Theory of Income · Part II: Overlapping Generations · Lesson 9 of 25

Equilibrium Prices and Interest Rates in the OLG Economy

Covers Class Note II, pp. 9–14 · About 35 min · Prerequisite: Lesson 8

By the end you can:

derive the equilibrium price sequence $p_t$ and interest rates $r_t$ of the no-trade equilibrium from first-order conditions;
solve the log example: $\bar r = (\alpha-\beta)/(\beta(1-\alpha))$ and $p_t = (1+\bar r)^{-(t-1)}$;
derive the savings function $s(r;\alpha,\beta)$, sign its derivatives, and use it to explain why $\bar r$ rises with $\alpha$ and falls with $\beta$.

Primary source Alvarez, Class Note II, pp. 9–14. Samuelson's 1958 paper §II derives the same "biological" interest rate. Ljungqvist & Sargent, Recursive Macroeconomic Theory, ch. "Overlapping Generations", treat the same endowment economy with "Samuelson" and "classical" cases.

1. Prices that support autarky

Lesson 8 told us the allocation, $x^i = e^i$. It did not tell us the prices. The proposition held for any positive prices at each step, so prices must be pinned down by something else: each generation must choose its endowment when facing them. In equilibrium, autarky is not imposed. It is optimal at the equilibrium prices.

Normalize $p_1 = 1$. If $v^i$ is differentiable and the endowment is interior, the first-order conditions are necessary at the optimum $x^i = e^i$ II p.9:

$$v^i_1(e^i_i,e^i_{i+1}) = \lambda^i p_i,\qquad v^i_2(e^i_i,e^i_{i+1}) = \lambda^i p_{i+1} \quad\Longrightarrow\quad \frac{p_{i+1}}{p_i} = \frac{v^i_2(e^i_i,e^i_{i+1})}{v^i_1(e^i_i,e^i_{i+1})}\qquad i\ge1.$$

Equilibrium relative prices equal each generation's marginal rate of substitution evaluated at its endowment. Geometrically, the budget line through the endowment is tangent to the indifference curve there.

Definition: interest rates from Arrow–Debreu prices II p.10

The net interest rate between $t$ and $t+1$ is defined by $$\frac{1}{1+r_t} = \frac{p_{t+1}}{p_t}.$$ In the no-trade equilibrium, for all $t\ge1$, $$r_t = \frac{v^t_1(e^t_t,e^t_{t+1})}{v^t_2(e^t_t,e^t_{t+1})} - 1,\qquad p_t = \frac{1}{(1+r_1)(1+r_2)\cdots(1+r_{t-1})} .$$

Interpretation: $p_t/p_{t+1} = 1+r_t$ is how many date-$t+1$ goods one date-$t$ good buys. That is exactly the gross return on a one-period loan (Exercise 8.4 makes the sequential-markets equivalence precise). The formula for $p_t$ chains the one-period rates. Each $r_t$ is determined by generation $t$ alone, the only generation with access to both dates $t$ and $t+1$.

Intuition: an "autarky interest rate"

$r_t$ is the interest rate at which generation $t$ is exactly content to neither borrow nor lend. If the young are impatient relative to their endowment profile, so that they would like to borrow against a large old-age endowment, the rate must be high enough to talk them out of it. If they would like to save, the rate must be low (possibly negative) enough to discourage saving. Since nobody can actually trade, it is the price that must adjust.

2. The log example

The baseline parametrization II p.11

All generations $i\ge1$ have $$v(c_y,c_o) = (1-\beta)\log c_y + \beta\log c_o,\qquad \beta\in(0,1),$$ and endowments $e^i_i = 1-\alpha$ (young), $e^i_{i+1} = \alpha$ (old), with $e^0_1=\alpha$ and $\alpha\in(0,1)$. The aggregate endowment is constant: $\bar e_t = (1-\alpha) + \alpha = 1$.

$\beta$ is the weight on old-age consumption: a "patience" or "taste for the future" parameter. $\alpha$ is the share of lifetime endowment received when old: a "pension-like" endowment. These two numbers alone determine everything.

