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# Classical Control with Linear Algebra¶

## Overview¶

In an earlier lecture Linear Quadratic Dynamic Programming Problems we have studied how to solve a special class of dynamic optimization and prediction problems by applying the method of dynamic programming. In this class of problems

• the objective function is quadratic in states and controls
• the one-step transition function is linear
• shocks are i.i.d. Gaussian or martingale differences

In this lecture and a companion lecture Classical Filtering with Linear Algebra, we study the classical theory of linear-quadratic (LQ) optimal control problems.

The classical approach does not use the two closely related methods – dynamic programming and Kalman filtering – that we describe in other lectures, namely, Linear Quadratic Dynamic Programming Problems and A First Look at the Kalman Filter

• $z$-transform and lag operator methods, or
• matrix decompositions applied to linear systems of first-order conditions for optimum problems.

In this lecture and the sequel Classical Filtering with Linear Algebra, we mostly rely on elementary linear algebra

The main tool from linear algebra we’ll put to work here is LU decomposition

We’ll begin with discrete horizon problems

Then we’ll view infinite horizon problems as appropriate limits of these finite horizon problems

Later, we will examine the close connection between LQ control and least squares prediction and filtering problems

These classes of problems are connected in the sense that to solve each, essentially the same mathematics is used

### References¶

Useful references include [Whi63], [HS80], [Orf88], [AP91], and [Mut60]

## A Control Problem¶

Let $L$ be the lag operator, so that, for sequence $\{x_t\}$ we have $L x_t = x_{t-1}$

More generally, let $L^k x_t = x_{t-k}$ with $L^0 x_t = x_t$ and

$$d(L) = d_0 + d_1 L+ \ldots + d_m L^m$$

where $d_0, d_1, \ldots, d_m$ is a given scalar sequence

Consider the discrete time control problem

$$\max_{\{y_t\}} \lim_{N \to \infty} \sum^N_{t=0} \beta^t\, \left\{ a_t y_t - {1 \over 2}\, hy^2_t - {1 \over 2} \, \left[ d(L)y_t \right]^2 \right\}, \tag{1}$$

where

• $h$ is a positive parameter and $\beta \in (0,1)$ is a discount factor
• $\{a_t\}_{t \geq 0}$ is a sequence of exponential order less than $\beta^{-1/2}$, by which we mean $\lim_{t \rightarrow \infty} \beta^{\frac{t}{2}} a_t = 0$

Maximization in (1) is subject to initial conditions for $y_{-1}, y_{-2} \ldots, y_{-m}$

Maximization is over infinite sequences $\{y_t\}_{t \geq 0}$

### Example¶

The formulation of the LQ problem given above is broad enough to encompass many useful models

As a simple illustration, recall that in LQ Dynamic Programming Problems we consider a monopolist facing stochastic demand shocks and adjustment costs

Let’s consider a deterministic version of this problem, where the monopolist maximizes the discounted sum

$$\sum_{t=0}^{\infty} \beta^t \pi_t$$

and

$$\pi_t = p_t q_t - c q_t - \gamma (q_{t+1} - q_t)^2 \quad \text{with} \quad p_t = \alpha_0 - \alpha_1 q_t + d_t$$

In this expression, $q_t$ is output, $c$ is average cost of production, and $d_t$ is a demand shock

The term $\gamma (q_{t+1} - q_t)^2$ represents adjustment costs

You will be able to confirm that the objective function can be rewritten as (1) when

• $a_t := \alpha_0 + d_t - c$
• $h := 2 \alpha_1$
• $d(L) := \sqrt{2 \gamma}(I - L)$

Further examples of this problem for factor demand, economic growth, and government policy problems are given in ch. IX of [Sar87]

## Finite Horizon Theory¶

We first study a finite $N$ version of the problem

Later we will study an infinite horizon problem solution as a limiting version of a finite horizon problem

(This will require being careful because the limits as $N \to \infty$ of the necessary and sufficient conditions for maximizing finite $N$ versions of (1) are not sufficient for maximizing (1))

We begin by

1. fixing $N > m$,
2. differentiating the finite version of (1) with respect to $y_0, y_1, \ldots, y_N$, and
3. setting these derivatives to zero

For $t=0, \ldots, N-m$ these first-order necessary conditions are the Euler equations

For $t = N-m + 1, \ldots, N$, the first-order conditions are a set of terminal conditions

Consider the term

\begin{aligned} J & = \sum^N_{t=0} \beta^t [d(L) y_t] [d(L) y_t] \\ & = \sum^N_{t=0} \beta^t \, (d_0 \, y_t + d_1 \, y_{t-1} + \cdots + d_m \, y_{t-m}) \, (d_0 \, y_t + d_1 \, y_{t-1} + \cdots + d_m\, y_{t-m}) \end{aligned}

Differentiating $J$ with respect to $y_t$ for $t=0,\ 1,\ \ldots,\ N-m$ gives

\begin{aligned} {\partial {J} \over \partial y_t} & = 2 \beta^t \, d_0 \, d(L)y_t + 2 \beta^{t+1} \, d_1\, d(L)y_{t+1} + \cdots + 2 \beta^{t+m}\, d_m\, d(L) y_{t+m} \\ & = 2\beta^t\, \bigl(d_0 + d_1 \, \beta L^{-1} + d_2 \, \beta^2\, L^{-2} + \cdots + d_m \, \beta^m \, L^{-m}\bigr)\, d (L) y_t\ \end{aligned}

We can write this more succinctly as

$${\partial {J} \over \partial y_t} = 2 \beta^t \, d(\beta L^{-1}) \, d (L) y_t \tag{2}$$

