The Solow Growth Model

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The Solow Growth Model Econ 504 Roberto Chang Rutgers University September 2012 R. Chang (Rutgers University) Solow Model September 2012 1 / 18

The Model The first ingredient of a dynamic model is the description of the time horizon and the set of commodities. In the original Solow model, time is continuous and the horizon is infinite. Without loss of generality assume that time is indexed by t in [0, ). At each point in time, there is only one final good R. Chang (Rutgers University) Solow Model September 2012 2 / 18

The Model The Production Function The good is produced via the aggregate production function: Y = F (K, AL) Here Y, A, K, and L denote output, labor productivity, capital, and labor, and are functions of time (i.e. we should really write Y (t) and so on, but we omit time arguments when not needed). The product AL is called effective labor. The function F has neoclassical properties: F is continuously differentiable, strictly increasing, strictly concave F has constant returns to scale R. Chang (Rutgers University) Solow Model September 2012 3 / 18

A key example is the Cobb Douglas function: F (K, AL) = K α (AL) 1 α, 0 < α < 1 Exercise: Show that the Cobb Douglas function enjoys the neoclassical properties listed above. R. Chang (Rutgers University) Solow Model September 2012 4 / 18

Once you assume the existence of the aggregate production function Y = F (K, AL), it is clear that the growth of output can be due to growth of A, K, or L. Solow assumed that A and L grow at exogenous rates: L = dl dt = nl Ȧ = γa At time t = 0, A, and L are given by history. Then one can describe the value of L and A at each point in time: L(t) = L(0)e nt A(t) = A(0)e γt So it remains to describe the motion of capital. R. Chang (Rutgers University) Solow Model September 2012 5 / 18

Assume K (0) is given by history. The accumulation of capital must follow: K = I δk where I denotes investment and δ the depreciation rate of capital. As usual, investment is assumed to equal savings. The key behavioral equation of the Solow model is that savings equal a constant fraction of output, so: I = sy This completes the description of the model. R. Chang (Rutgers University) Solow Model September 2012 6 / 18

The Solution A solution is a description of the values of L, A, Y, and K at each point in time. Now, from the capital accumulation equation: K = sy δk Here we have defined k = K /AL. = sf (K, AL) δk = ALsF (K /AL, 1) δk = AL[sF (k, 1) δk] R. Chang (Rutgers University) Solow Model September 2012 7 / 18

Define f (k) = F (k, 1), the intensive production function. Now we can rewrite the previous equation as: ( K /K ) k = sf (k) δk But k = K /AL implies k/k = K /K (n + γ). Combine the two to obtain: k = sf (k) (n + γ + δ)k This is the fundamental equation of growth. R. Chang (Rutgers University) Solow Model September 2012 8 / 18

k = sf (k) (n + γ + δ)k What is the intuition? Capital per effective labor grows faster when there is more investment, hence the role of the savings coeffi cient s. On the other hand, population growth, productivity growth, and faster capital depreciation tend to reduce capital per unit of effective labor. Obviously once we have solved for the path of k we can obtain K = kal, and Y. So we can focus on the solution for k. R. Chang (Rutgers University) Solow Model September 2012 9 / 18

The Solution The Steady State A steady state is a constant value of k, say k, such that k(t) = k is a solution of the growth equation. Hence k must solve: sf (k ) = (n + γ + δ)k Example: if the production function is CD, f (k) = k α, so: s(k ) α = (n + γ + δ)k that is [ k s = n + γ + δ ] 1/1 α R. Chang (Rutgers University) Solow Model September 2012 10 / 18

More generally, under the Inada conditions f (0) = 0, f (0) =, f ( ) = 0 there is a unique nontrivial steady state. This can be shown using the famous Solow diagram. R. Chang (Rutgers University) Solow Model September 2012 11 / 18

The Solution Convergence The same diagram can be employed to show that k must grow when it is between 0 and k, and falls if it is greater than k. Hence the model implies global convergence towards the steady state. R. Chang (Rutgers University) Solow Model September 2012 12 / 18

Interpreting the Solow Model What kind of questions can we answer with this model? In general, we can trace the adjustment path to changes in the parameters of the economy (s, δ, n, γ), or in initial conditions (K (0), L(0), A(0)). We can also modify the model in different ways. In this sense, it is an excellent example of a truly dynamic model. Example: Starting from steady state, analyze the consequences of a permanent increase in the savings propensity s. R. Chang (Rutgers University) Solow Model September 2012 13 / 18

The Solow Growth Model and Growth In the steady state, however, k is constant. Hence the extensive variables, such as K and Y, grow at the rate (n + γ). Per capita variables, such as Y /L, grow at rate γ. But n and γ are exogenous variables in the model. So, in this sense, the model does not explain long run growth. See Romer, chapter 1, for a discussion. To illustrate, as Romer also discusses: a permanent increase in the savings rate s has only transitory effects on the growth rate (e.g. a level effect, but not a growth effect). R. Chang (Rutgers University) Solow Model September 2012 14 / 18

The Golden Rule Another classic question leads to the Golden Rule. Suppose that one can choose s. What would an optimal value of s be? In growth theory, it was proposed to choose s so as to maximize steady state consumption. Now, in steady state, c = (1 s)f (k) so that the optimal rule would require dc ds = f (k) + (1 s)f (k) dk ds = 0 In steady state, the Solow equation is sf (k) = (n + γ + δ)k so dk ds = f (k) (n + γ + δ) sf (k) R. Chang (Rutgers University) Solow Model September 2012 15 / 18

Combining the preceding, we get or (1 s)f (k) = (n + γ + δ) sf (k) f (k) δ = n + γ The Golden Rule level of capital is such that the marginal product of capital, net of depreciation, is equal to the growth rate. If k is larger than this level in steady state there is overaccumulation. R. Chang (Rutgers University) Solow Model September 2012 16 / 18

Mathematical aside The Solow growth equation is an example of a first order ordinary differential equation. Such equations are expressed as: ẋ(t) = dx dt = g(x(t)), t T for some function g. Here x(t) X = R + (the state space) and T is the time interval ([0, ) in our Solow discussion) A solution is a function x : T X such that x(t) that satisfies the ODE for each t T. A steady state is a constant solution, i.e. x X such that 0 = g(x ) If x is a steady state, it is locally a.s. if g (x ) < 0, and unstable if the inequality is reversed. (Draw phase diagram to see why.) R. Chang (Rutgers University) Solow Model September 2012 17 / 18

Convergence: suppose g is linear: Any solution is of the form ẋ = g(x) = ax x(t) = x 0 e at which converges to x = 0 if a < 0. We then say that x is (globally) stable. For nonlinear g, global convergence is much more involved, but local convergence is simple: If x is a steady state, it is locally a.s. if g (x ) < 0, and unstable if the inequality is reversed. (Draw phase diagram to see why.) R. Chang (Rutgers University) Solow Model September 2012 18 / 18