14

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 characterized by constant per capita magnitudes. Verbally, the term "steady-state" seemed appropriate to describe such a situation. Then Romer and endogenous growth ...


9

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 $$ \kappa_{t+1} = \lambda \kappa_t \tag{1b} $$ This can be solved very easy by realizing that \begin{eqnarray} \kappa_1 &=&\lambda \kappa_0 \\ \...


9

There is quite a bit of work being done in that area. One very recent example is Straub and Werning's working paper "Positive Long Run Capital Taxation: Chamley-Juff Revisited." The point seems to be that we need to consider the rate of convergence to the steady state. Also, there is other literature that gives some competing results (e.g., "the new ...


6

Ljungqvist and Sargent (2004). Recursive macroeconomic theory 2n ed. (ch. 15) present and review the issue. In the Concluding Remarks section, they mention two environments, where the "zero-optimal-capital tax rate" does not hold: Aiyagari(1995) presents a model with heterogeneous agents, incomplete insurance markets and borrowing constraints (i.e. a "...


4

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 example here, figure 2) is not a phase diagram, since reasonably we call "phase diagrams" those that contain zero-change loci, identify the crossing points of ...


4

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. Let $I_{a,t}$ denote the indicator function for attack, taking the value $1$ when there is an attack, and the value $0$ when there isn't. Then, properly, $$\dot{...


4

Rearranging the steady state equation $$ \overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \alpha\overline{p}^{\alpha}-\frac{a+1}{\sigma} $$ we get $$ (1 + \alpha)\overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \frac{a+1}{\sigma}. $$ As $\alpha \in [0,1]$, the left hand side of the equation is increasing in $\overline{p}$ and the right hand side ...


4

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 in finding the BGP). If your system is explosive no equilibrium (and BGP) exists. In your example where you have the equality: $$k_t-k^* \simeq \lambda^t (k_0-...


4

A system is said to be in steady state if certain variables do not change over time (and where "certain variables" depend on the context and ought to have been clearly specified by the writer). Say we define a system to be in steady state if inflation does not change over time. Then if inflation is constant at 2%, we say that the system is in steady state ...


3

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: $$\ln (XZ)=\ln X + \ln Z$$ $$\ln X/Z= \ln X - \ln Z$$ $$\ln X^a = a \ln X$$ $$\ln 1 = 0$$ Also an important approximations that hold close to zero are applied here as well these are: $\ln(1+x) \approx x $ for $x$ ...


3

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 your case, the unit of measurement is "units of output", and both $(n+\delta)k$ and $sf(k)$ are measured in it (capital is measured in units of output, in case ...


3

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 strictly concave, we need $u''<0$. What does this actually mean? $$u'(\frac{}{})=\phi p\theta c_t^{p-1}e^{\theta c_t^p}>0$$ Since we know that $c_t^{p-1}e^{...


3

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 $K$.


3

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: $$ \frac{d}{dt} Y = \frac{d}{dt} C \cdot X = C \cdot \frac{dX}{dt} = C \cdot \dot{X} \Rightarrow$$ $$ \frac{\dot{Y}}{Y} = \frac{C \cdot \dot{X}}{ C \cdot X} = \frac{\dot{X}}{X} = g$$ Therefore, as you concluded, they both grow at rate $...


3

I'm not sure I follow the logic on that equation having infinitely many solutions and steady states. In any case, in what follows are some guidelines for equilibrium selection. It depends a lot on the context. Here are some criteria, in descending order of relevance Clean Ways Is any of the steady states unstable? If so, it's less likely to be the one ...


2

If you have are trying to discretize the continuous time model $$ \dot{\textbf{x}} = A\textbf{x}, $$ then in discrete time you will have $$ \textbf{x}_{t+1} = B \textbf{x}_t $$ but $A\neq B$, since $A$ describes the change in $\textbf{x}$ while $B$ describes the next value of $\textbf{x}$, not just the change. However $$ \Delta\textbf{x}_t = \textbf{x}_{t+1} ...


2

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 : $(1+g)(1+n)(k_{t+1} - k^*) = s((k_{t+1})^{\alpha} - (k^*)^{\alpha}) +(1 - \delta) (k_t - k^*)$ Now if you denote by $d_t = k_t - k^*$ the distance to steady-state, this gives : $(1+g)(...


