_{1}

The unit root can lead to major problems in economic time series analyses. I obtain the asymptotic distributions of the ordinary least squares (OLS) estimator when the true model is trend stationary for the following three cases: 1) the null model is a random walk without drift, and the auxiliary regression model does not contain a constant; 2) the null model is a random walk with drift, and the auxiliary regression model contains a constant; and 3) the null model is a random walk with drift, and the auxiliary regression model contains both a constant and a time trend. In the third case, the asymptotic distribution of the OLS estimator is determined by the first order of the autocorrelation, and we can distinguish between the random walk and trend stationary models, unlike in previous studies. Based on these results, the real US gross domestic product is analyzed. A time trend model with autoregressive error terms is chosen. The results suggest that the impacts of a shock can become larger than the original shock in some periods and then gradually decline. However, the impacts continue for a long period, and policy makers should account for this to design better economic policies.

The unit root can lead to major problems in economic time series analyses. The Dickey-Fuller (DF) test [

In this paper, following Schmidt and Phillips [

Based on these results, I analyze the real US gross domestic product (GDP). The problem of whether GDP is a random walk or trend stationary has been studied by various authors [

In this section, I obtained the asymptotic distributions of the OLS estimator for the three cases when the true model is the trend stationary model, given by

y t = y 0 + α t + u t , u t = ε t + ϕ 1 ε t − 1 + ϕ 2 ε t − 2 + ⋯ + ϕ q ε t − q , t = 1 , 2 , ⋯ , T . (1)

where ε 1 , ε 2 , ⋯ , ε T are independent and identically distributed (i.i.d.) random variables following N ( 0 , σ 2 ) . Let m s = E ( u t u t − s ) = ∑ k = 1 q ϕ k ϕ k − s σ 2 where ϕ 0 = 1 and ϕ j = 0 if j < 0 . Then m 0 = V ( u t ) . The value of q may be infinite, but the following conditions are satisfied:

∑ s = 1 q | m s | s 2 < ∞ and (2)

∑ j = 1 q E ( ξ i ξ j ) < ∞ for any i, where ξ i = u i u i − 1 − m 1 .

These conditions are satisfied if q is finite or u t is a stationary AR process. It is also assumed that the fourth moment of u t is finite. Phillips and Perron [

I first consider the case in which the null model is a random walk without drift. Secondly, I analyze the case in which the null model is a random walk with drift and the auxiliary regression model contains a constant. Finally, I consider the case in which the auxiliary regression model contains both a constant and a time trend.

I consider the case in which the null hypothesis is the random walk model given by

H 0 : y t = y t − 1 + ε t , t = 1 , 2 , ⋯ , T . (3)

We consider the auxiliary regression model,

y t = ρ y t − 1 + ω t , y 0 = 0 and ω 0 = 0 , t = 1 , 2 , ⋯ , T . (4)

Long and Herrera [

ρ ^ = ∑ t = 1 T y t y t − 1 ∑ t = 1 T y t − 1 2 . (5)

Under the null hypothesis, it is well known that (Hamilton [ [

T ( ρ ^ − 1 ) → D [ W ( 1 ) ] 2 − 1 2 ∫ [ W ( r ) ] 2 and τ = ( ρ ^ − 1 ) / s . e . ( ρ ^ ) → D [ W ( 1 ) ] 2 − 1 2 ∫ [ W ( r ) ] 2 .(6)

where ∫ = ∫ 0 1 , W ( ⋅ ) represents standard Brownian motion, and s . e . ( ρ ^ ) is the standard error.

When the true model is trend stationary as given by (1), the asymptotic distribution of the OLS estimator is given by the following theorem.

Theorem 2.1

When (1) is the true model, the OLS estimator becomes

T ( ρ ˜ − 1 ) → P 3 2 . (7)

The proof is given in Appendix A1. Although the OLS estimator is given by the same formula, I use ρ ^ under the null (random walk) model and ρ ˜ for the true (trend stationary) model to distinguish the distributions of the estimator.

