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Solow-Swan Growth Model Analytical Solution, Numerical Solution, Linearization, and Relaxation

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Solow-Swan Growth Model Analytical Solution, Numerical Solution, Linearization, and Relaxation Solow-Swan Growth Model Analytical Solution, Numerical Solution, Linearization, and Relaxation Production Functions Box-Cox Transformation Let us introduce symbols for output (Y), capital stock (K , labor (N), and technology (A). ClearAll t, yy, kk, nn, aa * do not use caps for Mma symbols! * Macroeconomists usually focus on production relationshipsL that can be written as t yy aa @ a * t kk + b * tDnnH L eq1 = %; yy t@ ê Da t kk +@ b tD nn @ D aa t Y A a t K + b t N (1) B F @ D @ D Usually a Î 0, 1 and b = 1 - a. AH particularlyê L H popularL H choiceL for t is the Box-Cox transformation: ClearAll tH, x,L b xb - 1 t x_ := b @ D eq1 @ D yy b -1 + -1 + kkb a -1 + nnb b aa + b b b I M H L H L Implied Production Technology Let us look at the implied production technology. We can determine the implied relationship between output and inputs by using Solve: ClearAll yy, kk, nn * do not use caps for Mma symbols! * Solve eq1, yy Flatten Simplify Quiet bc`yy = %; @ D H 1 L yy ® @aa 1 + -D1êê+ kkb a + êê-1 + nnb b b êê : I I M I M M > 2 Solow_continuous.nb 1 b Y = A 1 - a - b + aKb + bNb (2) You are most familiar with this whenê b = 1 - a. IH L M Constant Returns to Scale ces`yy = bc`yy . b ® 1 - a Simplify 1 yy ® aa -nnb -1 + a + kkb a b ê 8 < êê 1 b Y A a Kb + 1 - a Nb (3) : I H L M > This is known as the constantê elasticity of substitution (CES) production function. A H L E Linear Production Technology When b 1, the CES becomes the linear production function lpt`yy = bc`yy . b ® 1 Simplify yy ® aa 1 + -1 + kk a + -1 + nn b Y = A 1 - a - bê +8a K + b<Nêê (4) When8 b =H1 - aH, the linearL technologyH L becomesL< Y AHaHK + 1 - aL N L (5) Cobb-Douglas Production Technology H H L L When b ® 0, we get a log-linear technology. Clear x, b Limit t x , b ® 0 Log x @ D @ @ D D x (Here’s one way to see that this is the case. Let Y e and use L’Hopital’s rule to show t Y ® x.) So our@ D production relationship is linear in logs: + a + b ln Y = ln A ln K ln L H L (6) bc`yy yy . bc`yy b0`yy = Limit %, b ® 0 1 yyê® aa 1 + -1 + kkb a + -1 + nnb b b @ D 1 aa 1 + -1 + kkb a + -1 + nnb b b : I I M I M M > aa kka nnb I I M I M M Y = A Ka Nb (7) When b = 1 - a this gives us the famous Cobb and Douglas (1927) production function (previously used by Wicksell). Solow_continuous.nb 3 b0`yy . b ® 1 - a Y = A Ka N1-a (8) ê 8 < Application Growth Accounting Solow Model: Analytical Solution Population Growth but No Technical Change ClearAll s, aa, k, a, d, n Fundamental@ DynamicD Equation Define capital per capita k := K N. nt Assuming N = N0 e n ³ 0 , capital per capita evolves according to K N K k = - ê N N N H L (17) K = - n k N The accounting relationship K = S - d K makes this S k = - n + d k (18) N We assume the behavior S = s Y , and we define y = Y N, so we get the fundamental dynamic equa- tion of theH SolowL model: k = s y - d + n k (19) ê Production Technology H L Under constant returns to scale in capital and labor, Y = F K, N y = Y N = F K N, 1 = f k (20) We often assume a Cobb-Douglas production technology, so that @ D Y = AKa N1-a ê @ ê D @ D (21) y = A ka y k_ := aa * ka @ D 4 Solow_continuous.nb Production Technology: Graphical Illustration To illustrate this technology graphically, we need to assume values for our model parameters. * choose parameters to match NFR * prms`core = aa ® 5.0, a ® 0.36, b ® 0.9, d ® 0.075, n ® 0.01 ; Plot y k . prms`core, k, 0, 20 , HPlotLabel ® "Cobb-Douglas ProductionL Function", AxesLabel ® k, y 8 < @ @ D ê 8 < Cobb-Douglas Production Function y 8 <D 14 12 10 8 6 4 2 k 5 10 15 20 Steady State of the Solow-Swan Model: C-D Technology Solve vs. Reduce Recall our fundamental dynamic equation for the Solow-Swan model. With a