Why the First Welfare Theorem Fails in OLG

1. (a) Strict monotonicity implies LNS. As in Lesson 4, a weakly preferred bundle cannot be cheaper (else a nearby, still affordable bundle would be strictly better), and a strictly preferred bundle must cost strictly more (else it was affordable). This applies to each generation and to the initial old.
2. (b) Prices are positive (strict monotonicity) and consumptions are non-negative, so every term \(p_tx^s_t\ge0\). For non-negative double series the order of summation doesn't matter (Tonelli), so \(\sum_{s}p\cdot x^s=\sum_{t\ge1}p_t(x^{t-1}_t+x^t_t)\). Feasibility gives \(x^{t-1}_t+x^t_t\le\bar e_t\) for both allocations, so both sums are at most \(\sum_tp_t\bar e_t\lt \infty\). For \(\bar x\), feasibility holds with equality (LNS, so nothing is wasted in a CE), and its sum equals \(\sum_tp_t\bar e_t\).
3. (c) Sum the inequalities from (a). All sums are finite, so a strict inequality in one term survives: \(\sum_sp\cdot\hat x^s>\sum_sp\cdot\bar x^s=\sum_tp_t\bar e_t\). But by (b), \(\sum_sp\cdot\hat x^s=\sum_tp_t(\hat x^{t-1}_t+\hat x^t_t)\le\sum_tp_t\bar e_t\). Contradiction. ∎
4. (d) Step (c). With infinite value, both sides of the summed inequality are \(+\infty\), and "\(\infty>\infty\)" is not a contradiction. Equivalently, you cannot subtract \(\sum_sp\cdot\bar x^s\) from both sides. The agent-by-agent inequalities remain true; there is just no finite total to compare them with.

Exam tip: when a T/F mentions OLG and "finite value", the answer is almost always this argument in two lines: LNS gives the inequalities, finite value makes them summable, and the sum contradicts feasibility.

1. Prices are non-negative. If \(p_\ell<0\) for some \(\ell\), any agent with monotone preferences would demand unboundedly much of good \(\ell\), since buying it relaxes the budget. That contradicts feasibility with \(\bar e_\ell\lt \infty\). So \(p\ge0\). (Under the strict monotonicity used in class, \(p\gg0\).)
2. Swap the sums. All terms \(p_\ell\bar x^i_\ell\) are non-negative, so by Tonelli's theorem for series, \(\sum_i\sum_\ell p_\ell\bar x^i_\ell=\sum_\ell p_\ell\sum_i\bar x^i_\ell\), whether finite or not.
3. Use feasibility good by good. \(\sum_i\bar x^i_\ell=\bar e_\ell\lt \infty\). Hence \(p\cdot\sum_i\bar x^i=\sum_{\ell=1}^m p_\ell\bar e_\ell\), a finite sum of finite numbers, so it is finite.
4. 1WT. Any dominating feasible \(\{x^i\}\) also has \(\sum_i p\cdot x^i\le\sum_\ell p_\ell\bar e_\ell\lt \infty\). Summing \(p\cdot x^i\ge p\cdot\bar x^i\) (strict for \(i'\)) over agents is now legitimate and gives \(\sum_i p\cdot x^i>\sum_i p\cdot\bar x^i=\sum_\ell p_\ell\bar e_\ell\). Contradiction.

Why this matters: an infinite number of agents alone is harmless, because there are only \(m\) prices and \(m\) finite aggregate quantities to multiply.

1. Each agent's wealth is finite. Suppose \(p\cdot e^i=\infty\). Then the budget constraint \(p\cdot x\le p\cdot e^i\) constrains nothing. In particular, \(\bar x^i\) plus one more unit of any good is affordable and, by strict monotonicity, strictly better. So no utility maximum exists, contradicting the CE. Hence \(p\cdot e^i\lt \infty\).
2. Each agent's consumption has finite value. \(p\cdot\bar x^i\le p\cdot e^i\lt \infty\) by the budget constraint.
3. Finite sum. All terms are non-negative (\(p\ge0\) by monotonicity), so \(p\cdot\sum_{i\in I}\bar x^i=\sum_{i\in I}p\cdot\bar x^i\). This is a sum of \(|I|\lt \infty\) finite numbers, hence finite.

Contrast with OLG: there too each \(p\cdot\bar x^i\) is finite (two-period lives), but there are infinitely many of them. When \(\bar r\le0\) their sum diverges.

(a) Yes. \(r(0)=f'(k_0)-\delta\), and \(f'(k)\to0\) as \(k\to\infty\) (Inada). So for \(k_0\) large enough, \(f'(k_0)\lt \delta\) and \(r(0)<0\).
(b) No. There is one agent, so by Exercise 12.4 (finitely many agents, strictly increasing utility) the value of aggregate consumption is finite, and the 1WT applies whatever \(k_0\) is. More directly, with a single agent the CE allocation solves the planner's problem. A high \(k_0\) just makes the agent rich: she consumes a lot early and runs capital down, and \(r(t)\) rises toward its steady-state value \(\rho>0\). In OLG, a negative \(r\) is a symptom of inefficiency only because the double infinity can make the value infinite. In the representative-agent model, \(r<0\) for a while is harmless.

Grading note from the 2025 review: the points are for the reasoning ("which 1WT hypothesis is relevant?"), not for algebra.