Differentiating $J$ with respect to $y_t$ for $t = N-m + 1, \ldots, N$ gives

\begin{aligned} {\partial J \over \partial y_N} &= 2 \beta^N\, d_0 \, d(L) y_N \cr {\partial J \over \partial y_{N-1}} &= 2\beta^{N-1} \,\bigl[d_0 + \beta \, d_1\, L^{-1}\bigr] \, d(L)y_{N-1} \cr \vdots & \quad \quad \vdots \cr {\partial {J} \over \partial y_{N-m+1}} &= 2 \beta^{N-m+1}\,\bigl[d_0 + \beta L^{-1} \,d_1 + \cdots + \beta^{m-1}\, L^{-m+1}\, d_{m-1}\bigr] d(L)y_{N-m+1} \end{aligned} \tag{3}

With these preliminaries under our belts, we are ready to differentiate (1)

Differentiating (1) with respect to $y_t$ for $t=0, \ldots, N-m$ gives the Euler equations

$$\bigl[h+d\,(\beta L^{-1})\,d(L)\bigr] y_t = a_t, \quad t=0,\, 1,\, \ldots, N-m \tag{4}$$

The system of equations (4) form a $2 \times m$ order linear difference equation that must hold for the values of $t$ indicated.

Differentiating (1) with respect to $y_t$ for $t = N-m + 1, \ldots, N$ gives the terminal conditions

\begin{aligned} \beta^N (a_N - hy_N - d_0\,d(L)y_N) &= 0 \cr \beta^{N-1} \left(a_{N-1}-hy_{N-1}-\Bigl(d_0 + \beta \, d_1\, L^{-1}\Bigr)\, d(L)\, y_{N-1}\right) & = 0 \cr \vdots & \vdots\cr \beta^{N-m+1} \biggl(a_{N-m+1} - h y_{N-m+1} -(d_0+\beta L^{-1} d_1+\cdots\ +\beta^{m-1} L^{-m+1} d_{m-1}) d(L) y_{N-m+1}\biggr) & = 0 \end{aligned} \tag{5}

In the finite $N$ problem, we want simultaneously to solve (4) subject to the $m$ initial conditions $y_{-1}, \ldots, y_{-m}$ and the $m$ terminal conditions (5)

These conditions uniquely pin down the solution of the finite $N$ problem

That is, for the finite $N$ problem, conditions (4) and (5) are necessary and sufficient for a maximum, by concavity of the objective function

Next we describe how to obtain the solution using matrix methods

### Matrix Methods¶

Let’s look at how linear algebra can be used to tackle and shed light on the finite horizon LQ control problem

#### A Single Lag Term¶

Let’s begin with the special case in which $m=1$

We want to solve the system of $N+1$ linear equations

\begin{aligned} \bigl[h & + d\, (\beta L^{-1})\, d\, (L) ] y_t = a_t, \quad t = 0,\ 1,\ \ldots,\, N-1\cr \beta^N & \bigl[a_N-h\, y_N-d_0\, d\, (L) y_N\bigr] = 0 \end{aligned} \tag{6}

where $d(L) = d_0 + d_1 L$

These equations are to be solved for $y_0, y_1, \ldots, y_N$ as functions of $a_0, a_1, \ldots, a_N$ and $y_{-1}$

Let

$$\phi (L) = \phi_0 + \phi_1 L + \beta \phi_1 L^{-1} = h + d (\beta L^{-1}) d(L) = (h + d_0^2 + d_1^2) + d_1 d_0 L+ d_1 d_0 \beta L^{-1}$$

Then we can represent (6) as the matrix equation

$$\left[ \begin{matrix} (\phi_0-d_1^2) & \phi_1 & 0 & 0 & \ldots & \ldots & 0 \cr \beta \phi_1 & \phi_0 & \phi_1 & 0 & \ldots & \dots & 0 \cr 0 & \beta \phi_1 & \phi_0 & \phi_1 & \ldots & \ldots & 0 \cr \vdots &\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \cr 0 & \ldots & \ldots & \ldots & \beta \phi_1 & \phi_0 &\phi_1 \cr 0 & \ldots & \ldots & \ldots & 0 & \beta \phi_1 & \phi_0 \end{matrix} \right] \left [ \begin{matrix} y_N \cr y_{N-1} \cr y_{N-2} \cr \vdots \cr y_1 \cr y_0 \end{matrix} \right ] = \left[ \begin{matrix} a_N \cr a_{N-1} \cr a_{N-2} \cr \vdots \cr a_1 \cr a_0 - \phi_1 y_{-1} \end{matrix} \right] \tag{7}$$

or

$$W\bar y = \bar a \tag{8}$$

Notice how we have chosen to arrange the $y_t$’s in reverse time order.

The matrix $W$ on the left side of (7) is “almost” a Toeplitz matrix (where each descending diagonal is constant)

There are two sources of deviation from the form of a Toeplitz matrix

1. The first element differs from the remaining diagonal elements, reflecting the terminal condition
2. The subdiagonal elements equal $\beta$ time the superdiagonal elements

The solution of (8) can be expressed in the form

$$\bar y = W^{-1} \bar a \tag{9}$$

which represents each element $y_t$ of $\bar y$ as a function of the entire vector $\bar a$

That is, $y_t$ is a function of past, present, and future values of $a$‘s, as well as of the initial condition $y_{-1}$

#### An Alternative Representation¶

An alternative way to express the solution to (7) or (8) is in so called feedback-feedforward form

The idea here is to find a solution expressing $y_t$ as a function of past $y$‘s and current and future $a$‘s

To achieve this solution, one can use an LU decomposition of $W$

There always exists a decomposition of $W$ of the form $W= LU$ where

• $L$ is an $(N+1) \times (N+1)$ lower trangular matrix
• $U$ is an $(N+1) \times (N+1)$ upper trangular matrix.