When we perform unit root tests, it may be more reasonable to consider the random walk model with drift.

Here, the null hypothesis is given by

H 0 : y t = α + y t − 1 + ε t , t = 1 , 2 , ⋯ , T . (8)

The auxiliary regression model is given by

y t = α + ρ y t − 1 + ω t , t = 1 , 2 , ⋯ , T . (9)

The regression model contains a constant term, and we do not need the initial condition, unlike in (4). As described above, note that when | ρ | < 1 , the model becomes trend stationary and included as a term in (1). The OLS estimator of ρ is given by

ρ ^ = ∑ t = 1 T ( y t − 1 − y ¯ 1 ) ( y t − y ¯ 2 ) ∑ t = 1 T ( y t − 1 − y ¯ 1 ) 2 , y ¯ 1 = ∑ t = 1 T y t − 1 T and y ¯ = ∑ t = 1 T y t T . (10)

Under the null hypothesis, the asymptotic distribution is given by (Hamilton [ [

T 3 / 2 ( ρ ^ − 1 ) → D N ( 0 , 12 m 0 / α 2 ) (11)

When the true model is the trend stationary model given by (1), the asymptotic distribution of the OLS estimator is given by the following theorem.

Theorem 2.2

When (1) is the true model, the OLS estimator (9) becomes

T 2 ( ρ ˜ − 1 ) → D N ( 12 α 2 m 0 , 36 α 2 m 0 ) . (12)

The proof is given in Appendix A2. Theorem 2.2 means that the DF test based on the OLS estimator has no asymptotic power against the trend stationary alternatives as suggested by the previous studies. West [

but ( ρ ^ − 1 ) = O P ( T − 3 2 ) and his derivation is incomplete. It is necessary to prove that ( ρ ˜ − 1 ) = o p ( T − 3 2 ) to show that the asymptotic power of the DF test

is zero, as in this paper. It is impossible to distinguish between the two hypotheses, i.e. “ ρ = 1 ” and trend stationary, based on the asymptotic distributions of the OLS estimator. Schmidt and Phillips [

Finally, I consider the case in which the null hypothesis is the random walk model with drift as given by (8), and the auxiliary regression model contains both drift and the time trend:

y t = β + ρ y t − 1 + δ t + ω t , y 0 = 0 and ω 0 = 0 , t = 1 , 2 , ⋯ , T . (13)

Harvey et al. [

y t = β + δ t + z t , z t = ρ z t + ε t . (14)

Models (13) and (14) become the same model if ρ = 1 . The asymptotic distribution of the OLS estimator under (8) is given by (Hamilton [ [

T ( ρ ^ − 1 ) → D F G , (15)

F = 1 3 ( c 1 − 1 2 c 3 ) d 1 + 1 8 d 2 + ( c 3 − 1 2 c 1 ) d 3 ,

G = 1 12 c 2 − 1 3 c 1 2 − c 3 2 + c 1 c 3 ,

c 1 = ∫ W ( r ) d r , c 2 = ∫ W ( r ) 2 d r , c 3 = ∫ r W ( r ) d r ,

d 1 = W ( 1 ) , d 2 = W ( 1 ) 2 − 1 , and d 3 = W ( 1 ) − ∫ W ( r ) d r

The asymptotic distribution of τ = ( ρ ^ − 1 ) / s . e . ( ρ ^ ) is given by Hamilton [ [

Theorem 2.3

ρ ˜ → P ρ 0 = m 1 m 0 = r 1 , (16)

where r 1 is the first order autocorrelation of the error terms. This means that ρ ˜ does not converge to 1, and we can distinguish the random walk with drift from the trend stationary model if the first order correlation of error terms is not equal to 1. Harvey et al. [ [

Note that if ρ = 1 and δ ≠ 0 , (13) becomes

Δ y t = β + δ t + ω t . (17)

This means that y t must be a quadratic function of the time trend, and the model obtained from (17) becomes different from a random walk with drift. As Schmidt and Phillips [ [_{0}: ρ = 1 and δ = 0 is also used for the unit root tests [

In the first case in which the null and auxiliary regression models are given by (3) and (4), the estimated results areas follow (the standard error is given in parentheses):

y t * = 1.00437 y t − 1 * , R 2 = 0.9997 , ( 0.000444 ) (18)

T ( ρ ^ − 1 ) = 1.2226 and τ = ( ρ ^ − 1 ) / s . e . ( ρ ^ ) = 9.8423 .

