By the end you can:
- derive the two-equation \((K,C)\) system of the Blanchard–Yaari economy with capital from the aggregate consumption rule and factor prices;
- draw the phase diagram (both loci, the field, the saddle path) and explain why the equilibrium must sit on the saddle path;
- prove the comparative statics of the steady state in \(p\), \(\theta\), \(\alpha\), and say when the steady state is above or below the golden rule.
1. Where we are: one aggregate rule, three economies
Lesson 18 aggregated the households into three equations that hold for any path of interest rates \(r(t)\) and labour income \(Y(t)\) BY slide 16:
$$C = (p+\theta)(H+V),\qquad \dot V = rV + Y - C,\qquad \dot H = (r+p+\alpha)H - Y,$$
and, after eliminating \(H\),
$$\boxed{\;\dot C = [\,r + \alpha - \theta\,]\,C \;-\; (p+\alpha)(p+\theta)\,V\;}$$
To close the model we need to say what \(Y\), \(r\) and \(V\) are in general equilibrium. Alvarez does this three ways BY slide 19:
| Economy | Labour income \(Y\) | Interest rate | Non-human wealth \(V\) | Lesson |
|---|
| Pure endowment | \(Y = C\) | whatever clears goods | \(V=0\) | 18 |
| Capital | \(\mathcal F(K) - \mathcal F'(K)K\) | \(r=\mathcal F'(K)\) | \(V = K\) | this lesson |
| Capital + government debt | same, minus taxes | \(r=\mathcal F'(K)\) | \(V = K + B\) | 20 |
Notation: read before every Blanchard–Yaari formula
- \(p\) is the death rate (expected lifetime \(1/p\)), not a price. \(\theta\) is the pure discount rate, not the PAYG tax of Lesson 14.
- \(\alpha\) is the rate at which labour income declines with age, not the old-age endowment of Class Note II or a capital share. When we use Cobb–Douglas below, the capital share is called \(a\).
- \(\mathcal F(K) \equiv F(K,1) - \delta K\) is output net of depreciation. So the golden rule is \(\mathcal F'(K_{GR}) = 0\), and \(r = \mathcal F'(K)\) with no "\(-\delta\)" in sight. (The slides write \(\mathcal F\) as a calligraphic F; plain \(F\) is the gross technology.)
2. Technology in continuous time
Take a neoclassical, constant-returns \(F(K,L)\) and labour supplied inelastically at \(L=1\) BY slide 22. The continuous-time equations are limits of discrete time with period length \(\Delta\):
- Feasibility at \(t\) (flows per unit time): \(I(t) + C(t) = F(K(t),1)\).
- Capital accumulation over a period of length \(\Delta\): \(K_{t+\Delta} = \Delta I_t + K_t(1-\Delta\delta)\). Rearranging, \(\dfrac{K_{t+\Delta}-K_t}{\Delta} = I_t - \delta K_t\), and letting \(\Delta\downarrow 0\) gives \(\dot K = I - \delta K\).
- Substitute \(I\) from feasibility:
$$\dot K(t) = F(K(t),1) - \delta K(t) - C(t) \;=\; \mathcal F(K(t)) - C(t).$$
Factor prices
Competitive firms rent capital and hire labour BY slide 23:
- Wage bill (aggregate labour income, \(L=1\)): \(Y = w = F_L(K,1) = F(K,1) - F_K(K,1)K = \mathcal F(K) - \mathcal F'(K)K\). The middle equality is Euler's theorem for CRS. The last one holds because the \(\delta K\) terms cancel: \(\mathcal F - \mathcal F'K = (F - \delta K) - (F_K - \delta)K\).
- Interest rate. In discrete time, owning a unit of capital from \(t\) to \(t+\Delta\) pays the rental \(\nu_{t+\Delta}\Delta\) and loses \(\delta\Delta\) to depreciation. Arbitrage with a bond paying \(r_t\Delta\) gives \(\nu_{t+\Delta} = r_t + \delta\), and firms set \(F_K(K_{t+\Delta},1) = \nu_{t+\Delta}\). In the limit, \(F_K(K(t),1) = r(t)+\delta\), i.e.
$$r(t) = \mathcal F'(K(t)).$$
- Product exhaustion: \(rK + Y = \mathcal F'(K)K + \mathcal F(K) - \mathcal F'(K)K = \mathcal F(K)\). Net output is paid out entirely as interest and wages.
3. The equilibrium \((K,C)\) system
In a closed economy without government, the only asset households can hold in net terms is capital, so \(V(t) = \int_{-\infty}^t N(s,t)\,v(s,t)\,ds = K(t)\). Substituting \(V=K\), \(r=\mathcal F'(K)\) and \(Y = \mathcal F - \mathcal F'K\) into the two aggregate laws of motion BY slide 24:
- \(\dot V = rV + Y - C\) becomes \(\dot K = \mathcal F'(K)K + \mathcal F(K) - \mathcal F'(K)K - C = \mathcal F(K) - C\). This is exactly the technology equation, as it must be (Walras' law again).
- The \(\dot C\) equation becomes the "Euler equation" of this economy.
