Ramsey Model (continued)

# Ramsey Model (continued)
## Macroeconomics
### Kenji Sato
### day 11

---

$$\let\oldhat\hat
\renewcommand{\hat}[1]{\oldhat{\hspace{0pt} #1}}$$

## Exogenous variables

Population growth

`$$\dot{L} = n L$$`

Technical growth

`$$\dot{A} = g A$$`

---

---

## Consumer Behavior: Objective

Consumers act altruistically. They care not only about 
their own consumption but also about offspring's.

Each individual maximizes the total utility:

`$$\max \int_0^\infty e^{-\rho t} u(c(t)) e^{nt} dt$$`

Since there are `$L(0)$` consumers at `$t = 0$`, the economy 
as a whole miximizes the following objective.

`$$\max \int_0^\infty e^{-\rho t} u(c(t)) L(t) dt$$`

---

## Consumer Behavior: Budget constraint

Each consumer is subject to the following flow budget constraint:

`$$c(t) + \dot{s}(t) = \left(r(t) - n\right) s(t)  + w(t)$$`

We have `$-n s(t)$` in the interest term because of the population growth. Per-person 
share of the interest income is reduced by the factor of growth rate.

---

## Consumer Behavior: First-order condition

The representative consumer solves the following maximization problem:

`$$\max \int_0^\infty e^{-(\rho - n) t} u(c(t)) dt$$`
subject to

`$$c(t) + \dot{s}(t) = \left(r(t) - n\right) s(t)  + w(t)$$`

**Problem: Set up the Hamiltonian. (3min)**

---

## Consumer Behavior: Canonical Equations

`$$\mathcal H =
e^{-(\rho - n) t} u(c) + \lambda [(r - n) s + w - c]$$`

**Problem: Compute the canonical equations. (5min)**

1. `$\partial \mathcal H / \partial c = 0$`
2. `$\dot{\lambda} = - \partial \mathcal H / \partial s$`
3. `$\dot{s} = \partial \mathcal H / \partial \lambda$`

---

## Consumer Behavior: Euler equation

**Problem: Derive the Euler equation. (5min)**

`$$\begin{aligned}
  \frac{\dot{c}(t)}{c(t)} = 
  \frac{r(t) - \rho}{\Theta(c)} 
\end{aligned}$$`

Under the assumption of CRRA utility

`$$u(c) = \frac{c^{1 - \theta} - 1}{1 - \theta},\quad \theta > 0$$`

the Euler equation is expressed as

`$$\begin{aligned}
  \frac{\dot{c}(t)}{c(t)} = 
  \frac{r(t) - \rho}{\theta} 
\end{aligned}$$`

---

## Euler equation for `$C/(AL)$`

Define the consumption per unit of effective labor:

`$$\hat{c} = \frac{c}{A} = \frac{C}{AL}$$`

Since `$A$` grows exogenously with a constant growth rate `$g$`,

`$$\frac{\dot{\hat{c}}}{\hat{c}} = \frac{\dot{c}(t)}{c(t)} - g$$`

We therefore have

`$$\frac{\dot{\hat{c}}}{\hat{c}} =
\frac{r(t) - \rho - \theta g}{\theta}$$`

---

---

## Firm behavior: Production function

Let's describe the behavior of the firms.

This is mostly equivalent to the setup of the Solow model.

The firms have access to the CRS production technology

`$$Y(t) = F(K(t), A(t) L(t)).$$`

In per-`$AL$` term,

`$$\hat{y}(t) = f\left(\hat{k} (t)\right) = 
\frac{F(K, AL)}{AL}$$`

---

## Firm behavior: Profit maximization

Rent

`$$r(t) + \delta = f'\left( \hat{k}(t) \right)$$`

Wage

`$$w(t) = A(t) \left[
  f\left( \hat{k}(t) \right) - \hat{k}(t) f'\left( \hat{k}(t) \right)
\right]$$`

Wage per effective labor

`$$\hat{w}(t) = f\left( \hat{k}(t) \right) - \hat{k}(t) f'\left( \hat{k}(t) \right)$$`

