# The Gaussian AR(1) Case without Intercept: Part 1

2.1 Introduction

Consider the AR(1) model without intercept, rewritten as3

Ayt = a0yt-1 + ut, where ut is iid N(0, a2), (29.2)

and y t is observed for t = 1, 2,…, n. For convenience I will assume that

yt = 0 for t < 0. (29.3)

This assumption is, of course, quite unrealistic, but is made for the sake of trans­parency of the argument, and will appear to be innocent.

The OLS estimator of a 0 is:

X yt-iAyt X yt-iut

7 о = = a о + І=П

X y2-i X y2-i

t=i t=i

If -2 < a 0 < 0, so that yt is stationary, then it is a standard exercise to verify that л/n (a0 – a0) ^ N(0, 1 – (1 + a0)2) in distribution. On the other hand, if a0 = 0, so that yt is a unit root process, this result reads: a0 ^ N(0, 0) in distribution,

hence plimn^^/n a 0 = 0. However, we show now that a much stronger result holds, namely that p0 = na 0 converges in distribution, but the limiting distribu­tion involved is nonnormal. Thus, the presence of a unit root is actually advan­tageous for the efficiency of the OLS estimator a 0. The main problem is that the t-test of the null hypothesis that a 0 = 0 has no longer a standard normal asymp­totic null distribution, so that we cannot test for a unit root using standard methods. The same applies to more general unit root processes.

In the unit root case under review we have yt = yt-1 + ut = y0 + X tj=1uj = X tj=1uj for t > 0, where the last equality involved is due to assumption (29.3). Denoting

t

St = 0 for t < 0, St = X uj for t > 1. (29.5)

і=1

and 62 = (1/n)Xn=1 u2t, it follows that

Next, let

Wn(x) = S[nx}/(oVn) for x Є [0, 1], (29.8)

where [z] means truncation to the nearest integer < z. Then we have:4

– X utyt-1 =1(0 2Wn(1)2 – 62)

nT1 2

1 1 = -(o 2Wn(1)2 – о2 – Ov(1/4~n)) = о 2-^(Wn(1)2 – 1) + op(1), (29.9)

and

where the integral in (29.10) and below, unless otherwise indicated, is taken over the unit interval [0, 1]. The last equality in (29.9) follows from the law of large numbers, by which 62 = о2 + Op(1/Vn). The last equality in (29.10) follows from the fact that for any power m,

Moreover, observe from (29.11), with m = 1, that /Wn(x)dx is a linear combination of iid standard normal random variables, and therefore normal itself, with zero mean and variance

Thus, fWn(x)dx ^ N(0, 1/3) in distribution. Since fWn(x)2dx > (fWn(x)dx)2, it follows therefore that fWn(x)2dx is bounded away from zero:

-1

Wn(x)2 dx

V /

Combining (29.9), (29.10), and (29.13), we now have:

This result does not depend on assumption (29.3).