To remove the influence of the initial value, y t * = y t − y 0 is used in this case. First, H 0 : ρ = 1 against H 1 : ρ > 1 . (Since y t * is clearly an increasing function of t, the alternative becomes H 1 : ρ > 1 .) T ( ρ ^ − 1 ) = 1.223 does not exceed the 95% critical value of 1.28. However, τ exceeds the 99% critical value of 2.01, which suggests that ρ > 1 . Since the τ test has usually been used in previous studies, I use the τ test when the results of the two tests are different. The 95% confidence interval of ρ ^ based on the distribution of τ is [0.995,

1.011], and ρ 0 T + 1 = 1.5 280 + 1 = 1.0054 , obtained from Theorem 2.1, is included in this interval, so the trend stationary alternative is not rejected.

For the second case, the null and auxiliary regression models are given by (8) and (9). The estimation results are

y t = 0.03675 + 0.9967 y t − 1 , R 2 = 0.999796 , (19)

T 3 2 ( ρ ^ − 1 ) = − 0.9558 , V 1 = 12 m 0 / α 2 = 0.7203 and t = T 3 2 ( ρ ^ − 1 ) / V 1 = − 1.1262 .

Therefore, H 0 : ρ = 1 against H 1 : ρ < 1 is not rejected at the 5% level. (The 5% critical value of the standard normal distribution is −1.645). However, in the

model trend stationary, ρ ˜ − 1 = o P ( T − 3 2 ) as shown in Theorem 2.2, and we

cannot distinguish the random walk with drift model from the trend stationary model based on the distribution of ρ ^ − 1 . Therefore, the result may suggest that the model may be either a random walk with drift or trend stationary.

In the third case, the null, alternative and auxiliary regression models are given by (8) and (13). The estimation results are

y t = 0.1367 + 0.9840 y t − 1 + 0.000100 t , R 2 = 0.999798 , ( 0.0 6573 ) ( 0.00 8394 ) ( 0.0000 655 ) (20)

T ( ρ ^ − 1 ) = − 4.485 and τ = ( ρ ^ − 1 ) / s . e . ( ρ ^ ) = − 1.9083 .

The 5% critical values of the unit root tests based T ( ρ ^ − 1 ) and τ are −21.3 and −3.43, respectively, and we cannot reject H 0 : ρ = 1 against H 1 : ρ < 1 at the 5% level. However, the F-value of H 0 : ρ = 1 and δ = 0 is 8.3996, which is bigger than the 5% critical value of the F-test, 6.34, and the null hypothesis is rejected at the 5% level. The following are the results of the DF tests for the first and second cases for Δ y t .

Δ y t * = − 0.37149 Δ y t − 1 * , ( 0.0 46532 ) (21)

T ( ρ ^ − 1 ) = − 104.0 , τ = ( ρ ^ − 1 ) / s . e . ( ρ ^ ) = − 7.9837 ,

Δ y t = 0.005113 + 0.35586 Δ y t − 1 , R 2 = 0.1284 ,

T 3 2 ( ρ ^ − 1 ) = − 23.956 , V 1 = 12 m 0 / α 2 = 33.90 and t = T 3 2 ( ρ ^ − 1 ) / V 1 = − 4.114

In these cases, not only the null hypotheses of the unit root but also the trend stationary model are rejected with any reasonable significance level. These results suggest that the process cannot have a quadric trend. In other words, if ρ = 1 , δ must be 0. However, this contradicts the results of the F-test.

