This lecture uses matrices to solve some linear difference equations.
As a running example, we’ll study a second-order linear difference equation that was the key technical tool in Paul Samuelson’s 1939 article [Sam39] that introduced the multiplier-accelerator model.
This model became the workhorse that powered early econometric versions of Keynesian macroeconomic models in the United States.
You can read about the details of that model in this QuantEcon lecture.
(That lecture also describes some technicalities about second-order linear difference equations.)
In this lecture, we’ll also learn about an autoregressive representation and a moving average representation of a non-stationary univariate time series $ \{y_t\}_{t=0}^T $.
We’ll also study a “perfect foresight” model of stock prices that involves solving a “forward-looking” linear difference equation.
We will use the following imports:
import numpy as np
%matplotlib inline
import matplotlib.pyplot as plt
from matplotlib import cm
plt.rcParams["figure.figsize"] = (11, 5) #set default figure size
Let $ t = 0, \pm 1, \pm 2, \ldots $ index time.
For $ t = 1, 2, 3, \ldots, T $ suppose that
$$ y_{t} = \alpha_{0} + \alpha_{1} y_{t-1} + \alpha_{2} y_{t-2} \tag{11.1} $$
where we assume that $ y_0 $ and $ y_{-1} $ are given numbers that we take as initial conditions.
In Samuelson’s model, $ y_t $ stood for national income or perhaps a different measure of aggregate activity called gross domestic product (GDP) at time $ t $.
Equation (11.1) is called a second-order linear difference equation.
But actually, it is a collection of $ T $ simultaneous linear equations in the $ T $ variables $ y_1, y_2, \ldots, y_T $.
Note
To be able to solve a second-order linear difference
equation, we require two boundary conditions that can take the form either of two initial conditions or two terminal conditions or possibly one of each.
Let’s write our equations as a stacked system
$$ \underset{\equiv A}{\underbrace{\left[\begin{array}{cccccccc} 1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0\\ -\alpha_{1} & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ -\alpha_{2} & -\alpha_{1} & 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & -\alpha_{2} & -\alpha_{1} & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & -\alpha_{2} & -\alpha_{1} & 1 \end{array}\right]}}\left[\begin{array}{c} y_{1}\\ y_{2}\\ y_{3}\\ y_{4}\\ \vdots\\ y_{T} \end{array}\right]=\underset{\equiv b}{\underbrace{\left[\begin{array}{c} \alpha_{0}+\alpha_{1}y_{0}+\alpha_{2}y_{-1}\\ \alpha_{0}+\alpha_{2}y_{0}\\ \alpha_{0}\\ \alpha_{0}\\ \vdots\\ \alpha_{0} \end{array}\right]}} $$or
$$ A y = b $$where
$$ y = \begin{bmatrix} y_1 \cr y_2 \cr \vdots \cr y_T \end{bmatrix} $$Evidently $ y $ can be computed from
$$ y = A^{-1} b $$The vector $ y $ is a complete time path $ \{y_t\}_{t=1}^T $.
Let’s put Python to work on an example that captures the flavor of Samuelson’s multiplier-accelerator model.
We’ll set parameters equal to the same values we used in this QuantEcon lecture.
T = 80
# parameters
𝛼0 = 10.0
𝛼1 = 1.53
𝛼2 = -.9
y_1 = 28. # y_{-1}
y0 = 24.
Now we construct $ A $ and $ b $.
A = np.identity(T) # The T x T identity matrix
for i in range(T):
if i-1 >= 0:
A[i, i-1] = -𝛼1
if i-2 >= 0:
A[i, i-2] = -𝛼2
b = np.full(T, 𝛼0)
b[0] = 𝛼0 + 𝛼1 * y0 + 𝛼2 * y_1
b[1] = 𝛼0 + 𝛼2 * y0
Let’s look at the matrix $ A $ and the vector $ b $ for our example.
A, b
Now let’s solve for the path of $ y $.
If $ y_t $ is GNP at time $ t $, then we have a version of Samuelson’s model of the dynamics for GNP.