Notation alert

Class Note II puts $1-\alpha$ in youth and $\alpha$ in old age. Class Note III's privatization slides flip this (young $\alpha$, old $1-\alpha$); see Lesson 14. Here, and in Lessons 10–13, we follow Note II.

Equilibrium prices

With log utility, $v_1 = (1-\beta)/c_y$ and $v_2 = \beta/c_o$. Evaluating at the endowment $(1-\alpha,\alpha)$ II p.12:

$$\frac{1}{1+r_t} = \frac{p_{t+1}}{p_t} = \frac{\beta/\alpha}{(1-\beta)/(1-\alpha)} = \frac{\beta}{1-\beta}\cdot\frac{1-\alpha}{\alpha}.$$

This is the same for every $t$, so $r_t\equiv\bar r$ with

$$\boxed{\;1+\bar r = \frac{(1-\beta)\,\alpha}{\beta\,(1-\alpha)},\qquad \bar r = \frac{\alpha-\beta}{\beta(1-\alpha)},\qquad p_t = \left(\frac{\beta}{1-\beta}\cdot\frac{1-\alpha}{\alpha}\right)^{t-1} = (1+\bar r)^{-(t-1)}.\;}$$

Check the algebra of $\bar r$: $\frac{(1-\beta)\alpha - \beta(1-\alpha)}{\beta(1-\alpha)} = \frac{\alpha - \beta}{\beta(1-\alpha)}$. So $\bar r>0 \iff \alpha>\beta$, $\bar r=0\iff\alpha=\beta$ and $\bar r<0\iff\alpha\lt \beta$ II p.14.

3. The savings function

The savings of generation $t$ when young are $e^t_t - x^t_t$. Facing a constant rate $r$, a young agent who saves $s$ consumes $1-\alpha-s$ now and $\alpha + (1+r)s$ later. Its problem (the slide question on II p.13) is

$$s(r;\alpha,\beta) = \arg\max_s\;(1-\beta)\log(1-\alpha-s) + \beta\log\big(\alpha+s(1+r)\big).$$

Derivation (slide question, p. 13)

FOC. The objective is strictly concave in $s$, so the FOC is necessary and sufficient: $$\frac{1-\beta}{1-\alpha-s} = \frac{\beta(1+r)}{\alpha+s(1+r)} .$$
Cross-multiply. $(1-\beta)\alpha + (1-\beta)(1+r)s = \beta(1+r)(1-\alpha) - \beta(1+r)s$.
Collect $s$. $(1+r)s\,[(1-\beta)+\beta] = \beta(1+r)(1-\alpha) - (1-\beta)\alpha$. Divide by $1+r$: $$\boxed{\;s(r;\alpha,\beta) = \beta(1-\alpha) - \frac{(1-\beta)\,\alpha}{1+r}\;}$$
Match the slide's form. $(1-\alpha)-(1-\beta)\big[(1-\alpha)+\tfrac{\alpha}{1+r}\big] = (1-\alpha)\beta - \tfrac{(1-\beta)\alpha}{1+r}$. ✓ The slide's version reads as "endowment minus consumption": the young consume a share $1-\beta$ of lifetime wealth $(1-\alpha)+\alpha/(1+r)$ and save the rest of their young endowment.
Comparative statics. $$\frac{\partial s}{\partial r} = \frac{(1-\beta)\alpha}{(1+r)^2} > 0,\qquad \frac{\partial s}{\partial\beta} = (1-\alpha) + \frac{\alpha}{1+r} > 0,\qquad \frac{\partial s}{\partial\alpha} = -\beta - \frac{1-\beta}{1+r} < 0.$$ Savings are increasing in $r$, increasing in $\beta$, and decreasing in $\alpha$ (given $1+r>0$).

∎

Why savings rise with $r$ here (not always!)

A higher $r$ has a substitution effect (save more) and an income effect. Because the agent receives some endowment when old, a higher $r$ lowers the present value of lifetime wealth $(1-\alpha)+\alpha/(1+r)$, and this income effect also raises saving. With log utility the pure-substitution and pure-income effects on $c_y$ from $r$ cancel, leaving only this wealth effect. If $\alpha = 0$, saving $s=\beta$ would not depend on $r$ at all.

