<p>Acemoglu in his book &quot;Introduction to modern economic growth&quot; says &quot;we should have
a one-dimensional manifold (a curve) along which capital-investment pairs tend toward the
steady state. This manifold is also referred to as the stable arm&quot; your instructor asked you to make clear the old and the new stable arm so you need to draw these lines. This is my guess despite I'm not totally sure simply because I'm not a postgraduate student or a professor.</p>
<p>Anyway are you sure the Euler equation is correct? I obtained this</p>
<p><span class="math-container">$\displaystyle -f'(k)+\delta+\rho=\frac{\dot{c}}{c}\cdot\beta$</span></p>
<p>where <span class="math-container">$\beta$</span> is <span class="math-container">$\displaystyle \frac{u''\cdot c}{u'}$</span> the elasticity of marginal utility of consumption</p>
<p>P.S. I read all other contributions to the discussion @Atheneus said that my Euler equation is not correct. As a curious person I'd like to know why, because given the constraint <span class="math-container">$\displaystyle \dot{k}=Ak^{\alpha}-(\delta+\rho)k-c$</span></p>
<p>first condition is <span class="math-container">$\displaystyle u'=\mu$</span> second condition (costate variable law of motion) is <span class="math-container">$\displaystyle\frac{\dot{\mu}}{\mu}=f'(k)-\rho-\delta$</span> now if the second condition holds then also the felicity function can be derived respect to time so we can write <span class="math-container">$\displaystyle \frac{u''\dot{c}\cdot c}{u'\cdot c}=\frac{\dot{\mu}}{\mu}$</span> so in order to obtain the term <span class="math-container">$\displaystyle\frac{\dot{c}}{c}$</span> in the costate law of motion the step is the following <span class="math-container">$\displaystyle\frac{\dot{c}}{c}=\frac{u'}{u''c}\cdot(-f'(k)+\rho+\delta)$</span>. My euler equation was worng because of the <span class="math-container">$-$</span> sign but now I think is correct.</p>