---

## Equilibrium: Capital Accumulation

The capital market clearing condition:

`$$s(t) L(t) = K(t)$$`

The flow budget constraint

`$$c(t) + \dot{s}(t) = \left(r(t) - n\right) s(t)  + w(t)$$`

is transformed into

`$$\frac{\dot{\hat{k}}}{\hat{k}} =
\frac{\dot s}{s} - g = 
\frac{f\left(\hat{k}\right) - \hat{c} - (\delta + g + n) \hat{k}}{\hat{k}}$$`

---

## Equilibrium: First-order Dynamics

We have the following capital accumulation equation:

`$$\dot{\hat{k}} = f\left(\hat{k}\right) - \hat{c} - (\delta + g + n) \hat{k}$$`

The dynamics of the Ramsey model is determined by this capital 
accumulation equation together with 
the Euler equation:

`$$\frac{\dot{\hat{c}}}{\hat{c}} =
\frac{f'\left(\hat{k}\right) - \delta - \rho - \theta g}{\theta}$$`

---

## Locus for `$\dot{\hat{k}} = 0$`

Let's derive the condition under which `$\dot{\hat{k}} = 0$`.

From the capital accumulation equation,

`$$\dot{\hat{k}} = f\left(\hat{k}\right) - \hat{c} - (\delta + g + n) \hat{k},$$`
we obtain

`$$\begin{aligned}
  \dot{\hat{k}} = 0 
  \Leftrightarrow
  \hat{c} = f\left(\hat{k}\right) - (\delta + g + n) \hat{k}
\end{aligned}$$`

---

## Regions for `$\dot{\hat{k}} > 0$` and `$\dot{\hat{k}} < 0$`

Moreover,

`$$\begin{aligned}
  \dot{\hat{k}} > 0 
  \Leftrightarrow
  \hat{c} < f\left(\hat{k}\right) - (\delta + g + n) \hat{k}
\end{aligned}$$`
and

`$$\begin{aligned}
  \dot{\hat{k}} < 0 
  \Leftrightarrow
  \hat{c} > f\left(\hat{k}\right) - (\delta + g + n) \hat{k}
\end{aligned}$$`

`$\dot{\hat{k}}$` is increasing below the `$\dot{\hat{k}} = 0$` locus
and decreasing above the locus.

---

## Regions for `$\dot{\hat{k}} > 0$` and `$\dot{\hat{k}} < 0$` (cont'd)

You can rewrite these conditions as below:

`$$\begin{aligned}
  \dot{\hat{k}} \gtreqless 0 
  \Longleftrightarrow
  f\left(\hat{k}\right) - \hat{c} \gtreqless (\delta + g + n) \hat{k}
\end{aligned}$$`

Capital increases if and only if the investment, 
`$f(\hat{k}) - \hat{c}$`, is greater than the break-even level,
`$(\delta + g + n)\hat{k}$`.

This condition is exactly what you have already encountered in the 
analysis of the Solow model.

---

## Locus for `$\dot{\hat{c}} = 0$`

From

`$$\frac{\dot{\hat{c}}}{\hat{c}} =
\frac{f'\left(\hat{k}\right) - \delta - \rho - \theta g}{\theta}$$`

we can derive the locus on which `$\dot{\hat{c}} = 0$`.

`$$\dot{\hat{c}} = 0
\Longleftrightarrow
f'\left( \hat{k} \right) = \delta + \rho + \theta g$$`

---

## Regions for `$\dot{\hat{c}} > 0$` and `$\dot{\hat{c}} < 0$`

Obviously,

`$$\dot{\hat{c}} \gtreqless 0
\Longleftrightarrow
f'\left( \hat{k} \right) \gtreqless \delta + \rho + \theta g$$`

Since `$f'$` is a decreasing function of `$\hat{k}$`,

`$$\dot{\hat{c}} \gtreqless 0
\Longleftrightarrow
\hat{k} \lesseqgtr (f')^{-1}\left(\delta + \rho + \theta g\right)$$`

Let's define

`$$\hat{k}^* =  (f')^{-1}\left(\delta + \rho + \theta g\right)$$`

---

## Regions for `$\dot{\hat{c}} > 0$` and `$\dot{\hat{c}} < 0$` (cont'd)

Recall `$f'(\hat{k}) - \delta = r$`. The first-order condition for the 
consumption is easy to grasp when `$g = 0$`.

