Exercises

Chapter 3 — worked, faded, then on your own

Start with the worked example, fill in the faded twin, then work the graded exercises (🟢 → 🔵 → 🔴). Open a Solution only after a real attempt. Every numerical answer on this page is recomputed independently by the guide’s verification script. The chapter’s model has no depreciation, and we use its Cobb–Douglas technology $Y = A L^{\alpha} K^{1-\alpha}$ (book eq. 3.19), so per capita $f(k) = A k^{1-\alpha}$, where $\alpha$ is the labour share.

Worked example — the optimal steady state versus the golden rule

Problem. Let $f(k) = k^{1/3}$ (so $A = 1$, $\alpha = 2/3$), $\rho = 0.04$ and $n = 0.01$. Find the optimal steady state $(k^*, c^*)$ and the golden-rule pair $(k_G, c_G)$. Which has more capital, and why is the optimum not the golden rule?

Solution, step by step.

The optimal steady state is pinned down by $\dot c = 0$, so by the modified golden rule $f'(k^*) = \rho$ (book eq. 3.17). With $f'(k) = \tfrac13 k^{-2/3}$: $\tfrac13 k^{*\,-2/3} = 0.04 \Rightarrow k^{*} = \left(\tfrac{1/3}{0.04}\right)^{3/2} \approx 24.06$. Why can we ignore $\sigma$ here? Because $\dot c / c = \sigma[f'(k) - \rho]$ is zero exactly when $f'(k) = \rho$, whatever $\sigma$ is. $\sigma$ affects the speed of adjustment, not the destination.
Steady-state consumption comes from $\dot k = 0$ (book eq. 3.16): $c^* = f(k^*) - n k^* \approx 2.887 - 0.241 \approx 2.646$.
The golden rule maximises $c = f(k) - nk$, so $f'(k_G) = n$ (book eq. 3.18): $k_G = \left(\tfrac{1/3}{0.01}\right)^{3/2} \approx 192.45$ and $c_G = f(k_G) - n k_G \approx 3.849$.
Compare. $k^* < k_G$ because $\rho > n$ and $f'$ is decreasing. Why is less capital optimal when it gives less steady-state consumption? Because reaching $k_G$ would require years of extra saving, and those sacrifices are discounted at $\rho$. It is better to consume more along the way.

Answer: $k^* \approx 24.06$, $c^* \approx 2.646$; $k_G \approx 192.45$, $c_G \approx 3.849$. The optimum has far less capital than the golden rule.

Your turn — the faded twin

Problem. Now $f(k) = k^{1/2}$ ($A = 1$, $\alpha = 1/2$), $\rho = 0.05$, $n = 0.02$. Find $(k^*, c^*)$ and $(k_G, c_G)$.

Modified golden rule: $\tfrac12 k^{*\,-1/2} = 0.05$, so $k^* = $ ____.
$c^* = f(k^) - n k^ = $ ____. (Hint: $\sqrt{k^*}$ is a round number.)
Golden rule: $f'(k_G) = n$, so $k_G = $ ____ and $c_G = $ ____.

Solution

$k^{*\,-1/2} = 0.1 \Rightarrow k^* = 100$.
$c^* = 10 - 0.02 \times 100 = 8$.
$\tfrac12 k_G^{-1/2} = 0.02 \Rightarrow k_G^{1/2} = 25 \Rightarrow k_G = 625$, and $c_G = 25 - 0.02 \times 625 = 12.5$.

Answer: $k^* = 100$, $c^* = 8$; $k_G = 625$, $c_G = 12.5$.

🟢 Core

Exercise 1 — Reading the Euler equation

A household’s marginal product of capital is $f'(k) = 0.06$ and its rate of time preference is $\rho = 0.04$. (a) Will its consumption rise or fall over time? (b) At what rate if $\sigma = 0.5$? (c) And if $\sigma = 2$? Explain the role of $\sigma$ in one sentence.

Solution

The Euler equation (book eq. 3.15) gives $\dot c / c = \sigma[f'(k) - \rho] = \sigma \times 0.02$. (a) Saving earns more than impatience costs, so consumption rises. (b) $0.5 \times 0.02 = 0.01$, or 1% per year. (c) $2 \times 0.02 = 0.04$, or 4% per year. $\sigma$, the elasticity of intertemporal substitution, measures how willing the household is to tilt consumption over time in response to the same incentive.

Answer: consumption rises; 1% per year with $\sigma = 0.5$; 4% per year with $\sigma = 2$.

Exercise 2 — How far below the golden rule?

With $f(k) = A k^{1-\alpha}$, show that $k^*/k_G = (n/\rho)^{1/\alpha}$. Evaluate it for $\alpha = 2/3$, $n = 0.01$, $\rho = 0.04$, and interpret.

