Let $\vec{C}$ = $\{C_0,\dots, C_T\}$ and

$\vec{K}$ = $\{K_0,\dots,K_{T+1}\}$.

### Digression: an Aggregation Theory

### Digression: Aggregation Theory

We use a concept of a representative consumer to be thought of as follows.

There is a unit mass of identical consumers.

There is a unit mass of identical consumers indexed by $\omega \in [0,1]$.

For $\omega \in [0,1]$, consumption of consumer is $c(\omega)$.

Consumption of consumer $\omega$ is $c(\omega)$.

Aggregate consumption is

$$

C = \int_0^1 c(\omega) d \omega

$$

Consider the a welfare problem of choosing an allocation $\{c(\omega)\}$ across consumers to maximize

Consider a welfare problem that chooses an allocation $\{c(\omega)\}$ across consumers to maximize

$$

\int_0^1 u(c(\omega)) d \omega

Form a Lagrangian $L = \int_0^1 u(c(\omega)) d \omega + \lambda [C - \int_0^1 c(\omega) d \omega ] $.

Differentiate under the integral signs with respect to each $\omega$ to obtain the first-order

necessary condtions

necessary conditions

$$

u'(c(\omega)) = \lambda.

$$

This condition implies that $c(\omega)$ equals a constant $c$ that is independent

These conditions imply that $c(\omega)$ equals a constant $c$ that is independent

of $\omega$.

To find $c$, use the feasibility constraint {eq}`eq:feas200` to conclude that

To find $c$, use feasibility constraint {eq}`eq:feas200` to conclude that

$$

c(\omega) = c = C.

It appears often in aggregate economics.

We shall use it in this lecture and in {doc}`Cass-Koopmans Competitive Equilibrium <cass_koopmans_2>`.

We shall use this aggregation theory here and also in this lecture {doc}`Cass-Koopmans Competitive Equilibrium <cass_koopmans_2>`.

#### An Economy

The representative household inelastically supplies a single unit of

labor $N_t$ at each $t$, so that

$N_t =1 \text{ for all } t \in [0,T]$.

$N_t =1 \text{ for all } t \in \{0, 1, \ldots, T\}$.

The representative household has preferences over consumption bundles

ordered by the utility functional:

```

where $\beta \in (0,1)$ is a discount factor and $\gamma >0$

governs the curvature of the one-period utility function with larger $\gamma$ implying more curvature.

governs the curvature of the one-period utility function.

Larger $\gamma$'s imply more curvature.

Note that

```{math}

:label: allocation

C_t + K_{t+1} \leq F(K_t,N_t) + (1-\delta) K_t, \quad \text{for all } t \in [0, T]

C_t + K_{t+1} \leq F(K_t,N_t) + (1-\delta) K_t \quad \text{for all } t \in \{0, 1, \ldots, T\}

```

where $\delta \in (0,1)$ is a depreciation rate of capital.

\left(F(K_t,1) + (1-\delta) K_t- C_t - K_{t+1} \right)\right\}

$$ (eq:Lagrangian201)

and then pose the following min-max problem:

and pose the following min-max problem:

```{math}

:label: min-max-prob

maximization with respect to $\vec{C}, \vec{K}$ and

minimization with respect to $\vec{\mu}$.

- Our problem satisfies

conditions that assure that required second-order

conditions that assure that second-order

conditions are satisfied at an allocation that satisfies the

first-order conditions that we are about to compute.

first-order necessary conditions that we are about to compute.

Before computing first-order conditions, we present some handy formulas.

\end{aligned}

$$

(Here we are using that $N_t = 1$ for all $t$, so that $K_t = \frac{K_t}{N_t}$.)

### First-order necessary conditions

We now compute **first-order necessary conditions** for extremization of the Lagrangian {eq}`eq:Lagrangian201`:

We now compute **first-order necessary conditions** for extremization of Lagrangian {eq}`eq:Lagrangian201`:

```{math}

:label: constraint1

```

In computing {eq}`constraint3` we recognize that $K_t$ appears

in both the time $t$ and time $t-1$ feasibility constraints.

in both the time $t$ and time $t-1$ feasibility constraints {eq}`allocation`.

