# Intertemporal equilibrium

## Temporary equilibrium

Before turning to the intertemporal equilibrium and the analysis of the steady state, we study the temporary equilibrium that takes place every period.

We have not discussed the firms, but they follow the same setup as in Ramsey: use capital and labour in a perfectly competitive environment.

We shall work in intensive form: $k_{t} \equiv \frac{K_{t}}{N_{t}}.$

1. Labour market equilibrium: Only young individuals supply labour. Moreover, they do so inelastically. During period $t$ there are $N_{t}$ young agents and, hence, the supply of labour is $N_{t}$. Equating this to labour demand from firms $L_{t}$ gives the wage rate: $$w_{t} = \omega\left(\frac{K_{t}}{N_{t}}\right)=\omega(k_{t}).$$

2. Capital market: Only old individuals own capital. Since firms operate competitively, they make zero profits. Hence, $f^{\prime}(k_{t})K_{t}$ is distributed as interests. Old households receive $N_{t-1}R_{t}s_{t-1} = R_{t}K_{t}.$ So, what old households receive must equal what firms distribute In other terms: $R_{t}K_{t}= f^{\prime}(k_{t}) K_{t}$ and $R_t = f^\prime (k_t).$

3. Good market: Finding the equilibrium for this market departs from the otherwise similar Ramsey case. Remember that we have two types of agents: young and old. Total production is given by: $$Y_{t} = F(K_{t}, N_{t}) = \color{red}{N_{t}} f(k_{t}).$$ Total demand for goods combines the consumption of $d_{t}$, the old generation living in period $t$, and the demand for consumption and investment from the young: $c_{t}, s_{t}.$ Therefore, counting how many old and young individuals live during period $t$ we have that total demand equals: $$N_{t-1}d_{t} + N_{t}(c_{t}+s_{t}).$$ The equilibrium on the goods market implies: $$Y_{t} = N_{t-1}d_{t} + N_{t}(c_{t}+s_{t}).$$

Also, we can check that total production equals total consumption: $$N_{t}(c_{t}+s_{t}) = N_{t}w_{t}=N_{t}(f(k_{t})-k_{t}f^{\prime}(k_{t})) = Y_{t}-K_{t}f^{\prime}(k_{t}),$$ $$N_{t-1}d_{t} = N_{t-1}R_{t}s_{t-1} = R_{t}K_{t} = K_{t}f^{\prime}(k_{t}).$$

A temporary equilibrium is a set $\{w_{t}, R_{t}, K_{t}, L_{t}, Y_{t}, k_{t}, I_{t}, c_{t}, s_{t}, d_{t}\}$ that satisfies:

$$\begin{eqnarray} w_{t} = \omega(k_{t}), \\\ R_{t} = f^{\prime}(k_{t}), \\\ L_{t} = N_{t}, \\\ Y_{t} = N_{t}f(k_{t}), \\\ Y_{t} = N_{t-1}d_t + N_{t}(c_{t}+s_{t}) \\\ I_{t} = N_{t}s_{t}, \\\ c_{t} = w_{t} - s_{t}, \\\ s_{t} = s(\omega(k_{t}),R_{t+1}), \\\ d_{t} = R_{t}s_{t-1}. \end{eqnarray}$$

The existence of a temporary equilibrium is guaranteed because the functions are single-valued.

## Intertemporal equilibrium with perfect foresight

The equilibrium equation that links two consecutive periods is the capital accumulation equation. In particular, the savings of young individuals at period $t$ are transformed into productive capital at $t+1.$

Note: in this model, it is useful to write the equations first in aggregate terms and then convert them to its intensive-form representation.

$$K_{t+1} = N_{t}s_{t} = N_{t} s\left( \omega(k_{t}), R_{t+1} \right).$$

In intensive form:

$$k_{t+1} = \frac{K_{t+1}}{\color{red}{N_{t+1}}} = \frac{N_{t}}{\color{red}{N_{t+1}}} s\left( \omega(k_{t}), R_{t+1} \right) = \frac{1}{1+n}s\left( \omega(k_{t}), R_{t+1} \right) .$$