`$$\dot{\hat{c}} \gtreqless 0
\Longleftrightarrow
r \gtreqless \rho + \theta\cdot 0$$`

When benefit from waiting `$r$` is greater than utility cost of waiting `$\rho$`, it's best to defer consuming. Thus, increasing consumption (`$\dot{\hat c} > 0$`)

When `$g \neq 0$`, an additional term is needed to adjust
for technology change.

---

---

## Additional assumption

Recall the objective function:

`$$\max \int_0^\infty e^{-(\rho - n) t} u(c(t)) dt,$$`
where 
`$$u(c) = \frac{c^{1 - \theta} - 1}{1 - \theta}.$$`

Obviously, we have to assume `$\rho - n > 0$`. In the following analysis
we assume a slightly stronger condition:

`$$\rho - n > \max \{ 0, (1-\theta)g\}$$`

---

## Equilibrium: First-order Dynamics

The dynamics of the Ramsey model is determined by the following
two differential equations.

Capital accumulation equation:

`$$\dot{\hat{k}} = f\left(\hat{k}\right) - \hat{c} - (\delta + g + n) \hat{k}$$`

Euler equation:

`$$\frac{\dot{\hat{c}}}{\hat{c}} =
\frac{f'\left(\hat{k}\right) - \delta - \rho - \theta g}{\theta}$$`

---

## Locus of `$\dot{\hat{k}} = 0$`

Let's derive the condition under which `$\dot{\hat{k}} = 0$`.

From the capital accumulation equation,

`$$\dot{\hat{k}} = f\left(\hat{k}\right) - \hat{c} - (\delta + g + n) \hat{k},$$`
we obtain

`$$\begin{aligned}
  \dot{\hat{k}} = 0 
  \Leftrightarrow
  \hat{c} = f\left(\hat{k}\right) - (\delta + g + n) \hat{k}
\end{aligned}$$`

---

## Regions for `$\dot{\hat{k}} > 0$` and `$\dot{\hat{k}} < 0$`

Moreover,

`$$\begin{aligned}
  \dot{\hat{k}} > 0 
  \Leftrightarrow
  \hat{c} < f\left(\hat{k}\right) - (\delta + g + n) \hat{k}
\end{aligned}$$`
and

`$$\begin{aligned}
  \dot{\hat{k}} < 0 
  \Leftrightarrow
  \hat{c} > f\left(\hat{k}\right) - (\delta + g + n) \hat{k}
\end{aligned}$$`

`$\dot{\hat{k}}$` is increasing below the `$\dot{\hat{k}} = 0$` locus
and decreasing above the locus.

---

## Regions for `$\dot{\hat{k}} > 0$` and `$\dot{\hat{k}} < 0$` (cont'd)

You can rewrite these conditions as below:

`$$\begin{aligned}
  \dot{\hat{k}} \gtreqless 0 
  \Longleftrightarrow
  f\left(\hat{k}\right) - \hat{c} \gtreqless (\delta + g + n) \hat{k}
\end{aligned}$$`

Capital increases if and only if the investment, 
`$f(\hat{k}) - \hat{c}$`, is greater than the break-even level,
`$(\delta + g + n)\hat{k}$`.

This condition is exactly what you have already encountered in the 
analysis of the Solow model.

---

## Phase diagram

---

## Golden Rule

Notice that the curve has a unique maximum, `$\hat{k}_G$`, where

`$$f'\left(\hat{k}_G \right) = \delta + g + n$$`

must hold.

In the Solow model, the steady state that satisfies this 
condition is called the Golden Rule steady state, at which 
the consumption level is maximized over all steady states.

---

## Golden Rule (cont'd)

---

## Locus of `$\dot{\hat{c}} = 0$`

From

`$$\frac{\dot{\hat{c}}}{\hat{c}} =
\frac{f'\left(\hat{k}\right) - \delta - \rho - \theta g}{\theta}$$`

we can derive the locus on which `$\dot{\hat{c}} = 0$`.