Equilibrium is $s(\bar r) = 0$. Since only generation $t$ can be in the loan market at $t$, the equilibrium rate is the one at which the young choose zero savings. Setting $s=0$: $\beta(1-\alpha)(1+\bar r) = (1-\beta)\alpha$. This gives the same $\bar r$ as the FOC route in §2.

Interactive. Left: a young agent's budget line through the endowment $(1-\alpha,\alpha)$, with slope $-(1+r)$, and the chosen point (blue). Right: the savings function $s(r)$, with the current $r$ marked and the equilibrium $\bar r$ where it crosses zero (violet). Move $r$ until the chosen point sits on the endowment: that's $\bar r$, and the indifference curve is tangent at the endowment. Then raise $\beta$. The $s(r)$ curve shifts up, so $\bar r$ must fall to restore $s=0$. Raise $\alpha$ and the curve shifts down, so $\bar r$ rises.

4. Why $\bar r$ rises with $\alpha$ and falls with $\beta$

The slides give a "zero-savings" argument II p.14. Make it precise with the implicit function theorem applied to $s(\bar r;\alpha,\beta)=0$:

$$\frac{\partial\bar r}{\partial\beta} = -\frac{\partial s/\partial\beta}{\partial s/\partial r} < 0,\qquad \frac{\partial\bar r}{\partial\alpha} = -\frac{\partial s/\partial\alpha}{\partial s/\partial r} > 0 .$$

Higher $\beta$: the young want to shift consumption toward old age, so at the old rate they would save ($s>0$). Nobody can absorb those savings, so the rate must fall until they no longer want to.
Higher $\alpha$: more of the endowment arrives in old age, so the young want to borrow ($s<0$). The rate must rise until they no longer want to.

The same conclusions follow from differentiating $1+\bar r = \frac{(1-\beta)\alpha}{\beta(1-\alpha)}$ directly: it is increasing in $\alpha$ and decreasing in $\beta$.

5. The price path

Interactive. The equilibrium prices $p_t = (1+\bar r)^{-(t-1)}$, $t=1,\dots,15$ (with $p_1=1$), on a log scale, so a geometric path is a straight line. When $\alpha>\beta$ (so $\bar r>0$) prices fall geometrically and the sum $\sum_t p_t$ is finite. When $\alpha\le\beta$ (so $\bar r\le 0$) prices are constant or rising, and $\sum_t p_t = \infty$. That divergence is exactly what breaks the First Welfare Theorem in Lesson 12.

6. Check your understanding

1. In the no-trade equilibrium, what determines $p_{t+1}/p_t$?

The initial old's marginal utility at date one only
Aggregate endowment growth between dates t and t+1
Generation t's MRS evaluated at its own endowment
An arbitrary choice, since every price is consistent

Generation $t$ is the only one that values both date-$t$ and date-$(t+1)$ goods. Its FOC at the (forced) optimum $x^t=e^t$ gives $p_{t+1}/p_t = v^t_2/v^t_1$ at $e^t$. Only the level $p_1$ is arbitrary.

2. In the log example with $\alpha = 0.25$ and $\beta = 0.5$, what is the sign of $\bar r$?

Negative, because the old-age endowment is relatively small
Positive, because the young are relatively quite patient
Zero, because aggregate endowment is constant over time
Undetermined, because prices are only defined up to scale

$\bar r = (\alpha-\beta)/(\beta(1-\alpha)) = (-0.25)/(0.375) = -2/3$. The young want old-age consumption share $\beta=0.5$ but receive only $\alpha=0.25$ of their endowment when old. They want to save, and the rate must fall below zero to discourage them.

3. Holding $r$ fixed, why does saving $s(r;\alpha,\beta)$ fall when $\alpha$ rises?

Higher α raises patience, so agents want to consume later
Higher α lowers the interest rate that young agents face
Higher α makes agents value old-age consumption much less
Higher α moves endowment to old age, so less needs saving

$\alpha$ is an endowment parameter, not a preference parameter. Shifting endowment from youth to old age means the agent already has more old-age resources, so it saves less (and may borrow). Formally $\partial s/\partial\alpha = -\beta - (1-\beta)/(1+r) < 0$.