Assuming that ρ < 1 in (13), we obtain the trend stationary model. The estimates of the trend model are given by

y t = 7.8400 + 0.007738 t , R 2 = 0.9898 , ( 0.00 76 0 ) ( 0.00000 472 ) (22)

Note that, from Theorem 2.3, we get ρ ˜ → P ρ 0 = 0.9896 , very close to 1. Let e t be the residuals of Equation (22) (detrended values of y t ).

e t = γ 1 e t − 1 + γ 2 Δ e t − 1 = − 0.01813 e t − 1 + 0.3625 Δ e t − 1 , R 2 = 0.9821 , ( 0.0 5543 ) ( 0.0 5527 ) (23)

τ = ( γ ^ 1 − 1 ) / s . e . ( γ ^ 1 ) = − 2.2688.

The 5% critical value of τ is −1.95, and we can reject the null hypothesis of the unit root for the detrended process; the trend stationary assumption is accepted. The result is very close to the unit root process, and it is consistent with previous studies; some found a unit root and others did not. When the process is trend stationary, the estimator of α becomes a supper efficient estimator, and we can use the ADF test for the residuals.

Since the trend stationary model was supported in the previous section, we combine (22) and (23) and consider the model given by:

y ^ t = 7.8400 + 0.007738 t + 1.3443 e t − 1 − 0.3625 e t − 2 (24)

R^{2} obtained from y ^ t is 0.99979. The very large R^{2} indicates that y ^ t fits y t very well.

Using (24), we can simulate an economic shock to the economy in a given period (for example, the Lehman shock in 2008).

In this paper, I considered the asymptotic properties of the OLS estimator when the true model is trend stationary. The power of the DF depends on the null and auxiliary regression models. I considered the following three cases: 1) the null model is a random walk without drift and the auxiliary regression model does

not contain a constant; 2) the null model is a random walk with drift and the auxiliary regression model contains a constant; and 3) the null model is a random walk with drift and the auxiliary regression model contains both a constant and a time trend. Although the DF test has no asymptotic power for the second case, we can distinguish between the random walk and trend stationary models in first and third cases in contrast to what was suggested in previous studies. The asymptotic distribution of the OLS estimator depends on the first order of the autocorrelation of error terms in the third case.

Then I analyzed the real US GDP for 280 quarters or 70 years. The trend stationary model with AR (2) errors fits very well and gives a very large R^{2} value. However, the result is very close to the unit root process and is consistent with those of previous studies, some of which found a unit root while others did not. Then the impacts of the economic shock (such as the Lehman stock) were evaluated. The impacts were found to continue for a long period and to take 40 quarters or 10 years to decrease to half the original impact.

In this study, only the real US GDP was analyzed. Economic variables such as the GDP of various countries will be studied in the future to determine whether they follow the random walk model or trend model.

The author declares no conflicts of interest regarding the publication of this paper.

Nawata, K. (2021) The Random Walk and Trend Stationary Models with an Analysis of the US Real GDP: Can We Distinguish between the Two Models? Open Journal of Statistics, 11, 213-229. https://doi.org/10.4236/ojs.2021.111011

When the alternative model given by (1) is the true one, the OLS estimator becomes

ρ ˜ = ∑ t = 1 T y t y t − 1 ∑ t = 1 T y t − 1 2 = 1 + ∑ t = 1 T y t − 1 ( y t − y t − 1 ) ∑ t = 1 T y t − 1 2 = 1 + ∑ t = 1 T { α ( t − 1 ) + u t − 1 } ( α + u t − u t − 1 ) ∑ t = 1 T { α ( t − 1 ) + u t − 1 } 2 and (A.1)

ρ ˜ − 1 = ∑ t = 1 T { α ( t − 1 ) + u t − 1 } ( α + u t − u t − 1 ) ∑ t = 1 T { α ( t − 1 ) + u t − 1 } 2 = A T B T .

