.. index::
single: Autoregressive processesContents
In this lecture we are going to study a very simple class of stochastic models called AR(1) processes.
These simple models are used again and again in economic research to represent the dynamics of series such as
- labor income
- dividends
- productivity, etc.
AR(1) processes can take negative values but are easily converted into positive processes when necessary by a transformation such as exponentiation.
We are going to study AR(1) processes partly because they are useful and partly because they help us understand important concepts.
Let's start with some imports:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
The AR(1) model (autoregressive model of order 1) takes the form
X_{t+1} = a X_t + b + c W_{t+1}
where a, b, c are scalar-valued parameters.
This law of motion generates a time series \{ X_t\} as soon as we specify an initial condition X_0.
This is called the state process and the state space is \mathbb R.
To make things even simpler, we will assume that
- the process \{ W_t \} is IID and standard normal,
- the initial condition X_0 is drawn from the normal distribution N(\mu_0, v_0) and
- the initial condition X_0 is independent of \{ W_t \}.
Iterating backwards from time t, we obtain
X_t = a X_{t-1} + b + c W_t
= a^2 X_{t-2} + a b + a c W_{t-1} + b + c W_t
= \cdots
If we work all the way back to time zero, we get
X_t = a^t X_0 + b \sum_{j=0}^{t-1} a^j +
c \sum_{j=0}^{t-1} a^j W_{t-j}
Equation :eq:`ar1_ma` shows that X_t is a well defined random variable, the value of which depends on
- the parameters,
- the initial condition X_0 and
- the shocks W_1, \ldots W_t from time t=1 to the present.
Throughout, the symbol \psi_t will be used to refer to the density of this random variable X_t.
One of the nice things about this model is that it's so easy to trace out the sequence of distributions \{ \psi_t \} corresponding to the time series \{ X_t\}.
To see this, we first note that X_t is normally distributed for each t.
This is immediate from :eq:`ar1_ma`, since linear combinations of independent normal random variables are normal.
Given that X_t is normally distributed, we will know the full distribution \psi_t if we can pin down its first two moments.
Let \mu_t and v_t denote the mean and variance of X_t respectively.
We can pin down these values from :eq:`ar1_ma` or we can use the following recursive expressions:
\mu_{t+1} = a \mu_t + b
\quad \text{and} \quad
v_{t+1} = a^2 v_t + c^2
These expressions are obtained from :eq:`can_ar1` by taking, respectively, the expectation and variance of both sides of the equality.
In calculating the second expression, we are using the fact that X_t and W_{t+1} are independent.
(This follows from our assumptions and :eq:`ar1_ma`.)
Given the dynamics in :eq:`ar1_ma` and initial conditions \mu_0, v_0, we obtain \mu_t, v_t and hence
\psi_t = N(\mu_t, v_t)
The following code uses these facts to track the sequence of marginal distributions \{ \psi_t \}.
The parameters are
a, b, c = 0.9, 0.1, 0.5
mu, v = -3.0, 0.6 # initial conditions mu_0, v_0Here's the sequence of distributions:
from scipy.stats import norm
sim_length = 10
grid = np.linspace(-5, 7, 120)
fig, ax = plt.subplots()
for t in range(sim_length):
mu = a * mu + b
v = a**2 * v + c**2
ax.plot(grid, norm.pdf(grid, loc=mu, scale=np.sqrt(v)),
label=f"$\psi_{t}$",
alpha=0.7)
ax.legend(bbox_to_anchor=[1.05,1],loc=2,borderaxespad=1)
plt.show()Notice that, in the figure above, the sequence \{ \psi_t \} seems to be converging to a limiting distribution.
This is even clearer if we project forward further into the future:
def plot_density_seq(ax, mu_0=-3.0, v_0=0.6, sim_length=60):
mu, v = mu_0, v_0
for t in range(sim_length):
mu = a * mu + b
v = a**2 * v + c**2
ax.plot(grid,
norm.pdf(grid, loc=mu, scale=np.sqrt(v)),
alpha=0.5)
fig, ax = plt.subplots()
plot_density_seq(ax)
plt.show()Moreover, the limit does not depend on the initial condition.
For example, this alternative density sequence also converges to the same limit.
fig, ax = plt.subplots()
plot_density_seq(ax, mu_0=3.0)
plt.show()In fact it's easy to show that such convergence will occur, regardless of the initial condition, whenever |a| < 1.