4. When $\bar r<0$, how do the Arrow–Debreu prices $p_t$ behave as $t\to\infty$?

They fall geometrically and their sum is finite
They rise geometrically and their sum is infinite
They stay constant and their sum is finite overall
They oscillate and their sum is not well defined

$p_t = (1+\bar r)^{-(t-1)}$ with $1+\bar r\in(0,1)$, so $p_t$ grows like $\big(1/(1+\bar r)\big)^{t-1}\to\infty$. Future goods are increasingly expensive in today's units: the economy is "starved" of ways to save.

7. Exercises

Past exams test this material through short true/false justifications and through one long guided problem on an OLG economy with an asset (Exercise 9.2). Work with general parameters: the exams never ask you to plug in numbers.

Exercise 9.1 Exam-inspired True or false? Justify in 1–4 lines.

With preferences $(1-\beta)\log c_y+\beta\log c_o$ and endowments $(1-\alpha,\alpha)$, the equilibrium interest rate is negative if and only if $\alpha\lt \beta$.
Show answer
True. $1+\bar r = \frac{(1-\beta)\alpha}{\beta(1-\alpha)}<1 \iff (1-\beta)\alpha\lt \beta(1-\alpha)\iff\alpha\lt \beta$. The young want old-age consumption share $\beta$ but are endowed with only $\alpha$. Choking off their desire to save requires a negative rate.
With log utility, the savings of the young do not depend on the interest rate.
Show answer
False in general. Log utility fixes the expenditure share: $c_y=(1-\beta)W$ with $W=(1-\alpha)+\alpha/(1+r)$. The present value of the old-age endowment falls with $r$, so $s=(1-\alpha)-c_y$ rises with $r$. Savings are independent of $r$ only when there is no old-age endowment, e.g. $s_t=\beta w_t$ in the Diamond question on the Fall 2025 final (Lesson 15).
Since nobody trades in equilibrium, the OLG interest rate has no economic meaning.
Show answer
False. $\bar r$ is the shadow price at which the young are just willing not to trade: $1/(1+\bar r)=v_2/v_1$ at the endowment. Its sign decides welfare: $\bar r<0$ is exactly when transfers from young to old (social security, money, debt) can make everyone better off (Lessons 10–11).
In the OLG economy with a tree paying a constant dividend $\bar d>0$ (Exercise 8.2), the steady-state interest rate may be negative. (cf. Past exam: 2017/18 Midterm 1, part 6)
Show answer
False. In a steady state, no arbitrage gives $1+\bar r=(\bar q+\bar d)/\bar q$, so $\bar r=\bar d/\bar q>0$ for any positive price. An asset with a positive dividend can't be priced at $r\le0$, because its price would be infinite.
In the same economy, a permanent increase in the dividend raises the price of the tree and lowers the interest rate. (cf. 2017/18 Midterm 1, part 8)
Show answer
False. If savings $\bar\sigma$ increase strictly with the return, $\bar q$ rises, and so does $\bar r$: the young will absorb the more valuable tree only at a higher return. With log utility, $\bar q=\beta\bar e$ does not move at all and $\bar r=\bar d/(\beta\bar e)$ rises. Either way $\bar r$ rises.
The equilibrium interest rate between $t$ and $t+1$ depends on the endowments of all generations.
Show answer
False (pure endowment economy). Generation $t$ alone trades between $t$ and $t+1$, and it consumes its endowment. So $1/(1+r_t)$ is generation $t$'s MRS at $(e^t_t,e^t_{t+1})$ and nothing else (Exercise 9.5).
A more patient population (higher $\beta$) has a higher equilibrium interest rate.
Show answer
False. At the old rate, more patient young want to save more ($s_\beta>0$), but saving must stay at zero, so the rate must fall: $d\bar r/d\beta=-s_\beta/s_r<0$ (Exercise 9.3).