The Blanchard–Yaari dynamical system
$$\dot C = \big[\mathcal F'(K) + \alpha - \theta\big]\,C \;-\; (p+\alpha)(p+\theta)\,K, \qquad \dot K = \mathcal F(K) - C,$$
with \(K(0)\) given (predetermined) and \(C(0)\) free (a "jump" variable, since \(C(0) = (p+\theta)(K(0)+H(0))\) and \(H(0)\) depends on the whole future path of wages and interest rates).
Intuition: the extra term is generational turnover
Each individual obeys the Ramsey-like Euler equation \(\dot c/c = r - \theta\). Aggregate consumption grows more slowly, by the term \((p+\alpha)(p+\theta)K\). At every instant, a flow \(p\) of people die and are replaced by newborns who own no capital, so they consume less than the people they replace. With \(\alpha>0\) there is a second effect: the newborns' labour income is the highest anyone gets, while older people's income has decayed, so wealth shifts towards the old. Set \(p=\alpha=0\) and you get back the representative-agent Ramsey model, \(\dot C = (\mathcal F'(K)-\theta)C\).
4. The phase diagram
The recipe, from BY slide 25: draw the two loci, find the steady state, find the direction of motion everywhere, then discard the paths that eventually become infeasible.
The \(\dot K = 0\) locus: \(C = \mathcal F(K)\)
This curve is concave, starts at the origin with infinite slope (Inada), peaks at the golden rule \(\mathcal F'(K_{GR}) = 0\), and then falls, crossing zero at \(K_{\max}\) where depreciation eats all output BY slide 26. Above it, \(C > \mathcal F(K)\) and capital falls. Below it, capital rises.
The \(\dot C = 0\) locus: \(C = \dfrac{(p+\alpha)(p+\theta)}{\mathcal F'(K)+\alpha-\theta}\,K\)
This is defined where the denominator is positive, i.e. for \(K < \hat K\) with \(\mathcal F'(\hat K) = \theta - \alpha\). If \(\theta-\alpha \le -\delta\), no such \(\hat K\) exists and the locus is defined for all \(K\).
- It passes through the origin, since \(\mathcal F'(K)\to\infty\) as \(K\to 0\).
- It is increasing: the numerator \(K\) rises and the denominator falls because \(\mathcal F''<0\).
- It shoots to \(+\infty\) as \(K\uparrow\hat K\) (the denominator goes to \(0^+\)).
- It is convex. Alvarez states this (slide 26), and it holds for Cobb–Douglas; you can see it in the figure.
Direction: for \(K\lt \hat K\), the coefficient on \(C\) in \(\dot C\) is positive. So \(\dot C>0\) above the locus and \(\dot C<0\) below it. For \(K\ge\hat K\), both terms of \(\dot C\) are \(\le 0\), so consumption falls everywhere there.
The steady state is unique
There is exactly one \(\bar K\in(0,\min\{\hat K,K_{\max}\})\) with
$$\frac{\mathcal F(\bar K)}{\bar K} \;=\; \frac{(p+\alpha)(p+\theta)}{\mathcal F'(\bar K)+\alpha-\theta}, \qquad \bar C = \mathcal F(\bar K).$$
Proof
- A steady state needs \(C = \mathcal F(K)\) and \(C\) on the \(\dot C=0\) locus. Divide by \(K>0\) to get the displayed equation. Since the right side is positive, we also need \(\mathcal F(K)>0\), i.e. \(K\lt K_{\max}\), and \(K\lt \hat K\).
- The left side \(\mathcal F(K)/K\) (average product) is strictly decreasing. Its derivative is \((\mathcal F'K - \mathcal F)/K^2 < 0\), because strict concavity with \(\mathcal F(0)=0\) gives \(\mathcal F(K) > \mathcal F'(K)K\). It tends to \(+\infty\) as \(K\to0\) (Inada) and to \(0\) at \(K_{\max}\).
- The right side is strictly increasing in \(K\) (its denominator falls). It tends to \(0\) as \(K\to 0\) and to \(+\infty\) as \(K\to\hat K\).
- A strictly decreasing function that starts above a strictly increasing one and ends below it crosses it exactly once.
∎
The same picture (LHS falling, RHS rising) drives every comparative static in §6: anything that raises the right-hand side at a given \(K\) lowers \(\bar K\).
The saddle path
The two loci cut the plane into four regions. In the region to the left of the steady state and above both loci, \(K\) falls and \(C\) rises: paths shoot up-left and hit \(K=0\) in finite time with \(C>0\), which is infeasible. Below and to the right, \(C\) falls while \(K\) first rises. Such paths eventually reach \(C = 0\) in finite time, because on the axis \(\dot C = -(p+\alpha)(p+\theta)K < 0\), so \(C\) goes negative, which is also infeasible BY slide 27. Between these two families there is exactly one upward-sloping curve \(C = S(K)\), the stable manifold or saddle path, that converges to \((\bar K,\bar C)\).
Perfect-foresight equilibrium
Given \(K(0)>0\), the equilibrium sets \(C(0) = S(K(0))\), and then \(C(t) = S(K(t))\) for all \(t\ge 0\), with \(K(t)\to\bar K\) monotonically.