Solution

From $f'(k) = (1-\alpha) A k^{-\alpha}$: $k^* = [(1-\alpha)A/\rho]^{1/\alpha}$ and $k_G = [(1-\alpha)A/n]^{1/\alpha}$, so $k^*/k_G = (n/\rho)^{1/\alpha}$. With the numbers, $(0.25)^{3/2} = 0.125$. The optimal capital stock is one eighth of the golden-rule stock. Discounting ($\rho > n$) makes the long, costly build-up to $k_G$ not worth it.

Answer: $k^*/k_G = (n/\rho)^{1/\alpha} = 0.125$.

🔵 Advanced

Exercise 3 — Comparative statics of patience 🔵 Advanced

Using $k^* = [(1-\alpha)A/\rho]^{1/\alpha}$ with $\alpha = 2/3$: (a) find the elasticity of $k^*$ with respect to $\rho$; (b) compute the exact percentage change in $k^*$ when $\rho$ falls from 0.04 to 0.03, and when it falls by 10% (to 0.036). (c) Why does the elasticity understate the exact effect?

Solution

$\ln k^* = \tfrac1\alpha[\ln((1-\alpha)A) - \ln\rho]$, so $\dfrac{d\ln k^*}{d\ln\rho} = -\tfrac1\alpha = -1.5$.
From 0.04 to 0.03: $k^*_{\text{new}}/k^*_{\text{old}} = (0.04/0.03)^{1.5} \approx 1.540$, a rise of about 54%. A 10% fall: $(1/0.9)^{1.5} \approx 1.171$, a rise of about 17.1%. (c) The elasticity is a local, first-order approximation (it predicts 15% for a 10% fall); $k^*$ is convex in $\rho$, so larger falls raise it more than proportionally.

Answer: elasticity $-1.5$; $+54\%$ (0.04 → 0.03); $+17.1\%$ (10% fall).

Exercise 4 — A permanent fall in impatience 🔵 Advanced

The economy of the worked example ($A = 1$, $\alpha = 2/3$, $n = 0.01$) is at its steady state when $\rho$ falls unexpectedly and permanently from 0.04 to 0.03. (a) Which locus moves in the phase diagram? (b) What happens to $c$ and $k$ on impact? (c) Find the new steady state. (d) Is steady-state consumption higher or lower than before?

Solution

Only $\dot c = 0$ ($f'(k) = \rho$) moves, to the right. $\rho$ does not enter $\dot k = 0$.
$k$ is predetermined and cannot jump. $c$ jumps down onto the new saddle path: more patient households save more. (See the key graph.)
$k^*_{\text{new}} = (\tfrac{1/3}{0.03})^{1.5} \approx 37.04$ and $c^*_{\text{new}} = 37.04^{1/3} - 0.01 \times 37.04 \approx 2.963$.
Higher: up about 12.0% from 2.646, because the new steady state is closer to the golden rule.

Answer: $\dot c = 0$ shifts right; $c$ falls on impact, $k$ unchanged; new $k^* \approx 37.04$, $c^* \approx 2.963$ (higher).

Exercise 5 — Speed of convergence 🔵 Advanced

For the worked example with $\sigma = 1$: (a) write the matrix $\Omega$ of book eq. 3.33 numerically; (b) find its eigenvalues; (c) how many years does it take to close half of a small gap to the steady state?

Solution

$\rho - n = 0.03$ and $f''(k^*) = -\alpha(1-\alpha)k^{*\,-\alpha-1} \approx -0.001109$, so $\sigma c^* f''(k^*) \approx -0.002933$: $\Omega \approx \begin{bmatrix} 0.03 & -1 \\ -0.002933 & 0 \end{bmatrix}$.
$\lambda = \tfrac12\left[0.03 \pm \sqrt{0.03^2 + 4 \times 0.002933}\right]$, so $\lambda_s \approx -0.0412$ and $\lambda_u \approx 0.0712$ (opposite signs: a saddle).
Along the saddle path the gap shrinks like $e^{\lambda_s t}$. Half-life $= \ln 2 / 0.0412 \approx 16.8$ years.

Answer: $\lambda_s \approx -0.0412$, $\lambda_u \approx 0.0712$; half-life $\approx 16.8$ years.

🔴 Frontier

Exercise 6 — Derive the Euler equation 🔴 Frontier

With CRRA utility $u(c) = \dfrac{c^{1-1/\sigma}}{1 - 1/\sigma}$, set up the current-value Hamiltonian of book eq. 3.6. Derive the Keynes–Ramsey rule $\dot c / c = \sigma[f'(k) - \rho]$, and show that $\sigma$ is exactly the elasticity of book eq. 3.13 for this utility function.