Exercise 9.2 Past exam: 2017/18 Midterm 1, "OLG and trees", parts 4–12

Continue the economy of Exercise 8.2, where you showed $a_t=1$, $s_t=0$ and $1+r_t=(q_{t+1}+d_{t+1})/q_t$. Define the savings function $$\sigma_t(R)\equiv\arg\max_x\; u^t\big(e^t_t-x,\;xR\big),$$ the amount a young agent saves at gross return $R$.

Using market clearing and the young's optimality, write an implicit difference equation in $q_t$, $q_{t+1}$ and $d_{t+1}$ that the equilibrium price sequence must satisfy.
From now on $e^t_t=\bar e$, $u^t=\bar u$ and $d_t=\bar d\ge0$ are constant; let $\bar\sigma(R)$ be the corresponding savings function. Write an implicit equation for a constant price $\bar q$. Hint: one side is $\bar q$; the other is a function of $\bar d/\bar q$.
Assume $\bar u$ satisfies Inada conditions and $\bar d>0$. Is the implied interest rate $\bar r$ strictly positive? Is the equilibrium allocation Pareto optimal?
Assume $\bar\sigma$ is strictly increasing. Is the steady state unique? Argue with a graph.
$\bar d$ increases. What happens to $\bar q$ and $\bar r$?
Preferences shift so that, at any return, the young want to save more. What happens to $\bar q$ and $\bar r$?
Consumption when young and when old are both normal goods, and $\bar e$ increases. What happens to $\bar q$ and $\bar r$?
$\bar u$ is homothetic, and $\bar e$ and $\bar d$ both scale up by $\lambda>0$. What happens to $\bar q$ and $\bar r$?
Let $\bar u=(1-\beta)\log c^t_t+\beta\log c^t_{t+1}$. Solve for $\bar q$ and $\bar r$, and check your answers to (e)–(h).

Hint

Total saving of the young is $a_tq_t+s_t=q_t$. For (d)–(h), plot the 45° line $y=q$ against $y=\bar\sigma(1+\bar d/q)$ and ask which curve shifts.

Show solution

(a) The young save $q_t$ in total, at gross return $(q_{t+1}+d_{t+1})/q_t$. So $$q_t=\sigma_t\!\left(\frac{q_{t+1}+d_{t+1}}{q_t}\right).$$

(b) $\bar q=\bar\sigma\big(1+\bar d/\bar q\big)$, with $\bar r=\bar d/\bar q$.

(c) Yes and yes. Inada conditions make the young save a positive amount, so $\bar q>0$ and $\bar r=\bar d/\bar q>0$. With $p_t=(1+\bar r)^{-(t-1)}$, the value of the aggregate endowment is $\sum_t p_t(\bar e+\bar d)=(\bar e+\bar d)(1+\bar r)/\bar r\lt \infty$. By the First Welfare Theorem for OLG economies (Lesson 12), the equilibrium is PO. Exam warning: writing "$\bar r>0$, so the 1WT applies" is not enough. The theorem needs a finite value of the aggregate endowment, and you must say so.

(d) Yes. The left side $y=q$ is strictly increasing. The right side $y=\bar\sigma(1+\bar d/q)$ is strictly decreasing in $q$, because the return $1+\bar d/q$ falls with $q$ and $\bar\sigma$ is increasing. So the curves cross at most once. They do cross: near $q=0$ the right side is positive while the left side is near zero, and at $q=\bar e$ the right side is below $\bar e$, because Inada keeps $c_y>0$.

(e) The right-hand curve shifts up at every $q$, so $\bar q$ rises. Since $\bar q=\bar\sigma(1+\bar r)$ and $\bar\sigma$ is strictly increasing, a higher $\bar q$ needs a higher return: $\bar r$ rises too, so $\bar q$ rises less than proportionally to $\bar d$.

(f) The new savings function $\tilde\sigma\ge\bar\sigma$ shifts the curve up, so $\bar q$ rises, and with $\bar d$ fixed, $\bar r=\bar d/\bar q$ falls. (Write the inequality as $\tilde\sigma(R)\ge\bar\sigma(R)$. A circulated answer key has it backwards.)