To solve $ y = A^{-1} b $ we can either invert $ A $ directly, as in
A_inv = np.linalg.inv(A)
y = A_inv @ b
or we can use np.linalg.solve:
y_second_method = np.linalg.solve(A, b)
Here make sure the two methods give the same result, at least up to floating point precision:
np.allclose(y, y_second_method)
Note
In general,
np.linalg.solveis more numerically stable than using
np.linalg.inv directly.
However, stability is not an issue for this small example. Moreover, we will
repeatedly use A_inv in what follows, so there is added value in computing
it directly.
Now we can plot.
plt.plot(np.arange(T)+1, y)
plt.xlabel('t')
plt.ylabel('y')
plt.show()
The steady state value $ y^* $ of $ y_t $ is obtained by setting $ y_t = y_{t-1} = y_{t-2} = y^* $ in (11.1), which yields
$$ y^* = \frac{\alpha_{0}}{1 - \alpha_{1} - \alpha_{2}} $$If we set the initial values to $ y_{0} = y_{-1} = y^* $, then $ y_{t} $ will be constant:
y_star = 𝛼0 / (1 - 𝛼1 - 𝛼2)
y_1_steady = y_star # y_{-1}
y0_steady = y_star
b_steady = np.full(T, 𝛼0)
b_steady[0] = 𝛼0 + 𝛼1 * y0_steady + 𝛼2 * y_1_steady
b_steady[1] = 𝛼0 + 𝛼2 * y0_steady
y_steady = A_inv @ b_steady
plt.plot(np.arange(T)+1, y_steady)
plt.xlabel('t')
plt.ylabel('y')
plt.show()
To generate some excitement, we’ll follow in the spirit of the great economists Eugen Slutsky and Ragnar Frisch and replace our original second-order difference equation with the following second-order stochastic linear difference equation:
$$ y_{t} = \alpha_{0} + \alpha_{1} y_{t-1} + \alpha_{2} y_{t-2} + u_t \tag{11.2} $$
where $ u_{t} \sim N\left(0, \sigma_{u}^{2}\right) $ and is IID, meaning independent and identically distributed.
We’ll stack these $ T $ equations into a system cast in terms of matrix algebra.
Let’s define the random vector
$$ u=\left[\begin{array}{c} u_{1}\\ u_{2}\\ \vdots\\ u_{T} \end{array}\right] $$Where $ A, b, y $ are defined as above, now assume that $ y $ is governed by the system
The solution for $ y $ becomes
$$ y = A^{-1} \left(b + u\right) \tag{11.4} $$
Let’s try it out in Python.
𝜎u = 2.
u = np.random.normal(0, 𝜎u, size=T)
y = A_inv @ (b + u)
plt.plot(np.arange(T)+1, y)
plt.xlabel('t')
plt.ylabel('y')
plt.show()
The above time series looks a lot like (detrended) GDP series for a number of advanced countries in recent decades.
We can simulate $ N $ paths.
N = 100
for i in range(N):
col = cm.viridis(np.random.rand()) # Choose a random color from viridis
u = np.random.normal(0, 𝜎u, size=T)
y = A_inv @ (b + u)
plt.plot(np.arange(T)+1, y, lw=0.5, color=col)
plt.xlabel('t')
plt.ylabel('y')
plt.show()
Also consider the case when $ y_{0} $ and $ y_{-1} $ are at steady state.
N = 100
for i in range(N):
col = cm.viridis(np.random.rand()) # Choose a random color from viridis
u = np.random.normal(0, 𝜎u, size=T)
y_steady = A_inv @ (b_steady + u)
plt.plot(np.arange(T)+1, y_steady, lw=0.5, color=col)
plt.xlabel('t')
plt.ylabel('y')
plt.show()
We can apply standard formulas for multivariate normal distributions to compute the mean vector and covariance matrix for our time series model
$$ y = A^{-1} (b + u) . $$You can read about multivariate normal distributions in this lecture Multivariate Normal Distribution.
Let’s write our model as
$$ y = \tilde A (b + u) $$where $ \tilde A = A^{-1} $.