5. Adjustment to the steady state
BY slide 29 asks: suppose \(0 < K(0) < \bar K\). Describe the paths of \(C\), \(K\), \(Y\), \(r\), \(H\). Are they monotone?
Answers, with reasons
- \(K(t)\) and \(C(t)\) both rise monotonically. The saddle path is upward sloping and lies below the \(\dot K = 0\) curve to the left of \(\bar K\), so \(\dot K>0\) all along it. \(K\) rises monotonically, and \(C = S(K)\) rises with it.
- \(Y(t)\) rises monotonically. \(Y = \mathcal F(K) - \mathcal F'(K)K\) has \(dY/dK = -\mathcal F''(K)K > 0\), and \(K\) is rising. (More capital per worker means a higher wage.)
- \(r(t)\) falls monotonically. \(r = \mathcal F'(K)\) with \(\mathcal F''<0\). Capital is scarce early, so its return is high.
- \(H(t)\) rises monotonically. Write \(H(t) = \int_0^\infty Y(t+u)\exp\!\big(-\!\int_0^u [r(t+v)+p+\alpha]\,dv\big)du\). Moving \(t\) later raises every future wage \(Y(t+u)\) and lowers every future interest rate \(r(t+v)\), so the integrand rises pointwise. (This needs \(\alpha\) and \(p\) constant, which they are.)
- Relation between \(S(K)\) and \(C = (\theta+p)(K+H)\) slide 27. The consumption rule \(C=(\theta+p)(K+H)\) holds at every date for any price path. In equilibrium the price path is itself a function of the current state, so human wealth is a function of \(K\) too: \(H(t) = h(K(t))\). The saddle path is just the consumption rule with this equilibrium human wealth plugged in:
$$S(K) = (\theta+p)\,\big(K + h(K)\big).$$
- Is \(S\) linear? No, not in general. \(h(K)\) is the present value of the wages the economy will pay starting from \(K\), discounted at interest rates that depend on \(K\) nonlinearly through \(\mathcal F\). Numerically (\(a=0.33\), \(\delta=0.08\), \(\theta=0.03\), \(p=0.05\), \(\alpha=0\)), \(C/K\) falls from about \(0.48\) at \(K=0.3\bar K\) to \(0.29\) at \(\bar K\), so \(S\) is not a ray. It is close to linear only near \(\bar K\), where its slope is the stable eigenvector \(\mathcal F'(\bar K) - \lambda_s\).
6. Comparative statics of the steady state
Recall the steady-state equation \(\dfrac{\mathcal F(\bar K)}{\bar K} = \dfrac{(p+\alpha)(p+\theta)}{\mathcal F'(\bar K)+\alpha-\theta}\). Its left side is decreasing in \(K\), its right side increasing, and neither side depends on parameters except through the right side. So a parameter change that raises the right side at the old \(\bar K\) lowers \(\bar K\), and vice versa BY slide 30.
| Change | Effect on RHS at fixed \(K\) | \(\bar K\) | \(\bar r\) | Why, economically |
|---|
| \(p\uparrow\) | numerator \((p+\alpha)(p+\theta)\uparrow\) | ↓ | ↑ | Shorter horizons: more of the capital stock is "lost" to newborns each instant, so there is less aggregate saving. |
| \(\theta\uparrow\) | numerator ↑, denominator ↓ | ↓ | ↑ | More impatience. |
| \(\alpha\uparrow\) | ↓ if \(\mathcal F'(\bar K) < \theta+p\) (always true, see below) | ↑ | ↓ | Income is front-loaded, so agents save when young to finance old age, even with flat consumption. |
Proof that \(\bar K\) is increasing in \(\alpha\)
BY slide 32
- Since \((p+\theta)\) does not involve \(\alpha\), it suffices to show \(\dfrac{p+\alpha}{\mathcal F'(\bar K)+\alpha-\theta}\) is decreasing in \(\alpha\) at the steady state.
- Differentiate, holding \(K\) fixed: \(\dfrac{\partial}{\partial\alpha}\dfrac{p+\alpha}{\mathcal F'+\alpha-\theta} = \dfrac{(\mathcal F'+\alpha-\theta)-(p+\alpha)}{(\mathcal F'+\alpha-\theta)^2} = \dfrac{\mathcal F'(\bar K)-\theta-p}{(\mathcal F'+\alpha-\theta)^2}.\) This is negative iff \(\mathcal F'(\bar K) < \theta+p\).
- At \(\alpha=0\) we have \(\mathcal F'(\bar K) < \theta + p\) (proved next). So at \(\alpha=0\), raising \(\alpha\) lowers the RHS and raises \(\bar K\).
- A higher \(\bar K\) means an even lower \(\mathcal F'(\bar K)\), so \(\mathcal F'(\bar K)\lt \theta+p\) continues to hold as \(\alpha\) rises. Therefore the derivative stays negative and \(\bar K\) is increasing for all \(\alpha\ge0\).
