Deterministic Simulations

The Stacked Time Method

Pablo Winant

Introduction

Deterministic Simulations

What are they?
- Simulations where the entire path of future shocks is known at \(t=0\).
- Known as Perfect Foresight or MIT Shocks.
Why use them?
- Essential for examining large, structural deviations from steady states.
- Analyzing gradual, anticipated policy changes.
- Solving models with “lumpy” transitions.
Contrast with Stochastic Simulations:
- No uncertainty about the future after \(t=0\).
- Agents do not form expectations over distributions, saving us from the “curse of dimensionality.”

The Stacked Time Method

The Core Concept

Dynamic models consist of difference equations linking periods: \[ f(x_{t-1}, x_t, x_{t+1}) = 0 \]

Instead of time-stepping (which can be unstable), the Stacked Time Method (Laffargue 1990; Boucekkine 1995; Juillard 1996) solves the entire path simultaneously.
We “stack” the variables for all \(T\) periods into one giant vector: \[ X = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_T \end{bmatrix} \]
We define a stacked function \(F(X) = 0\) encompassing the equations for every time step.

Boundary Conditions

To solve the system \(F(X) = 0\), we need to anchor the start and the end of the simulation:

Initial Condition: \(x_0\) is given (usually the economy’s state before the shock).
Terminal Condition: \(x_{T+1}\) is assumed to reach a new steady state (\(x^*\)).
- We must choose \(T\) large enough so that the transition is fully complete before the terminal date.

\[ \begin{align*} f(x_0, x_1, x_2) &= 0 \quad \text{(Initial boundary)} \\ f(x_1, x_2, x_3) &= 0 \\ \vdots \\ f(x_{T-1}, x_T, x^*) &= 0 \quad \text{(Terminal boundary)} \end{align*} \]

Solving the System

The system \(F(X) = 0\) is highly non-linear and very large (\(N \times T\) equations, where \(N\) is the number of variables per period).
We solve it using exactly the same tools as a static non-linear problem: Newton’s Method. \[ X^{(k+1)} = X^{(k)} - [J_F(X^{(k)})]^{-1} F(X^{(k)}) \]
Sparsity: The Jacobian \(J_F\) is massive but banded (sparse).
- An equation at time \(t\) only depends on variables at \(t-1, t, t+1\).
- Modern sparse linear algebra solvers invert this matrix efficiently.

Example: The Ecological Transition

An Ecological Transition Model

Consider an economy forced to transition from polluting to clean energy.

Two Types of Capital:
- \(K^b_t\): “Brown” (carbon-intensive) capital.
- \(K^g_t\): “Green” (clean) capital.
Production: Output requires both types of energy.
The Policy: The government introduces a carbon tax \(\tau_t\) on the use of Brown capital.
- The tax is announced at \(t=0\), ramps up linearly for 20 years, and remains permanently high.
- The entire path of the tax is perfectly foreseeable.

The Dynamics of Transition

Firms face adjustment costs to changing capital. How will they react to the announcement?

Euler Equations & Arbitrage:
- Firms immediately foresee that brown capital will be unprofitable in the future.
- Even before the tax is fully implemented, they stop investing in \(K^b_t\).
Stranded Assets:
- The shadow value (Tobin’s q) of brown capital drops sharply at \(t=0\).
Green Investment Boom:
- Massive investments scale up \(K^g_t\) to replace the lost energy capacity.

Applying Stacked Time to the Transition

How do we compute this?

Initial SS (\(t=0\)): Solve the steady state with \(\tau=0\). This yields initial capital \(K^b_0, K^g_0\).
Final SS (\(t=T\)): Solve a new steady state with \(\tau = \tau_{max}\). This yields terminal capital \(K^{b*}, K^{g*}\).
Stack the Equations: For \(t=1 \dots T\), stack the Euler equations, budget constraints, and capital accumulation rules.
Solve: Use a sparse Newton solver to find the optimal sequences \((K^b_t, K^g_t, C_t, \dots)_{t=1}^T\).

This provides a deterministic trajectory showing the macro-financial effects of climate policy!

Conclusion

Summary

Deterministic Simulations are crucial for studying structured transitions over time (like climate change or demographics).
Stacked Time converts a dynamic problem into a large, sparse static root-finding problem.
Leveraging sparse Jacobians allows us to rapidly simulate hundreds of variables over hundreds of periods simultaneously.

Appendix

Notation Conventions: \(f\) and \(g\)

In other parts of this course, we split models into:

Transition functions (backward-looking): \(s_t = g(s_{t-1}, x_{t-1}, z_t)\)
Optimality conditions (forward-looking): \(0 = f(s_t, x_t, x_{t+1}, s_{t+1})\)

(Where \(s\) are states, \(x\) are controls, and \(z\) are exogenous variables.)

Translating \(f\)-\(g\) to \(F\)

How do we take this two-part system and bridge it back to the general difference equation form \(F(y_{t-1}, y_t, y_{t+1}) = 0\) from earlier?

We define a combined variable vector \(y_t = \begin{bmatrix} s_t \\ x_t \end{bmatrix}\). We can then stack the rules into a single generic function \(F\):

\[ F(y_{t-1}, y_t, y_{t+1}, z_t) = \begin{bmatrix} s_t - g(s_{t-1}, x_{t-1}, z_t) \\ f(s_t, x_t, x_{t+1}, s_{t+1}) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

Notice how the top block is backward-looking (depends on \(y_{t-1}, y_t\)) and the bottom block is forward-looking (depends on \(y_t, y_{t+1}\)), but the combined system exactly matches \(F(y_{t-1}, y_t, y_{t+1}) = 0\).

Implications for Stacked Time

How does this explicit separation affect the Stacked Time approach compared to an arbitrary system \(F(y_{t-1}, y_t, y_{t+1}) = 0\)?

Variable Condensation: Sometimes, you can evaluate the transition function \(g\) recursively or substitute it algebraically into \(f\), significantly reducing the dimension of the stacked Jacobian.
Boundary Condition Clarity: The formal separation clearly allocates boundary conditions: the initial period requires a fixed state \(s_0\), while the terminal date \(T+1\) targets a steady state condition for \((s, x)\).
Block Solving Structure: Advanced Newton solvers can exploit the explicit state / control division to reorder the Jacobian elements into a beautifully banded structure, making the inversion step even faster.