The factorization can be normalized so that the diagonal elements of $U$ are unity

Using the LU representation in (9), we obtain

$$U \bar y = L^{-1} \bar a \tag{10}$$

Since $L^{-1}$ is lower trangular, this representation expresses $y_t$ as a function of

• lagged $y$‘s (via the term $U \bar y$), and
• current and future $a$’s (via the term $L^{-1} \bar a$)

Because there are zeros everywhere in the matrix on the left of (7) except on the diagonal, superdiagonal, and subdiagonal, the $LU$ decomposition takes

• $L$ to be zero except in the diagional and the leading subdiagonal
• $U$ to be zero except on the diagonal and the superdiagional

Thus, (10) has the form

$$\left[ \begin{matrix} 1& U_{12} & 0 & 0 & \ldots & 0 & 0 \cr 0 & 1 & U_{23} & 0 & \ldots & 0 & 0 \cr 0 & 0 & 1 & U_{34} & \ldots & 0 & 0 \cr 0 & 0 & 0 & 1 & \ldots & 0 & 0\cr \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\cr 0 & 0 & 0 & 0 & \ldots & 1 & U_{N,N+1} \cr 0 & 0 & 0 & 0 & \ldots & 0 & 1 \end{matrix} \right] \ \ \ \left[ \begin{matrix} y_N \cr y_{N-1} \cr y_{N-2} \cr y_{N-3} \cr \vdots \cr y_1 \cr y_0 \end{matrix} \right] =$$$$\quad \left[ \begin{matrix} L^{-1}_{11} & 0 & 0 & \ldots & 0 \cr L^{-1}_{21} & L^{-1}_{22} & 0 & \ldots & 0 \cr L^{-1}_{31} & L^{-1}_{32} & L^{-1}_{33}& \ldots & 0 \cr \vdots & \vdots & \vdots & \ddots & \vdots\cr L^{-1}_{N,1} & L^{-1}_{N,2} & L^{-1}_{N,3} & \ldots & 0 \cr L^{-1}_{N+1,1} & L^{-1}_{N+1,2} & L^{-1}_{N+1,3} & \ldots & L^{-1}_{N+1\, N+1} \end{matrix} \right] \left[ \begin{matrix} a_N \cr a_{N-1} \cr a_{N-2} \cr \vdots \cr a_1 \cr a_0 - \phi_1 y_{-1} \end{matrix} \right ]$$

where $L^{-1}_{ij}$ is the $(i,j)$ element of $L^{-1}$ and $U_{ij}$ is the $(i,j)$ element of $U$

Note how the left side for a given $t$ involves $y_t$ and one lagged value $y_{t-1}$ while the right side involves all future values of the forcing process $a_t, a_{t+1}, \ldots, a_N$

We briefly indicate how this approach extends to the problem with $m > 1$

Assume that $\beta = 1$ and let $D_{m+1}$ be the $(m+1) \times (m+1)$ symmetric matrix whose elements are determined from the following formula:

$$D_{jk} = d_0 d_{k-j} + d_1 d_{k-j+1} + \ldots + d_{j-1} d_{k-1}, \qquad k \geq j$$

Let $I_{m+1}$ be the $(m+1) \times (m+1)$ identity matrix

Let $\phi_j$ be the coefficients in the expansion $\phi (L) = h + d (L^{-1}) d (L)$

Then the first order conditions (4) and (5) can be expressed as:

$$(D_{m+1} + hI_{m+1})\ \ \left[ \begin{matrix} y_N \cr y_{N-1} \cr \vdots \cr y_{N-m} \end{matrix} \right]\ = \ \left[ \begin{matrix} a_N \cr a_{N-1} \cr \vdots \cr a_{N-m} \end{matrix} \right] + M\ \left[ \begin{matrix} y_{N-m+1}\cr y_{N-m-2}\cr \vdots\cr y_{N-2m} \end{matrix} \right]$$

where $M$ is $(m+1)\times m$ and

$$M_{ij} = \begin{cases} D_{i-j,\,m+1} \textrm{ for } i>j \\ 0 \textrm{ for } i\leq j\end{cases}$$\begin{aligned} \phi_m y_{N-1} &+ \phi_{m-1} y_{N-2} + \ldots + \phi_0 y_{N-m-1} + \phi_1 y_{N-m-2} +\cr &\hskip.75in \ldots + \phi_m y_{N-2m-1} = a_{N-m-1} \cr \phi_m y_{N-2} &+ \phi_{m-1} y_{N-3} + \ldots + \phi_0 y_{N-m-2} + \phi_1 y_{N-m-3} +\cr &\hskip.75in \ldots + \phi_m y_{N-2m-2} = a_{N-m-2} \cr &\qquad \vdots \cr \phi_m y_{m+1} &+ \phi_{m-1} y_m + + \ldots + \phi_0 y_1 + \phi_1 y_0 + \phi_m y_{-m+1} = a_1 \cr \phi_m y_m + \phi_{m-1}& y_{m-1} + \phi_{m-2} + \ldots + \phi_0 y_0 + \phi_1 y_{-1} + \ldots + \phi_m y_{-m} = a_0 \end{aligned}

As before, we can express this equation as $W \bar y = \bar a$

The matrix on the left of this equation is “almost” Toeplitz, the exception being the leading $m \times m$ sub matrix in the upper left hand corner

We can represent the solution in feedback-feedforward form by obtaining a decomposition $LU = W$, and obtain

$$U \bar y = L^{-1} \bar a \tag{11}$$

\begin{aligned} \sum^t_{j=0}\, U_{-t+N+1,\,-t+N+j+1}\,y_{t-j} &= \sum^{N-t}_{j=0}\, L_{-t+N+1,\, -t+N+1-j}\, \bar a_{t+j}\ ,\cr &\qquad t=0,1,\ldots, N \end{aligned}

where $L^{-1}_{t,s}$ is the element in the $(t,s)$ position of $L$, and similarly for $U$