CC BY-SA 4.0
2022-11-12T22:03:53.107Z
Your equation considers the general case. In the question of the OP there is a a particular utility function, $ln (c)$. This is of the family of iso-elastic utility function, in particular with $u(x)=ln (n)$ we have $\beta=1$
BakerStreet
CC BY-SA 4.0
2022-11-12T22:15:11.417Z
thank you very much @BakerStreet
Pepus
CC BY-SA 4.0
2022-11-12T22:15:45.567Z
Your observation is interesting, anyway. I gave you an upvote .
BakerStreet
CC BY-SA 4.0
2022-11-13T16:38:22.410Z
Thank you. In regard to your answer could it be said that consumption shifts to $B$ because the scenario is analogue to the first one in which in order not to remain in the steady state $(k=0)$ it is assumed that consumer start from a specific point in the unique stable arm with $k(0)>0$? @BakerStreet
Pepus
CC BY-SA 4.0
2022-11-13T16:52:42.420Z
Yes, I think it is correct to say that the scenario is analogue to the first one, the difference is a new level of $A$. The initial condition is $k_0^*$, and the consumer solves the new optimization problem setting her consumption on the new stable arm  in correspondence of $k_0^*$. Then  the system  goes, spontaneously,  to the new steady state $E_n$ .
BakerStreet
CC BY-SA 4.0
2022-11-02T18:22:56.250Z
2022-11-12T22:12:36.680Z
2022-11-12T22:12:36.680Z
Pepus
<p>It is really difficult to answer your question, because one has to guess what are the requests of your teacher in this exercise, namely if she wants just a graph, or she wants you to explain in words, or you to explain something analytically, for instance the shifts of the <span class="math-container">$\dot c=0$</span> and <span class="math-container">$\dot k=0$</span>  loci.</p>
<p>But I try to tell you my impression, and to interpret the words of your teacher, hoping it could help you.</p>
<p>I think that the equations for <span class="math-container">$\dot c$</span> and <span class="math-container">$\dot k$</span> are correct. They are the equations coming from the form of the utility function as in your image,  <span class="math-container">$u(c)=ln (c)$</span>.</p>
<p>As for the graph, the only thing that seems to me incorrect in your drawing is the ‘old’ stable arm. The arrows on the right of the point <span class="math-container">$E_0$</span>, of the arm passing through <span class="math-container">$E_0$</span>,  should be in the opposite direction, that is pointing to the left, toward <span class="math-container">$E_0$</span>.</p>
<p>In other respects, the graph seems right, <span class="math-container">$\dot c=0$</span> shifts to the right, and the <span class="math-container">$\dot k$</span>=0 locus shifts upward. The point <span class="math-container">$E_N$</span> is the new path and you showed the new stable arm passing through <span class="math-container">$E_N$</span>.</p>
<p>Consumption jumps to point <span class="math-container">$B$</span>.</p>
<p><span class="math-container">$$***$$</span></p>
<p>What seems to me unclear and also not very correct is your explanation, in words, of the fact  <span class="math-container">$c$</span> jumps upward to <span class="math-container">$B$</span>.  It is the formulation of the problem, not the graph in itself, that seems  incorrect.</p>
<p>You said:</p>
<blockquote>
<p>the ˙=0 locus shifts to right. As capital accumulates, consumption
jumps up and so, consumption rises to point B.</p>
</blockquote>
<p>This is unsatisfactory. Why just <span class="math-container">$B$</span>, for example, and not another point above <span class="math-container">$E_0$</span>?</p>
<p>You explain the jump in consumption only through an increase of saving and capital, before analyzing what happens to the <span class="math-container">$\dot k=0$</span> locus and without considering the new stable arm.</p>
<p>Instead, it should be said, before considering point <span class="math-container">$B$</span>,  that the loci <span class="math-container">$\dot c=0$</span> and <span class="math-container">$\dot k=0$</span> <strong>shift both,</strong> and we have a <strong>new stable arm</strong> passing through <span class="math-container">$E_N$</span>.</p>
<p>So, the <strong>new optimal solution</strong> of the control problem, with <span class="math-container">$A^*&gt;A$</span>, is on the new stable arm. This is the reason why  <span class="math-container">$c$</span> jumps to point <span class="math-container">$B$</span>:  the optimizing consumer sets the control variable <span class="math-container">$c(t)$</span> on the stable arm, in correspondence of the old value of <span class="math-container">$k_0$</span>, because optimal points are always on the stable arm of the steady state point.</p>
<p>At this point begins the transitional dynamics, along the new stable arm,  leading to point <span class="math-container">$E_N$</span>, where the transitional dynamics stops, and we are in the new steady state <span class="math-container">$E_N$</span>.</p>
<p>The difference with what you said is not irrelevant, because the jump in consumption is the result of the optimum  problem of the consumer:  the consumer chooses the control function <span class="math-container">$c(t)$</span> and sets <span class="math-container">$c$</span>  on the stable arm.  It is not the result, as you seemed to suggest, of an endogenous, spontaneous adjustment of the system after the technological shock.</p>
<p>For this reason, there is a 'jump' of <span class="math-container">$c$</span>, a sudden increase, and not a ‘smooth’ movement toward <span class="math-container">$B$</span>.</p>
<p><span class="math-container">$$***$$</span></p>
<p>Just another observation: The new <span class="math-container">$\dot c=0$</span> locus in the graph should be a little further to the left. This is because the <span class="math-container">$\dot c=0$</span> line cannot be on the right of the <em>golden rule</em>,  that is the point of maximum of the <span class="math-container">$\dot k=0$</span> locus: in the Ramsey-Cass-Koopman model, a balanced growth path cannot be in correspondence of a level of <span class="math-container">$k$</span> above the golden rule.</p>
<p>A  graph with a <span class="math-container">$\dot c=0$</span> line on the right of the golden rule is not ‘nice’,  it doesn’t make a good impression.</p>

