\(p\) = death rate, \(\theta\) = discount rate, \(\alpha\) = rate of decline of labour income with age (as in Lessons 17–19). \(T\) = aggregate taxes, \(G\) = government purchases, \(B\) = government debt (so \(B>0\) means the government owes the private sector). This is the opposite of Class Note III (Lesson 14), where \(B_t\) denotes government assets and \(B_t<0\) is debt.
BY slide 35 The sequential constraint says debt grows with the primary deficit plus interest:
$$\dot B(z) = G(z) - T(z) + r(z)B(z).$$Given the sequential constraint, the boundary condition \(\lim_{T'\to\infty}B(T')e^{-\int_t^{T'}r(\mu)d\mu}=0\) holds if and only if $$B(t) = \int_t^\infty [T(z)-G(z)]\;e^{-\int_t^z r(\mu)d\mu}\,dz .$$ Today's debt equals the present value of future primary surpluses.
This is the same algebra as the household's \(v\) and \(h\) in Lesson 17, but with discount rate \(r\) instead of \(r+p\). The government does not die.
BY slide 36 To keep the algebra short, set \(\alpha=0\) and let taxes be lump sum, the same for everyone alive. Labour income becomes after-tax labour income \(Y-T\). Private non-human wealth now includes government bonds: \(V = K + B\).
$$C = (p+\theta)(H+V),\qquad H(t) = \int_t^\infty [Y(z)-T(z)]\,e^{-\int_t^z (r+p)}\,dz,\qquad \dot V = rV + Y - T - C .$$Households discount future taxes at \(r+p\) because there is a probability \(1-e^{-p(z-t)}\) that they will be dead (and so will not pay) by date \(z\). The government's budget discounts at \(r\): someone will pay those taxes, but partly people who are not yet born. That wedge is the entire story of this lesson.
BY slide 37 Cut taxes at \(t\) by \(dT(t)<0\), finance the cut with debt, and repay it with a tax increase at \(t+s\). Hold the path of \(r\) fixed (partial equilibrium).
The people alive at \(t\) receive the whole cut. Only a fraction \(e^{-ps}\) of them are still alive at \(t+s\) to pay the higher tax. The rest is paid by people born after \(t\), whose welfare does not enter anyone's budget today. So the current population is richer, even though the government's present-value constraint (discounting at \(r\)) is unchanged. Two limits:
Example: \(p=0.05\), \(s=10\) years. Each dollar of tax cut raises human wealth by \(1-e^{-0.5}\approx 0.39\) dollars.
BY slide 38 Now allow \(\alpha\ge0\). To keep the aggregate \(\dot C\) equation unchanged, assume taxes are levied in proportion to labour income, so after-tax income still declines with age at rate \(\alpha\). (With \(\alpha=0\) this is automatic.) Then the derivation of Lesson 18 goes through with \(Y\) replaced by \(Y-T\), and \(\dot C = (r+\alpha-\theta)C - (p+\alpha)(p+\theta)V\) still holds.
It is convenient to scale by \(Y-G\): let \(b \equiv B/(Y-G)\), \(k\equiv(p+\alpha)(p+\theta)\) and \(\sigma \equiv (T-G)/(Y-G)\) (the primary surplus ratio). Then $$\dot b = k b^2 + (\theta-\alpha)\,b - \sigma \;=\; r(b)\,b - \sigma, \qquad r(b) = kb + \theta - \alpha .$$ The second form makes steady states transparent: \(\dot b = 0 \iff r^* b^* = \sigma\). In a steady state, the interest bill is exactly covered by the primary surplus.
The parabola \(\dot b = kb^2+(\theta-\alpha)b-\sigma\) shifts vertically by \(-\sigma\). Its roots are \(b_\pm = \dfrac{-(\theta-\alpha) \pm\sqrt{(\theta-\alpha)^2 + 4k\sigma}}{2k}\), and \(\tfrac{d\dot b}{db}\big|_{b_\pm} = \pm\sqrt{(\theta-\alpha)^2+4k\sigma}\). So the smaller root is always the stable one and the larger root the unstable one.
