both homework and self study
1muflon1
CC BY-SA 4.0
2021-12-02T11:21:17.367Z
Thank you for the feedback, I will adapt the question and my future comments.
Denis Dinand
CC BY-SA 4.0
2021-12-02T12:46:08.063Z
Dear contributors, could you check my edited submission? Is it better in terms of the policies?
Denis Dinand
<macroeconomics><mathematical-economics><economic-growth>
Comparative statics of the Ramsey model, the effect of a change in population growth on the steady state capital-labor ratio
Denis Dinand
# Comparative statics of the Ramsey model, the effect of a change in population growth on the steady state capital-labor ratio

(Asked by: username_0 on 2021-12-02)

In Acemoglu's book "Introduction to Modern Economic Growth" there is Proposition 8.3 which states that the impact of a change in the population growth rate on the steady state capital-labour ratio is zero. However, I was not able to prove this, which is asked in Exercise 8.17 as well. It makes more sense that this effect is negative. Can anyone offer an explanation to why this effect should be zero?

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

Edit:

Adding my attempt:

We have two differential equations:

$\frac{\dot c\_t}{c\_t}=\frac{1}{\epsilon\_u}(A\tilde f'(k(t))-\rho-\delta)$

$\dot k\_t = f(k(t)) -(n+\delta)k(t) -c(t)$

For the steady state $\frac{\dot c\_t}{c\_t}=0$, since $\epsilon\_u<0$ it has to be the case that $\tilde f'(k^\*)=\frac{\rho+\delta}{A}$. Then,

$k^\* = \tilde f'^{-1}(\frac{\rho+\delta}{A})$ .

Since $\tilde f(k(t))$ is concave in $k(t)$ then $\tilde f'(k(t))$ is strictly decreasing. Hence by the inverse function theorem, we have the result that $\frac{\partial k^{\*}}{\partial A} >0,\frac{\partial k^{\*}}{\partial \rho} <0$ and $\frac{\partial k^{\*}}{\partial \delta} < 0$. Moreover $\frac{\partial k^{\*}}{\partial n} = 0$ since $n$ is not an argument of $k^{\*}$.

So I am confused here because then I do not understand why Acemoglu wrote $k^\* = k^\*(A,\rho,n,\delta)$ if $n$ is not an argument of $k^{\*}$.

Furthermore we have to show that $\frac{\partial c^{\*}}{\partial n}<0$. For this I use the law motion of capital and set it equal to zero and re-arrange in terms of $c^\*$,

$c^\* = f(k^\*) -(n+\delta)k^\*$.

Differentiating w.r.t $n$,

$\frac{\partial c^\*}{\partial n}= f'(k^\*)\frac{\partial k^\*}{\partial n} -k^\* - n\frac{\partial k^\*}{\partial n}$

If $\frac{\partial k^\*}{\partial n}=0$ then the above reduces to,

$\frac{\partial c^\*}{\partial n}= -k^\* <0$.

So my question would be, am I correct in using the inverse function theorem? Are my derivations accurate?

**Tags:** macroeconomics, mathematical-economics, economic-growth
> **Comments:**
> **username_0**: Sure, will edit asap with what I have tried so far. This is not a homework though, it is a Proposition in the book with the proof postulated as an exercise. Fairly typical for such books but quite annoying if the proposition is not typical.  
> **username_1**: @DenisDinand the policy applies to both homework and self study  
> **username_0**: Thank you for the feedback, I will adapt the question and my future comments.  
> **username_0**: Dear contributors, could you check my edited submission? Is it better in terms of the policies?  

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## 1 answers

### Answer by username_2 on 2021-12-02. Score: -1

1. $n$ has no effect on $K/L$ ratio simply because $n$ does not show up in the Euler. This is however, dependent on how your write the objective function. In your formulation, you have $e^{-\rho t}U(c)L$ under the integral.
2. You can (almost) avoid derivatives. Diminishing marginal product of $f(k)$ implies the following:

When $\rho$ increases, $f’(k^\*)$ rises, which then must be consistent with lower $k^\*$.

> **Comments:**
> **username_2**: It might be constructive to say why you (the downvoter), think the answer is not good. :)  

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