CC BY-SA 4.0
2022-11-09T15:58:30.683Z
2022-11-12T15:18:51.947Z
2022-11-12T15:18:51.947Z
BakerStreet
<p>My instructor said that the answer I wrote below is not correct without any explanation. But I dont know why. I need to learn its correct answer. Please share your ideas with me.</p>
<p>I asked the same question before. But i could not start bounty. The system doesnt allow to start bounty. Thus, I re-post <a href="https://economics.stackexchange.com/questions/52119/question-on-ramsey-model-with-shock-on-technological-level">the question</a>.</p>
<p>All helps will be appreciated. Many thanks.</p>
<p>I have a question on Ramsey Model.</p>
<p>Consider the following one-sector, closed, representative household economy.</p>
<p>I have the following Cobb Douglas production function in intensive form</p>
<p><span class="math-container">$$f(k)=Ak(t)^{\alpha}$$</span></p>
<p>A is the constant technology level: There is no technological progress: Let  denote n the population growth rate. This production function displays constant returns to scale in both Capital and labor ; hence each factor is paid its marginal product.</p>
<p><span class="math-container">$\delta$</span> is depreciation rate for capital. And <span class="math-container">$p$</span> is time preference rate.</p>
<p>I found the characteristics of the equilibrium in this model as follows</p>
<p>Euler equation is</p>
<p><span class="math-container">$$\frac{\dot{c}}{c}=A\alpha k^{\alpha -1}- \delta -p$$</span></p>
<p>Resource constraint is</p>
<p><span class="math-container">$$\dot{k} = Ak^{\alpha} - (\delta +n) k - c $$</span></p>
<p>The question asks that</p>
<p>suppose at some t = 0, a permanent shock raises technological level A to <span class="math-container">$A&gt;A^*$</span>. Show the effect of this shock on the phase diagram of the model. In your answer, be sure to compare the equilibrium paths for capital stock and consumption in the economies with and without technological shock.</p>
<p><strong>My answer</strong>
The economy is initially at <span class="math-container">$E_0$</span>.</p>
<p>If <span class="math-container">$A$</span> increases, then the return to saving <span class="math-container">$A\alpha k^{\alpha -1}- \delta $</span> increases, so saving increases, this implies that capital accumulates. So the <span class="math-container">$\dot{c}=0$</span> locus shifts to right.  As capital accumulates, consumption jumps up and so, consumption rises to point B.</p>
<p>If <span class="math-container">$A$</span> increases, then the Output <span class="math-container">$f(k)$</span> is much higher than the break even investment <span class="math-container">$ (\delta +n)k$</span>, so c will be also increasing such that <span class="math-container">$\dot{k}=0$</span> holds. Then the locus <span class="math-container">$\dot{c}=0$</span> shifts up.</p>
<p>At the end, the economy reaches the new equilibrium point <span class="math-container">$E_N$</span>.</p>
<p>The graph is as follows.</p>
<p><a href="https://i.sstatic.net/ujNru.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ujNru.jpg" alt="enter image description here" /></a></p>
<p>However, my instructor gives such a feedback about my answer. And my answer is not enough for her.</p>
<blockquote>
<blockquote>
<p>you need to show the old and new stable arms and also show that at the time of the shock the c jumps up and the new path remains above the old path.</p>
</blockquote>
</blockquote>
<p><strong>How can I correct my answer according to this feedback?</strong></p>
<p>Please help me to answer this question correctly. Thank you.</p>
<p>Edit: I add the original question.
<a href="https://i.sstatic.net/Vy2Gi.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Vy2Gi.jpg" alt="enter image description here" /></a></p>
<p>Now suppose at some t = 0, a permanent shock raises technological level A to
<span class="math-container">$A^* &gt; A$</span>. Show the effects of this shock on the phase diagram of the model. Please compare the equilibrium paths for capital stock and consumption in the economies with and without the technological shock on the phase diagram.</p>

CC BY-SA 4.0
2022-11-05T12:25:49.127Z
Since you have "Hicks-neutral" technology in your production function, does it mean your intensive form is just: $k = K/L$? Please define.
Athaeneus
CC BY-SA 4.0
2022-11-05T14:23:34.837Z
You’re right! @Athaeneus what is your idea about this question?
studentp
CC BY-SA 4.0
2022-11-05T15:02:41.673Z
It's way trickier than I initially thought... :D I think you have correctly deduced Euler eq. and capital dynamics (contrary to @Pepus). I am trying to approach the problem as analogic to the Ramsey-Cass with "Harrod-neutral" technology $k = K/(AL)$, where the change in $A$ changes only the initial point. But since you have it in this form, I think the increase in $A$ should be similar to the situation of growth rate of technology decreasing... I think you have the right solution... Maybe your teacher did not read it correctly (due to "thickness" of your illustration)?
Athaeneus
<macroeconomics><economic-growth><self-study><ramsey>
Effect of shock to technology level in the Ramsey model
studentp
# Effect of shock to technology level in the Ramsey model

(Asked by: username_0 on 2022-10-31)

My instructor said that the answer I wrote below is not correct without any explanation. But I dont know why. I need to learn its correct answer. Please share your ideas with me.

I asked the same question before. But i could not start bounty. The system doesnt allow to start bounty. Thus, I re-post [the question](https://economics.stackexchange.com/questions/52119/question-on-ramsey-model-with-shock-on-technological-level).

All helps will be appreciated. Many thanks.