`$$\dot{\hat{c}} = 0
\Longleftrightarrow
f'\left( \hat{k} \right) = \delta + \rho + \theta g$$`

---

## Regions for `$\dot{\hat{c}} > 0$` and `$\dot{\hat{c}} < 0$`

Obviously,

`$$\dot{\hat{c}} \gtreqless 0
\Longleftrightarrow
f'\left( \hat{k} \right) \gtreqless \delta + \rho + \theta g$$`

Since `$f'$` is a decreasing function of `$\hat{k}$`,

`$$\dot{\hat{c}} \gtreqless 0
\Longleftrightarrow
\hat{k} \lesseqgtr (f')^{-1}\left(\delta + \rho + \theta g\right)$$`

Let's define

`$$\hat{k}^* =  (f')^{-1}\left(\delta + \rho + \theta g\right)$$`

---

## Regions for `$\dot{\hat{c}} > 0$` and `$\dot{\hat{c}} < 0$` (cont'd)

Recall `$f'(\hat{k}) - \delta = r$`. The first-order condition for the 
consumption is easy to grasp when `$g = 0$`.

`$$\dot{\hat{c}} \gtreqless 0
\Longleftrightarrow
r \gtreqless \rho + \theta\cdot 0$$`

Consumption is growing when benefit from waiting `$\rho$`
is greater than utility cost of waiting `$\rho$`.

When `$g \neq 0$`, an additional term is needed to adjust
for technology change.

---

## Regions for `$\dot{\hat{c}} > 0$` and `$\dot{\hat{c}} < 0$` (cont'd)

---

## `$\hat{k}^*$` and `$\hat{k}_G$`

Notice how I arranged the locus, `$\hat{k} = \hat{k}^*$`, and 
the Golden Rule level, `$\hat{k} = \hat{k}_G$`.

By the assumption that `$\rho - n > \max \{ 0, (1-\theta)g\}$`,

`$$f'\left(\hat{k}^* \right) = \delta + \rho + \theta g  > 
\delta + n + g = f'\left(\hat{k}_G\right)$$`

which implies

$$\hat{k}^* < \hat{k}_G $$

---

## `$\hat{k}^*$` and `$\hat{k}_G$`

The intersection, `$\left(\hat{k}^*, \hat{c}^*\right)$`, of the loci for `$\dot{\hat{k}} = 0$` and 
`$\dot{\hat{c}} = 0$` corresponds to the steady state 
level of `$\left(\hat{k}, \hat{c}\right)$`.

In the steady state, the level of capital stock per 
unit of effective labor is necessarily smaller than the 
golden rule level.

`$\hat{k}^*$` is sometimes called, modified golden rule.

---

## Phase Diagram

---

## A trajectory from `$\hat{k}(0) < k^*$`: 1

---

## A trajectory from `$\hat{k}(0) < k^*$`: 2

---

## A trajectory from `$\hat{k}(0) < k^*$`: 3

---

## Optimal trajectory

The problem yet to be solved is the determination of 
`$\hat{c}(0)$`. Given `$\hat{k}(0)$`, depending on the level 
of candidates for `$\hat{c}(0)$`, the trajectories for the 
differential equations exibit either of the following.

1. `$\hat{c}(t) \to 0$`
2. `$\hat{k}(t) \to 0$`
3. `$\hat{k}(t) \to \hat{k}^*$` and `$\hat{c}(t) \to \hat{c}^*$`

---

## Optimal trajectory (cont'd)

1. `$\hat{c}(t) \to 0$`
2. `$\hat{k}(t) \to 0$`

The first trajectory doesn't solve the maximization problem 
because by raising `$\hat{c}(0)$` by a small amount we can 
make `$\hat{c}(t)$` larger for any `$t$`.

The second trajectory will eventually violate the feasibility 
condition.

