Lesson 4 proved that every CE with locally non-satiated preferences is Pareto optimal. Log utility is strictly increasing, hence LNS. So the First Welfare Theorem seems to say the autarky equilibrium of Lessons 8–9 is PO. Here we test that claim directly, by searching for a feasible allocation that everybody weakly prefers. To keep the search tractable, we restrict attention to allocations that treat every generation the same way.
An allocation is symmetric if consumption depends only on age, not on the generation. It is described by two numbers \((c_y,c_o)\): $$x^i_i = c_y\ \ (i\ge1),\qquad x^i_{i+1} = c_o\ \ (i\ge0).$$ Note that the initial old also receive \(c_o\). Since \(\bar e_t=1\), feasibility at each date reads \(x^{t-1}_t + x^t_t = c_o + c_y = 1\): one condition, the same at every date.
The CE \((1-\alpha,\alpha)\) is itself a feasible symmetric allocation, since \(1-\alpha+\alpha = 1\). So the set we are searching contains the equilibrium, which is what makes the comparison clean.
"Best" here means best from the point of view of a generation \(i\ge1\), all of whom have the same \(v\) II p.16:
$$\max_{c_y,c_o}\;(1-\beta)\log c_y + \beta\log c_o\quad\text{s.t.}\quad c_y + c_o = 1 .$$Along the symmetric frontier, giving the old one more unit costs the young exactly one unit, because there is one old person per young person and no growth. The best allocation equates the MRS to this biological rate of return of 1. Equivalently, the implied interest rate at \((1-\beta,\beta)\) is \(r = v_1/v_2 - 1 = 0\), the population growth rate. In Lesson 13, with growth, the biological rate becomes \((1+n)(1+g)\).
There are two kinds of agent to check II p.17:
| CE | Best symmetric | |
|---|---|---|
| Generations \(i\ge1\) | \(v(1-\alpha,\alpha)\) | \(v(1-\beta,\beta)\) |
| Initial old \(i=0\) | \(\alpha\) | \(\beta\) |
| Case | Parameters | \(\bar r\) | Initial old | Conclusion |
|---|---|---|---|---|
| 1 | \(\beta>\alpha\) | \(<0\) | prefer best | CE is Pareto-dominated |
| 2 | \(\beta=\alpha\) | \(=0\) | indifferent | CE = best symmetric |
| 3 | \(\beta\lt \alpha\) | \(>0\) | prefer CE | not comparable |
Hence: the CE is Pareto-dominated by the best symmetric allocation if and only if \(\bar r<0\). (Recall \(\bar r = \frac{\alpha-\beta}{\beta(1-\alpha)}\).)
Read the move from CE to best symmetric in Case 1 as a transfer. Each young person hands \(\beta-\alpha>0\) units to the contemporaneous old person:
If the transfers stopped at some date \(T\), generation \(T-1\) would give when young and receive nothing when old. They would be worse off, so the scheme would no longer be a Pareto improvement. The gain to the initial old is financed by an infinite sequence of future generations, each passing the bill forward. This is possible only because the economy has infinitely many agents and dates. That double infinity is the heart of Lesson 12.
When \(\bar r>0\), the young would rather borrow against their large old-age endowment. Moving toward \((1-\beta,\beta)\) means transferring from the old to the young. Generations \(i\ge1\) like it (they receive when young, repay when old at rate \(0\lt \bar r\)), but the initial old pay without being repaid. Someone always loses, so the CE is not dominated by any symmetric allocation (Exercise 10.3). In fact it is Pareto optimal among all allocations, as Lesson 12 will show using the finiteness of \(\sum_t p_t\).
"Is the OLG equilibrium Pareto optimal?" is one of the most frequently examined questions in this course. It has appeared as a true/false item on Fall 2005, Spring 2023 and Fall 2025 exams, and as the last part of the Fall 2025 final's Diamond question. The grade goes to the reasoning: name the condition you are using.