I have a question on Ramsey Model.

Consider the following one-sector, closed, representative household economy.

I have the following Cobb Douglas production function in intensive form

$$f(k)=Ak(t)^{\alpha}$$

A is the constant technology level: There is no technological progress: Let denote n the population growth rate. This production function displays constant returns to scale in both Capital and labor ; hence each factor is paid its marginal product.

$\delta$ is depreciation rate for capital. And $p$ is time preference rate.

I found the characteristics of the equilibrium in this model as follows

Euler equation is

$$\frac{\dot{c}}{c}=A\alpha k^{\alpha -1}- \delta -p$$

Resource constraint is

$$\dot{k} = Ak^{\alpha} - (\delta +n) k - c $$

The question asks that

suppose at some t = 0, a permanent shock raises technological level A to $A>A^\*$. Show the effect of this shock on the phase diagram of the model. In your answer, be sure to compare the equilibrium paths for capital stock and consumption in the economies with and without technological shock.

**My answer**
The economy is initially at $E\_0$.

If $A$ increases, then the return to saving $A\alpha k^{\alpha -1}- \delta $ increases, so saving increases, this implies that capital accumulates. So the $\dot{c}=0$ locus shifts to right. As capital accumulates, consumption jumps up and so, consumption rises to point B.

If $A$ increases, then the Output $f(k)$ is much higher than the break even investment $ (\delta +n)k$, so c will be also increasing such that $\dot{k}=0$ holds. Then the locus $\dot{c}=0$ shifts up.

At the end, the economy reaches the new equilibrium point $E\_N$.

The graph is as follows.

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

However, my instructor gives such a feedback about my answer. And my answer is not enough for her.

> > you need to show the old and new stable arms and also show that at the time of the shock the c jumps up and the new path remains above the old path.

**How can I correct my answer according to this feedback?**

Please help me to answer this question correctly. Thank you.

Edit: I add the original question.
[![enter image description here](https://i.sstatic.net/Vy2Gi.jpg)](https://i.sstatic.net/Vy2Gi.jpg)

Now suppose at some t = 0, a permanent shock raises technological level A to
$A^\* > A$. Show the effects of this shock on the phase diagram of the model. Please compare the equilibrium paths for capital stock and consumption in the economies with and without the technological shock on the phase diagram.

**Tags:** macroeconomics, economic-growth, self-study, ramsey
> **Comments:**
> **username_1**: Since you have "Hicks-neutral" technology in your production function, does it mean your intensive form is just: $k = K/L$? Please define.  
> **username_0**: You’re right! @Athaeneus what is your idea about this question?  
> **username_1**: It's way trickier than I initially thought... :D I think you have correctly deduced Euler eq. and capital dynamics (contrary to @Pepus). I am trying to approach the problem as analogic to the Ramsey-Cass with "Harrod-neutral" technology $k = K/(AL)$, where the change in $A$ changes only the initial point. But since you have it in this form, I think the increase in $A$ should be similar to the situation of growth rate of technology decreasing... I think you have the right solution... Maybe your teacher did not read it correctly (due to "thickness" of your illustration)?  

---

## 2 answers

### Answer by username_3 on 2022-11-09. Score: 2 | Accepted answer

It is really difficult to answer your question, because one has to guess what are the requests of your teacher in this exercise, namely if she wants just a graph, or she wants you to explain in words, or you to explain something analytically, for instance the shifts of the $\dot c=0$ and $\dot k=0$ loci.

But I try to tell you my impression, and to interpret the words of your teacher, hoping it could help you.

I think that the equations for $\dot c$ and $\dot k$ are correct. They are the equations coming from the form of the utility function as in your image, $u(c)=ln (c)$.

As for the graph, the only thing that seems to me incorrect in your drawing is the ‘old’ stable arm. The arrows on the right of the point $E\_0$, of the arm passing through $E\_0$, should be in the opposite direction, that is pointing to the left, toward $E\_0$.

In other respects, the graph seems right, $\dot c=0$ shifts to the right, and the $\dot k$=0 locus shifts upward. The point $E\_N$ is the new path and you showed the new stable arm passing through $E\_N$.

Consumption jumps to point $B$.

$$\*\*\*$$

What seems to me unclear and also not very correct is your explanation, in words, of the fact $c$ jumps upward to $B$. It is the formulation of the problem, not the graph in itself, that seems incorrect.

You said:

> the ˙=0 locus shifts to right. As capital accumulates, consumption
> jumps up and so, consumption rises to point B.

