Questions: Steady-State Analysis in Growth Models

Question 1

A linearized growth model has a Jacobian matrix with eigenvalues −0.3 and +0.5. What does this imply about the steady state?

• A. The steady state is globally stable — both eigenvalues are non-zero, so all paths converge
• B. The steady state is a saddle point — only the saddle path converges; all other trajectories diverge
• C. The steady state is unstable — any perturbation causes permanent divergence regardless of initial conditions
• D. The positive eigenvalue indicates oscillatory cycling around the steady state

Answer: (B)

One negative and one positive eigenvalue defines saddle-point stability. The stable manifold (saddle path) is a single converging trajectory. Paths not on it diverge — one eigenvalue negative means there is a direction of attraction, but the positive eigenvalue means there is also a direction of repulsion. This is the typical result in two-variable growth models like Ramsey-Cass-Koopmans.

Question 2

In the Solow model, which mathematical condition defines the steady-state capital-per-worker ratio k*?

• A. The capital stock is at its maximum possible level given technology
• B. Per-capita output growth equals the population growth rate
• C. New investment exactly replaces depreciated capital: sf(k*) = (n + δ)k*
• D. The marginal product of capital equals zero

Answer: (C)

The steady state is where the capital-per-worker ratio stops changing — the 'bathtub' analogy where inflow equals outflow. sf(k) represents new investment per worker; (n + δ)k represents capital needed to replace depreciation and equip new workers. At k*, these are equal, so Δk = 0. Finding k* reduces a dynamic system to a single algebraic equation.

Question 3

In the Solow model, a higher savings rate leads to a higher steady-state capital stock per worker.

• T. True
• F. False

Answer: True

A higher savings rate s shifts up the investment curve sf(k), raising its intersection with the breakeven investment line (n + δ)k. This produces a higher k*. Raising s doesn't change the breakeven line — it only changes how much saving is available to accumulate capital — so the new steady state has more capital per worker and correspondingly higher output and consumption (up to the Golden Rule level).

Question 4

If a steady state has a positive eigenvalue, the economy is very likely to converge to it eventually as long as the initial conditions are close enough.

• T. True
• F. False

Answer: False

A positive eigenvalue means the system is locally unstable in that direction — perturbations grow, not shrink, along the corresponding eigenvector. Convergence requires all eigenvalues to have negative real parts. A saddle point (one negative, one positive) has only a one-dimensional stable manifold; initial conditions not on it diverge. 'Close to steady state' is not sufficient for convergence when a positive eigenvalue exists.

Question 5

In a model with saddle-point stability, why must forward-looking agents immediately 'jump' to the saddle path after a shock rather than adjusting gradually?

Model answer: Paths not on the saddle path lead to economically impossible outcomes: capital eventually goes to zero (no production) or consumption goes to zero or infinity. Since rational forward-looking agents anticipate these outcomes and rule them out, they must jump to the unique path that converges — the saddle path. The 'jump' is not arbitrary; it is the only choice consistent with both optimality and feasibility.

This is what makes the saddle path a prediction, not just a mathematical artifact. In the Ramsey model, consumption is the 'jumping' variable — it can be chosen freely in each period. Capital is 'predetermined' — it can't jump. So after a shock, the agent immediately sets consumption to the saddle path value. Any other choice leads to a path the agent knows will eventually violate a feasibility constraint, so rational agents never choose it.