∎
Proof that with \(\alpha = 0\): \(\theta < \mathcal F'(\bar K) < \theta + p\)
BY slide 31
- With \(\alpha=0\), the steady state is \(\mathcal F(\bar K)\,[\mathcal F'(\bar K)-\theta] = p(p+\theta)\bar K\). The right side is positive and \(\mathcal F(\bar K)>0\), so \(\mathcal F'(\bar K) > \theta\). Lower bound done.
- Define \(K^{**}\) by \(\mathcal F'(K^{**}) = \theta+p\) and \(K^*\) by \(\mathcal F'(K^*) = \theta\). Then \(K^{**}\lt K^* = \hat K\).
- Let \(g(K) \equiv \mathcal F(K)[\mathcal F'(K)-\theta] - p(p+\theta)K\). By the uniqueness argument, \(g>0\) to the left of \(\bar K\) and \(g<0\) between \(\bar K\) and \(\hat K\).
- Evaluate at \(K^{**}\): \(g(K^{**}) = p\,\mathcal F(K^{**}) - p(p+\theta)K^{**} = p\,[\mathcal F(K^{**}) - \mathcal F'(K^{**})K^{**}]\).
- Concavity with \(\mathcal F(0)=0\) gives \(\mathcal F(K) > \mathcal F'(K)K\) (average product exceeds marginal product), so \(g(K^{**})>0\). Hence \(K^{**}\lt \bar K\), i.e. \(\mathcal F'(\bar K) < \mathcal F'(K^{**}) = \theta+p\). ∎
Watch the direction of the concavity inequality
Slide 31 states the concavity step as a comparison between \(\mathcal F(K^{**})\) and \((p+\theta)K^{**} = \mathcal F'(K^{**})K^{**}\). The direction that concavity delivers, and that the argument needs, is \(\mathcal F(K^{**}) > \mathcal F'(K^{**})K^{**}\). If your copy shows "\(<\)", read it as "\(>\)". You can check: for \(\mathcal F = \sqrt K\), \(\sqrt K > \tfrac{1}{2\sqrt K}K = \tfrac12\sqrt K\).
This is Blanchard's Figure 2 (BY slide 28): with \(\alpha=0\) the steady state lies strictly between \(K^{**}\) (\(\mathcal F'=\theta+p\)) and \(K^*\) (\(\mathcal F'=\theta\)). The individual consumption profile has \(\dot c/c = \bar r - \theta > 0\), so people consume more as they age and therefore save when young. That saving is what builds the capital stock.
Golden rule and dynamic (in)efficiency
- \(\alpha = 0\): \(\bar r = \mathcal F'(\bar K) > \theta > 0\), so \(\bar K < K_{GR}\). The economy is dynamically efficient; compare the positive-interest case of Lesson 10.
- \(\alpha>\theta\): \(\hat K\) satisfies \(\mathcal F'(\hat K) = \theta-\alpha<0\), so \(\hat K > K_{GR}\) and \(\bar K\) can exceed \(K_{GR}\), giving \(\bar r<0\). This is Blanchard's Figure 4 (BY slide 33). Agents save heavily for the years when their labour income has decayed.
Exact condition for \(\bar r < 0\)
\(\bar K > K_{GR}\) iff \(g\) is still positive at \(K_{GR}\), where now \(g(K) = \mathcal F(K)[\mathcal F'(K)+\alpha-\theta] - (p+\alpha)(p+\theta)K\). Since \(\mathcal F'(K_{GR})=0\),
$$\bar r < 0 \iff (\alpha-\theta)\,\frac{\mathcal F(K_{GR})}{K_{GR}} \;>\; (p+\alpha)(p+\theta).$$
This requires \(\alpha>\theta\). Because the right side is increasing in \(p\), it holds when \(p\) is small enough relative to \(\alpha-\theta\).
A slide bullet to read carefully
Slide 31 says "if \(\alpha>\theta\), for \(p\) large enough, \(\bar K\) larger than golden rule." Since \(\bar K\) is decreasing in \(p\) (and \(\bar K\to0\) as \(p\to\infty\)), the condition above shows it is small \(p\) (long horizons) together with \(\alpha>\theta\) that pushes \(\bar K\) past the golden rule. The figure confirms this: with \(a=0.33\), \(\delta=0.08\), \(\theta=0.01\), \(\alpha=0.06\), you get \(\bar r\approx-0.024\) at \(p=0.02\) but \(\bar r\approx+0.082\) at \(p=0.2\). The bullet probably means "for \(\alpha\) large enough". Ask Alvarez or a TA if in doubt. Also remember that \(p\to0\) forces \(\alpha=0\) (§7), so the inefficient region needs \(p>0\).
7. The infinite-lifetime limit \(p\downarrow 0\)
BY slide 34 With \(p=0\) nobody dies and nobody is born. If individual labour income decayed at rate \(\alpha>0\), aggregate labour income would decay too, so a stationary economy requires \(\alpha=0\). The system becomes the Ramsey–Cass–Koopmans model:
$$\dot C = [\mathcal F'(K)-\theta]\,C, \qquad \dot K = \mathcal F(K) - C.$$
- The \(\dot C=0\) locus is the vertical line \(\mathcal F'(\bar K)=\theta\) (the modified golden rule). In the figure, the blue curve steepens towards this line as \(p,\alpha\to0\).