The left side of equation (11) is the “feedback” part of the optimal control law for $y_t$, while the right-hand side is the “feedforward” part

We note that there is a different control law for each $t$

Thus, in the finite horizon case, the optimal control law is time dependent

It is natural to suspect that as $N \rightarrow\infty$, (11) becomes equivalent to the solution of our infinite horizon problem, which below we shall show can be expressed as

$$c(L) y_t = c (\beta L^{-1})^{-1} a_t\ ,$$

so that as $N \rightarrow \infty$ we expect that for each fixed $t, U^{-1}_{t, t-j} \rightarrow c_j$ and $L_{t,t+j}$ approaches the coefficient on $L^{-j}$ in the expansion of $c(\beta L^{-1})^{-1}$

This suspicion is true under general conditions that we shall study later

For now, we note that by creating the matrix $W$ for large $N$ and factoring it into the $LU$ form, good approximations to $c(L)$ and $c(\beta L^{-1})^{-1}$ can be obtained

## The Infinite Horizon Limit¶

For the infinite horizon problem, we propose to discover first-order necessary conditions by taking the limits of (4) and (5) as $N \to \infty$

This approach is valid, and the limits of (4) and (5) as $N$ approaches infinity are first-order necessary conditions for a maximum

However, for the infinite horizon problem with $\beta < 1$, the limits of (4) and (5) are, in general, not sufficient for a maximum

That is, the limits of (5) do not provide enough information uniquely to determine the solution of the Euler equation (4) that maximizes (1)

As we shall see below, a side condition on the path of $y_t$ that together with (4) is sufficient for an optimum is

$$\sum^\infty_{t=0}\ \beta^t\, hy^2_t < \infty \tag{12}$$

All paths that satisfy the Euler equations, except the one that we shall select below, violate this condition and, therefore, evidently lead to (much) lower values of (1) than does the optimal path selected by the solution procedure below

Consider the characteristic equation associated with the Euler equation

$$h+d \, (\beta z^{-1})\, d \, (z) = 0 \tag{13}$$

Notice that if $\tilde z$ is a root of equation (13), then so is $\beta \tilde z^{-1}$

Thus, the roots of (13) come in “$\beta$-reciprocal” pairs

Assume that the roots of (13) are distinct

Let the roots be, in descending order according to their moduli, $z_1, z_2, \ldots, z_{2m}$

From the reciprocal pairs property and the assumption of distinct roots, it follows that $\vert z_j \vert > \sqrt \beta\ \hbox{ for } j\leq m \hbox { and } \vert z_j \vert < \sqrt\beta\ \hbox { for } j > m$

It also follows that $z_{2m-j} = \beta z^{-1}_{j+1}, j=0, 1, \ldots, m-1$

Therefore, the characteristic polynomial on the left side of (13) can be expressed as

\begin{aligned} h+d(\beta z^{-1})d(z) &= z^{-m} z_0(z-z_1)\cdots (z-z_m)(z-z_{m+1}) \cdots (z-z_{2m}) \cr &= z^{-m} z_0 (z-z_1)(z-z_2)\cdots (z-z_m)(z-\beta z_m^{-1}) \cdots (z-\beta z^{-1}_2)(z-\beta z_1^{-1}) \end{aligned} \tag{14}

where $z_0$ is a constant

In (14), we substitute $(z-z_j) = -z_j (1- {1 \over z_j}z)$ and $(z-\beta z_j^{-1}) = z(1 - {\beta \over z_j} z^{-1})$ for $j = 1, \ldots, m$ to get

$$h+d(\beta z^{-1})d(z) = (-1)^m(z_0z_1\cdots z_m) (1- {1\over z_1} z) \cdots (1-{1\over z_m} z)(1- {1\over z_1} \beta z^{-1}) \cdots(1-{1\over z_m} \beta z^{-1})$$

Now define $c(z) = \sum^m_{j=0} c_j \, z^j$ as

$$c\,(z)=\Bigl[(-1)^m z_0\, z_1 \cdots z_m\Bigr]^{1/2} (1-{z\over z_1}) \, (1-{z\over z_2}) \cdots (1- {z\over z_m}) \tag{15}$$

Notice that (14) can be written

$$h + d \ (\beta z^{-1})\ d\ (z) = c\,(\beta z^{-1})\,c\,(z) \tag{16}$$

It is useful to write (15) as

$$c(z) = c_0(1-\lambda_1\, z) \ldots (1-\lambda_m z) \tag{17}$$

where

$$c_0 = \left[(-1)^m\, z_0\, z_1 \cdots z_m\right]^{1/2}; \quad \lambda_j={1 \over z_j},\,\ j=1, \ldots, m$$

Since $\vert z_j \vert > \sqrt \beta \hbox { for } j = 1, \ldots, m$ it follows that $\vert \lambda_j \vert < 1/\sqrt \beta$ for $j = 1, \ldots, m$

Using (17), we can express the factorization (16) as

$$h+d (\beta z^{-1})d(z) = c^2_0 (1-\lambda_1 z) \cdots (1 - \lambda_m z) (1-\lambda_1 \beta z^{-1}) \cdots (1 - \lambda_m \beta z^{-1})$$

In sum, we have constructed a factorization (16) of the characteristic polynomial for the Euler equation in which the zeros of $c(z)$ exceed $\beta^{1/2}$ in modulus, and the zeros of $c\,(\beta z^{-1})$ are less than $\beta^{1/2}$ in modulus

Using (16), we now write the Euler equation as

$$c(\beta L^{-1})\,c\,(L)\, y_t = a_t$$

The unique solution of the Euler equation that satisfies condition (12) is

$$c(L)\,y_t = c\,(\beta L^{-1})^{-1}a_t \tag{18}$$

This can be established by using an argument paralleling that in chapter IX of [Sar87]