Because linear combinations of normal random variables are normal, we know that
$$ y \sim {\mathcal N}(\mu_y, \Sigma_y) $$where
$$ \mu_y = \tilde A b $$and
$$ \Sigma_y = \tilde A (\sigma_u^2 I_{T \times T} ) \tilde A^T $$Let’s write a Python class that computes the mean vector $ \mu_y $ and covariance matrix $ \Sigma_y $.
class population_moments:
"""
Compute population moments mu_y, Sigma_y.
---------
Parameters:
alpha0, alpha1, alpha2, T, y_1, y0
"""
def __init__(self, alpha0, alpha1, alpha2, T, y_1, y0, sigma_u):
# compute A
A = np.identity(T)
for i in range(T):
if i-1 >= 0:
A[i, i-1] = -alpha1
if i-2 >= 0:
A[i, i-2] = -alpha2
# compute b
b = np.full(T, alpha0)
b[0] = alpha0 + alpha1 * y0 + alpha2 * y_1
b[1] = alpha0 + alpha2 * y0
# compute A inverse
A_inv = np.linalg.inv(A)
self.A, self.b, self.A_inv, self.sigma_u, self.T = A, b, A_inv, sigma_u, T
def sample_y(self, n):
"""
Give a sample of size n of y.
"""
A_inv, sigma_u, b, T = self.A_inv, self.sigma_u, self.b, self.T
us = np.random.normal(0, sigma_u, size=[n, T])
ys = np.vstack([A_inv @ (b + u) for u in us])
return ys
def get_moments(self):
"""
Compute the population moments of y.
"""
A_inv, sigma_u, b = self.A_inv, self.sigma_u, self.b
# compute mu_y
self.mu_y = A_inv @ b
self.Sigma_y = sigma_u**2 * (A_inv @ A_inv.T)
return self.mu_y, self.Sigma_y
my_process = population_moments(
alpha0=10.0, alpha1=1.53, alpha2=-.9, T=80, y_1=28., y0=24., sigma_u=1)
mu_y, Sigma_y = my_process.get_moments()
A_inv = my_process.A_inv
It is enlightening to study the $ \mu_y, \Sigma_y $’s implied by various parameter values.
Among other things, we can use the class to exhibit how statistical stationarity of $ y $ prevails only for very special initial conditions.
Let’s begin by generating $ N $ time realizations of $ y $ plotting them together with population mean $ \mu_y $ .
# plot mean
N = 100
for i in range(N):
col = cm.viridis(np.random.rand()) # Choose a random color from viridis
ys = my_process.sample_y(N)
plt.plot(ys[i,:], lw=0.5, color=col)
plt.plot(mu_y, color='red')
plt.xlabel('t')
plt.ylabel('y')
plt.show()
Visually, notice how the variance across realizations of $ y_t $ decreases as $ t $ increases.
Let’s plot the population variance of $ y_t $ against $ t $.
# plot variance
plt.plot(Sigma_y.diagonal())
plt.show()
Notice how the population variance increases and asymptotes
Let’s print out the covariance matrix $ \Sigma_y $ for a time series $ y $
my_process = population_moments(alpha0=0, alpha1=.8, alpha2=0, T=6, y_1=0., y0=0., sigma_u=1)
mu_y, Sigma_y = my_process.get_moments()
print("mu_y = ",mu_y)
print("Sigma_y = ", Sigma_y)
Notice that the covariance between $ y_t $ and $ y_{t-1} $ – the elements on the superdiagonal – are not identical.
This is is an indication that the time series respresented by our $ y $ vector is not stationary.
To make it stationary, we’d have to alter our system so that our initial conditions $ (y_1, y_0) $ are not fixed numbers but instead a jointly normally distributed random vector with a particular mean and covariance matrix.
We describe how to do that in another lecture in this lecture Linear State Space Models.
But just to set the stage for that analysis, let’s print out the bottom right corner of $ \Sigma_y $.
mu_y, Sigma_y = my_process.get_moments()
print("bottom right corner of Sigma_y = \n", Sigma_y[72:,72:])
Please notice how the sub diagonal and super diagonal elements seem to have converged.