- A slightly different argument rules out the non-saddle paths. Paths below the saddle path now have \(C\to0\) only asymptotically: \(\dot C/C = \mathcal F'-\theta\) is bounded once \(K>K^*\), so \(C\) never actually hits zero. These paths are feasible, so infeasibility cannot rule them out. Instead they violate the household's transversality condition: they converge to \(K_{\max}\) with \(r=\mathcal F'(K_{\max})<0\), so wealth discounted at \(r\) explodes. With \(p>0\), the extra term \(-(p+\alpha)(p+\theta)K\) makes \(C\) hit zero in finite time, which is simpler.
- The adjustment from \(K(0)\lt \bar K\) looks the same: \(K, C, Y\) rise and \(r\) falls.
8. Check your understanding
1. Why does aggregate consumption grow more slowly than individual consumption \(\dot c/c = r-\theta\)?
- Old agents face a higher interest rate on their annuities
- Dying agents are replaced by newborns who hold no wealth
- Newborns discount the future at a higher rate than adults
- Firms retain part of profits instead of paying dividends
The term \(-(p+\alpha)(p+\theta)V\) is generational turnover. Wealthy dying cohorts are replaced by newborns with \(v(t,t)=0\). With \(\alpha>0\), wealth also tilts towards the old, whose labour income has decayed. Everyone faces the same \(r+p\) on annuities and has the same \(\theta\).
2. With \(\alpha=0\) and \(p>0\), where does the steady-state interest rate lie?
- Exactly at θ, as in the Ramsey model
- Below zero, past the golden-rule level
- Between zero and θ, below impatience
- Strictly between θ and θ plus p
\(\theta\lt \mathcal F'(\bar K)\lt \theta+p\). The lower bound comes from the sign of the steady-state equation; the upper bound from concavity (\(\mathcal F > \mathcal F'K\)) evaluated at \(K^{**}\). So the economy is below the golden rule and dynamically efficient.
3. Along the transition from \(K(0)\lt \bar K\), which variable falls?
- The interest rate r
- Human wealth H
- Labour income Y
- Consumption C
\(K\) rises along the saddle path, so \(r=\mathcal F'(K)\) falls while \(Y = \mathcal F - \mathcal F'K\) rises. \(H\) rises because future wages rise and future discount rates fall. \(C=S(K)\) rises.
4. Which parameter change raises steady-state capital \(\bar K\)?
- A higher death rate p
- A higher discount rate θ
- A faster income decline α
- A higher depreciation δ
Raising \(\alpha\) lowers the right side \((p+\alpha)(p+\theta)/(\mathcal F'+\alpha-\theta)\) whenever \(\mathcal F'(\bar K)\lt \theta+p\), which always holds. So the intersection with the falling average-product curve moves right. Higher \(p\) or \(\theta\) raise the right side. Higher \(\delta\) lowers net average product and \(\mathcal F'\), which shrinks \(\bar K\).
5. Why must \(C(0)\) lie on the saddle path when \(p>0\)?
- Paths elsewhere violate the government budget constraint
- Paths elsewhere reach zero capital or consumption in finite time
- Paths elsewhere converge to a second, unstable steady state
- Paths elsewhere imply negative wages in the long run
Above \(S\), \(C\) keeps rising while \(K\) falls, so \(K\) hits 0 with \(C>0\). Below \(S\), \(\dot C<0\) with \(\dot C = -(p+\alpha)(p+\theta)K<0\) even at \(C=0\), so \(C\) turns negative in finite time. Both are infeasible. (With \(p=0\) we need the transversality condition instead.)
9. Exercises
Problem Set 2 (2025) works with this system extended to include government debt; those parts are in Lesson 20. The exercises below follow Alvarez's formats: true/false with a short justification, and guided derivations with hints and point weights. Work with a general concave \(\mathcal F\) unless told otherwise.
Exercise 19.1 Exam-inspired True or false? Justify. (7 × 10 points)
- In the perpetual-youth model with capital and flat income profiles (\(\alpha=0\)), the steady state is dynamically efficient. exam-inspired
Show solution
True. With \(\alpha=0\) the steady state satisfies \(\mathcal F(\bar K)[\mathcal F'(\bar K)-\theta]=p(p+\theta)\bar K>0\), so \(\bar r=\mathcal F'(\bar K)>\theta>0\). Hence \(\bar K<K_{GR}\) (where \(\mathcal F'=0\)): no overaccumulation.
- In the same model, the steady-state interest rate equals the rate of time preference, as in the Ramsey model. exam-inspired
Show solution
False unless \(p=0\). With \(\alpha=0\), \(\theta<\bar r<\theta+p\). Generational turnover keeps transferring capital to wealthless newborns, so aggregate saving is lower than in Ramsey and \(\bar K\) is smaller. Individual consumption grows at \(\bar r-\theta>0\) over the life cycle.