To exhibit the solution in a form paralleling that of [Sar87], we use (17) to write (18) as

$$(1-\lambda_1 L) \cdots (1 - \lambda_mL)y_t = {c^{-2}_0 a_t \over (1-\beta \lambda_1 L^{-1}) \cdots (1 - \beta \lambda_m L^{-1})} \tag{19}$$

Using partial fractions, we can write the characteristic polynomial on the right side of (19) as

$$\sum^m_{j=1} {A_j \over 1 - \lambda_j \, \beta L^{-1}} \quad \text{where} \quad A_j := {c^{-2}_0 \over \prod_{i \not= j}(1-{\lambda_i \over \lambda_j})}$$

Then (19) can be written

$$(1-\lambda_1 L) \cdots (1-\lambda_m L) y_t = \sum^m_{j=1} \, {A_j \over 1 - \lambda_j \, \beta L^{-1}} a_t$$

or

$$(1 - \lambda_1 L) \cdots (1 - \lambda_m L) y_t = \sum^m_{j=1}\, A_j \sum^\infty_{k=0}\, (\lambda_j\beta)^k\, a_{t+k} \tag{20}$$

Equation (20) expresses the optimum sequence for $y_t$ in terms of $m$ lagged $y$‘s, and $m$ weighted infinite geometric sums of future $a_t$‘s

Furthermore, (20) is the unique solution of the Euler equation that satisfies the initial conditions and condition (12)

In effect, condition (12) compels us to solve the “unstable” roots of $h+d (\beta z^{-1})d(z)$ forward (see [Sar87])

The step of factoring the polynomial $h+d (\beta z^{-1})\, d(z)$ into $c\, (\beta z^{-1})c\,(z)$, where the zeros of $c\,(z)$ all have modulus exceeding $\sqrt\beta$, is central to solving the problem

We note two features of the solution (20)

• Since $\vert \lambda_j \vert < 1/\sqrt \beta$ for all $j$, it follows that $(\lambda_j \ \beta) < \sqrt \beta$
• The assumption that $\{ a_t \}$ is of exponential order less than $1 /\sqrt \beta$ is sufficient to guarantee that the geometric sums of future $a_t$‘s on the right side of (20) converge

We immediately see that those sums will converge under the weaker condition that $\{ a_t\}$ is of exponential order less than $\phi^{-1}$ where $\phi = \max \, \{\beta \lambda_i, i=1,\ldots,m\}$

Note that with $a_t$ identically zero, (20) implies that in general $\vert y_t \vert$ eventually grows exponentially at a rate given by $\max_i \vert \lambda_i \vert$

The condition $\max_i \vert \lambda_i \vert <1 /\sqrt \beta$ guarantees that condition (12) is satisfied

In fact, $\max_i \vert \lambda_i \vert < 1 /\sqrt \beta$ is a necessary condition for (12) to hold

Were (12) not satisfied, the objective function would diverge to $- \infty$, implying that the $y_t$ path could not be optimal

For example, with $a_t = 0$, for all $t$, it is easy to describe a naive (nonoptimal) policy for $\{y_t, t\geq 0\}$ that gives a finite value of (17)

We can simply let $y_t = 0 \hbox { for } t\geq 0$

This policy involves at most $m$ nonzero values of $hy^2_t$ and $[d(L)y_t]^2$, and so yields a finite value of (1)

Therefore it is easy to dominate a path that violates (12)

## Undiscounted Problems¶

It is worthwhile focusing on a special case of the LQ problems above: the undiscounted problem that emerges when $\beta = 1$

In this case, the Euler equation is

$$\Bigl( h + d(L^{-1})d(L) \Bigr)\, y_t = a_t$$

The factorization of the characteristic polynomial (16) becomes

$$\Bigl(h+d \, (z^{-1})d(z)\Bigr) = c\,(z^{-1})\, c\,(z)$$

where

\begin{aligned} c\,(z) &= c_0 (1 - \lambda_1 z) \ldots (1 - \lambda_m z) \cr c_0 &= \Bigl[(-1)^m z_0 z_1 \ldots z_m\Bigr ] \cr \vert \lambda_j \vert &< 1 \, \hbox { for } \, j = 1, \ldots, m\cr \lambda_j &= \frac{1}{z_j} \hbox{ for } j=1,\ldots, m\cr z_0 &= \hbox{ constant} \end{aligned}

The solution of the problem becomes

$$(1 - \lambda_1 L) \cdots (1 - \lambda_m L) y_t = \sum^m_{j=1} A_j \sum^\infty_{k=0} \lambda^k_j a_{t+k}$$

### Transforming discounted to undiscounted problem¶

Discounted problems can always be converted into undiscounted problems via a simple transformation

Consider problem (1) with $0 < \beta < 1$

Define the transformed variables

$$\tilde a_t = \beta^{t/2} a_t,\ \tilde y_t = \beta^{t/2} y_t \tag{21}$$

Then notice that $\beta^t\,[d\, (L) y_t ]^2=[\tilde d\,(L)\tilde y_t]^2$ with $\tilde d \,(L)=\sum^m_{j=0} \tilde d_j\, L^j$ and $\tilde d_j = \beta^{j/2} d_j$

Then the original criterion function (1) is equivalent to

$$\lim_{N \rightarrow \infty} \sum^N_{t=0} \{\tilde a_t\, \tilde y_t - {1 \over 2} h\,\tilde y^2_t - {1\over 2} [ \tilde d\,(L)\, \tilde y_t]^2 \} \tag{22}$$

which is to be maximized over sequences $\{\tilde y_t,\ t=0, \ldots\}$ subject to $\tilde y_{-1}, \cdots, \tilde y_{-m}$ given and $\{\tilde a_t,\ t=1, \ldots\}$ a known bounded sequence

