Class Note II has four parts II p.2:
This lesson builds the environment and solves for the equilibrium allocation. It is the foundation for all the welfare results that follow.
The point to keep in mind is that nothing here is new machinery. The OLG economy is an instance of the general economy of Lesson 1, with \(L=\mathbb{R}^\infty\), countably many households and no firms. Everything from Part I (budget constraints, CE, Pareto optimality) applies word for word. What is unusual is only that both the number of goods and the number of agents are infinite.
Notation: a superscript is who (the generation), a subscript is when (the date). So \(x^i_{i+1}\) is generation \(i\)'s consumption when old. The slides note that in any CE or efficient allocation, \(x^i_t = 0\) for \(t\notin\{i,i+1\}\) II p.4. Goods at dates when you are dead give no utility, so giving them to you wastes resources (inefficient) or wastes budget (not optimal at positive prices).
Material balance from Lesson 1, \(\sum_{i\in I}x^i = \sum_{i\in I}e^i\), is an equation in \(\mathbb{R}^\infty\): one equation per date. At date \(t\) only generations \(t-1\) (old) and \(t\) (young) consume or hold endowments. So the infinite sum over \(i\) collapses to two terms II p.6:
$$x^{t-1}_t + x^t_t \;=\; e^{t-1}_t + e^t_t \;\equiv\; \bar e_t \qquad t = 1,2,\dots$$Here \(\bar e_t\) is the aggregate endowment at date \(t\). Written out:
$$\begin{aligned} t=1:&\quad x^0_1 + x^1_1 = e^0_1 + e^1_1\\ t=2:&\quad x^1_2 + x^2_2 = e^1_2 + e^2_2\\ t=3:&\quad x^2_3 + x^3_3 = e^2_3 + e^3_3\\ &\quad\vdots \end{aligned}$$Prices are a sequence \(p=(p_1,p_2,\dots)\in\mathbb{R}^\infty\): \(p_t\) is the price, in a common unit of account, of one unit of the date-\(t\) good. Think of all trades as happening in one big Arrow–Debreu market "before time starts" (or, equivalently, think of a sequence of markets with one-period loans; see Exercise 8.4). Generation \(i\)'s problem is the general household problem \(\max u^i(x)\) s.t. \(p\cdot x\le p\cdot e^i\). Since \(x^i_t=e^i_t=0\) outside \(\{i,i+1\}\), it specializes to II p.7:
For \(i\ge1\): $$\max_{x_i,\,x_{i+1}}\; v^i(x_i,x_{i+1})\quad\text{s.t.}\quad p_i x_i + p_{i+1}x_{i+1} \;=\; p_i e^i_i + p_{i+1}e^i_{i+1}.$$ For \(i=0\): $$\max_{x_1}\; x_1 \quad\text{s.t.}\quad p_1 x_1 = p_1 e^0_1 .$$ A CE is a price sequence \(p\) and an allocation \(\{\bar x^i\}_{i\ge0}\) that solve these problems and satisfy \(\bar x^{t-1}_t+\bar x^t_t = \bar e_t\) for every \(t\).
The slides write the budgets with equality. That is legitimate whenever \(v^i\) is strictly increasing (so LNS holds and budgets bind; see Lesson 3).
In any competitive equilibrium, \(\bar x^i = e^i\) for every \(i\ge0\). There is no trade.
Steps 1 and 3 divided by \(p_t\). If some \(p_t\le 0\), the date-\(t\) good would be free (or better than free) to a generation that values it with strictly increasing \(v\). Demand would be unbounded and the market could not clear. So in any CE with strictly increasing utilities, \(p_t>0\) for all \(t\).
Trade needs a double coincidence across time. The young at \(t\) would like to save, i.e. give up goods today for goods tomorrow. The only people they can trade with today are the current old, and the old will be dead tomorrow, so they cannot repay. The young at \(t\) will meet the young of \(t+1\) tomorrow, but those people are not yet born today. So every mutually beneficial intertemporal loan has a counterparty who is either dead or unborn.
The induction shows that this logic is airtight: the initial old are stuck with their endowment, and that "locks" every subsequent link of the chain. Prices still exist and are well defined (Lesson 9). They are simply the prices at which nobody wants to trade.
Exams on this material come in two formats: rapid true/false questions with a short justification, and long guided derivations that walk you through one economy step by step (see the past exams cited below). Attempt each part before opening its solution.
False. Every agent in a cohort is identical and the only other agents alive are the old, who will be dead when the loan is repaid. Loan-market clearing at \(t\) involves generation \(t\) alone, so \(s_t=0\): the unique CE is autarky (the proposition in §4).
