16 votes
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Solow Model: Steady State v Balanced Growth Path

This is when the attempt at accuracy creates confusion and misunderstanding. Back in the day, growth models were not incorporating technological progress, and led to a long-run equilibrium ...
Alecos Papadopoulos's user avatar
10 votes
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"If $\lambda$ is greater than than 1, the system explodes." Why does the system explode?

Eq. (2.64) can be written as (at first order) $$ k_{t+1} - k^* = \lambda (k_t - k^*) \tag{1a} $$ Define the quantity $\kappa_t$ as $$ \kappa_t \stackrel{\rm def}{=} k_t - k^* \tag{2} $$ So that ...
caverac's user avatar
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5 votes

"If $\lambda$ is greater than than 1, the system explodes." Why does the system explode?

A system is explosive if its coefficients are non-stationary. Stationary is an important property to have in dynamic models as it tells us that an equilibrium value is obtainable (which is important ...
EconJohn's user avatar
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5 votes

Solow Model: Steady State v Balanced Growth Path

Following the conversation with user @denesp at the comments of my previous answer, I have to clarify the following: the usual graphical device we use related to the basic Solow growth model (see for ...
Alecos Papadopoulos's user avatar
5 votes
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Prove the uniqueness of steady state

Rearranging the steady state equation $$ \overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \alpha\overline{p}^{\alpha}-\frac{a+1}{\sigma} $$ we get $$ (1 + \alpha)\overline{p}^{\alpha}=\alpha y\...
Giskard's user avatar
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4 votes
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Solow Growth Model. Steady State. Can someone explain please?

An axis is characterized not by the variable it measures, but by the unit of measurement. So an axis can measure any number of variables, as long as they are measured in the same measurement unit. In ...
Alecos Papadopoulos's user avatar
4 votes

An Optimal Control Model: A Ridiculous Result for a Steady State

One general issue I see is that you try to include uncertainty in a framework developed for a deterministic setup. What you do is to use expected income in the equation of motion for human capital. ...
Alecos Papadopoulos's user avatar
4 votes
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Solution Method for Infinite-Horizon Maximization Problem

Your first question (regarding constraints on the parameters) can be answered through first and second derivative analysis. In order to satisfy strictly increasing, we need $u'>0$ and to satisfy ...
DornerA's user avatar
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4 votes

Solow model with population growth - proof of steady state level of capital per worker

Part 1 of the solution The fundamental equation of Solow model is (neglecting the $t$ subscripts): $$ \Delta k= sf(k) -(n+d),$$ where $k= K/N$, $\Delta k= k_{t+1}-k_t$ and $f(k)$ is the intensive ...
BakerStreet's user avatar
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3 votes
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Constant to the power of t in steady state

If the variables are constant, everything in the equation is time-invariant, while $a^t$ will still grow or fall over time (unless $a=1$). This is a contradiction and no steady state exists, unless $a=...
jpfeifer's user avatar
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3 votes
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Log linearising EUler equation

As you say the first step is to take log of both sides after that you are just applying the rules for logarithms and rearrange. For example: $$\ln (XZ)=\ln X + \ln Z$$ $$\ln X/Z= \ln X - \ln Z$$ $$\...
1muflon1's user avatar
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3 votes
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Steady state equilibrium in Solow model with a convex production function

You only provide partial information. E.g., this production function is unusual; is anything else unusual? Is depreciation still linear in $k$? Is the rate of population growth constant? etc. If ...
Giskard's user avatar
  • 28.6k
3 votes

Steady-state savings rate

$\delta = 0.02$ is depreciation. $p = 0.02$ is population growth. $g = 0.03$ is technological growth. $s = 0.14$ is the savings rate. $Y=0.5\cdot K^{\frac{1}{3}}\left(AN\right)^{\frac{2}{3}}$ is the ...
BKay's user avatar
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3 votes
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Derive the Demographic Structure in the Steady State

I don't know the paper nor the notation, so I am just guessing here. I gues $N(a,t)$ is the number of agents of age $a$ at time period $t$. Let's follow the number of age $B$ accross generations: $$ \...
tdm's user avatar
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3 votes

Marginal Product of Capital in the Solow Model

According to your calculations MPK is not increasing in $K$. The Solow model assumes $0< \alpha < 1$, thus $\alpha - 1 < 0$ and $K^{\alpha - 1}$ is decreasing in $K$.
Giskard's user avatar
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3 votes
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Solow Model, Growth rate of K/L and Y/L in steady state