The Euler equation for this problem is $[h+\tilde d \,(L^{-1}) \, \tilde d\, (L) ]\, \tilde y_t = \tilde a_t$

The solution is

$$(1 - \tilde \lambda_1 L) \cdots (1 - \tilde \lambda_m L)\,\tilde y_t = \sum^m_{j=1} \tilde A_j \sum^\infty_{k=0} \tilde \lambda^k_j \, \tilde a_{t+k}$$

or

$$\tilde y_t = \tilde f_1 \, \tilde y_{t-1} + \cdots + \tilde f_m\, \tilde y_{t-m} + \sum^m_{j=1} \tilde A_j \sum^\infty_{k=0} \tilde \lambda^k_j \, \tilde a_{t+k}, \tag{23}$$

where $\tilde c \,(z^{-1}) \tilde c\,(z) = h + \tilde d\,(z^{-1}) \tilde d \,(z)$, and where

$$\bigl[(-1)^m\, \tilde z_0 \tilde z_1 \ldots \tilde z_m \bigr]^{1/2} (1 - \tilde \lambda_1\, z) \ldots (1 - \tilde \lambda_m\, z) = \tilde c\,(z), \hbox { where } \ \vert \tilde \lambda_j \vert < 1$$

We leave it to the reader to show that (23) implies the equivalent form of the solution

$$y_t = f_1\, y_{t-1} + \cdots + f_m\, y_{t-m} + \sum^m_{j=1} A_j \sum^\infty_{k=0} \, (\lambda_j\, \beta)^k \, a_{t+k}$$

where

$$f_j = \tilde f_j\, \beta^{-j/2},\ A_j = \tilde A_j,\ \lambda_j = \tilde \lambda_j \, \beta^{-1/2} \tag{24}$$

The transformations (21) and the inverse formulas (24) allow us to solve a discounted problem by first solving a related undiscounted problem

## Implementation¶

Code that computes solutions to the LQ problem using the methods described above can be found in file control_and_filter.py

Here’s how it looks

In [1]:
"""

Authors: Balint Skoze, Tom Sargent, John Stachurski

"""

import numpy as np
import scipy.stats as spst
import scipy.linalg as la

class LQFilter:

def __init__(self, d, h, y_m, r=None, h_eps=None, β=None):
"""

Parameters
----------
d : list or numpy.array (1-D or a 2-D column vector)
The order of the coefficients: [d_0, d_1, ..., d_m]
h : scalar
Parameter of the objective function (corresponding to the
y_m : list or numpy.array (1-D or a 2-D column vector)
Initial conditions for y
r : list or numpy.array (1-D or a 2-D column vector)
The order of the coefficients: [r_0, r_1, ..., r_k]
(optional, if not defined -> deterministic problem)
β : scalar
Discount factor (optional, default value is one)
"""

self.h = h
self.d = np.asarray(d)
self.m = self.d.shape[0] - 1

self.y_m = np.asarray(y_m)

if self.m == self.y_m.shape[0]:
self.y_m = self.y_m.reshape(self.m, 1)
else:
raise ValueError("y_m must be of length m = {self.m:d}")

#---------------------------------------------
# Define the coefficients of ϕ up front
#---------------------------------------------
ϕ = np.zeros(2 * self.m + 1)
for i in range(- self.m, self.m + 1):
ϕ[self.m - i] = np.sum(np.diag(self.d.reshape(self.m + 1, 1) @ \
self.d.reshape(1, self.m + 1), k=-i))
ϕ[self.m] = ϕ[self.m] + self.h
self.ϕ = ϕ

#-----------------------------------------------------
# If r is given calculate the vector ϕ_r
#-----------------------------------------------------
if r is None:
pass
else:
self.r = np.asarray(r)
self.k = self.r.shape[0] - 1
ϕ_r = np.zeros(2 * self.k + 1)
for i in range(- self.k, self.k + 1):
ϕ_r[self.k - i] = np.sum(np.diag(self.r.reshape(self.k + 1, 1) @ \
self.r.reshape(1, self.k + 1), k=-i))
if h_eps is None:
self.ϕ_r = ϕ_r
else:
ϕ_r[self.k] = ϕ_r[self.k] + h_eps
self.ϕ_r = ϕ_r

#-----------------------------------------------------
# If β is given, define the transformed variables
#-----------------------------------------------------
if β is None:
self.β = 1
else:
self.β = β
self.d = self.β**(np.arange(self.m + 1)/2) * self.d
self.y_m = self.y_m * (self.β**(- np.arange(1, self.m + 1)/2)).reshape(self.m, 1)

def construct_W_and_Wm(self, N):
"""
This constructs the matrices W and W_m for a given number of periods N
"""

m = self.m
d = self.d

W = np.zeros((N + 1, N + 1))
W_m = np.zeros((N + 1, m))

#---------------------------------------
# Terminal conditions
#---------------------------------------

D_m1 = np.zeros((m + 1, m + 1))
M = np.zeros((m + 1, m))

# (1) Constuct the D_{m+1} matrix using the formula

for j in range(m + 1):
for k in range(j, m + 1):
D_m1[j, k] = d[:j + 1] @ d[k - j: k + 1]

# Make the matrix symmetric
D_m1 = D_m1 + D_m1.T - np.diag(np.diag(D_m1))

# (2) Construct the M matrix using the entries of D_m1

for j in range(m):
for i in range(j + 1, m + 1):
M[i, j] = D_m1[i - j - 1, m]

#----------------------------------------------
# Euler equations for t = 0, 1, ..., N-(m+1)
#----------------------------------------------
ϕ = self.ϕ

