# The Gaussian AR(1) Case with Intercept under the Alternative of Stationary

If under the stationarity hypothesis the AR(1) process has an intercept, but not under the unit root hypothesis, the AR(1) model that covers both the null and the alternative is:

Ayt = a 0 + a 1 yt-1 + ut, where a 0 = – ca1. (29.28) If -2 < a 1 < 0, then the process yt is stationary around the constant c:

hence E( yt) = c2 + (1 – (1 + a1)2) 1o2, E(ytyt-1) = c2 + (1 + a 1)(1 – (1 + a 1)2) 1o2, and  which approaches zero if c2/о2 ^ «>. Therefore, the power of the test p0 will be low if the variance of ut is small relative to [E(yf)]2. The same applies to the f-test x0. We should therefore use the OLS estimator of a1 and the corresponding f-value in the regression of Ayt on yf-1 with intercept.

Denoting y_1 = (1/п)Щ=1 У f-1, й = (1/„)’L„=1u f, the OLS estimator of a 1 is:

X иУм _ пйЕ_1

71 = a 1 + ^ . (29.31)

X y2_1 _ nE_1

t = 1     Since by (29.8), л[пй = оWn(1), and under the null hypothesis a 1 = 0 and the maintained hypothesis (29.3),

where the last equality follows from (29.11) with m = 1, it follows from Lemma 1, similarly to (29.14) and (29.22) that (1/2)(W„(1)2 _ 1) _ Wn(1)/W„(x)dx + o (1) /Wn(x)2dx _ (fWn(x)dx)2 p( ) ^ p1 s (1/2)(W(1)2 _ !) _ W(1)/W(x)dx in distribution.

/W(x)2 dx _ (fW(x)dx)2

The density of p1 is displayed in Figure 29.3. Comparing Figures 29.1 and 29.3, we see that the density of p1 is farther left of zero than the density of p0, and has a fatter left tail. As to the f-value t of a1 in this case, it follows similarly to (29.24) and (29.33) that under the unit root hypothesis,

(1/2) (W(1)2 – 1) – W(1)/W(x)dx

t1 ^ т1 = in distribution. (29.34)

/W(x)2 dx – (fW(x)dx)2

Again, the results (29.33) and (29.34) do not hinge on assumption (29.3).

The distribution of т1 is even farther away from the normal distribution than the distribution of т0, as follows from comparison of (29.26) with

P(t1 < -2.86) = 0.05, P(t1 < -2.57) = 0.1 (29.35)

See again Fuller (1996, p. 642). This is corroborated by Figure 29.4, where the density of т1 is compared with the standard normal density.

We see that the density of т1 is shifted even more to the left of the standard normal density than in Figure 29.2, hence the left-sided standard normal test would result in a dramatically higher type 1 error than in the case without an intercept: compare

P(t1 < -1.64) = 0.46, P(t1 < -1.28) = 0.64 (29.36)

with (29.26) and (29.27).