---

## Optimal trajectory (cont'd)

3. `$\hat{k}(t) \to \hat{k}^*$` and `$\hat{c}(t) \to \hat{c}^*$`

The third trajectory (which lies in the middlle of 1 and 2) 
converges to the steady state, 
`$\left(\hat{k}^*, \hat{c}^*\right)$`, where 
`$\dot{\hat{k}} = 0$` and `$\dot{\hat{c}} = 0$` hold simultaneously.

This trajectory is the optimal path.

To obtain the convergence, the trajectory must start from 
an appropriate `$\hat{c}(0)$`. The steady state 
is said to be **saddle-point stable**.

---

## Optimal Consumption

You have different (optimal) `$\hat{c}(0)$`'s for different `$\hat{k}(0)$`'s. 
This mapping gives the optimal consumption function. (The Solow model's consumption 
function was `$\hat{c} = (1 - s) f\left( \hat{k} \right)$`)

---

## Balanced Growth

In the long-run, the Ramsey model converges to the 
balanced growth path, in which all variables 
grow at constat rates.

Growth rates for

`$$y, k, c, Y, K, C$$`

are identical to those in the Solow model.

---

## In the long run (comparative statics)

What will happen to the steady state and the steady state growth rates,
when one of the above parameters changes?

Sample questions

* Will the steady-state _growth rate_ of aggregate consumption, `$C$`, be increased, decreased,   or unchanged by an increase in `$\delta$`?
* What policy measure might be effective when you (as a policy maker) try to increase 
  the _level_ of output per capita, `$y = Y / L$`, in the steady state?

---

## On transition (comparative dynamics)

As in the Solow model, the adjustment to the new steady state is not (necessarily) immediate.

You can analyze the transition to the new steady state by the phase diagram.

---

## Example: Increase in `$n$`

`$\dot{\hat{c}} = 0$` locus is unchanged because the Euler equation does not have `$n$`.

An increase in `$n$` brings `$\dot{\hat{k}} = 0$` locus down:

`$$\hat{c} = f\left(\hat{k}\right) - (\delta + g + n) \hat{k}$$`

The economy jumps down to the new steady state immediately.

Since `$\hat{c}$` is a variable that the consumer can choose, discontinuity is allowed.

---

## Example: Increase in `$n$`

---

## Example: Increase in `$\rho$`

`$\dot{\hat{k}} = 0$` locus is unchanged because the capital accumulation equation does not have `$\rho$`.

An increase in `$\rho$` shifts `$\dot{\hat{c}} = 0$` locus to the left.

Recall that

* `$\hat{c}$` can be discontinuous
* `$\hat{k}$` must be continuous

Describe the transition to the new steady state using the following diagram.

---

## Example: Increase in `$\rho$`

---

## Comparative statics/dynamics

You need to understand:

* How steady states move;
* How the economy adjusts to the new steady state.

As we discussed earlier in this course, anticipation for future 
is an important component of the Ramsey model.

You can analyze implications of such a policy change as

* Permanent increase in `$n$` due to an improvement of adoption policy
  or immigration policy, which will be enforced at a known point 
  in future.

---

## Example: Anticipated increase in `$n$`

---

## Example: Anticipated increase in `$n$`

---

## Example: Anticipated increase in `$n$`

---

## Example: Anticipated increase in `$n$`

---

## Anticipated increase in `$n$`

---

## Why this path is optimal

Mathematical proof is based on Kemp and Long (1977, Economic Record): 
The costate variable `$\lambda(t) = e^{- \rho t} u'(c(t))$` cannot jump 
at the time of policy change.

If there is anticipated discontinuous drop of `$c(t)$` at `$t = t_0$`, 
the agents anticipate a hike in `$\lambda(t) = e^{- \rho t} u'(c(t))$`.

If the price of asset, `$\lambda$`, is known to jump in the future, 
economic agents all try to buy the asset and the _present price_ would 
rise.

What you observe thus is an immediate drop of `$c$`.

---

## Summary

When you solve a problem of comparative dynamics, keep in mind the following points:

* There is no discontinuity in `$k$`
* There is no anticipated discontinuity in `$u'(c)$` or `$c$`
* The path must converge to the steady state at `$t = \infty$`.

That is, in the Ramsey model, consumers change their behavior at the time of 
arrival of new information.