Finally, we can incorporate the equilibrium in the capital market (together with perfect foresight) and replace $R_{t+1} = f^{\prime}(k_{t+1}):$

$$k_{t+1} =\frac{1}{1+n}s \left( \omega(k_{t}), f^{\prime}(k_{t+1}) \right).$$

Intertemporal equilibrium (for perfect foresigt): Given an initial capital stock $k_{0} = K_{0} \big{/} N_{-1}$, an intertemporal equilibrium (for perfect foresight) is a sequence of temporary equilibria that satisfies for all $t>0$ the equation:

$$k_{t+1} =\frac{1}{1+n}s \left( \omega(k_{t}), f^{\prime}(k_{t+1}) \right).$$

Note: de la Croix and Michel discuss on pp. 20–27 the existence and uniqueness of the intertemporal equilibrium.

## Existence of an intertemporal equilibrium

The existence of at least one temporary equilibrium is guaranteed by the properties of the functions. The proof is quite involved, though, as is presented below. Having an intertemporal equilibrium means having a solution for $k_{t+1}$ in the equation

$$k_{t+1} = \frac{1}{1+n}s(\omega(k_t), f^\prime (k_{t+1})),$$

where $k_t$ is predetermined at $t.$

Proof

The proof uses the following equation about savings: $$0< s\left( w, f^\prime(k) \right) < w.$$ In words, it indicates that individuals have positive savings, and they save only part of their total income.

Next, define $$H(k,w) = (1+n)k - s(w, f^\prime(k))= 0.$$ Basically, here we are using the definition of an intertemporal equilibrium. Having an intertemporal equilibrium is then equivalent to finding a $k$ satisfying the previous equation.

We use the intermediate value theorem to show that at least one solution exists. First, we analyse the behaviour of $H(k,w)$ when $k$ tends to $+\infty.$

From $0< s\left( w, f^\prime(k) \right) < w,$ we have

$$0 < \frac{s(w, f^\prime (k))}{k} < \frac{w}{k}.$$

Limit when $k \rightarrow +\infty$

Keeping $w$ fixed, the limit of $\frac{w}{k}$ when $k \rightarrow +\infty$ is 0. Then,

$$\lim_{k \rightarrow +\infty} \frac{s(w, f^\prime (k))}{k} = 0.$$

Consequently,

$$\lim_{k \rightarrow +\infty}\frac{H(w,k)}{k} = \lim_{k \rightarrow +\infty} (1+n) - \frac{s(w, f^\prime (k))}{k} = 1+n > 0.$$

Limit when $k \rightarrow 0$

When $k$ tends to zero, we shall distinguish two cases regarding $f^\prime (k)$. It can be that either

• Case $\lim_{k \rightarrow 0} f^\prime (k) = f^\prime (0)> 0$ and finite. Then, the function $s(w, f^\prime(k))$ is well defined and positive and we have:

$$\lim_{k \rightarrow 0} H(w,k) = \lim_{k \rightarrow 0} \left[(1+n)k - s(w,f^\prime (k)) \right] = -s(w, f^\prime (k)) < 0.$$

• Case $\lim_{k \rightarrow 0} f^\prime (k) = +\infty.$ Two things can occur in that case.

• Savings are positive: $\lim_{k \rightarrow 0} s(w,f^\prime (k))>0.$ This case is analogous to the previous one:

$$\lim_{k \rightarrow 0} H(w,k) = \lim_{k \rightarrow 0} \left[(1+n)k - s(w,f^\prime (k)) \right] = -s(w, f^\prime (k)) < 0.$$

• Savings tend to zero: $\lim_{k \rightarrow 0} s(w, f^\prime (k)) = 0.$ In this case, since the interest rate goes to infinity individuals tend to save zero. This implies that consumption in the second period, $d$, tends to infinity. To see this, first notice that $d = f^\prime (k) s(w, f^\prime (k)).$ Moreover, from the first order conditions we know that: $$u^\prime (d) = \frac {u^\prime ( w - s)}{\beta f^\prime (k)}$$ and hence $$\lim_{k \rightarrow 0}u^\prime (d) = \lim_{k \rightarrow 0}\frac {u^\prime ( w - s)}{\beta f^\prime (k)}=0.$$ Hence, $u^\prime (d)$ tends to zero which means that $d$ goes to infinity. Therefore: $$\lim_{k \rightarrow 0} f^\prime (k) s(w, f^\prime (k)) = +\infty.$$

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