W[:(m + 1), :(m + 1)] = D_m1 + self.h * np.eye(m + 1)
W[:(m + 1), (m + 1):(2 * m + 1)] = M

for i, row in enumerate(np.arange(m + 1, N + 1 - m)):
W[row, (i + 1):(2 * m + 2 + i)] = ϕ

for i in range(1, m + 1):
W[N - m + i, -(2 * m + 1 - i):] = ϕ[:-i]

for i in range(m):
W_m[N - i, :(m - i)] = ϕ[(m + 1 + i):]

return W, W_m

def roots_of_characteristic(self):
"""
This function calculates z_0 and the 2m roots of the characteristic equation
associated with the Euler equation (1.7)

Note:
------
numpy.poly1d(roots, True) defines a polynomial using its roots that can be
evaluated at any point. If x_1, x_2, ... , x_m are the roots then
p(x) = (x - x_1)(x - x_2)...(x - x_m)
"""
m = self.m
ϕ = self.ϕ

# Calculate the roots of the 2m-polynomial
roots = np.roots(ϕ)
# sort the roots according to their length (in descending order)
roots_sorted = roots[np.argsort(abs(roots))[::-1]]

z_0 = ϕ.sum() / np.poly1d(roots, True)(1)
z_1_to_m = roots_sorted[:m]     # we need only those outside the unit circle

λ = 1 / z_1_to_m

return z_1_to_m, z_0, λ

def coeffs_of_c(self):
'''
This function computes the coefficients {c_j, j = 0, 1, ..., m} for
c(z) = sum_{j = 0}^{m} c_j z^j

Based on the expression (1.9). The order is
c_coeffs = [c_0, c_1, ..., c_{m-1}, c_m]
'''
z_1_to_m, z_0 = self.roots_of_characteristic()[:2]

c_0 = (z_0 * np.prod(z_1_to_m).real * (- 1)**self.m)**(.5)
c_coeffs = np.poly1d(z_1_to_m, True).c * z_0 / c_0

return c_coeffs[::-1]

def solution(self):
"""
This function calculates {λ_j, j=1,...,m} and {A_j, j=1,...,m}
of the expression (1.15)
"""
λ = self.roots_of_characteristic()[2]
c_0 = self.coeffs_of_c()[-1]

A = np.zeros(self.m, dtype=complex)
for j in range(self.m):
denom = 1 - λ/λ[j]
A[j] = c_0**(-2) / np.prod(denom[np.arange(self.m) != j])

return λ, A

def construct_V(self, N):
'''
This function constructs the covariance matrix for x^N (see section 6)
for a given period N
'''
V = np.zeros((N, N))
ϕ_r = self.ϕ_r

for i in range(N):
for j in range(N):
if abs(i-j) <= self.k:
V[i, j] = ϕ_r[self.k + abs(i-j)]

return V

def simulate_a(self, N):
"""
Assuming that the u's are normal, this method draws a random path
for x^N
"""
V = self.construct_V(N + 1)
d = spst.multivariate_normal(np.zeros(N + 1), V)

return d.rvs()

def predict(self, a_hist, t):
"""
This function implements the prediction formula discussed is section 6 (1.59)
It takes a realization for a^N, and the period in which the prediciton is formed

Output:  E[abar | a_t, a_{t-1}, ..., a_1, a_0]
"""

N = np.asarray(a_hist).shape[0] - 1
a_hist = np.asarray(a_hist).reshape(N + 1, 1)
V = self.construct_V(N + 1)

aux_matrix = np.zeros((N + 1, N + 1))
aux_matrix[:(t + 1), :(t + 1)] = np.eye(t + 1)
L = la.cholesky(V).T
Ea_hist = la.inv(L) @ aux_matrix @ L @ a_hist

return Ea_hist

def optimal_y(self, a_hist, t=None):
"""
- if t is NOT given it takes a_hist (list or numpy.array) as a deterministic a_t
- if t is given, it solves the combined control prediction problem (section 7)
(by default, t == None -> deterministic)

for a given sequence of a_t (either determinstic or a particular realization),
it calculates the optimal y_t sequence using the method of the lecture

Note:
------
scipy.linalg.lu normalizes L, U so that L has unit diagonal elements
To make things cosistent with the lecture, we need an auxiliary diagonal
matrix D which renormalizes L and U
"""

N = np.asarray(a_hist).shape[0] - 1
W, W_m = self.construct_W_and_Wm(N)

L, U = la.lu(W, permute_l=True)
D = np.diag(1 / np.diag(U))
U = D @ U
L = L @ np.diag(1 / np.diag(D))

J = np.fliplr(np.eye(N + 1))

if t is None:   # if the problem is deterministic

a_hist = J @ np.asarray(a_hist).reshape(N + 1, 1)

#--------------------------------------------
# Transform the a sequence if β is given
#--------------------------------------------
if self.β != 1:
a_hist =  a_hist * (self.β**(np.arange(N + 1) / 2))[::-1].reshape(N + 1, 1)

a_bar = a_hist - W_m @ self.y_m               # a_bar from the lecture
Uy = np.linalg.solve(L, a_bar)                # U @ y_bar = L^{-1}
y_bar = np.linalg.solve(U, Uy)                # y_bar = U^{-1}L^{-1}

# Reverse the order of y_bar with the matrix J
J = np.fliplr(np.eye(N + self.m + 1))
y_hist = J @ np.vstack([y_bar, self.y_m])     # y_hist : concatenated y_m and y_bar

#--------------------------------------------
# Transform the optimal sequence back if β is given
#--------------------------------------------
if self.β != 1:
y_hist = y_hist * (self.β**(- np.arange(-self.m, N + 1)/2)).reshape(N + 1 + self.m, 1)

return y_hist, L, U, y_bar

else:           # if the problem is stochastic and we look at it

Ea_hist = self.predict(a_hist, t).reshape(N + 1, 1)
Ea_hist = J @ Ea_hist

a_bar = Ea_hist - W_m @ self.y_m              # a_bar from the lecture
Uy = np.linalg.solve(L, a_bar)                # U @ y_bar = L^{-1}
y_bar = np.linalg.solve(U, Uy)                # y_bar = U^{-1}L^{-1}