For the numerator,

A T = ∑ t = 1 T α 2 ( t − 1 ) + α ∑ t = 1 T u t − 1 + ∑ t = 1 T α ( t − 1 ) ( u t − u t − 1 ) + ∑ t = 1 T u t − 1 u t − ∑ t = 1 T u t − 1 2 ≡ α 2 T ( T − 1 ) 2 + A 1 + A 2 + A 3 + A 4 . (A.2)

It is easy to show that

A 1 = O p ( T ) and A 4 = O p ( T ) . (A.3)

Here,

A 2 / α = ( u 2 − u 1 ) + 2 ( u 3 − u 2 ) + ⋯ + ( T − 2 ) ( u T − 1 − u T − 2 ) + ( T − 1 ) ( u T − u T − 1 ) = − ( u 1 + u 2 + ⋯ + u t − 1 ) + ( T − 1 ) u T = ( T − 1 ) u T + O p ( T 1 2 ) = O p ( T ) . (A.4)

Since E ( A 3 ) = σ 2 m 1 T and the fourth moment of u t is finite,

A 3 = O P ( T ) . (A.5)

From (A.2)-(A.5),

A T T 2 → P α 2 2 . (A.6)

For the denominator,

B T = ∑ t = 1 T { α 2 ( t − 1 ) 2 + 2 α ( t − 1 ) u t − 1 + u t − 1 2 } = α 2 T ( T − 1 ) ( 2 T − 1 ) 6 + 2 ∑ t = 1 T α ( t − 1 ) u t − 1 + ∑ t = 1 T u t − 1 2 ≡ α 2 T ( T − 1 ) ( 2 T − 1 ) 6 + B 1 − A 4 . (A.7)

Here,

E { ∑ t = 1 T ( t − 1 ) u t − 1 } = 0 and ∑ s = 1 ∞ | m s | s n < ∞ , (A.8)

and

E { ∑ t = 1 T ( t − 1 ) u t − 1 } 2 = ∑ t = 1 T ( t − 1 ) 2 + 2 ∑ i = 1 T ∑ j = i + 1 T ( i − 1 ) ( j − 1 ) E ( u i − 1 u j − 1 ) = m 0 ∑ t = 1 T ( t − 1 ) 2 + ∑ i = 1 T ∑ j = i + 1 T ( i − 1 ) ( i − 1 + j − i ) m j − i = m 0 ∑ t = 1 T ( t − 1 ) 2 + ∑ i = 1 T ( i − 1 ) 2 ∑ s = 1 T − i m s + ∑ t = 1 T ( i − 1 ) ∑ s = 1 T − i s m s < m 0 T ( T − 1 ) ( 2 T − 1 ) 6 + φ 1 T ( T − 1 ) ( 2 T − 1 ) 6 + φ 2 T ( T − 1 ) 2 = O ( T 3 ) , (A.9)

where φ 1 = ∑ s = 1 ∞ | m s | and φ 2 = ∑ s = 1 ∞ | s m s |

From (A.8) and (A.9), we get

B 1 = O p ( T 3 / 2 ) . (A.10)

From (A.3), (A.7), and (A.10),

B T T 3 → P α 2 3 . (A.11)

From (A.1), (A.7), (A.11), and Slutsky’s theorem,

T ( ρ ˜ − 1 ) → P 3 2 . (A.12)

Here, y t − y t − 1 = α + u t − u t − 1 and the OLS estimator of ρ becomes

ρ ˜ = ∑ t = 1 T ( y t − 1 − y ¯ 1 ) ( y t − y ¯ ) ∑ t = 1 T ( y t − 1 − y ¯ 1 ) 2 , (A.13)