If $Y = C \cdot X$ where $C$ is constant and $\frac{\dot{X}}{X} = g$ then we can solve for $\frac{\dot{Y}}{Y}$ as follows: $$ \frac{d}{dt} Y = \frac{d}{dt} C \cdot X = C \cdot \frac{dX}{dt} = C \cdot \...
BKay's user avatar
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3 votes

Solution Method for Infinite-Horizon Maximization Problem

Your first question, if it's literally correct, is easy: The only way for $u'$ to be positive for c=0 is for p=1. if p =1 then sign($\phi$)=sign($\theta$) so that the product is positive. But, since $...
Fix.B.'s user avatar
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3 votes

The effect of saving rate on steady state

I suppose you are speaking of the standard Solow Growth model. Yes, in this case, the economic system will come back to the original steady state equilibrium. This is a consequence of the stability of ...
BakerStreet's user avatar
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3 votes
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Log-linearizing a second order term around the steady-state

If $\Pi_t$ is gross inflation then indeed $(\Pi_t-1)^2$ is a second-order term and is approximately zero. For example, for a reasonable quarterly steady state gross inflation rate of 1.005, the term ...
BrsG's user avatar
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2 votes

Taylor Series Approximation around steady state in Solow

The steady-state value $k^*$ must be a fixed point : $(1+g)(1+n)k^* = s(k^*)^{\alpha} +(1 - \delta) k^*$ Taking the difference between this equation and the dynamic one : $(1+g)(1+n)(k_{t+1} - k^*) ...
user11629's user avatar
  • 170
2 votes

Solow model, time and steady state

In the model with technological progress the capital per effective worker remains constant, implies that capital per worker grows at the rate of exogenous rate of technological progress. See Barro and ...
Kshitiz Mishra's user avatar
2 votes
Accepted

Evaluation around steady state for a specific DSGE Model

I found the error in my derivation: I mistakenly supposet that the steady state of $p_t+1$ equals $\bar{\rho}$. I did not recognize that p was different from $\rho$ because of the poor quality of my ...
Ralle Kalle's user avatar
2 votes

What does steady state mean?

Usually the term steady state is derived from the Solow Model and its derivatives that seek to explain long-term economic growth. The steady state is a state in which the growth rate of the economy is ...
E. Sommer's user avatar
  • 1,307
2 votes
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Non-trivial steady state

Let's guess that the value function is of the form $a + b \ln(k)$. Then substituting for $V(k) = a + b \ln(k)$ in the Bellman equation gives: $$ a + b \ln(k) = \max_{k'}\left(\ln(k^\alpha - k') + \...
tdm's user avatar
  • 10.3k
2 votes

Would a zero-growth economy with zero-growth population still have the same GDP?

Smith pointed out that as wealth was growing in any nation, the rate of profit would tend to fall and investment opportunities would diminish. Source The first half of the sentence makes it ...
Giskard's user avatar
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2 votes
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Solow model with population growth - proof of steady state level of capital per worker

Part 2 of the solution I post a step by step solution, slow passages are the best way to avoid mistakes (I hope). In the previuos answer we calculated the intensive production function: $$f(k)=z[\...
BakerStreet's user avatar
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2 votes
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Capital-Output Ratio using Nominal GDP and Nominal GFCF

Dividing GFCF by GDP is a standard way to approximate $K/Y$. Also, if I am not mistaken, K/Y = (s / (g + δ)) only holds in steady state when $K/Y$ is constant. In real life economies are typically not ...
csilvia's user avatar
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1 vote
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Question on overlapping generations

Is that derivation for $k(t+1)$ correct? Technically, you never reach the steady state, but only asimptotically as $t\rightarrow\infty$, but at infinity the $A(t)$ will also be infinite because it ...
Pekisch's user avatar
  • 130
1 vote
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Difference between long run coefficient and non stochastic steady state coefficient ARDL model

yes, the term that you showed for the ALDR non-stochastic steady state: $$\frac{ \beta_1 + \beta_2 }{1- \rho_1 -\rho_2}$$ is long-run multiplier or sometimes also called long run equilibrium ...
1muflon1's user avatar
  • 54.4k

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