True. With \(r_t<0\), prices \(p_{t+1}=p_t/(1+r_t)>p_t\) never fall, so the value of the aggregate endowment is infinite and the 1WT proof breaks. An explicit improvement exists: transfer a small \(\varepsilon\) from young to old at every date, starting with the initial old. Each generation values the extra old-age unit at \(1/(1+r_t)>1\) young-age units (Exercise 10.4; Lesson 16 for \(k\) periods). Explain it by the infinite value of the endowment, not by "over-accumulation of capital": this economy has no capital.
True (with local non-satiation). Suppose a feasible allocation dominates. LNS and optimality give each generation's dominating bundle a weakly higher value, and strictly higher for someone. Because the sums are finite, they can be regrouped by date, which gives \(\sum_tp_t(\hat x^{t-1}_t+\hat x^t_t)>\sum_tp_t\bar e_t\). That contradicts feasibility. Lesson 12 proves this in full.
False as stated. (A circulated study guide marks this "always true", but it isn't.) With growth, what matters is \(1+r\) versus \((1+g)(1+n)\) (Lesson 13): a positive but low \(r\) can be inefficient. Even without growth, \(r_t>0\) with \(r_t\to0\) fast enough (e.g. \(r_t\sim 1/t^2\)) keeps \(\sum_tp_t\) infinite. The safe claim is: stationary \(\bar r>0\) with constant aggregate endowment ⇒ PO.
False. The first half is true but the conclusion does not follow: the initial old consume \(\beta\lt \alpha\) and lose. In fact the CE is PO, because \(\bar r>0\) makes \(\sum_tp_t\lt \infty\) (Lesson 12). For symmetric allocations, see Exercise 10.3.
False. A finite value is sufficient for PO, not necessary. In Case 2 the CE equals the best symmetric allocation, which is PO among all allocations (Exercise 10.6).
True. From Exercise 10.5(d), \(1/(1+r)=\pi_o/\pi_y<1\), so \(r>0\), constant. With a constant aggregate endowment, \(\sum_tp_t\bar e_t\lt \infty\), so the OLG version of the 1WT applies.
False. In a stochastic economy the relevant test is whether the value of aggregate consumption is finite, i.e. the return on a claim to aggregate output versus growth. In the problem set's tree economy the safe rate can be below growth, yet the tree's expected return exceeds growth and the equilibrium is PO.
False. Generation 1 pays when young and gets nothing back. The same is true of any scheme that stops at a finite date: the last payer loses. Every improvement over the Case-1 CE must transfer from young to old at every date, forever.
Let \(v\) be strictly increasing, strictly quasi-concave and differentiable, with stationary endowments \((1-\alpha,\alpha)\).
(a) \(\max v(c_y,1-c_y)\): the FOC is \(v_1 = v_2\), so the MRS is 1 and the implied rate is \(r=0\).
(b) Move along the line: \(c_o = \alpha+\tau\), \(c_y = 1-\alpha-\tau\). Then \(\frac{d}{d\tau}v = v_2-v_1 = v_1\big(\tfrac{1}{1+\bar r}-1\big) > 0\) at \(\tau=0\) iff \(\bar r<0\). So a small \(\tau>0\) raises the utility of every generation \(i\ge1\) and the initial old's consumption: a Pareto improvement.
(c) Yes. If \(\bar r\ge0\), the initial old need \(\tau\ge0\). Along the line, \(v\) is quasi-concave in \(\tau\) with \(dv/d\tau\le0\) at \(\tau=0\), so it is non-increasing for \(\tau\ge0\). Then no generation \(i\ge1\) gains, and in Case 2 nobody gains strictly.
Show that in Case 3 (\(\beta\lt \alpha\)) no feasible symmetric allocation Pareto-dominates the CE.