(g) If $c_o=xR$ is normal, savings at any $R$ increase with $\bar e$. This is the same shift as (f): $\bar q$ rises and $\bar r$ falls.

(h) Homotheticity gives $\bar\sigma(R;\lambda\bar e)=\lambda\bar\sigma(R;\bar e)$. Try $\lambda\bar q$: the return $1+\lambda\bar d/(\lambda\bar q)$ is unchanged and saving scales by $\lambda$, so it solves the equation. By uniqueness, $\bar q$ scales by $\lambda$ and $\bar r$ is unchanged.

(i) The FOC $(1-\beta)/(\bar e-x)=\beta/x$ gives $x=\beta\bar e$ at every $R$. So $\bar q=\beta\bar e$ and $\bar r=\bar d/(\beta\bar e)$. Then $\bar r$ rises with $\bar d$, falls with $\beta$ and with $\bar e$, and is invariant to scaling $\bar e,\bar d$ together. $\bar q$ does not respond to $\bar d$: log savings are flat in $R$, the boundary case of (e).

Exercise 9.3 Slides II p.13 Exam-inspired The savings function

Let $s(r;\alpha,\beta)=\arg\max_s (1-\beta)\log(1-\alpha-s)+\beta\log(\alpha+s(1+r))$.

[10 pts] Show that $s(r;\alpha,\beta)=\beta(1-\alpha)-(1-\beta)\dfrac{\alpha}{1+r}$, and that this equals the slide's form $(1-\alpha)-(1-\beta)\big[(1-\alpha)+\frac{\alpha}{1+r}\big]$.
[10 pts] Show that $s$ is increasing in $r$ and $\beta$ and decreasing in $\alpha$. Interpret each sign in one line.
[10 pts] Without using the closed form for $\bar r$, use $s(\bar r;\alpha,\beta)=0$ and the implicit function theorem to sign $d\bar r/d\alpha$ and $d\bar r/d\beta$.
[10 pts] Now let the young have no old-age endowment, so $e=(w,0)$. Show that savings no longer depend on $r$. Which effect has disappeared?

Show solution

(a) The FOC is $\frac{1-\beta}{1-\alpha-s}=\frac{\beta(1+r)}{\alpha+s(1+r)}$, so $(1-\beta)(\alpha+s(1+r))=\beta(1+r)(1-\alpha-s)$. Collecting $s$: $s(1+r)=\beta(1+r)(1-\alpha)-(1-\beta)\alpha$, which is the stated formula. Expanding the slide's form gives $(1-\alpha)\beta-(1-\beta)\alpha/(1+r)$, the same thing.

(b) $s_r=\frac{(1-\beta)\alpha}{(1+r)^2}>0$: a higher rate lowers the present value of the old-age endowment, so the young are poorer and consume less today. $s_\beta=(1-\alpha)+\frac{\alpha}{1+r}>0$: more patience moves consumption toward old age. $s_\alpha=-\beta-\frac{1-\beta}{1+r}<0$: more old-age endowment means less need to save.

(c) $d\bar r/d\alpha=-s_\alpha/s_r>0$ and $d\bar r/d\beta=-s_\beta/s_r<0$. A higher $\alpha$ lowers desired saving at the old rate, so the rate must rise to restore $s=0$. A higher $\beta$ works the other way. Check against $\bar r=\frac{\alpha-\beta}{\beta(1-\alpha)}$: $\partial\bar r/\partial\alpha=\frac{1-\beta}{\beta(1-\alpha)^2}>0$ and $\partial\bar r/\partial\beta=-\frac{\alpha}{\beta^2(1-\alpha)}<0$.

(d) With $e=(w,0)$ the FOC is $\frac{1-\beta}{w-s}=\frac{\beta}{s}$, so $s=\beta w$. The term in $r$ came only from the present value of the old-age endowment (a wealth effect); with log utility, the income and substitution effects of $r$ on a pure saver cancel exactly.