This is an indication that our process is asymptotically stationary.
You can read about stationarity of more general linear time series models in this lecture Linear State Space Models.
There is a lot to be learned about the process by staring at the off diagonal elements of $ \Sigma_y $ corresponding to different time periods $ t $, but we resist the temptation to do so here.
Let’s print out $ A^{-1} $ and stare at its structure
To study the structure of $ A^{-1} $, we shall print just up to $ 3 $ decimals.
Let’s begin by printing out just the upper left hand corner of $ A^{-1} $
with np.printoptions(precision=3, suppress=True):
print(A_inv[0:7,0:7])
Evidently, $ A^{-1} $ is a lower triangular matrix.
Let’s print out the lower right hand corner of $ A^{-1} $ and stare at it.
with np.printoptions(precision=3, suppress=True):
print(A_inv[72:,72:])
Notice how every row ends with the previous row’s pre-diagonal entries.
Since $ A^{-1} $ is lower triangular, each row represents $ y_t $ for a particular $ t $ as the sum of
Thus, let $ \tilde{A}=A^{-1} $.
Evidently, for $ t\geq0 $,
$$ y_{t+1}=\sum_{i=1}^{t+1}\tilde{A}_{t+1,i}b_{i}+\sum_{i=1}^{t}\tilde{A}_{t+1,i}u_{i}+u_{t+1} $$This is a moving average representation with time-varying coefficients.
Just as system (11.4) constitutes a moving average representation for $ y $, system (11.3) constitutes an autoregressive representation for $ y $.
Samuelson’s model is backwards looking in the sense that we give it initial conditions and let it run.
Let’s now turn to model that is forward looking.
We apply similar linear algebra machinery to study a perfect foresight model widely used as a benchmark in macroeconomics and finance.
As an example, we suppose that $ p_t $ is the price of a stock and that $ y_t $ is its dividend.
We assume that $ y_t $ is determined by second-order difference equation that we analyzed just above, so that
$$ y = A^{-1} \left(b + u\right) $$Our perfect foresight model of stock prices is
$$ p_{t} = \sum_{j=0}^{T-t} \beta^{j} y_{t+j}, \quad \beta \in (0,1) $$where $ \beta $ is a discount factor.
The model asserts that the price of the stock at $ t $ equals the discounted present values of the (perfectly foreseen) future dividends.
Form
$$ \underset{\equiv p}{\underbrace{\left[\begin{array}{c} p_{1}\\ p_{2}\\ p_{3}\\ \vdots\\ p_{T} \end{array}\right]}}=\underset{\equiv B}{\underbrace{\left[\begin{array}{ccccc} 1 & \beta & \beta^{2} & \cdots & \beta^{T-1}\\ 0 & 1 & \beta & \cdots & \beta^{T-2}\\ 0 & 0 & 1 & \cdots & \beta^{T-3}\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{array}\right]}}\left[\begin{array}{c} y_{1}\\ y_{2}\\ y_{3}\\ \vdots\\ y_{T} \end{array}\right] $$𝛽 = .96
# construct B
B = np.zeros((T, T))
for i in range(T):
B[i, i:] = 𝛽 ** np.arange(0, T-i)
B
𝜎u = 0.
u = np.random.normal(0, 𝜎u, size=T)
y = A_inv @ (b + u)
y_steady = A_inv @ (b_steady + u)
p = B @ y
plt.plot(np.arange(0, T)+1, y, label='y')
plt.plot(np.arange(0, T)+1, p, label='p')
plt.xlabel('t')
plt.ylabel('y/p')
plt.legend()
plt.show()
Can you explain why the trend of the price is downward over time?
Also consider the case when $ y_{0} $ and $ y_{-1} $ are at the steady state.
p_steady = B @ y_steady
plt.plot(np.arange(0, T)+1, y_steady, label='y')
plt.plot(np.arange(0, T)+1, p_steady, label='p')
plt.xlabel('t')
plt.ylabel('y/p')
plt.legend()
plt.show()