Assume that we have the following over-determined linear system \begin{cases} z_{1}=c + \phi z_0\\ z_{2}=c + \phi z_{1}\\ \dots\\ z_{n} = c + \phi z_{n-1} \end{cases} with $n>2$ and all $z_{0}, \dots, z_{n}$ are different numbers. Then, using ordinary least squares (OLS), i.e. we can get the solution $(\hat{c}_{1}, \hat{\phi}_{2})$ (which is, actually, the estimate) of the system.

Next, let us consider the following system \begin{cases} z_{2} - z_{1} = \phi(z_{1} - z_{0})\\ (z_{3} - z_{2})= \phi (z_{2} - z_{1})\\ \dots\\ (z_{n} - z_{n-1})=\phi (z_{n-1} - z_{n-2}) \end{cases} and using OLS again let us get the solution for $\hat{\phi}_{2}$.

Why the solution in the second case is very different compare to the first one?

From what I can see from the simulations below, one can not just subtract rows in the system, which is correct for consistent system.

The python code for simulations is below.

import numpy as np
import scipy as sp
import statsmodels.api as sm

function for simulation of z_{t} (the previous value plus some noise)
def simulate_z(nSample, phi, sigma_e, fVal, c):
    noise_e = sp.random.normal(0, sigma_e, nSample)
    z = np.zeros(nSample)
    z[0] = fVal
    for period in range(1, nSample):
        z[period] = c + phi * z[(period - 1)] + noise_e[period]
    return z

OLS estimation
def est_c_ph(z):
    x = z[0:-1]
    y = z[1:]
    p = sp.polyfit(x, y, 1)
    # Estimate phi
    phi_est = p[0]
    # Estimate c
    c_est = sp.mean(z) * (1 - phi_est)
    return [c_est, phi_est]

values of the parameters for simulation
phi = 0.95  # slope
c = 0.5  # intercept
sigma_e = 0.08  # standard deviation of observation noise
nSample = 500  # sample size
E = c / (1 - phi)  # mean value
fVal = E  # first value of the simulated process
simulation of AR(1)
z = simulate_z(nSample, phi, sigma_e, fVal, c)

c_est, phi_est = est_c_ph(z)
print("OLS [c, phi]: ", [c_est, phi_est])

differencing of data
z_dif = z[1:] - z[0:-1]

c_d_est, phi_d_est = est_c_ph(z_dif)
print("diff z OLS [c, phi]: ", [c_d_est, phi_d_est])
  • 1
    $\begingroup$ Your second system is not equivalent to the first. To make it equivalent, you have to add one more constraint $$z_1 = c + \phi z_0$$ By dropping this constraint, you allow a wider range of possible solutions, so it is not surprising that the best fit without it is different from the best fit with it. $\endgroup$ – Paul Sinclair Nov 15 '19 at 2:59
  • $\begingroup$ Dear @Paul Sinclair, what if system is consistent (n=2)? In this case we would be able to make this transformation and $z_{2} - z_{1} = \phi(z_{1} - z_{0})$ would give us correct solution for $\theta$. $\endgroup$ – ABK Nov 15 '19 at 9:24
  • $\begingroup$ Dear @Paul Sinclair, system itself is not equivalent, we do not have $c$ in the second one. But why the solution for $\phi$ is very different? $\endgroup$ – ABK Nov 15 '19 at 9:47
  • 1
    $\begingroup$ Read the part of my previous comment that comes after the formula. $\endgroup$ – Paul Sinclair Nov 15 '19 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.