Solution

$H = u(c)e^{nt} + \lambda[f(k) - nk - c]$. The FOCs are $u'(c)e^{nt} = \lambda$ (eq. 3.7) and $\dot\lambda = -\lambda[f'(k) - n] + \rho\lambda$ (eq. 3.8). Differentiate the first in time: $u''(c)\dot c\, e^{nt} + n u'(c)e^{nt} = \dot\lambda$. Divide by $\lambda = u'(c)e^{nt}$: $\dfrac{u''(c)c}{u'(c)}\dfrac{\dot c}{c} + n = \dfrac{\dot\lambda}{\lambda} = \rho - f'(k) + n$. For CRRA, $u'(c) = c^{-1/\sigma}$ and $u''(c)c/u'(c) = -1/\sigma$, so $-\tfrac1\sigma\dfrac{\dot c}{c} = \rho - f'(k)$, which is $\dot c / c = \sigma[f'(k) - \rho]$. And $-u'(c)/(u''(c)c) = \sigma$ for every $c$, so the elasticity is constant.

Answer: $\dot c / c = \sigma[f'(k) - \rho]$, with $-u'/(u''c) = \sigma$ constant.

Exercise 7 — Why the saddle path slopes up 🔴 Frontier

Using $\text{tr}\,\Omega$ and $\det\Omega$, prove that $\Omega$ has one negative and one positive eigenvalue. (b) Show that, near the steady state, the stable arm satisfies $c - c^* = (\rho - n - \lambda_s)(k - k^*)$, and that its slope is positive. (c) Evaluate the slope for the worked example ($\sigma = 1$).

Solution

$\det\Omega = \sigma c^* f''(k^*) < 0$ since $f'' < 0$. The determinant is the product of the eigenvalues, so they have opposite signs.
An eigenvector $(\Delta k, \Delta c)$ for $\lambda_s$ satisfies the first row, $\lambda_s \Delta k = (\rho - n)\Delta k - \Delta c$, so $\Delta c = (\rho - n - \lambda_s)\Delta k$. Since $\rho > n$ and $\lambda_s < 0$, the slope $\rho - n - \lambda_s > 0$. (It equals $\lambda_u$, because $\lambda_s + \lambda_u = \text{tr}\,\Omega = \rho - n$.)
$0.03 + 0.0412 \approx 0.0712$.

Answer: opposite-sign eigenvalues; slope $\rho - n - \lambda_s \approx 0.0712 > 0$.

Exercise 8 — The market reproduces the planner 🔴 Frontier

In the decentralised economy, firms pay $r = f'(k)$ (book eq. 3.38), and households face $\dot a = (r - n)a + w - c$. (a) Show that, with CRRA utility, the household’s Euler equation is $\dot c / c = \sigma(r - \rho)$. (b) Explain why $a = k$ in equilibrium. (c) Conclude that the decentralised paths of $c$ and $k$ coincide with the planner’s.

Solution

The household Hamiltonian (book eq. 3.48) gives, as in Exercise 6, $\dfrac{u''(c)c}{u'(c)}\dfrac{\dot c}{c} = \rho - r$, and with CRRA, $\dot c / c = \sigma(r - \rho)$ (book eq. 3.49).
Assets are capital minus debt, $a = k - b$. Households are identical, so if one wanted to borrow, no one would lend: $b = 0$ and $a = k$.
Substituting $r = f'(k)$ gives the planner’s Euler equation. Substituting $w = f(k) - f'(k)k$ and $a = k$ into the budget constraint gives $\dot k = f(k) - nk - c$, the planner’s resource constraint. The no-Ponzi condition binds and matches the transversality condition (book eq. 3.53). Same equations, same initial $k_0$: the same paths.

Answer: $\dot c / c = \sigma(f'(k) - \rho)$ and $\dot k = f(k) - nk - c$, the planner’s system.

Answer key

Exercise	Tier	Final answer
Worked example	🟢	$k^* \approx 24.06$, $c^* \approx 2.646$; $k_G \approx 192.45$, $c_G \approx 3.849$
Faded twin	🟢	$k^* = 100$, $c^* = 8$; $k_G = 625$, $c_G = 12.5$
1	🟢	rises; 1% per year ($\sigma = 0.5$); 4% per year ($\sigma = 2$)
2	🟢	$k^*/k_G = (n/\rho)^{1/\alpha} = 0.125$
3	🔵	elasticity $-1.5$; $+54\%$; $+17.1\%$
4	🔵	$c$ falls on impact; new $k^* \approx 37.04$, $c^* \approx 2.963$
5	🔵	$\lambda_s \approx -0.0412$, $\lambda_u \approx 0.0712$; half-life $\approx 16.8$ years
6	🔴	$\dot c / c = \sigma[f'(k) - \rho]$
7	🔴	slope $\rho - n - \lambda_s \approx 0.0712$
8	🔴	the planner’s system