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Lecture

# solow.pdf

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Economics
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ECON 2450
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Neoclassical Growth Model - Solow September 30, 2004 Environment • Production Function : describes how, given the level of technology (E t, the economy’s pro- ductive resources, capital ( tandlbr( L t are used to produce output, Y = F (K ,E L ) (0.1) t t t t We assume that the production function has the following properties: (a) it exhibits constant returns to scale (if you double inputs, you double output), (b) it is increasing in capital ( > 0), (c) exhibits diminishing returns to capital (as capital increases MPK falls). • Savings/Investment : Consumers are assumed to consume a constant fraction, s< 1 ,ofeir income (real GDP). Therefore total investment in this economy is, It= sY t (0.2) • Law of Motion for Capital :Acnft δ < 1 of the capital stock depreciates each period. The total capital stock next period is then, K t+1= K +tI −tδK t (0.3) • Labor Force : The labor force is assumed to grow at a constant proportional rate per period. Thus the labor force next period is, Lt+1 =(1+ n)L t (0.4) • Consumption: Whatever is not saved in this economy is consumed. Therefore investment and consumption exhaust total output, Ct+ I t Y t (0.5) Equations (0.1) - (0.5) fully describe the environment of the neoclassical growth model. The variables Y ,K ,I ,C are all aggregate or total variables. We are interested however in per t t t t worker variables because these re ﬂect the well being of the population. We use lower case letters to denote the per worker variables: yt = Lt,k t= L , it= Lt, ct= Lt. Now we want to express equations (0.1) - (0.5) in per worker terms. We simply divide both sides of these equations by .L t Then we have the following, µ ¶ Y t K t = F ,Et L t Lt It = s Yt L t L t L K K I t+1 t+1 =(1 − δ) t+ t L t+1 L t L t L t 1 Lt+1 =(1+ n) Lt C t It Y t + = L t L t L t Notice that in the law of motion I have multiplied and divided the term Kt+1 by Lt+1.he Kt+1 Lt Lt+1 reason is that I want to create the variable Lt+1. But notice that I am also left with a term L t which is equal to (1 + n). Substituting in this set of equations the lower case variables to which they are equal, we have the following, y = F (k ,E ) t t t i = sy t t (1 + n)k =(1 − δ)k + i t+1 t t c + i = y t t t We can substitute t = sy t into the last two equation to be left with the following three equations, yt= F (k tE )t (0.6) (1 + n)k =(1 − δ)k + sy (0.7) t+1 t t ct=(1 − s)y t (0.8) Consider the transformed law of motion, equato in (0.7). W
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