ρ ˜ − 1 = ∑ t = 1 T ( y t − 1 − y ¯ 1 ) ( y t − y ¯ − y t − 1 + y ¯ 1 ) ∑ t = 1 T ( y t − 1 − y ¯ 1 ) 2 = ∑ t = 1 T ( y t − 1 − y ¯ 1 ) ( y t − y t − 1 + y ¯ 1 − y ¯ 2 ) ∑ t = 1 T ( y t − 1 − y ¯ 1 ) 2 = ∑ t = 1 T ( y t − 1 − y ¯ 1 ) ( α + u t − u t − 1 + y ¯ 1 − y ¯ ) ∑ t = 1 T ( y t − 1 − y ¯ 1 ) 2 ≡ C T D T ,

y ¯ = 1 T ∑ t = 1 T y t , and y ¯ 1 = 1 T ∑ t = 1 T y t − 1 .

For the numerator, since ∑ t = 1 T ( y t − 1 − y ¯ 1 ) = 0 and y ¯ 1 − y ¯ = y T T , we get

C T = ∑ t = 1 T ( y t − 1 − y ¯ 1 ) ( u t − u t − 1 ) = ∑ t = 1 T y t − 1 ( u t − u t − 1 ) + y ¯ 1 ∑ t = 1 T ( u t − u t − 1 ) = ∑ t = 1 T y t − 1 ( u t − u t − 1 ) + y ¯ 1 u T

= ∑ t = 1 T ( α ( t − 1 ) + u t − 1 ) ( u t − u t − 1 ) + y ¯ 1 u T = ∑ t = 1 T α ( t − 1 ) ( u t − u t − 1 ) + ∑ t = 1 T u t − 1 u t − ∑ t = 1 T u t − 1 2 + y ¯ 1 u T = A 2 + A 3 + A 4 + y ¯ 1 u T . (A.14)

As before, A 2 = O p ( T ) , A 3 = O P ( T ) and A 4 = O p ( T ) .

Here,

y ¯ 1 = 1 T ∑ t = 1 T α ( t − 1 ) + 1 T ∑ t = 1 T u t = α T − 1 2 + 1 T ∑ t = 1 T u t = α T − 1 2 + O p ( T − 1 2 ) , (A.15)

and y ¯ 1 u T = O p ( T ) .

From (A.14) and (A.15),

C T T P → P α 2 u T + m 0 . (A.16)

For the denominator,

D T = ∑ t = 1 T y t − 1 2 − T y ¯ 1 2 ≡ D 1 + D 2 . (A.17)

From (A.10)

D 1 = ∑ t = 1 T { α ( t − 1 ) + u t − 1 } 2 = α 2 ∑ t = 1 T ( t − 1 ) 2 + 2 ∑ t = 1 T ( t − 1 ) u t − 1 + ∑ t = 1 T u t − 1 2 = α 2 T ( T − 1 ) ( 2 T − 1 ) 6 + O P ( T 3 / 2 ) . (A.18)

From (A.18),

D 2 = − α 2 T ( T − 1 ) 2 4 + O P ( T 1 2 ) . (A.19)

From (A.17)-(A.19),

D T = α 2 T 3 12 + O P ( T 2 ) . (A.20)

From (A13), (A16) and (A.20), we get

T 2 ( ρ ˜ − 1 ) → p 6 α u T + 12 α 2 m 0 and T 2 ( ρ ˜ − 1 ) → D N ( 12 α 2 m 0 , 36 α 2 m 0 ) . (A.21)

A.3 Proof of Theorem 2.3

The OLS estimator of ρ is equivalent to the OLS estimator of

y t * = ρ y t − 1 + + δ t * + u t * (A.22)

y t * = y t − y ¯ , y t − 1 + = y t − 1 − y ¯ 1 , t * = t − T ¯ , T ¯ = 1 T ∑ t = 1 T t = T + 1 2 ,

u t * = u t − u ¯ and u ¯ = 1 T ∑ t = 1 T u t .