Let \((c_y,c_o)\) be feasible (\(c_y+c_o\le1\)) and weakly preferred by everyone. The initial old need \(c_o\ge\alpha\), so \(c_y\le1-\alpha\). Since \(v\) is increasing, it suffices to check \(c_o = 1-c_y\). The function \(f(c_y) = v(c_y,1-c_y)\) is strictly concave with its maximum at \(1-\beta > 1-\alpha\), so it is strictly increasing on \((0,1-\alpha]\). Hence \(f(c_y)\le f(1-\alpha)\) for \(c_y\le1-\alpha\), with equality only at the CE. Nobody can be strictly better off.
The exam asked only for "true or false". Here you prove the two-period case without stationarity. Generation \(t\) has strictly increasing, strictly concave, twice continuously differentiable utility \(v\) and endowment \((e^t_t,e^t_{t+1})\) lying in a compact set \(E\) bounded away from zero. Suppose the CE interest rates satisfy \(r_t\le\underline r<0\) for all \(t\).
In the pure-endowment CE each generation consumes its endowment, so \(v_2/v_1\) at \(e^t\) equals \(1/(1+r_t)\). For (c)–(d), continuous functions are bounded on compact sets: bound \(v_1\) below and \(|\Delta''_t|\) above, uniformly in \(t\).
(a) \(p_{t+1}/p_t=1/(1+r_t)\ge1/(1+\underline r)>1\), so \(p_t\ge p_1\) for all \(t\). With \(\bar e_t\) bounded away from zero, the series diverges.
(b) At each date the young give \(\varepsilon\) and the old receive \(\varepsilon\), so totals are unchanged. Consumption stays positive for \(\varepsilon\) below the lower bound on endowments. The initial old consume \(e^0_1+\varepsilon>e^0_1\).
(c) \(\Delta_t'(0)=-v_1+v_2=v_1(v_2/v_1-1)=v_1\big(\frac{1}{1+r_t}-1\big)\), using the FOC at the endowment. On \(E\), \(v_1\ge m>0\), and \(\frac1{1+r_t}-1\ge\frac1{1+\underline r}-1>0\). So \(\Delta_t'(0)\ge c\equiv m\big(\frac{1}{1+\underline r}-1\big)>0\).
(d) \(\Delta''_t(\varepsilon)=v_{11}-2v_{12}+v_{22}\), evaluated along the segment. On a compact neighbourhood of \(E\) this is bounded: \(|\Delta''_t|\le K\). By Taylor's theorem, \(\Delta_t(\varepsilon)\ge\varepsilon c-\tfrac12K\varepsilon^2>0\) for \(0\lt \varepsilon\lt \bar\varepsilon\equiv\min\{2c/K,\ \underline e\}\). Every generation \(t\ge1\) strictly gains and so do the initial old, so the CE is not Pareto optimal.
(e) If \(r_t\uparrow0\), the first-order gain \(\Delta'_t(0)\) shrinks to zero along the sequence. Then no single \(\varepsilon\) need work for all generations: a constant transfer may hurt late generations, and you would need a transfer path that adapts. The general criterion (Cass; Balasko–Shell) is more delicate. On an exam, state the uniform bound or assume stationarity.
Periods \(t=1,2,\dots\). Each cohort contains types \(i=1,\dots,I\). Type \(i\) of cohort \(t\) has endowment \(\alpha^i_y\) when young and \(\alpha^i_o\) when old, and preferences $$-\pi_y\frac{(\theta^i_y-x^{i,t}_t)^{1+\sigma}}{1+\sigma}-\pi_o\frac{(\theta^i_o-x^{i,t}_{t+1})^{1+\sigma}}{1+\sigma},\qquad \sigma>0,$$ defined for consumption below the satiation points \(\theta^i\). The initial old of type \(i\) have endowment \(\alpha^i_o\) and utility \(x^{0,i}_1\). Let \(\bar\alpha_y=\sum_i\alpha^i_y\), \(\bar\alpha_o=\sum_i\alpha^i_o\), \(\bar\theta_y=\sum_i\theta^i_y\), \(\bar\theta_o=\sum_i\theta^i_o\).