- A higher death rate \(p\) raises the steady-state capital stock, because people save more against an uncertain lifetime. exam-inspired
Show solution
False. \(\bar K\) is decreasing in \(p\). At fixed \(K\), the right side of \(\mathcal F/K=(p+\alpha)(p+\theta)/(\mathcal F'+\alpha-\theta)\) rises with \(p\), shifting the intersection left. A higher \(p\) raises the MPC \(\theta+p\) and speeds up turnover. With fair annuities there is no precautionary motive against death.
- If labour income declines with age fast enough, the perpetual-youth economy can accumulate capital beyond the golden rule. exam-inspired
Show solution
True. \(\bar r<0\) iff \((\alpha-\theta)\,\mathcal F(K_{GR})/K_{GR}>(p+\alpha)(p+\theta)\). This needs \(\alpha>\theta\), and given \(\alpha>\theta\) it holds when \(p\) is small enough (Exercise 19.4). Agents save heavily for the part of life when their earnings have decayed.
- As \(p\downarrow0\) (with \(\alpha=0\)), the steady state converges to the modified golden rule \(\mathcal F'(K)=\theta\). exam-inspired
Show solution
True. The bounds \(\theta<\mathcal F'(\bar K)<\theta+p\) squeeze \(\mathcal F'(\bar K)\) to θ. In the limit the \(\dot C=0\) locus becomes the vertical line \(\mathcal F'(K)=\theta\) of the Ramsey model.
- Because \(C=(\theta+p)(K+H)\) holds at every date, the saddle path \(C=S(K)\) is a straight line in the \((K,C)\) plane. exam-inspired (slide 27)
Show solution
False. Along the equilibrium path, \(H\) is itself a function of the state, \(H=\mathcal H(K)\), because wages and interest rates depend on the future path of \(K\). So \(S(K)=(\theta+p)[K+\mathcal H(K)]\), which is generally nonlinear. It is linear only in special cases where \(\mathcal H\) is affine in \(K\).
- Starting from \(K(0)<\bar K\), the interest rate falls monotonically along the equilibrium path while consumption rises. exam-inspired (slide 29)
Show solution
True. On the upward-sloping saddle path below the steady state, \(K\) and \(C\) rise monotonically. Then \(r=\mathcal F'(K)\) falls monotonically and the wage \(Y=\mathcal F-\mathcal F'K\) rises.
Exercise 19.2 Proof The steady state is a saddle (40 points)
- (10) Linearize \(\dot C=(\mathcal F'(K)+\alpha-\theta)C-(p+\alpha)(p+\theta)K\), \(\dot K=\mathcal F(K)-C\) around \((\bar K,\bar C)\).
- (20) Show that the Jacobian has a negative determinant. [Hint: use the steady-state condition to replace \((p+\alpha)(p+\theta)\) by \(\mathcal F(\bar K)(\mathcal F'(\bar K)+\alpha-\theta)/\bar K\), and use that average product exceeds marginal product.]
- (10) Conclude that there is one stable and one unstable eigenvalue, and find the slope of the saddle path at the steady state.
Show solution
(a) In the order \((K,C)\): \(J = \begin{pmatrix} \mathcal F' & -1 \\ \mathcal F''\bar C - (p+\alpha)(p+\theta) & \mathcal F'+\alpha-\theta \end{pmatrix}\), evaluated at \(\bar K\).
(b) \(\det J = \mathcal F'(\mathcal F'+\alpha-\theta) + \mathcal F''\bar C - (p+\alpha)(p+\theta)\). Substituting, \(\det J = (\mathcal F'+\alpha-\theta)\big(\mathcal F' - \tfrac{\mathcal F}{\bar K}\big) + \mathcal F''\bar C\). The first factor is positive (we are left of \(\hat K\)), the second is negative by concavity with \(\mathcal F(0)=0\), and \(\mathcal F''\bar C<0\). So \(\det J<0\).
(c) The product of the eigenvalues is \(\det J<0\), so they are real with opposite signs, \(\lambda_s<0<\lambda_u\): a saddle. The first row gives \((\mathcal F'-\lambda_s)dK-dC=0\), so the saddle path has slope \(dC/dK=\mathcal F'(\bar K)-\lambda_s\). This is positive when \(\mathcal F'(\bar K)\ge0\).
Exercise 19.3 Slides pp.30–32 Comparative statics of the steady state (40 points)
The steady state solves \(\dfrac{\mathcal F(\bar K)}{\bar K}=\dfrac{(p+\alpha)(p+\theta)}{\mathcal F'(\bar K)+\alpha-\theta}\) with \(\bar K<\hat K\), where \(\mathcal F'(\hat K)=\theta-\alpha\).
- (10) Show that the left side is decreasing and the right side increasing in \(K\) on \((0,\hat K)\), so the steady state is unique.
- (10) Show that \(\bar K\) is decreasing in \(p\) and in \(\theta\).
- (10) With \(\alpha=0\), show \(\theta<\mathcal F'(\bar K)<\theta+p\). [Hint: evaluate \(g(K)=\mathcal F(K)[\mathcal F'(K)-\theta]-p(p+\theta)K\) at \(K^{**}\), where \(\mathcal F'(K^{**})=\theta+p\). Watch the direction of the concavity inequality; see the note in §6.]