False. The induction uses only that the initial old value only date-1 goods, that budgets bind at positive prices (strictly increasing \(v^i\)), and that just two generations overlap. Preferences matter for the prices that support autarky (Lesson 9), not for the allocation.
False. Types within a cohort have different MRSs at their endowments, so they borrow from and lend to each other. What survives is aggregate no trade across generations: each cohort's total consumption equals its total endowment in each period of life (Exercise 8.7). Beware: one circulated problem-set solution says "each agent consumes its endowment". That is wrong.
True. The old will not be alive next period, so they sell the tree. The young must hold it for the tree market to clear, \(a_t=1\). Bonds are in zero net supply and only the young can be on either side, so \(s_t=0\). See Exercise 8.2.
False. Generation \(t\) must choose its endowment. With differentiable \(v^t\) and an interior optimum this requires \(p_{t+1}/p_t = v^t_2/v^t_1\) evaluated at \((e^t_t,e^t_{t+1})\). With \(p_1=1\), that determines every \(p_t\) (Lesson 9). Prices would be indeterminate only at kinks, e.g. Leontief preferences.
False in general. The young and the middle-aged are both alive next period, so one can lend to the other and be repaid. Market clearing at \(t=1\) now has two unknowns, and the induction does not start (Exercise 8.3). This is the setting of the Fall 2025 true/false item on agents who live \(k\ge2\) periods (Lesson 10).
Time is discrete, \(t=1,2,\dots\). A measure one of agents is born each period and lives two periods. Generation \(t\) has endowment \(e^t_t>0\) when young and nothing when old, with strictly increasing, concave utility \(u^t(c^t_t,c^t_{t+1})\). There is one tree that pays a dividend \(d_t\ge0\) of the good each period, and at \(t=1\) each initial old agent owns one tree. Feasibility is \(c^{t-1}_t+c^t_t = d_t + e^t_t\). Whoever owns the tree collects the dividend and may sell the tree at the ex-dividend price \(q_t\). Agents may also borrow or lend for one period at net rate \(r_t\). Generation \(t\) buys \(a_t\) trees and saves \(s_t\) in bonds: $$c^t_t + a_tq_t + s_t = e^t_t,\qquad c^t_{t+1} = a_t(q_{t+1}+d_{t+1}) + s_t(1+r_t),$$ and the initial old consume \(c^0_1 = d_1+q_1\).
The problem continues in Lesson 9 (Exercise 9.2), where you derive the equilibrium price of the tree.
For (b), start with the initial old: what do they consume? Then use market clearing at \(t=1\) and generation 1's first budget constraint with \(a_1=1\).
(a) \(a_t = 1\). The supply of trees is one. An old agent values the tree only for what it sells for today, so the old sell everything and the young must hold the entire stock.
(b) \(s_t = 0\). Economics: all young are identical, and the only other agents are the old, who cannot repay. So aggregate lending must be zero. Induction: the initial old consume \(d_1+q_1\). Market clearing at \(t=1\) gives \(c^1_1 = e^1_1 - q_1\). Generation 1's budget with \(a_1=1\) then gives \(s_1 = e^1_1 - c^1_1 - q_1 = 0\). If \(s_t=0\), generation \(t\) consumes \(q_{t+1}+d_{t+1}\) when old, so market clearing gives \(c^{t+1}_{t+1} = e^{t+1}_{t+1}-q_{t+1}\), and the budget gives \(s_{t+1}=0\).
(c) No arbitrage: \(1+r_t = (q_{t+1}+d_{t+1})/q_t\). Trees and bonds are both ways to carry resources forward. If the tree paid more, the young would borrow without limit to buy trees. If it paid less, they would sell trees short. Either way \(a_t=1\), \(s_t=0\) could not be an optimum.
Suppose each generation lives three periods (young, middle-aged, old), with \(u^i = v(x_i,x_{i+1},x_{i+2})\) and strictly positive endowments in all three periods. At \(t=1\) there are an initial old (alive only at \(t=1\)) and an initial middle-aged (alive at \(t=1,2\)).
(a) \(x^{t-2}_t + x^{t-1}_t + x^t_t = e^{t-2}_t + e^{t-1}_t + e^t_t\). Three generations are alive.
(b) The initial old are still stuck with their endowment. But market clearing at \(t=1\) is now one equation in two unknowns, the consumption of the initial middle-aged and of generation 1. So it no longer pins down anyone's consumption, and the chain of single-unknown equations breaks.
(c) At date \(t\) the young (born \(t\)) and the middle-aged (born \(t-1\)) are both alive at \(t+1\). The young can lend to the middle-aged at \(t\) and be repaid at \(t+1\), when the borrower is old and the lender middle-aged. Both counterparties are alive at both ends of the loan, so equilibrium interest rates now reflect actual borrowing and lending. Inefficiency is still possible; see the \(N\)-period argument in Lesson 16.