You may use the result of the exam's previous question (Gorman aggregation, Lesson 25): in a pure-exchange economy with these preferences and aggregate endowment \(\bar e\), interior equilibrium relative prices are $$\frac{p_\ell}{p_k}=\frac{\pi_\ell(\bar\theta_\ell-\bar e_\ell)^\sigma}{\pi_k(\bar\theta_k-\bar e_k)^\sigma},$$ whatever the distribution of endowments.
(a) No. Types differ in MRS at their endowments, so they trade within the cohort.
(b) Yes. There is no aggregate trade across cohorts (Exercise 8.7(b)): the initial old just eat their endowment, and induction pins each cohort's totals.
(c) Yes. Each cohort is the same two-good exchange economy, with goods "young-date" and "old-date" and aggregate endowment \((\bar\alpha_y,\bar\alpha_o)\). So its relative price is the same for every \(t\).
(d) Apply the aggregation formula with \(\ell=o\), \(k=y\) and \(\bar e=(\bar\alpha_y,\bar\alpha_o)\): $$\frac{1}{1+r}=\frac{\pi_o}{\pi_y}\left(\frac{\bar\theta_o-\bar\alpha_o}{\bar\theta_y-\bar\alpha_y}\right)^{\sigma}.$$
(e) The ratio in brackets is 1, so \(1/(1+r)=\pi_o/\pi_y<1\) and \(r>0\). PO: prices fall geometrically and the aggregate endowment is constant, so its value is finite and the OLG 1WT applies.
(f) \(r<0\iff\pi_o(\bar\theta_o-\bar\alpha_o)^\sigma>\pi_y(\bar\theta_y-\bar\alpha_y)^\sigma\): old-age consumption is weighted more heavily, or the cohort is further below its satiation point when old. In an interior equilibrium, every type's MRS equals \(1/(1+r)>1\). Take \(\varepsilon\) from every young agent of every type and give \(\varepsilon\) to every old agent, at every date. The initial old gain \(\varepsilon\), since their utility is linear. Each type of every cohort gains to first order \(u_y\,\varepsilon\,(\tfrac{1}{1+r}-1)>0\), and stationarity makes the second-order terms uniformly small (Exercise 10.4). So the CE is not PO.
Show that the best symmetric allocation \((1-\beta,\beta)\) is Pareto optimal among all feasible allocations, not just symmetric ones. In Case 2 this is the CE, even though its value is infinite.
Write any feasible allocation that doesn't waste goods as \(x^t_t = 1-\beta-\delta_t\) and \(x^{t-1}_t = \beta+\delta_t\), where \(\delta_t\) is the extra amount given to the old at date \(t\). Use \(\log(1+z)\le z\), and \(\log(1-z)\le -z - z^2/2\) for \(z\in[0,1)\), to bound each generation's utility gain.
Wasting goods never helps, so impose feasibility with equality and parametrize as in the hint. The initial old need \(\delta_1\ge0\). Generation \(t\)'s utility change is $$\Delta_t = (1-\beta)\log\Big(1-\tfrac{\delta_t}{1-\beta}\Big) + \beta\log\Big(1+\tfrac{\delta_{t+1}}{\beta}\Big).$$ Suppose \(\delta_t\ge0\), which is true for \(t=1\). The two inequalities give $$\Delta_t \;\le\; -\delta_t - \kappa\,\delta_t^2 + \delta_{t+1},\qquad \kappa = \tfrac{1}{2(1-\beta)} > 0 .$$ So \(\Delta_t\ge0\) requires \(\delta_{t+1}\ge\delta_t + \kappa\delta_t^2\ge\delta_t\ge0\). By induction every \(\delta_t\ge0\) and the sequence is non-decreasing. If some \(\delta_k>0\), then \(\delta_{t+1}\ge\delta_t(1+\kappa\delta_k)\) for \(t\ge k\). The sequence grows geometrically and eventually exceeds \(1-\beta\), which is infeasible. Hence all \(\delta_t = 0\). ∎
Remark: "transfers must grow geometrically to keep everyone happy" is the essence of the Balasko–Shell/Cass characterization of OLG efficiency, and it reappears in Lesson 16.