Exercise 9.4 Exam-inspired CRRA preferences

Replace log utility with $v(c_y,c_o) = (1-\beta)\frac{c_y^{1-\sigma}}{1-\sigma} + \beta\frac{c_o^{1-\sigma}}{1-\sigma}$, $\sigma>0$, keeping endowments $(1-\alpha,\alpha)$.

Find $1+\bar r$.
For which parameters is $\bar r<0$? Does the condition depend on $\sigma$?
How does $\bar r$ respond to $\sigma$ when $\alpha<1/2$? Interpret.

Show solution

(a) $1+\bar r = \dfrac{v_1}{v_2}\Big|_e = \dfrac{1-\beta}{\beta}\Big(\dfrac{\alpha}{1-\alpha}\Big)^{\sigma}$.

(b) $\bar r<0 \iff (1-\beta)\alpha^\sigma < \beta(1-\alpha)^\sigma$. For $\sigma = 1$ this is $\alpha\lt \beta$. In general it depends on $\sigma$, unless $\alpha=\tfrac12$.

(c) If $\alpha<1/2$, then $\alpha/(1-\alpha)<1$ and $1+\bar r$ is decreasing in $\sigma$. With more curvature the young care more about smoothing toward their poor old age, so the rate must fall further to stop them saving.

Exercise 9.5 Exam-inspired Non-stationary endowments

Keep log preferences with weight $\beta$, but let generation $t$'s endowments be $e^t_t = w_t$ and $e^t_{t+1}=a_t$, where $\{w_t,a_t\}$ are positive sequences bounded above and away from zero.

Find $r_t$ and $p_t$ (with $p_1=1$).
Explain in one line why nothing links $r_t$ to $r_{t+1}$.
Suppose the ratio $a_t/w_t$ alternates between two values, $\rho_1$ at odd dates and $\rho_2$ at even dates. Give a necessary and sufficient condition, in terms of the two interest rates, for the value of the aggregate endowment $\sum_tp_t\bar e_t$ to be finite. Can it be finite even if $r_t<0$ at every odd date?

Show solution

(a) $1+r_t = \dfrac{(1-\beta)/w_t}{\beta/a_t} = \dfrac{1-\beta}{\beta}\cdot\dfrac{a_t}{w_t}$ and $p_t = \prod_{s=1}^{t-1}\dfrac{1}{1+r_s}$.

(b) Each $r_t$ is generation $t$'s MRS at its own endowment, and no one trades across generations.

(c) Over each two-period cycle prices are multiplied by $\kappa\equiv\frac{1}{(1+r_{\text{odd}})(1+r_{\text{even}})}$. Since $\bar e_t$ is bounded above and away from zero, $\sum_tp_t\bar e_t$ behaves like a geometric series in $\kappa$. So it is finite iff $\kappa<1$, i.e. $(1+r_{\text{odd}})(1+r_{\text{even}})>1$. Yes: a negative rate at odd dates is fine if it is outweighed by a high enough rate at even dates. Efficiency depends on the cumulative discounting, not on the sign of any single rate (Lesson 12).

Exercise 9.6 Proof

Show directly, without FOCs, that at the prices $p_t = (1+\bar r)^{-(t-1)}$ the young of generation $t$ find their endowment optimal: the optimal $c_y$ at rate $\bar r$ equals $1-\alpha$.

Show solution

Log utility spends a share $1-\beta$ of lifetime wealth on youth: $c_y = (1-\beta)\big[(1-\alpha) + \alpha/(1+\bar r)\big]$. Since $1/(1+\bar r) = \beta(1-\alpha)/((1-\beta)\alpha)$, we get $\alpha/(1+\bar r) = \beta(1-\alpha)/(1-\beta)$. So $c_y = (1-\beta)(1-\alpha) + \beta(1-\alpha) = 1-\alpha$. ✓ Equivalently, $s(\bar r) = 0$.