To see this, we just have to look at the dynamics of the first two moments, as given in :eq:`dyn_tm`.
When |a| < 1, these sequences converge to the respective limits
\mu^* := \frac{b}{1-a}
\quad \text{and} \quad
v^* = \frac{c^2}{1 - a^2}
(See our :doc:`lecture on one dimensional dynamics <scalar_dynam>` for background on deterministic convergence.)
Hence
\psi_t \to \psi^* = N(\mu^*, v^*)
\quad \text{as }
t \to \infty
We can confirm this is valid for the sequence above using the following code.
fig, ax = plt.subplots()
plot_density_seq(ax, mu_0=3.0)
mu_star = b / (1 - a)
std_star = np.sqrt(c**2 / (1 - a**2)) # square root of v_star
psi_star = norm.pdf(grid, loc=mu_star, scale=std_star)
ax.plot(grid, psi_star, 'k-', lw=2, label="$\psi^*$")
ax.legend()
plt.show()As claimed, the sequence \{ \psi_t \} converges to \psi^*.
A stationary distribution is a distribution that is a fixed point of the update rule for distributions.
In other words, if \psi_t is stationary, then \psi_{t+j} = \psi_t for all j in \mathbb N.
A different way to put this, specialized to the current setting, is as follows: a density \psi on \mathbb R is stationary for the AR(1) process if
X_t \sim \psi
\quad \implies \quad
a X_t + b + c W_{t+1} \sim \psi
The distribution \psi^* in :eq:`ar1_psi_star` has this property --- checking this is an exercise.
(Of course, we are assuming that |a| < 1 so that \psi^* is well defined.)
In fact, it can be shown that no other distribution on \mathbb R has this property.
Thus, when |a| < 1, the AR(1) model has exactly one stationary density and that density is given by \psi^*.
The concept of ergodicity is used in different ways by different authors.
One way to understand it in the present setting is that a version of the Law of Large Numbers is valid for \{X_t\}, even though it is not IID.
In particular, averages over time series converge to expectations under the stationary distribution.
Indeed, it can be proved that, whenever |a| < 1, we have
\frac{1}{m} \sum_{t = 1}^m h(X_t) \to
\int h(x) \psi^*(x) dx
\quad \text{as } m \to \infty
whenever the integral on the right hand side is finite and well defined.
Notes:
- In :eq:`ar1_ergo`, convergence holds with probability one.
- The textbook by :cite:`MeynTweedie2009` is a classic reference on ergodicity.
For example, if we consider the identity function h(x) = x, we get
\frac{1}{m} \sum_{t = 1}^m X_t \to
\int x \psi^*(x) dx
\quad \text{as } m \to \infty
In other words, the time series sample mean converges to the mean of the stationary distribution.
As will become clear over the next few lectures, ergodicity is a very important concept for statistics and simulation.
Let k be a natural number.
The k-th central moment of a random variable is defined as
M_k := \mathbb E [ (X - \mathbb E X )^k ]
When that random variable is N(\mu, \sigma^2), it is known that
M_k =
\begin{cases}
0 & \text{ if } k \text{ is odd} \\
\sigma^k (k-1)!! & \text{ if } k \text{ is even}
\end{cases}
Here n!! is the double factorial.
According to :eq:`ar1_ergo`, we should have, for any k \in \mathbb N,
\frac{1}{m} \sum_{t = 1}^m
(X_t - \mu^* )^k
\approx M_k
when m is large.
Confirm this by simulation at a range of k using the default parameters from the lecture.
Write your own version of a one dimensional kernel density estimator, which estimates a density from a sample.
Write it as a class that takes the data X and bandwidth h when initialized and provides a method f such that
f(x) = \frac{1}{hn} \sum_{i=1}^n
K \left( \frac{x-X_i}{h} \right)
For K use the Gaussian kernel (K is the standard normal density).
Write the class so that the bandwidth defaults to Silverman’s rule (see the “rule of thumb” discussion on this page). Test the class you have written by going through the steps
- simulate data X_1, \ldots, X_n from distribution \phi
- plot the kernel density estimate over a suitable range
- plot the density of \phi on the same figure
for distributions \phi of the following types
- beta distribution with \alpha = \beta = 2
- beta distribution with \alpha = 2 and \beta = 5
- beta distribution with \alpha = \beta = 0.5
Use n=500.