2

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 Martin book, Chapter 1.


2

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 $exp()>0$ and $p=1$, the first term of $u''$ cancels out and the only way for u'' to be negative is for $\phi$ to be negative. Therefore $\theta<0$ too. ...


2

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 clear that this is not about steady state economies, as wealth is still growing. my question can be rephrased in a simpler way: If the economy was/became a ...


1

Start by writing the Lagrangian (as it sounds like you have already done): $$L=\sum_{t=0}^{\infty} \{\beta^t(\ln C_t^y + \ln C_t^o) + \lambda_t (C_t^y +\frac{C_t^o}{1+n} +k_{t+1}(n+1)-k_t-f(k_t))\}.$$ First order conditions are given by $$\frac{\partial L}{\partial C_t^y} = \frac{\beta^t}{C_t^y} +\lambda_t = 0$$ $$\frac{\partial L}{\partial C_t^o} = \frac{...


1

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 constant (but positive!). In the Solow model, the growth rate is more or less a function of the saving rate. An economy might deviate from this because it's ...


1

Let production function $$F(K_t,L_t)=K_t^aL_t^{1-a}$$ $$max[c^*=f(k^*)-(δ+n)k^*]$$ with respect to $k^*$ Then $MPK=a(k^*)^{a-1}$ So $k_G= (\frac{a}{(δ+n)})^{1/1-a}$ which is golden rule level of capital stock per capita. Now calculate consumption at golden rule level $$c_G=f(k_G)-(δ+n)k_G$$ $$c_G=(k_G)^a-(δ+n)k_G$$ $$c_G=(\frac{a}{(δ+n)})^{a/1-a}-(δ+...


1

Ok, first try for your first question. From $u'>0\Rightarrow \phi\theta p>0 \Rightarrow \phi,\theta, p\ne 0$. From $u''<0\Leftrightarrow \phi \theta p c_t^{p-1}\exp(\theta c_t^p)(p-1)c_t^{-1}+\phi \theta p c_t^{p-1}c_t^{p-1}\exp(\theta c_t^p)c_t^{-1}p\theta<0\Leftrightarrow \phi \theta p c_t^{p-1}\exp(\theta c_t^p)(p-1)c_t^{-1}+\phi \theta p c_t^{...


1

To stick to the simpler deterministic case, it is perfect foresight and the Transvarsality condition (TVC)that guarantees that an optimizing agent will stay on the saddle-path, because all other paths lead to the violation of TVC (corner solutions violate the TVC). In a deterministic environment and with perfect foresight the agent knows this and so he ...


1

In general, a steady state in discrete time is that the first difference is zero at some $t=t_0$, and that it remains so: $$y_t-y_{t-1}=x_t-x_{t-1}=0$$ for all $t\geq t_0$. Hence, in a steady state, $y_t$ and $x_t$ are constant at some values $y$ and $x$. Thus, $y=\beta(y+\gamma z)$, which implies $$y=\frac{\beta\gamma z}{1-\beta},$$ and $x=\rho x+y$, which ...


1

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 printout. Reading the paper again on a screen I immediately spotted the mistake.


1

Normally, in the Solow model, there are not increasing marginal returns because by construction, Solow model assumes that marginal return of capital is decreasing. In fact this feature ensures the catch up between countries? which is the main insight of this model. There is another branch of literature on endogeneous growth models which has taken into ...


1

Upgrading an exchange of comments, a critical point in the question is the expression \begin{align} \frac{\partial u(\tilde x)}{\partial \rho}=\frac{1}{(\rho + u'(\tilde x) + 1)^2} \end{align} which is mistaken , because from \begin{align} u(\tilde x) = \frac{\rho + u'(\tilde x)}{\rho + u'(\tilde x) + 1}. \end{align} we obtain \begin{align} \frac{\...


1

I find that the system is saddle-path stable. Setting $z\equiv 1/k$ we get \begin{eqnarray*} \dot{z} & = & -z^2\left(\frac{\rho}{1+\rho} - \frac{1}{1+\lambda}\right) \\ \\ \dot{\lambda} & = & \rho \lambda - z \end{eqnarray*} The fixed point is $E=\{z^*, \lambda^*\} = \{1, 1/\rho\}$ The Jacobian of this system evaluated at the steady ...


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