- (10) Show that \(\bar K\) is increasing in α, first near \(\alpha=0\) and then for all \(\alpha\ge0\).
Show solution
(a) \(\frac{d}{dK}\frac{\mathcal F}{K}=\frac{\mathcal F'K-\mathcal F}{K^2}<0\). The right side's numerator is constant, and its denominator \(\mathcal F'(K)+\alpha-\theta\) is positive and decreasing on \((0,\hat K)\), so the right side is increasing. It goes to \(+\infty\) as \(K\uparrow\hat K\). With Inada conditions, the left side exceeds the right side near 0, so there is exactly one crossing.
(b) At fixed \(K\), raising \(p\) raises the numerator. Raising θ raises the numerator and lowers the denominator. Either way the right-side curve shifts up, so it crosses the decreasing left side at a smaller \(K\).
(c) The right side is positive, so \(\mathcal F'(\bar K)>\theta\). At \(K^{**}\), \(g(K^{**})=p[\mathcal F(K^{**})-\mathcal F'(K^{**})K^{**}]>0\) by concavity. Since \(g>0\) exactly to the left of \(\bar K\), we get \(K^{**}<\bar K\), i.e. \(\mathcal F'(\bar K)<\theta+p\).
(d) It suffices that \((p+\alpha)/(\mathcal F'+\alpha-\theta)\) falls with α at fixed \(K\). Its derivative is \((\mathcal F'-\theta-p)/(\mathcal F'+\alpha-\theta)^2\), negative iff \(\mathcal F'(\bar K)<\theta+p\). That holds at \(\alpha=0\) by (c). As α rises, \(\bar K\) rises and \(\mathcal F'(\bar K)\) falls, so the inequality keeps holding: \(\bar K\) is increasing for all \(\alpha\ge0\).
Exercise 19.4 Exam-inspired When is the economy dynamically inefficient? (40 points)
- (15) Show that \(\bar r<0\) (equivalently \(\bar K>K_{GR}\)) if and only if \((\alpha-\theta)\,\mathcal F(K_{GR})/K_{GR}>(p+\alpha)(p+\theta)\). [Hint: \(\bar K>K_{GR}\) iff the \(\dot C=0\) locus lies below \(C=\mathcal F(K)\) at \(K_{GR}\).]
- (10) Let \(F(K,1)=K^a\) with depreciation δ, so \(\mathcal F(K)=K^a-\delta K\). Show that \(\mathcal F(K_{GR})/K_{GR}=\delta(1-a)/a\).
- (10) For the Cobb–Douglas case with \(\alpha>\theta\), find the threshold \(\bar p\) in closed form such that \(\bar r<0\) iff \(p<\bar p\).
- (5) Why do longer expected lifetimes make overaccumulation possible when \(\alpha>\theta\), but impossible when \(\alpha=0\)?
Show solution
(a) The \(\dot C=0\) locus \(C=(p+\alpha)(p+\theta)K/(\mathcal F'+\alpha-\theta)\) starts below \(\mathcal F\) near 0, is increasing and convex, and crosses \(\mathcal F\) once at \(\bar K\). So \(\bar K>K_{GR}\) iff the locus is still below \(\mathcal F\) at \(K_{GR}\). For the locus to be defined at \(K_{GR}\) with \(\mathcal F'(K_{GR})=0\), we need \(\alpha>\theta\). The condition is then \((p+\alpha)(p+\theta)K_{GR}/(\alpha-\theta)<\mathcal F(K_{GR})\), which rearranges to the stated inequality. If \(\alpha\le\theta\), then \(\hat K\le K_{GR}\), so \(\bar K<K_{GR}\) automatically.
(b) \(aK_{GR}^{a-1}=\delta\), so \(\mathcal F/K=K_{GR}^{a-1}-\delta=\delta/a-\delta\).
(c) We need \(p^2+(\alpha+\theta)p+\alpha\theta<(\alpha-\theta)\delta(1-a)/a\equiv M\). The left side is increasing in \(p\ge0\) and equals \(\alpha\theta\) at \(p=0\). So \(\bar p=\tfrac12\big[-(\alpha+\theta)+\sqrt{(\alpha-\theta)^2+4M}\big]\), which is positive iff \(M>\alpha\theta\).
(d) With \(\alpha>\theta\), agents save for the period when their earnings have decayed. The longer they expect to live, the more they save, and the slower turnover redistributes capital to newborns. With \(\alpha=0\), the only saving motive is the tilt \(\bar r>\theta\), which requires \(\bar r>0\). As \(p\to0\) the steady state approaches \(\bar r=\theta>0\), never past the golden rule.
Exercise 19.5 Exam-inspired An unexpected rise in mortality (30 points)
The economy (\(\alpha=0\)) is in steady state. At \(t=0\) the death rate rises unexpectedly and permanently from \(p\) to \(p'>p\).
- (10) Which locus shifts, and in which direction? Compare the new steady state with the old.
- (15) Does \(C\) jump up or down on impact? Describe the subsequent paths of \(K\), \(C\) and \(r\).
- (5) Give the economic intuition for the impact effect.