Make a comment on your results. (Do you think this is a good estimator of these distributions?)
In the lecture we discussed the following fact: for the AR(1) process
X_{t+1} = a X_t + b + c W_{t+1}
with \{ W_t \} iid and standard normal,
\psi_t = N(\mu, s^2) \implies \psi_{t+1}
= N(a \mu + b, a^2 s^2 + c^2)
Confirm this, at least approximately, by simulation. Let
- a = 0.9
- b = 0.0
- c = 0.1
- \mu = -3
- s = 0.2
First, plot \psi_t and \psi_{t+1} using the true distributions described above.
Second, plot \psi_{t+1} on the same figure (in a different color) as follows:
- Generate n draws of X_t from the N(\mu, s^2) distribution
- Update them all using the rule X_{t+1} = a X_t + b + c W_{t+1}
- Use the resulting sample of X_{t+1} values to produce a density estimate via kernel density estimation.
Try this for n=2000 and confirm that the simulation based estimate of \psi_{t+1} does converge to the theoretical distribution.
from numba import njit
from scipy.special import factorial2
@njit
def sample_moments_ar1(k, m=100_000, mu_0=0.0, sigma_0=1.0, seed=1234):
np.random.seed(seed)
sample_sum = 0.0
x = mu_0 + sigma_0 * np.random.randn()
for t in range(m):
sample_sum += (x - mu_star)**k
x = a * x + b + c * np.random.randn()
return sample_sum / m
def true_moments_ar1(k):
if k % 2 == 0:
return std_star**k * factorial2(k - 1)
else:
return 0
k_vals = np.arange(6) + 1
sample_moments = np.empty_like(k_vals)
true_moments = np.empty_like(k_vals)
for k_idx, k in enumerate(k_vals):
sample_moments[k_idx] = sample_moments_ar1(k)
true_moments[k_idx] = true_moments_ar1(k)
fig, ax = plt.subplots()
ax.plot(k_vals, true_moments, label="true moments")
ax.plot(k_vals, sample_moments, label="sample moments")
ax.legend()
plt.show()Here is one solution:
K = norm.pdf
class KDE:
def __init__(self, x_data, h=None):
if h is None:
c = x_data.std()
n = len(x_data)
h = 1.06 * c * n**(-1/5)
self.h = h
self.x_data = x_data
def f(self, x):
if np.isscalar(x):
return K((x - self.x_data) / self.h).mean() * (1/self.h)
else:
y = np.empty_like(x)
for i, x_val in enumerate(x):
y[i] = K((x_val - self.x_data) / self.h).mean() * (1/self.h)
return y
def plot_kde(ϕ, x_min=-0.2, x_max=1.2):
x_data = ϕ.rvs(n)
kde = KDE(x_data)
x_grid = np.linspace(-0.2, 1.2, 100)
fig, ax = plt.subplots()
ax.plot(x_grid, kde.f(x_grid), label="estimate")
ax.plot(x_grid, ϕ.pdf(x_grid), label="true density")
ax.legend()
plt.show()
from scipy.stats import beta
n = 500
parameter_pairs= (2, 2), (2, 5), (0.5, 0.5)
for α, β in parameter_pairs:
plot_kde(beta(α, β))
We see that the kernel density estimator is effective when the underlying distribution is smooth but less so otherwise.
Here is our solution
a = 0.9
b = 0.0
c = 0.1
μ = -3
s = 0.2
μ_next = a * μ + b
s_next = np.sqrt(a**2 * s**2 + c**2)
ψ = lambda x: K((x - μ) / s)
ψ_next = lambda x: K((x - μ_next) / s_next)
ψ = norm(μ, s)
ψ_next = norm(μ_next, s_next)
n = 2000
x_draws = ψ.rvs(n)
x_draws_next = a * x_draws + b + c * np.random.randn(n)
kde = KDE(x_draws_next)
x_grid = np.linspace(μ - 1, μ + 1, 100)
fig, ax = plt.subplots()
ax.plot(x_grid, ψ.pdf(x_grid), label="$\psi_t$")
ax.plot(x_grid, ψ_next.pdf(x_grid), label="$\psi_{t+1}$")
ax.plot(x_grid, kde.f(x_grid), label="estimate of $\psi_{t+1}$")
ax.legend()
plt.show()
The simulated distribution approximately coincides with the theoretical distribution, as predicted.