Exercise 9.7 Problem set 3 (2022), P4 Q1–Q7 A growing tree under uncertainty (harder)

Two-period lives, a mass one per cohort, labour income only when young. Output is $Y_t$, with $Y_{t+1}=Y_tz_{t+1}$, where $z>0$ is i.i.d. The young earn $w_t=(1-\delta)Y_t$, and one tree, owned initially by the old, pays fruit $d_t=\delta Y_t$. Preferences are $(1-\beta)\log c^t_t+\beta E_t\log c^t_{t+1}$. The young buy $s_t$ trees at price $q^e_t$ and $b_t$ one-period discount bonds at price $q^b_t$, each paying one unit next period: $$c^t_t+s_tq^e_t+b_tq^b_t=w_t,\qquad c^t_{t+1}=s_t(q^e_{t+1}+d_{t+1})+b_t .$$

Write the first-order conditions for $s_t$ and $b_t$.
What are the equilibrium $s_t$ and $b_t$? Why?
Guess $q^e_t=Q^eY_t$ and $q^b_t=Q^b$. Show that $Q^e=\beta(1-\delta)$. Explain in one line why $Q^e$ rises with $\beta$ and falls with $\delta$.
Find the expected gross return on the tree and the gross return on the bond, $R^b=1/Q^b$.

The efficiency half of this problem (is the equilibrium PO even though the safe rate can be below growth?) appears in Lessons 12–13.

Show solution

(a) Tree: $\frac{(1-\beta)q^e_t}{c^t_t}=\beta E_t\Big[\frac{q^e_{t+1}+d_{t+1}}{c^t_{t+1}}\Big]$. Bond: $\frac{(1-\beta)q^b_t}{c^t_t}=\beta E_t\Big[\frac{1}{c^t_{t+1}}\Big]$.

(b) $s_t=1$: the old sell the tree and the young must hold it. $b_t=0$: bonds are in zero net supply and all young are identical.

(c) In equilibrium $c^t_{t+1}=q^e_{t+1}+d_{t+1}$, so the right side of the tree FOC is $\beta$, and $(1-\beta)q^e_t=\beta c^t_t=\beta(w_t-q^e_t)$. Hence $q^e_t=\beta w_t=\beta(1-\delta)Y_t$. (Log utility: the young save a fraction $\beta$ of income, and the tree is the only thing to save in.) A higher $\beta$ means more saving chasing one tree; a higher $\delta$ means less labour income to save.

(d) Old consumption is $c^t_{t+1}=[\beta(1-\delta)+\delta]Y_tz_{t+1}$. The tree return is $R^e=c^t_{t+1}/q^e_t$, so $E_tR^e=\Big(1+\frac{\delta}{\beta(1-\delta)}\Big)E[z]$. The bond FOC gives $Q^b=\frac{\beta(1-\delta)}{\beta(1-\delta)+\delta}E[1/z]$, so $R^b=\Big(1+\frac{\delta}{\beta(1-\delta)}\Big)\big/E[1/z]$. Since $E[1/z]\ge1/E[z]$ (Jensen), $R^b\le E R^e$: the gap is a risk premium.

Key takeaways

Equilibrium prices make autarky optimal: $p_{t+1}/p_t = v^t_2/v^t_1$ at the endowment, and $1+r_t = p_t/p_{t+1}$.
Log example: $1+\bar r = \frac{(1-\beta)\alpha}{\beta(1-\alpha)}$, so $\bar r \gtrless 0 \iff \alpha\gtrless\beta$, and $p_t = (1+\bar r)^{-(t-1)}$.
Savings: $s(r) = \beta(1-\alpha) - (1-\beta)\alpha/(1+r)$, increasing in $r$ and $\beta$, decreasing in $\alpha$. $\bar r$ solves $s(\bar r)=0$.
When $\bar r\le0$, the prices $p_t$ don't fall, so $\sum_t p_t=\infty$. Keep this in mind for Lesson 12.

💬 Stuck or curious? Ask me (your Claude teacher) anything. Try "why are the FOCs necessary even though agents don't trade?" or "decompose the effect of r on savings into income and substitution effects."

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Equilibrium Prices and Interest Rates in the OLG Economy

1. Prices that support autarky

2. The log example

Equilibrium prices

3. The savings function

4. Why \(\bar r\) rises with \(\alpha\) and falls with \(\beta\)

5. The price path

6. Check your understanding

7. Exercises

Key takeaways