1. (a) \(a_t=1\). There is one tree, and the old (who don't value anything after \(t\)) sell it, so the young must hold it all. \(s_t=0\). Members of a generation are identical, so net lending within the cohort is zero, and the old cannot borrow (they won't be alive to repay). Formally, by induction: the initial old eat \(d_1+q_1\); market clearing gives \(c^1_1=e^1_1-q_1\), so \(s_1=0\); and so on.
2. (b) No arbitrage: \(1+r_t=(q_{t+1}+d_{t+1})/q_t\). If the tree paid more, the young would borrow without limit to buy trees. If it paid less, nobody would hold it. Either way the tree market could not clear at \(a_t=1\).
3. (c) The young's total saving is \(a_tq_t+s_t=q_t\), and optimal saving is \(\sigma_t(1+r_t)\). So \(\sigma_t\big((q_{t+1}+d_{t+1})/q_t\big)=q_t\). In a steady state, \(\bar q=\bar\sigma(1+\bar d/\bar q)\) and \(\bar r=\bar d/\bar q\).
4. (d) Inada conditions imply \(\bar\sigma>0\), so \(\bar q>0\), and \(\bar r=\bar d/\bar q>0\). The equilibrium is PO: prices are \(p_t=(1+\bar r)^{-(t-1)}\), and the value of aggregate consumption is \(\sum_tp_t(\bar e+\bar d)=(\bar e+\bar d)(1+\bar r)/\bar r\lt \infty\). The OLG 1WT (Exercise 12.2) applies. Common mistake (in a circulating set of answers): writing "\(\bar r>0\), so by the 1WT it is PO" without checking finite value. The standard 1WT does not apply to OLG without that hypothesis.
5. (e) Log utility saves a fixed share: \(\bar\sigma=\beta\bar e\), independent of \(R\). Hence \(\bar q=\beta\bar e\) and \(\bar r=\bar d/(\beta\bar e)\). The rate rises with \(\bar d\), falls with \(\beta\) (more saving) and falls with \(\bar e\).
6. (f) With \(\bar d=0\), no arbitrage reads \(1+r=q_{t+1}/q_t\). One steady state is \(\bar q=0\): the tree is worthless, there is autarky, and generation \(t\) consumes \((\bar e,0)\). The other is \(\bar q=\bar\sigma(1)>0\) with \(\bar r=0\). Under log utility this is \(\bar q=\beta\bar e\), and each generation consumes \(((1-\beta)\bar e,\beta\bar e)\). That is the best symmetric allocation, so the bubble equilibrium is PO even though its value is infinite (compare Exercise 12.7). Autarky is Pareto-dominated by it, and the initial old gain \(\beta\bar e\). The valued bubble does exactly what PAYG does in Lesson 11 and rolled-over debt does in Lesson 14 (Exercise 14.7).

Intuition for (d): an asset paying a positive dividend forever cannot be priced at \(r\le g\), since its price would be infinite. So a positive-dividend asset rules out dynamic inefficiency.

1. WLOG feasibility holds with equality (throwing goods away cannot help). Describe any allocation by \(d_t\equiv x^{t-1}_t-\beta\), so \(x^t_t=1-\beta-d_t\). The initial old are weakly better off iff \(d_1\ge0\).
2. Generation \(t\) gets \((1-\beta-d_t,\ \beta+d_{t+1})\). At \(\bar c\) the MRS is 1, so the indifference curve is tangent to a line of slope \(-1\). By strict quasi-concavity it lies strictly above that line away from \(\bar c\). Hence weakly preferring requires \(d_{t+1}\ge d_t\), strictly if \(d_t\ne0\).
3. For a strict improvement, some \(d_k>0\): either the initial old gain (\(d_1>0\)), or some generation strictly gains, which by step 2 forces some later \(d>0\). From then on \(d\) is strictly increasing. Strict concavity gives more. Log has curvature bounded away from zero on compact sets, so \(d_{t+1}-d_t\ge\kappa d_t^2\) for some \(\kappa>0\). An increasing sequence satisfying this exceeds any bound in finitely many steps.
4. But \(d_t\le 1-\beta\) (the young cannot consume negative amounts). Contradiction. ∎

Compare with Case 1: there the MRS at the CE exceeds 1. Generation \(t\) accepts \(d_{t+1}=d_t=\varepsilon\), a constant transfer, so no explosion is needed.

(a) \(q=p_{t+1}/p_t=v_2/v_1\) at the endowment \(=\frac{\beta/\alpha}{(1-\beta)/(1-\alpha)}\). It exceeds 1 iff \(\beta(1-\alpha)>\alpha(1-\beta)\iff\beta>\alpha\).
(b) \(p\cdot x^{*t}-p\cdot\bar x^t=p_t[(1-\beta)-(1-\alpha)]+p_{t+1}[\beta-\alpha]=(\beta-\alpha)(p_{t+1}-p_t)=p_t(\beta-\alpha)(q-1)\). This is positive, and since \(p_t=q^{t-1}\to\infty\), the gap explodes. No contradiction: both allocations have total value \(\sum_tp_t=\infty\), so summing the agent-level strict inequalities gives "\(\infty>\infty\)". Economically, generation \(t\)'s extra cost is paid by generation \(t+1\)'s transfer, forever.
(c) The forward induction of Lesson 8 works as before, and at \(T\) the last generation eats what is left, \(e^T_T\). Moving to \((1-\beta,\beta)\) now requires generation \(T\) to give \(\beta-\alpha>0\) to generation \(T-1\) and get nothing back, so it loses. With \(T+1\) agents and \(T\) goods, the value of the aggregate endowment is finite, and the 1WT says autarky is PO. It is the absence of a last generation that makes the improvement possible.