| Primary balance | Steady states | Stable root | Interpretation |
|---|---|---|---|
| Surplus \(\sigma>0\) | always two, one each side of 0 (product \(=-\sigma/k<0\)) | \(b_-<0\) | \(b_+>0\) is the "Ricardian" debt level, the present value of surpluses at \(r^*=\sigma/b_+>0\). It is unstable: start above it and debt explodes; start below it and the surpluses pay off the debt and the government accumulates assets. |
| Deficit \(\sigma<0\), \(\alpha>\theta\) (\(\bar r<0\)) | two positive roots if \(|\sigma|\le(\theta-\alpha)^2/4k\), none otherwise | \(b_->0\) | A permanent primary deficit is sustainable with positive debt, at \(r^* = \sigma/b_-<0\). This is Blanchard's point: when the economy is dynamically inefficient, the government can run deficits forever, up to a maximum size. |
| Deficit \(\sigma<0\), \(\theta>\alpha\) (\(\bar r>0\)) | two negative roots, or none | \(b_-<0\) | A permanent deficit is sustainable only if the government owns enough assets that their interest covers it. Positive debt plus deficits explode. |
How it depends on \(\theta-\alpha\): the sign of \(\theta-\alpha\) (the sign of the no-debt interest rate) decides whether the stable steady state with a deficit involves positive debt (\(\alpha>\theta\)) or government assets (\(\theta>\alpha\)). Its size decides how large a deficit is sustainable: \(|\sigma|_{\max} = (\theta-\alpha)^2/4k\).
Take Class Note II's economy (young endowment \(1-\alpha\), old endowment \(\alpha\), utility \((1-\beta)\log c_y + \beta\log c_o\); note that this \(\alpha\) is the Note II endowment, not BY's \(\alpha\)). The young hold government debt: \(B_t = s(r_t)\) with \(s(r) = \beta(1-\alpha) - (1-\beta)\alpha/(1+r)\) (Lesson 9). With \(T=G=0\), \(B_{t+1} = (1+r_t)B_t\). Solving \(B_t = s(r_t)\) for \(1+r_t\) and substituting: $$B_{t+1} = \phi(B_t) \equiv \frac{(1-\beta)\alpha\,B_t}{\beta(1-\alpha) - B_t}.$$
BY slide 40 Set \(\alpha=0\) and hold \(T\) and \(G\) constant. Now \(V=B+K\), and the system has three equations:
$$\dot C = (r-\theta)C - p(p+\theta)(B+K),\qquad \dot K = \mathcal F(K) - C - G,\qquad \dot B = rB + G - T,\qquad r=\mathcal F'(K).$$Steady states satisfy \([\mathcal F'(K^*)-\theta]C^* = p(p+\theta)(B^*+K^*)\), \(C^* = \mathcal F(K^*)-G\), and \(\mathcal F'(K^*)B^* = T-G\). Eliminating \(C^*\) and \(B^*\) BY slide 41: $$\mathcal F(K^*) - G = \frac{p(p+\theta)}{\mathcal F'(K^*)-\theta}\left[\frac{T-G}{\mathcal F'(K^*)} + K^*\right].$$
| Numerical illustration: \(F(K,1)=K^{0.33}\), \(\delta=0.08\), \(\theta=0.03\) (net output ≈ 1.28) | |||||
|---|---|---|---|---|---|
| \(p\) | \((G, T)\) | \(K^*\) | \(r^*\) | \(B^*\) | \(C^*\) |
| 0.05 | (0, 0) | 4.332 | 0.0436 | 0 | 1.276 |
| 0.05 | (0, 0.05): tax ↑ | 4.180 | 0.0466 | 1.074 | 1.269 |
| 0.05 | (0, −0.05): tax ↓ | 4.515 | 0.0402 | −1.244 | 1.283 |
| 0.05 | (0.05, 0): spending ↑ | 4.493 | 0.0406 | −1.232 | 1.232 |
| 0.01 | (0, 0.05): tax ↑ | 5.016 | 0.0320 | 1.561 | 1.301 |
| 0.001 | (0, 0.05): tax ↑ | 5.142 | 0.0302 | 1.658 | 1.305 |
The table shows each answer: higher \(T\) means more debt and less capital; a tax cut and a spending increase of the same size give almost the same \(B^*\), but the spending increase leaves less for consumption and raises \(K^*\) slightly less; and as \(p\to0\), \(r^*\to\theta=0.03\) whatever the fiscal policy.
Now the government keeps \(B\) constant and adjusts taxes, \(T = \mathcal F'(K)B + G\). The steady state solves $$[\mathcal F'(K^*)-\theta]\,[\mathcal F(K^*)-G] = p(p+\theta)\,(B+K^*).$$
Exercise 20.2 is Problem Set 2 (2025), which asks you to linearize the debt-and-capital system and count its eigenvalues. No past exam has asked about Blanchard–Yaari directly. The other exercises follow the formats Alvarez does use: true/false with a short justification, and guided derivations with hints and point weights. Use the notation of §4: \(k\equiv(p+\alpha)(p+\theta)\), \(b\equiv B/(Y-G)\), \(\sigma\equiv(T-G)/(Y-G)\).