This is unsatisfactory. Why just $B$, for example, and not another point above $E\_0$?

You explain the jump in consumption only through an increase of saving and capital, before analyzing what happens to the $\dot k=0$ locus and without considering the new stable arm.

Instead, it should be said, before considering point $B$, that the loci $\dot c=0$ and $\dot k=0$ **shift both,** and we have a **new stable arm** passing through $E\_N$.

So, the **new optimal solution** of the control problem, with $A^\*>A$, is on the new stable arm. This is the reason why $c$ jumps to point $B$: the optimizing consumer sets the control variable $c(t)$ on the stable arm, in correspondence of the old value of $k\_0$, because optimal points are always on the stable arm of the steady state point.

At this point begins the transitional dynamics, along the new stable arm, leading to point $E\_N$, where the transitional dynamics stops, and we are in the new steady state $E\_N$.

The difference with what you said is not irrelevant, because the jump in consumption is the result of the optimum problem of the consumer: the consumer chooses the control function $c(t)$ and sets $c$ on the stable arm. It is not the result, as you seemed to suggest, of an endogenous, spontaneous adjustment of the system after the technological shock.

For this reason, there is a 'jump' of $c$, a sudden increase, and not a ‘smooth’ movement toward $B$.

$$\*\*\*$$

Just another observation: The new $\dot c=0$ locus in the graph should be a little further to the left. This is because the $\dot c=0$ line cannot be on the right of the *golden rule*, that is the point of maximum of the $\dot k=0$ locus: in the Ramsey-Cass-Koopman model, a balanced growth path cannot be in correspondence of a level of $k$ above the golden rule.

A graph with a $\dot c=0$ line on the right of the golden rule is not ‘nice’, it doesn’t make a good impression.


---

### Answer by username_2 on 2022-11-02. Score: 1

Acemoglu in his book "Introduction to modern economic growth" says "we should have
a one-dimensional manifold (a curve) along which capital-investment pairs tend toward the
steady state. This manifold is also referred to as the stable arm" your instructor asked you to make clear the old and the new stable arm so you need to draw these lines. This is my guess despite I'm not totally sure simply because I'm not a postgraduate student or a professor.

Anyway are you sure the Euler equation is correct? I obtained this

$\displaystyle -f'(k)+\delta+\rho=\frac{\dot{c}}{c}\cdot\beta$

where $\beta$ is $\displaystyle \frac{u''\cdot c}{u'}$ the elasticity of marginal utility of consumption

P.S. I read all other contributions to the discussion @Atheneus said that my Euler equation is not correct. As a curious person I'd like to know why, because given the constraint $\displaystyle \dot{k}=Ak^{\alpha}-(\delta+\rho)k-c$

first condition is $\displaystyle u'=\mu$ second condition (costate variable law of motion) is $\displaystyle\frac{\dot{\mu}}{\mu}=f'(k)-\rho-\delta$ now if the second condition holds then also the felicity function can be derived respect to time so we can write $\displaystyle \frac{u''\dot{c}\cdot c}{u'\cdot c}=\frac{\dot{\mu}}{\mu}$ so in order to obtain the term $\displaystyle\frac{\dot{c}}{c}$ in the costate law of motion the step is the following $\displaystyle\frac{\dot{c}}{c}=\frac{u'}{u''c}\cdot(-f'(k)+\rho+\delta)$. My euler equation was worng because of the $-$ sign but now I think is correct.

> **Comments:**
> **username_3**: Your equation considers the general case. In the question of the OP there is a a particular utility function, $ln (c)$. This is of the family of iso-elastic utility function, in particular with $u(x)=ln (n)$ we have $\beta=1$  
> **username_2**: thank you very much @BakerStreet  
> **username_3**: Your observation is interesting, anyway. I gave you an upvote .  
> **username_2**: Thank you. In regard to your answer could it be said that consumption shifts to $B$ because the scenario is analogue to the first one in which in order not to remain in the steady state $(k=0)$ it is assumed that consumer start from a specific point in the unique stable arm with $k(0)>0$? @BakerStreet  
> **username_3**: Yes, I think it is correct to say that the scenario is analogue to the first one, the difference is a new level of $A$. The initial condition is $k_0^*$, and the consumer solves the new optimization problem setting her consumption on the new stable arm  in correspondence of $k_0^*$. Then  the system  goes, spontaneously,  to the new steady state $E_n$ .  

---