Cobb-Douglas production technology this becomes k = s A ka - d + n k (22) kdot k_ := s * aa * ka - d + n * k At the steadyH stateL we have k = 0, so we can see that @ D 1 H L s A k = 1-a (23) ss d+n To get this solution from Mathematica, we need to make our assumptions explicit. But because of the non-integerI M exponent on k, this is not quite enough. Solve primarily handles polynomial (or even ratio- nal) functions. $Assumptions = 1 > s > 0 && aa > 0 && k ³ 0 && k0 > 0 && 1 > a > 0 && 1 > d > 0 && 1 > n > 0 && g ³ 0; Solve kdot k 0, k Solve::nsmet: This system cannot be solved with the methods available to Solve. a Solve@aa k @s D- k n + dD 0, k When Solve fails, we can sometimes succeed by turning to Reduce, which also looks for a fuller charac- terization@ of the possibleH Lsolutions.D Solow_continuous.nb 5 Reduce s * aa * ka - d + n * k 0 && k ³ 0, k, Reals Simplify 1 n + d -1+a k 0 k @ aa s H L D êê Note that Reduce characterizes solutions as equalities. ÈÈ The first solution is a good reminder that there is more than one steady state. The second solution yields the result we were looking for. s * aa 1 ss`k = Power , n + d 1 - a 1 aa s 1-a n + d B F In sum, the non-trivial steady state of the model can be characterized in terms of this steady-state value ofK k: O 1 s A k = 1-a (24) ss n+d Transformations Can Help Solve I M To help Mathematica along, look for ways to rewrite the equality we are trying to solve. Solve kdot k k 0, k Simplify Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. @ @ D ê D êê 1 n + d -1+a k ® aa s Note that Solve returns the solutions it finds as rules. :: >> The other obvious transformation does not work immediately (in Mathematica 9). Solve kdot k ka 0, k Simplify Solve::nsmet: This system cannot be solved with the methods available to Solve. Solve@k-a aa@ Dkêa s - k n + Dd êê 0, k However, we can again push things along by giving Mathematica a some guidance. Here is one approa@ch: H H LL D FullSimplify kdot k ka Solve % 0, k Simplify - 1-a + d aa s k n @ @ D ê D Solve::ifun@ : Inverse functionsD êê are being used by Solve, so someHsolutionsL may not be found; use Reduce for complete solution information. 1 n + d -1+a k ® aa s A related approach is to recognize that if we define z = k1-a we can solve a linear equation in z: :: >> 6 Solow_continuous.nb Clear a, s, aa, d, n, z Solve s * aa - d + n * z 0, z ss`z = z . % 1 @aa s D z ® @ H L D n + d ê P T aa s :n:+ d >> We then reverse the transformation to find the implied soln for kss. 1 ss`z 1-a % ss`k 1 aa s 1-a n + d True K O Graphical Illustration of the Steady State prms`s20 = Append prms`core, s ® 0.20 ; Plot Evaluate s * aa * ka, d + n * k . prms`s20 , k, 0, 60 , AxesLabel ® k, "s y\n d+n k" , PlotLegends ® "s y", " d+n k" , Epilog ® Gray, Dashed@ , Line ss`k,D0 , ss`k, s * y ss`k . prms`s20 @ @8 H L < ê D 8 < s y d+n k 8 H L < 8 H L < 8 @88 < 8 @ D<< ê D<D 5 H L 4 s y 3 d+n k 2 H L 1 k 10 20 30 40 50 60 Solutions for y and c Our steady-state value of k implies values for y and c. Clear ss`y, ss`c ss`y, ss`c = y ss`k Simplify, 1 - s * ss`y aa s a aa s a aa @ 1-a , aa D1 - s 1-a 8 n + d < 8 @ D êên + d H L < : K O H L K O > Solow_continuous.nb 7 Comparative Statics ss`kyc = ss`k, ss`y, ss`c ; ss`exogs = s, aa, d, n ; ss`dkdx = D ss`k, ss`exogs Simplify ss`dkdx 8 Sign Simplify< ss`k8 1 1< -1 -1 ss`dkdx @ 8, , <D êê, Simplify 1 - a s aa n + d n + d êê êê 1 1 1 1 aa s aa s aa s aa s 1-a 1-a 1-a 1-a n+d n+d n+d n+d - , - : , >,êê s -1 + a aa -1 + a H -1L+ a H n +Ld -1 + a n + d 1,I1, -M1, -1 I M I M I M : > TrueH L H L H L H L H L H L 8 < ss`dydx = D ss`y, ss`exogs Simplify ss`dydx Sign Simplify a ss`y 1 1 -1 -1 ss`dydx @ 8 , <,D êê , Simplify 1 - a s a * aa n + d n + d êê êê a a 1 1 aa s aa s aa s aa s aa a 1-a 1-a a 1-a a 1-a n+d n+d n+d n+d - , : , , > êê s -1 + a 1 - a s -H1 + a L sH -1 +La 1, 1, -I1, -M1 I M I M I M : > True H L H L H L 8 < Exercise: explain why the following is false.
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