Show solution
(a) \(\dot K=0\) does not involve \(p\). The \(\dot C=0\) locus shifts up, because its numerator \(p(p+\theta)\) is bigger. So \(\bar K'<\bar K\) and \(\bar C'=\mathcal F(\bar K')<\bar C\), since we are left of the golden rule.
(b) \(K\) is predetermined, so \(C\) jumps onto the new saddle path at the old \(\bar K\). To the right of the new steady state, the saddle path lies above \(C=\mathcal F(K)\), where \(\dot K<0\). So \(C(0^+)>\bar C\): consumption jumps up. Then \(K\) and \(C\) fall monotonically to the new steady state, and \(r\) rises monotonically.
(c) Higher mortality raises everyone's MPC out of wealth, \(\theta+p\), so consumption rises and saving falls on impact.
Exercise 19.6 Slides pp.27, 34 Proof Ruling out the other paths (30 points)
- (20) Show that any path starting below the saddle path reaches \(C=0\) in finite time when \(p>0\).
- (10) Explain why this argument fails when \(p=\alpha=0\), and which condition replaces it.
Show solution
(a) A path below \(S\) eventually enters \(\mathcal R=\{C<\mathcal F(K),\ C<\text{the }\dot C=0\text{ locus}\}\), where \(\dot K>0\) and \(\dot C<0\). \(\mathcal R\) is invariant: the path moves right and down while the \(\dot C=0\) locus is increasing, and on the \(\dot K=0\) boundary the path moves straight down. Inside \(\mathcal R\), \(K\) is increasing and bounded and \(C\) is decreasing. If \(C\) stayed positive, the path would converge to a rest point in the closure of \(\mathcal R\). But \((\bar K,\bar C)\) is not in that closure, and at \(C=0\), \(\dot C=-(p+\alpha)(p+\theta)K<0\), so no point with \(C=0\) is a rest point. Hence \(C\) hits 0 in finite time, which is infeasible.
(b) With \(p=\alpha=0\), \(\dot C=(\mathcal F'-\theta)C\) is proportional to \(C\), so \(C\) decays only exponentially and never reaches 0. The path converges to \((K_{\max},0)\), which is feasible but violates the transversality condition \(\lim_{T\to\infty}K(T)e^{-\int_0^Tr}=0\), since \(r=\mathcal F'(K_{\max})<0\). That is the "slightly different argument" of slide 34.
Exercise 19.7 Slides pp.27, 29 Adjustment and the saddle path (30 points)
- (15) Starting from \(K(0)>\bar K\), describe the paths of \(C\), \(K\), \(Y\), \(r\) and \(H\). Are they monotone?
- (10) What is the relationship between \(C=S(K)\) and \(C=(\theta+p)(K+H)\)? Is \(S\) necessarily linear?
- (5) Explain why aggregate human wealth \(H\) can move even though no individual's labour-services profile changes.
Show solution
(a) Everything reverses relative to \(K(0)<\bar K\): \(K\), \(C\) and \(Y=\mathcal F-\mathcal F'K\) fall monotonically, and \(r=\mathcal F'(K)\) rises monotonically. \(H\), the PV of wages discounted at \(r+p+\alpha\), falls because wages fall and discount rates rise.
(b) The consumption function holds on every equilibrium path. On the saddle path, \(H\) is a function of the state, \(H=\mathcal H(K)\), so \(S(K)=(\theta+p)[K+\mathcal H(K)]\). It is not linear in general, because \(\mathcal H\) inherits the curvature of \(\mathcal F\) through future wages and rates.
(c) \(H\) is price-weighted: a present value of market wages at market interest rates. Both prices move with \(K\) in general equilibrium.
Key takeaways
- With capital: \(\dot C = [\mathcal F'(K)+\alpha-\theta]C - (p+\alpha)(p+\theta)K\) and \(\dot K = \mathcal F(K)-C\). It is Ramsey plus a generational-turnover term.
- \(\dot K=0\): \(C=\mathcal F(K)\), hump-shaped with its peak at the golden rule. \(\dot C=0\): increasing and convex, with an asymptote at \(\mathcal F'(\hat K)=\theta-\alpha\). The steady state is unique (falling average product vs rising RHS), and it is a saddle.
- From \(K(0)\lt \bar K\): \(K,C,Y,H\) rise and \(r\) falls, monotonically. \(S(K) = (\theta+p)(K+h(K))\) is nonlinear in general.
- \(\bar K\) falls with \(p\) and \(\theta\) and rises with \(\alpha\). With \(\alpha=0\): \(\theta\lt \bar r\lt \theta+p\), so the economy is dynamically efficient. With \(\alpha>\theta\) and long enough lifetimes, \(\bar r<0\) is possible.
- As \(p\to0\) (forcing \(\alpha=0\)), we get Ramsey: a vertical \(\dot C=0\) locus at \(\mathcal F'=\theta\), and the transversality condition rules out the bad paths.
💬 Ask your teacher. Try: "derive the convexity of the \(\dot C=0\) locus for Cobb–Douglas", "why does \(H\) use \(r+p+\alpha\) but \(V\) grow at \(r\)?", or "redo Exercise 19.4 for an increase in \(\theta\)".