False when \(p>0\). Human wealth of those alive rises by \(dH=-dT(t)[1-e^{-ps}]>0\) for a tax cut \(dT(t)<0\), so \(C\) rises by \((\theta+p)dH\). Only the fraction \(e^{-ps}\) of today's population is still alive when the repayment comes; the rest of the bill falls on people not yet born. It is true when \(p=0\).
False. Fair annuities are already in the model. The failure comes from the arrival of new agents who will pay part of future taxes and are not linked to current agents by altruism. In the aggregate, the death rate \(p\) is the rate at which taxpayers are replaced by newcomers.
False. \(\dot B=kB^2/Y+(\theta-\alpha)B\) has slope \(\alpha-\theta>0\) at \(\bar B\), so it is unstable. The stable steady state is \(B=0\), whose slope is \(\theta-\alpha<0\). Starting below \(\bar B\), debt shrinks to zero; starting above, it explodes and no equilibrium exists.
True. With no primary surplus, a constant \(B>0\) requires \(rB=0\), so \(r(\bar B)=k\bar B/Y+\theta-\alpha=0\). This is the continuous-time counterpart of Note II's steady state with valued debt, which reproduces the PAYG allocation (Exercise 20.4).
False. A steady state needs \(r^*b^*=\sigma<0\), so \(b^*>0\) requires \(r^*<0\). But \(r(b)=kb+\theta-\alpha>0\) for every \(b\ge0\) when \(\theta>\alpha\). So the steady states (if any) have \(b^*<0\): the government must own assets. Positive debt plus a deficit explodes.
False. In the steady state \(B^*=(T-G)/r^*\) rises, so households hold more bonds. Bonds are net wealth for finite-lived households, which raises desired consumption. Bonds therefore crowd capital out: \(K^*\) falls and \(r^*\) rises. (The table in §5 shows \(K^*\) falling from 4.33 to 4.18.)
True. The steady state solves \([\mathcal F'(K)-\theta][\mathcal F(K)-G]=p(p+\theta)(B+K)\). A higher \(B\) raises the right side, lowering \(K^*\). As \(p\to0\) the right side vanishes, \(\mathcal F'(K^*)=\theta\), and debt is neutral (Ricardian).
False. The linearized system has two positive eigenvalues and one negative eigenvalue (Exercise 20.2). With two predetermined states \((K,B)\) and one jump variable \(C\), convergence would need two stable roots. Generically no path converges: \(b\) grows at rate \(r=\mathcal F'(\bar K)>0\). Debt is explosive unless taxes respond to it.
Take \(\alpha=0\) and constant \(T,G\). \(B\) is government debt. $$\dot C=(\mathcal F'(K)-\theta)C-p(p+\theta)(B+K),\qquad \dot K=\mathcal F(K)-C-G,\qquad \dot B=\mathcal F'(K)B+G-T.$$ \(K\) and \(B\) are predetermined; \(C\) can jump.
(a) $$J=\begin{pmatrix}\mathcal F'-\theta & \mathcal F''C-p(p+\theta) & -p(p+\theta)\\ -1 & \mathcal F' & 0\\ 0 & \mathcal F''B & \mathcal F'\end{pmatrix}.$$ (The official solution's text says the \((1,2)\) term multiplies \(c\); it multiplies \(k\). The matrix itself is right.)
(b) \(\dot B=0\) gives \(\mathcal F'(K)B=0\), and \(\mathcal F'>0\), so \(B=0\). Then \(C=\mathcal F(K)\) and \([\mathcal F'(K)-\theta]\mathcal F(K)=p(p+\theta)K\): exactly the steady state of Lesson 19 with \(\alpha=0\).
(c) With \(B=0\), the \(\dot b\) row of \(J-\lambda I\) is \((0,0,\mathcal F'-\lambda)\). Expanding along it gives \(\det(J-\lambda I)=(\mathcal F'-\lambda)\,g(\lambda)\), where \(g\) is the characteristic polynomial of the \((C,K)\) block. So one root is \(\mathcal F'\) and the other two are the roots of \(g\).