# Reverse the order of y_bar with the matrix J
J = np.fliplr(np.eye(N + self.m + 1))
y_hist = J @ np.vstack([y_bar, self.y_m])     # y_hist : concatenated y_m and y_bar

return y_hist, L, U, y_bar


### Example¶

In this application we’ll have one lag, with

$$d(L) y_t = \gamma(I - L) y_t = \gamma (y_t - y_{t-1})$$

Suppose for the moment that $\gamma = 0$

Then the intertemporal component of the LQ problem disappears, and the agent simply wants to maximize $a_t y_t - hy^2_t / 2$ in each period

This means that the agent chooses $y_t = a_t / h$

In the following we’ll set $h = 1$, so that the agent just wants to track the $\{a_t\}$ process

However, as we increase $\gamma$, the agent gives greater weight to a smooth time path

Hence $\{y_t\}$ evolves as a smoothed version of $\{a_t\}$

The $\{a_t\}$ sequence we’ll choose as a stationary cyclic process plus some white noise

Here’s some code that generates a plot when $\gamma = 0.8$

In [2]:
import matplotlib.pyplot as plt
%matplotlib inline

# == Set seed and generate a_t sequence == #
np.random.seed(123)
n = 100
a_seq = np.sin(np.linspace(0, 5 * np.pi, n)) + 2 + 0.1 * np.random.randn(n)

def plot_simulation(γ=0.8, m=1, h=1, y_m=2):

d = γ * np.asarray([1, -1])
y_m = np.asarray(y_m).reshape(m, 1)

testlq = LQFilter(d, h, y_m)
y_hist, L, U, y = testlq.optimal_y(a_seq)
y = y[::-1]  # reverse y

# == Plot simulation results == #

fig, ax = plt.subplots(figsize=(10, 6))
p_args = {'lw' : 2, 'alpha' : 0.6}
time = range(len(y))
ax.plot(time, a_seq / h, 'k-o', ms=4, lw=2, alpha=0.6, label='$a_t$')
ax.plot(time, y, 'b-o', ms=4, lw=2, alpha=0.6, label='$y_t$')
ax.set(title=rf'Dynamics with $\gamma = {γ}$', xlabel='Time', xlim=(0, max(time)))
ax.legend()
ax.grid()
plt.show()

plot_simulation()


Here’s what happens when we change $\gamma$ to 5.0

In [3]:
plot_simulation(γ=5)


And here’s $\gamma = 10$

In [4]:
plot_simulation(γ=10)


## Exercises¶

### Exercise 1¶

Consider solving a discounted version $(\beta < 1)$ of problem (1), as follows

Convert (1) to the undiscounted problem (22)

Let the solution of (22) in feedback form be

$$(1 - \tilde \lambda_1 L)\, \cdots\, (1 - \tilde \lambda_m L) \tilde y_t = \sum^m_{j=1} \tilde A_j \sum^\infty_{k=0} \tilde \lambda^k_j \tilde a_{t+k}$$

or

$$\tilde y_t = \tilde f_1 \tilde y_{t-1} + \cdots + \tilde f_m \tilde y_{t-m} + \sum^m_{j=1} \tilde A_j \sum^\infty_{k=0} \tilde \lambda^k_j \tilde a_{t+k} \tag{25}$$

Here

• $h + \tilde d (z^{-1}) \tilde d (z) = \tilde c (z^{-1}) \tilde c (z)$
• $\tilde c (z) = [(-1)^m \tilde z_0 \tilde z_1 \cdots \tilde z_m ]^{1/2} (1 - \tilde \lambda_1 z) \cdots (1 - \tilde \lambda_m z)$

where the $\tilde z_j$ are the zeros of $h +\tilde d (z^{-1})\, \tilde d(z)$

Prove that (25) implies that the solution for $y_t$ in feedback form is

$$y_t = f_1 y_{t-1} + \ldots + f_m y_{t-m} + \sum^m_{j=1} A_j \sum^\infty_{k=0} \beta^k \lambda^k_j a_{t+k}$$

where $f_j = \tilde f_j \beta^{-j/2}, A_j = \tilde A_j$, and $\lambda_j = \tilde \lambda_j \beta^{-1/2}$

### Exercise 2¶

Solve the optimal control problem, maximize

$$\sum^2_{t=0}\ \Bigl\{a_t y_t - {1 \over 2} [(1 - 2 L) y_t]^2\Bigr\}$$

subject to $y_{-1}$ given, and $\{ a_t\}$ a known bounded sequence

Express the solution in the “feedback form” (20), giving numerical values for the coefficients

Make sure that the boundary conditions (5) are satisfied

(Note: this problem differs from the problem in the text in one important way: instead of $h > 0$ in (1), $h = 0$. This has an important influence on the solution.)

### Exercise 3¶

Solve the infinite time optimal control problem to maximize

$$\lim_{N \rightarrow \infty} \sum^N_{t=0}\, -\, {1 \over 2} [(1 -2 L) y_t]^2,$$

subject to $y_{-1}$ given. Prove that the solution is

$$y_t = 2y_{t-1} = 2^{t+1} y_{-1} \qquad t > 0$$

### Exercise 4¶

Solve the infinite time problem, to maximize

$$\lim_{N \rightarrow \infty}\ \sum^N_{t=0}\ (.0000001)\, y^2_t - {1 \over 2} [(1 - 2 L) y_t]^2$$

subject to $y_{-1}$ given. Prove that the solution $y_t = 2y_{t-1}$ violates condition (12), and so is not optimal

Prove that the optimal solution is approximately $y_t = .5 y_{t-1}$

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