Instead of one Arrow–Debreu market at the beginning of time, suppose that at each date \(t\) there is a spot market for the date-\(t\) good and a market for one-period loans at net interest rate \(r_t\). Generation \(t\) chooses savings \(s_t\): \(x^t_t + s_t = e^t_t\) and \(x^t_{t+1} = e^t_{t+1} + (1+r_t)s_t\).
(a) Eliminate \(s_t\): \(x^t_t + x^t_{t+1}/(1+r_t) = e^t_t + e^t_{t+1}/(1+r_t)\). Multiply by \(p_t\) and set \(p_{t+1} = p_t/(1+r_t)\) to get \(p_tx^t_t + p_{t+1}x^t_{t+1} = p_te^t_t + p_{t+1}e^t_{t+1}\).
(b) Only generation \(t\) can be in the loan market at \(t\), because the old will not be around to repay or collect. So loan-market clearing is \(s_t = 0\).
(c) \(s_t = 0\) for all \(t\) gives \(x^t_t=e^t_t\) and \(x^t_{t+1}=e^t_{t+1}\). The interest rate is whatever rate makes generation \(t\) choose \(s_t=0\). That is the subject of Lesson 9.
Prove that in any CE of the OLG economy with strictly increasing \(v^i\), \(p_t>0\) for all \(t\). Where exactly did the proof of the proposition need this?
Suppose \(p_{t}\le0\) for some \(t\ge1\). Consider generation \(t\), or generation \(0\) if \(t=1\). Raising \(x_t\) by any amount weakly lowers expenditure \(p_tx_t\). So the budget set contains bundles with arbitrarily large \(x_t\) and the other components unchanged. Since utility is strictly increasing in \(x_t\), no maximizer exists, which contradicts \(\bar x^t\) solving the problem. So \(p_t>0\). The proposition used this when dividing by \(p_1\) (step 1) and by \(p_{t+1}\) (the budget steps).
Suppose the initial old have preferences \(u^0(x)=w(x_1)\) for some strictly increasing \(w\). Does the proposition still hold? What if the initial old also held an endowment \(e^0_2>0\) of the date-2 good? Prove it, or give a counterexample.
Step 1 of the proof is the only place where generation 0's preferences enter. In the second case, ask whether generation 0 can now sell something.
Strictly increasing \(w\): the proposition still holds. At positive prices, generation 0 never buys goods dated \(t\ge2\), so \(p_1x_1 = p_1 e^0_1\) still forces \(x_1^0=e^0_1\).
With \(e^0_2>0\): generation 0 sells its date-2 claim, so its budget becomes \(p_1 x_1 = p_1 e^0_1 + p_2 e^0_2\) and \(x^0_1 > e^0_1\). The induction no longer starts and trade occurs: generation 1 gives up date-1 goods for date-2 goods. The old now hold an asset that pays off after they die. This is the logic of the tree in Exercise 8.2, and of money and government debt in Lessons 11 and 14.
Keep two-period lives and one good per period.
For (b), sum generation \(t\)'s Arrow–Debreu budgets across types. If you know the cohort's total young-age consumption, the summed budget pins its total old-age consumption.
(a) \(N_tc^t_t + N_{t-1}c^{t-1}_t = N_te^t_t+N_{t-1}e^{t-1}_t\), or per young, \(c^t_t + c^{t-1}_t/(1+n) = e^t_t + e^{t-1}_t/(1+n)\). The initial old eat their endowment. Market clearing at \(t=1\) then gives \(c^1_1 = e^1_1\), generation 1's budget gives \(c^1_2=e^1_2\), and so on by induction. Population size only rescales the equations.
(b) Each initial old type consumes \(e^{0,n}_1\). Market clearing at \(t=1\) gives \(\sum_n c^{1,n}_1 = \sum_n e^{1,n}_1\). Summing generation 1's budgets over types, $$p_1\textstyle\sum_n c^{1,n}_1 + p_2\sum_n c^{1,n}_2 = p_1\sum_n e^{1,n}_1 + p_2\sum_n e^{1,n}_2,$$ so with \(p_2>0\), \(\sum_n c^{1,n}_2=\sum_n e^{1,n}_2\). Market clearing at \(t=2\) then gives \(\sum_n c^{2,n}_2=\sum_n e^{2,n}_2\), and induction finishes the proof.
(c) No. Types with high old-age endowments borrow from types with low ones, within the cohort. Only the cohort's net position against other cohorts is zero. Lesson 16 solves this economy.