(d) From the steady state, \(\mathcal F'-\theta=p(p+\theta)K/\mathcal F\). So \((\mathcal F'-\theta)\mathcal F'-p(p+\theta)=p(p+\theta)(\mathcal F'K-\mathcal F)/\mathcal F<0\), by concavity with \(\mathcal F(0)=0\). Also \(\mathcal F''C<0\), so \(g(0)<0\). The roots of \(g\) multiply to \(g(0)<0\): one positive, one negative (the familiar saddle of Lesson 19). Together with \(\lambda=\mathcal F'(\bar K)=\bar r>0\), that makes two positive and one negative. (The official solution stops at "\(g\) has a positive root"; this completes the count.)
(e) No. With two predetermined variables and only one stable root, the stable manifold is one-dimensional, while the initial conditions span two dimensions. For generic \(B(0)\ne0\), no choice of \(C(0)\) puts the system on a convergent path. Linearized debt obeys \(\dot b=\bar r\,b\) and explodes at rate \(\bar r>0\) when taxes don't respond. Only \(b(0)=0\) works, with \(C(0)\) on the \((C,K)\) saddle path.
(f) \(B=T/\mathcal F'(K)>0\): positive debt serviced by the surplus. The \((3,2)\) entry \(\mathcal F''B\) is now nonzero. The characteristic equation is $$(\mathcal F'-\lambda)\,g(\lambda)+\mathcal F''B\,p(p+\theta)=0,$$ since the cofactor of the \((3,2)\) entry is \((-1)^{3+2}\) times the minor \(-p(p+\theta)\), i.e. \(+p(p+\theta)\). (The official solution writes this term with a minus sign. That is a sign slip, but it is immaterial for the argument.) As \(T\downarrow0\), \(B\to0\) and the extra term vanishes. The roots are continuous in the coefficients, so for small \(T\) there are still two positive roots and one negative. Debt dynamics remain explosive: a stable debt path requires taxes that respond to the level of debt.
Set \(\alpha=0\). At date \(t\) the government cuts taxes by \(dT(t)<0\) and raises them at \(t+s\) by enough to repay the extra debt with interest.
(a) The extra debt accumulates at \(r\), so \(dT(t+s)=-e^{\int_t^{t+s}r}dT(t)>0\).
(b) \(H(t)=\int_t^\infty[Y-T]e^{-\int_t^z(r+p)}dz\), so \(dH(t)=-dT(t)-dT(t+s)e^{-\int_t^{t+s}(r+p)} = -dT(t)\big[1-e^{-ps}\big]\). Hence \(dC(t)=(\theta+p)dH(t)=-(\theta+p)(1-e^{-ps})dT(t)>0\).
(c) \(p\to0\): \(dC\to0\), Ricardian equivalence holds with infinitely lived agents. \(s\to0\): \(dC\to0\), because the same people repay almost immediately. \(s\to\infty\): \(dH\to-dT\) and \(dC\to-(\theta+p)dT\). Repayment falls entirely on the unborn, so the tax cut is pure net wealth for those alive.
(d) No. Goods-market clearing forces \(C=Y-G\), so aggregate consumption cannot rise. The interest rate adjusts instead: with \(B>0\), \(r=kB/(Y-G)+\theta-\alpha\) rises, enough that households willingly hold the new debt and keep \(C=Y-G\). The consumption profiles of different cohorts shift too. Prices and the distribution of consumption change, so Ricardian equivalence fails even though aggregate \(C\) cannot move.
Take the Class Note II economy (young endowment \(1-\alpha\), old \(\alpha\); utility \((1-\beta)\log c_y+\beta\log c_o\)) and add one-period government bonds, with no taxes and no spending. The young at \(t\) buy the whole debt \(B_t\), which is repaid with interest at \(t+1\) by issuing new debt.
(a) Asset-market clearing gives \(B_t=s(r_t)\), so \(1+r_t=(1-\beta)\alpha/(\beta(1-\alpha)-B_t)\). The debt rolls over: \(B_{t+1}=(1+r_t)B_t\). Substituting gives \(\phi\).
(b) The steady states are \(B=0\) and the solution of \(\beta(1-\alpha)-\bar B=(1-\beta)\alpha\), i.e. \(\bar B=\beta-\alpha\). This is positive iff \(\beta>\alpha\), iff \(\bar r=(\alpha-\beta)/(\beta(1-\alpha))<0\). At \(\bar B\) we have \(r=0\): each young agent lends \(\beta-\alpha\) and gets back \(\beta-\alpha\) when old, exactly the transfers of PAYG with \(\tau=\beta-\alpha\).
