# Tag Info

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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 ...

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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 $c \in (0, \infty)$: $$F(cK,cN) = (cK)^{\alpha}(cN)^{1-\alpha} = c^{\alpha}c^{1-\alpha} K^{\alpha}N^... 4 We have that:$$ \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} $$Equivalently:... 4 Since, I was urged to present an answer for didactic reasons to which I totally agree, I will provide the full set of corrections to avoid any ambiguity. % Model parameters alpha = 1/3; % s = 0.2; % Investment rate delta = 0.2; % Depreciation rate n = 0.02; % Growth of labor force g = 0.01; % technological progress eps = 1; k(1) = ... 3 The story in the other answer is not fundamentally wrong but incomplete and bit inaccurate. Saving does actually affect capital stock through investment in Solow model (assuming based on the Solow tag that is the model you want intuition for), but there is more complexity to it. What is saving Investment is actually equal not just to private saving but both ... 2 Romer (1986): Increasing Returns to Long Run Growth Lucas (1988): On the Mechanics of Economic Development Romer (1990): Endogenous Technological Change Jones (1995): Time Series Tests of Endogenous Growth Models These are all classic papers in this vein of endogenous growth and questions of cross-country convergence/divergence. 2 When you are using implicit differentiation along a level curve you will treat the variable with respect to which you are differentiating as a single variable, rather than function. This is because the formula for implicit differentiation along level curve is already based previous derivation where you already solve for y'. For example, for general ... 2 You're right, since in basic Solow model (with population growth and no technological progress) macroeconomic closure condition (in aggregate terms) is:$$Y(t) = C(t) + I(t)$$where$$I(t) = sY(t)$$Now replacing the second in the first equation:$$Y(t) = C(t) + sY(t)$$Factorizing we arrive at the equation you stated:$$C(t) = (1-s)Y(t)$$Taking the first ... 2 I don't know how this can be answered without any recourse to at least basic mathematics, since the question itself refers to movement along some curves which is mathematical concept. Hence I will go over some math and then provide intuition at the end. Explanation Including Math The investment curve in Solow model is defined as sf(k) where f(k) = Y and ... 2 Savings in macro are equal to investment (S=I). Investment is how you get new capital stock. When people save more they also by definition invest more and when they invest more there is more capital. 1 The direction of \frac{\partial c^{*}}{\partial n} is not ambiguous. An easy way to show this is taking derivative of c^*=(1-s)f(k^*) so that \frac{\partial c^{*}}{\partial n}=(1-s)f'\frac{\partial k^{*}}{\partial n} and because f'>0 and we can prove \frac{\partial k^{*}}{\partial n}<0 we thus have \frac{\partial c^{*}}{\partial n}<0. ... 1 They used a product rule, quotient rule and chain rule, I won't be discussing the rules themselves as such topics belong to Mathematics.SE, but the reason why you might not realize this outright is the particular notation in a growth theory, which would be on topic here so I will focus on that. In Solow growth model with labor augmenting technological ... 1 This is just standard derivative. The minus sign there is because you are taking derivative of a quotient and the quotient rule for derivatives is:$$ \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$$In this case:$$\dot{k} = \frac{d}{dt} \left(\frac{K(t)}{A(t)L(t)} \right)\\ = \frac{\dot{K}(t)A(t)L(t)-K(t)(\dot{A}L(t)+A(t)...

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Overall it is almost correct but there are some small mistakes and it is incomplete. 1st Mistake: in step (1) you write: $\frac{Y_t}{L_t}=\frac{K^{\alpha}_t}{L_t}\frac{L^{1-\alpha}_t}{L_t}$ But it should actually be $$\frac{Y_t}{L_t}=\frac{K^{\alpha}_t L^{1-\alpha}_t}{L_t}$$ You cannot divide LHS of the equation by $L$ and RHS by $L L \implies L^2$ as that ...

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There are some mistakes in the steps you took although they went in right direction. The correct steps are below. First we can start by expressing output in per capita terms by dividing first two equations by $L_t$: $$\frac{Y_t}{L_t}=\frac{K^{\alpha}_t L^{1−\alpha}_t}{L_t} \implies y_t = k_t^{\alpha}$$ and \frac{Y_t}{L_t}=\frac{C_t}{L_t}+\frac{I_t}{L_t}+\...

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