Restrictions {eq}`constraint4` come from differentiating with respect

to $K_{T+1}$ and applying the following **Karush-Kuhn-Tucker condition** (KKT)

u'\left(C_{t}\right) \quad \text{ for all } t=0,1,\dots, T

```

Applying the inverse of the utility function on both sides of the above

Applying the inverse marginal utility of consumption function on both sides of the above

equation gives

$$

(1-\delta)]\right)^{1/\gamma} \end{aligned}

$$

This is a non-linear first-order difference equation that an optimal sequence $\vec C$ must satisfy.

Below we define a `jitclass` that stores parameters and functions

that define our economy.

$\vec{C}, \vec{K}$ and an associated Lagrange multiplier sequence

$\vec{\mu}$.

The first-order necessary conditions

First-order necessary conditions

{eq}`constraint1`, {eq}`constraint2`, and

{eq}`constraint3` for the planning problem form a system of **difference equations** with

two boundary conditions:

- We could continue in this way to compute the remaining elements of

$\vec{C}, \vec{K}, \vec{\mu}$.

But we don't have an initial condition for $\mu_0$, so this

won't work.

However, we woujld not be assured that the Kuhn-Tucker condition {eq}`kkt` would be satisfied.

Furthermore, we don't have an initial condition for $\mu_0$.

So this won't work.

Indeed, part of our task is to compute the optimal value of $\mu_0$.

Indeed, part of our task is to compute the **optimal** value of $\mu_0$.

To compute $\mu_0$ and the other objects we want, a simple modification of the above procedure will work.

- Guess an initial Lagrange multiplier $\mu_0$.

- Apply the **simple algorithm** described above.

- Compute $k_{T+1}$ and check whether it

- Compute $K_{T+1}$ and check whether it

equals zero.

- If $K_{T+1} =0$, we have solved the problem.

- If $K_{T+1} > 0$, lower $\mu_0$ and try again.

The following Python code implements the shooting algorithm for the

planning problem.

We actually modify the algorithm slightly by starting with a guess for

$c_0$ instead of $\mu_0$ in the following code.

(Actually, we modified the preceding algorithm slightly by starting with a guess for

$c_0$ instead of $\mu_0$ in the following code.)

```{code-cell} python3

@njit

$\mu_0$ because $C_0$ is an exact function of

$\mu_0$).

We know that the lowest $C_0$ can ever be is $0$ and the

We know that the lowest $C_0$ can ever be is $0$ and that the

largest it can be is initial output $f(K_0)$.

Guess $C_0$ and shoot forward to $T+1$.

In a steady state $K_{t+1} = K_t=\bar{K}$ for all very

large $t$.

Evalauating the feasibility constraint {eq}`allocation` at $\bar K$ gives

Evalauating feasibility constraint {eq}`allocation` at $\bar K$ gives

```{math}

:label: feasibility-constraint

\bar{K} = f'^{-1}(\rho+\delta)

$$

For the production function {eq}`production-function` this becomes

For production function {eq}`production-function`, this becomes

$$

\alpha \bar{K}^{\alpha-1} = \rho + \delta

plot_paths(pp, 0.3, k_ss/3, [250, 150, 50, 25], k_ss=k_ss);

```

Different colors in the above graphs are associated with

In the above graphs, different colors are associated with

different horizons $T$.

Notice that as the horizon increases, the planner puts $K_t$

Notice that as the horizon increases, the planner keeps $K_t$

closer to the steady state value $\bar K$ for longer.

This pattern reflects a **turnpike** property of the steady state.

The planner chooses a positive saving rate that is higher than the steady state

saving rate.

Note, $f''(K)<0$, so as $K$ rises, $f'(K)$ declines.

Note that $f''(K)<0$, so as $K$ rises, $f'(K)$ declines.

The planner slowly lowers the saving rate until reaching a steady

state in which $f'(K)=\rho +\delta$.

In that lecture, we replace the planner of this lecture with Adam Smith's **invisible hand**.

In place of quantity choices made by the planner, there are market prices that are set by a mechanism outside the model, a so-called invisible hand.

In place of quantity choices made by the planner, there are market prices that are set by a *deus ex machina* from outside the model, a so-called invisible hand.

Equilibrium market prices must reconcile distinct decisions that are made independently

by a representative household and a representative firm.