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5

In this example, "mass" means the same as "length". Let me try to explain it with a simple example. If you work in a continuous setting, you have an infinite number of agents, which lie on a line between $0$ and $M$. Let's say individual consumption is given by the function $C(i)$, for all agents $i \in [0,M]$. For total consumption, you ...


5

We have that: $$ \dot K = s K^\alpha L_0^b e^{nbt} $$ Rewriting the differential equation gives: $$ K^{-\alpha} \frac{dK}{dt} = s L_0^b e^{nbt} $$ Integrate both sides with respect to $t$ from $0$ to $T$ gives: $$ \frac{1}{b} [K^b]^T_0 = s L_0^b \frac{1}{nb}[e^{nbt}]^T_0 $$ So: $$ K^b_T = K^b_0 - \frac{s}{n} L_0^b + \frac{s}{n} L_0^b e^{nbT} $$ Equivalently:...


4

Yes, there are DSGE models that can be used for forecasting. These models typically have a particular kind of steady-state, which is, more precisely, called balanced growth path (BGP). On the BGP (in the absence of shocks), key indicators growth at the same constant rate. For example, GDP, household consumption, investment all grow at 2% a year. This is ...


4

These two assumptions are not necessarily contradictory. Just check whether the assumptions are satisfied by any candidate function. For example, take $F(K,N) = K^{\alpha}N^{1-\alpha}$, with $\alpha \in (0,1)$. Constant returns to scale: For any scaling factor $c \in (0, \infty)$: $$F(cK,cN) = (cK)^{\alpha}(cN)^{1-\alpha} = c^{\alpha}c^{1-\alpha} K^{\alpha}N^...


3

Romer (1986): Increasing Returns to Long Run Growth Lucas (1988): On the Mechanics of Economic Development Romer (1990): Endogenous Technological Change Jones (1995): Time Series Tests of Endogenous Growth Models These are all classic papers in this vein of endogenous growth and questions of cross-country convergence/divergence.


2

When you are using implicit differentiation along a level curve you will treat the variable with respect to which you are differentiating as a single variable, rather than function. This is because the formula for implicit differentiation along level curve is already based previous derivation where you already solve for $y'$. For example, for general ...


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 $\frac{\partial c^{*}}{\partial n}<0$. ...


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