4

a) Your calculations are correct, but in order for consumption to be positive, so for $$ c_t=\bigg(\frac{R}{(\beta R)^{\frac{1}{\gamma}}}-1\bigg)k_t > 0, $$ you will need to additional conditions. The first is the obvious $k_t > 0$. If there is nothing to gain interest on, there will be no growth and no consumption. The second is the more nuanced $$ \...


3

We know that $g_A = \dot{A}/A$ and thus $g_A = \delta L_A A^{\phi-1}$. Take logs of both sides of $g_A = \delta L_A A^{\phi-1}$ and differentiating with respect to time gives us $$ \frac{\dot{g_A}}{g_A} = n + (\phi-1)g_A.$$ Multiply both sides of the equation by $g_A$: $$ \dot{g_A} = n g_A + (\phi-1) g_A^2$$ Given that $\phi<1$, and in the steady-...


3

There is more than one way to derive the formula for $g_A$ with constant growth: the following is a way I find conceptually simple. Starting from your $g_A=\delta L_A A^{\phi-1}$, differentiate with respect to time (using the product and chain rules) and (for constant growth) set the result to zero: $$\dot{g_A}=0=\delta[A^{\phi-1}\dot{L_A}+L_A(\phi-1)A^{\...


3

I guess you already know that $\exp(-\rho t)$ is the the discount factor. In particular, it is obtained by continuously compounding the discount factor $$\exp(-\rho t)=\lim_{m \to \infty} \left(1+ \frac{\rho}{m}\right)^{mt}$$


2

(That one can use the log-difference approximation for the growth rates, can be glimpsed by the fact that while the model apparently is set in discrete time, the log-evolution of technology is expressed using the exponential, which is how we express a constant growth rate in continuous time). From $$\frac{y_{t+1}}{y_t}=e^\mu\bigg(\frac{k_{t+1}}{k_t}\bigg)^...


1

The direction of $\frac{\partial c^{*}}{\partial n}$ is not ambiguous. An easy way to show this is taking derivative of $c^*=(1-s)f(k^*)$ so that $\frac{\partial c^{*}}{\partial n}=(1-s)f'\frac{\partial k^{*}}{\partial n}$ and because $f'>0$ and we can prove $\frac{\partial k^{*}}{\partial n}<0$ we thus have $\frac{\partial c^{*}}{\partial n}<0$. ...


1

He is just saying that there exists some exponent, reflecting either diminishing or increasing returns to scale, theta. Research can have nonlinear marginal returns. Given that, as theta increases the growth rate must increase linearly since the exponential returns on knowledge translate to a linear increase in exponential growth rate. So it's just how ...


1

income is total profit plus wage income, and the two are fixed in proportions I.e. $y_t = w_t + \pi_t$ and $w_t = \alpha y_t$ and $\pi_t = (1-\alpha)y_t$, where $\alpha$ and $(1-\alpha)$ are the fixed proportions. For $y_{t+1}$ we have $y_{t+1}=(1+g_y)y_t$ and $w_{t+1} = \alpha y_{t+1} = \alpha (1+g_y)y_t = (1+g_y)\alpha y_t = (1+g_y) w_t$ and $\pi_{t+1} ...


1

I guess you already went trough the algebra below, but just for context, the problem you're trying to solve is $$ \max_{c}\sum_{t=0}^{+\infty}\beta^t u(c_t) \\ \text{s.t.}~~ f(k_t) + (1- \delta)k_t = c_t + k_{t+1} \tag{1} $$ where $f(k_t) = k_t^\alpha$ and $$ u(c_t) = \frac{c_t^{1-\gamma}}{1-\gamma} - 1 \tag{2} $$ The problem in (1) can be cast into the ...


1

The approach is not correct, because $\mu^{BGP}$ is a result of underlying structural parameters. So to say "when $\mu^{BGP}$ increases..." immediately begs the question why it increases, which underlying parameter(s) has changed to cause such an increase... assume it was $\epsilon$ that decreased. But in this case $g_k$ is not affected at all, so you see ...


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