# Effect of shock to technology level in the Ramsey model

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.

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

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

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. 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?

Edit: I add the original question. 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.

• Since you have "Hicks-neutral" technology in your production function, does it mean your intensive form is just: $k = K/L$? Please define. Nov 5, 2022 at 12:25
• 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)? Nov 5, 2022 at 15:02

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 𝑐˙=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.

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.

• 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 $\beta=1$ Nov 12, 2022 at 22:03
• thank you very much @BakerStreet Nov 12, 2022 at 22:15
• Your observation is interesting, anyway. I gave you an upvote . Nov 12, 2022 at 22:15
• 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 Nov 13, 2022 at 16:38
• 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 $E_n$ . Nov 13, 2022 at 16:52