2.1. Kim's Tests

In the context of Kim's tests (2000), the null hypothesis can be described as

for t = 1,…,t, where r_t = r₀, or r_t = r₀ + γt, where r₀ is a constant and t is a linear time trend. The variable z_t is a stationary variable satisfying some regularity conditions. The alternative hypothesis implies that the variable z_t becomes a more persistent process at some moment [τT]. For simplicity, we assume that the null model corresponds to r_t = r₀.

Now, let us define the residual series³

where y₀ and y₁ are the sample means of each subsample. To test the null hypothesis, Kim (2000) designs the statistic

where

, for t = [τT] + 1,…,T.

However, because the true τ value is unknown a priori, Kim (2000) proposes three alternative ways for testing structural change. Following the notation of Busetti and Taylor (2004):

which are called maximum-Chow type, mean score, and mean-exponential test, respectively. In the actual computations, the integrals are replaced by sample averages, where

is a compact subset of [0,1],

.

Kim (2000) derives the asymptotic distribution of these statistics under the null hypothesis, and he shows that the tests are consistent against the alternative hypothesis of change in persistence.

2.2. Stationarity and Structural Breaks

The properties of the proposed tests under neglected structural changes are unknown, although we would expect the effects to be similar to the effects on the KPSS test (see Lee, Huang, and Shin, 1997).

Let us assume that the true data generating process is defined by the following equation:

where D_t = 1 for t ≥ λT + 1 and zero otherwise, ε_t is a stationary process of constant persistence, and λ is the fraction of the sample where the structural change occurs (0 < λ < 1).

The main results can be summarized in the following theorem.

THEOREM. If the time series is generated by process (7), then H_j(Ξ_T(·)) = O_p(T), j = 1,2,3. Therefore, Kim tests (2000) would reject the null asymptotically with probability one.

Proof. To prove the theorem, we analyze Kim's tests under two different assumptions. First, let us assume that τ = λ, i.e., the fraction of the sample where the process presumably changes persistence equals the fraction of the sample where the structural break occurs. The residuals for each subsample can be written as

Then,

where (see Kim, Belaire-Franch, and Badillo Amador, 2002)

and where W denotes a standard Wiener process. This result directly follows from Theorem 3.1 in Kim (2000).

Next, let us assume that τ < λ < 1. Then, if we neglect the occurrence of a structural break, the residual series for each subsample is

for j = 1,…,[τT] and j = [τT] + 1,…,T, respectively. Let us compute the partial sum residuals processes for the numerator in (3).

where ε₁ and D₁ are the sample means of ε_t and D_t, respectively, computed for the second subsample. The first two terms are of order O_p(T^1/2), whereas the third term is of order O_p(T). Thus,

.

Again, the whole expression is asymptotically dominated by the deterministic terms involved. Hence,

, as before.

In both cases,

where W denotes a standard Wiener process. Then, the denominator of the Kim (2000) statistic is O_p(1). It follows that (Ξ_T(τ))⁻¹ = O_p(T⁻¹) and, hence, (H_j(Ξ_T(·)))⁻¹ = O_p(T⁻¹), j = 1,2,3.⁴

Therefore, in both cases, if the tests were actually computed in the set

, e.g.,

, the Kim (2000) H_j(Ξ_T(·)), j = 1,2,3, tests would reject the null asymptotically with probability one, unless the fraction of the sample where the structural change is located is equal to or smaller than the lower end of the prespecified range (i.e., unless λ ≤ τ₁).

Finally, let us assume that λ < τ < 1. The corresponding residual series are

for j = 1,…,[τT] and j = [τT] + 1,…,T, respectively. The computation of the partial sum residuals processes is established as previously. But under the present assumption,

for 0 < b < λ and λ < b < τ, whereas, in both cases,

Therefore, Ξ_T(τ) → 0 as T → ∞. Then, if λ < τ₁, the one-tailed tests H_j(Ξ_T(·)), j = 1,2,3 would have zero power, asymptotically.