(c) \(\phi'(B)=(1-\beta)\alpha\beta(1-\alpha)/(\beta(1-\alpha)-B)^2\). At 0 this is \((1-\beta)\alpha/(\beta(1-\alpha))=1+\bar r\). At \(\bar B\) it is \(\beta(1-\alpha)/((1-\beta)\alpha)=1/(1+\bar r)\). When \(\bar r<0\), \(0\) is stable and \(\bar B\) is unstable: the same picture as Blanchard–Yaari with \(\alpha>\theta\).
(d) The initial old, who sell the first bonds and consume the proceeds. At \(\bar B\), every later generation also gains relative to autarky, as under PAYG.
In the endowment economy, \(\dot b=kb^2+(\theta-\alpha)b-\sigma\), with \(r(b)=kb+\theta-\alpha\).
(a) \(\dot b=b\,r(b)-\sigma=0\iff r^*b^*=\sigma\). In a steady state the interest bill on the debt equals the primary surplus.
(b) The roots are \(b_\pm=\big[(\alpha-\theta)\pm\sqrt{(\theta-\alpha)^2+4k\sigma}\big]/(2k)\). They are real iff \((\theta-\alpha)^2\ge-4k\sigma\), i.e. \(|\sigma|\le(\theta-\alpha)^2/(4k)\).
(c) With \(\sigma<0\), the product of the roots is \(-\sigma/k>0\) and their sum is \((\alpha-\theta)/k>0\), so both are positive. The slope \(d\dot b/db=2kb+\theta-\alpha\) equals \(\mp\sqrt{(\theta-\alpha)^2+4k\sigma}\) at \(b_\mp\), so \(b_-\) is stable and \(b_+\) unstable. From (a), \(r^*=\sigma/b^*<0\) at both: the deficit is financed by rolling over debt at a negative interest rate.
(d) Above \(b_+\), \(\dot b>0\) and the gap grows: debt and \(r\) explode and no equilibrium exists without a fiscal adjustment. If \(\theta>\alpha\), the sum of the roots \((\alpha-\theta)/k\) is negative, so any steady states are negative (government assets). A deficit with positive debt is unsustainable.
Take the capital economy of §5 (\(\alpha=0\)).
(a) From \(\dot B=0\), \(B^*=(T-G)/\mathcal F'(K^*)\). From \(\dot K=0\), \(C^*=\mathcal F(K^*)-G\). Substitute both into \([\mathcal F'-\theta]C^*=p(p+\theta)(B^*+K^*)\).
(b) \(B^*=(T-G)/r^*\) rises. More bonds mean more private wealth at a given \(K\), which raises the right side, so \(K^*\) falls (the same crossing argument as in Lesson 19). Then \(r^*\) rises and \(C^*=\mathcal F(K^*)-G\) falls, since we are below the golden rule. Bonds are net wealth for finite-lived households and displace capital in their portfolios.
(c) Both lower \(T-G\) by the same amount, so \(B^*\) falls by about the same amount: the government owes less (or owns more assets), which crowds capital in. But \(G\) also absorbs output, so \(C^*=\mathcal F(K^*)-G\) is lower and the left side of the steady-state equation falls. As a result, \(K^*\) rises by less under the spending increase than under the equal tax cut (§5 table: 4.493 vs 4.515), and consumption falls with \(G\) but rises with the tax cut.
(d) Both lower \(K^*\) and raise \(r^*\) when \(p>0\). As \(p\to0\), \(\mathcal F'(K^*)=\theta\) pins down \(K^*\), so neither affects capital. \(G\) still crowds out consumption one-for-one, while \(B\) becomes neutral. Debt matters only through finite horizons; government spending matters through resources in any model.
In the endowment economy with \(T=G=0\) and \(\alpha>\theta\), show that every path with \(0<B(0)<\bar B\) converges to \(B=0\). How does the interest rate move along such a path? Who gains from the initial debt?
On \((0,\bar B)\), \(\dot B=B\,r(B)\) with \(r(B)=kB/Y+\theta-\alpha<r(\bar B)=0\), so \(\dot B<0\). \(B\) is decreasing and bounded below by 0, so it converges to a steady state, which must be 0. Since \(r(B)\) is increasing in \(B\) and \(B\) falls, \(r\) falls monotonically, from below zero towards the autarky rate \(\theta-\alpha<0\). The initial bondholders gain: they sell an asset that the economy rolls over at an interest rate that is negative but above autarky's.