Let

y t * = a t * + v t . (A.23)

Then the OLS estimator of a is given by

a ^ = ∑ t = 1 T t * y t * ∑ t T t * 2 = ∑ t = 1 T t * y t ∑ t T t * 2 = ∑ t = 1 T t * ( α t + u t ) ∑ t T t * 2 = α + ∑ t = 1 T t * u t ∑ t T t * 2 (A.24)

ς = a ^ − α ~ N ( 0 , m 0 ∑ t = 1 T t * 2 ) , ∑ t = 1 T t * 2 = T ( T + 1 ) ( T − 1 ) 12 , and

T 3 2 ς D → D N ( 0 , 12 m 0 ) .

In the same way, let

y t − 1 + = b t * + w t . (A.25)

Then the OLS estimator of is given by

b ^ = ∑ t = 1 T t * y t + ∑ t T t * 2 = ∑ t = 1 T t * y t − 1 ∑ t T t * 2 = ∑ t = 1 T t * { α ( t − 1 ) + u t − 1 } ∑ t T t * 2 = α + − ∑ t = 1 T t * α + ∑ t = 1 T t * u t − 1 ∑ t T t * 2 , (A.26)

η = b ^ − α = − ∑ t = 1 T t * α + ∑ t = 1 T t * u t − 1 ∑ t T t * 2 ,

E ( η ) = E ( b ^ ) − α = − α ∑ t = 1 T t * ∑ t = 1 T t * 2 = 0 , and

V ( η ) = σ 2 ∑ t = 2 T t * 2 { ∑ t = 1 T t * 2 } 2 = T − 3 12 + O ( T − 4 ) , η = O p ( T − 3 / 2 ) .

Let

v ^ t = y t * − a ^ t * = y t − y ¯ − ( α + ς ) ( t − T ¯ ) = u t − u ¯ − ζ t * , and (A.27)

w ^ t = y t − 1 + − b ^ t * = y t − 1 − y ¯ 1 − b ^ t * = α ( t − 1 ) + u t − 1 − α T ¯ 1 − u ¯ 1 − ( α + η ) ( t − T ¯ ) = − α + u t − 1 + α ( T ¯ − T ¯ 1 ) − u ¯ 1 − η ( t − T ¯ ) = u t − 1 − u ¯ 1 − η t * .

Then the OLS estimator of ρ in (A.22) becomes

ρ ˜ = ∑ t = 1 T v ^ t w ^ t ∑ t = 1 T w ^ 2 ≡ H T I T . (A.28)

For the numerator,

H T = ∑ t = 1 T v ^ t w ^ t = ∑ t = 1 T ( u t − u ¯ − ζ t * ) ( u t − 1 − u ¯ 1 − η t * ) . (A.29)

Since u ¯ = O p ( T − 1 2 ) , u ¯ 1 = O p ( T − 1 2 ) and ζ = O p ( T − 3 / 2 ) ,

H T = ∑ t = 1 T v ^ t w ^ t = ∑ t = 1 T ( u t u t − 1 ) − T η ∑ t = 1 T 1 T t * u t + O p ( T 1 / 2 ) , and (A.30)

T − 1 H T → D T − 1 ∑ t = 1 T ( u t u t − 1 + 3 α 2 λ t u t ) , λ t = t * T .

From the assumption of (2) V ( ∑ t = 1 T u t u t − 1 ) = O ( T ) . Therefore,

T − 1 ∑ t = 1 T u t u t − 1 → P m 1 and T − 1 ∑ t = 1 T u t λ t → P 0 . (A.31)

For the denominator,

I T = ∑ t = 1 T w ^ 2 = ∑ t = 1 T ( u t − 1 − u ¯ 1 − η t * ) 2 = ∑ t = 1 T u t − 1 2 + η 2 ∑ t = 1 T t * 2 + T u ¯ 1 2 − 2 η ∑ t = 1 T t * u t − 1 − 2 u ¯ 1 ∑ t = 1 T u t − 1 + 2 η u ¯ 1 ∑ t = 1 T t * . (A.32)

Since ∑ t = 1 T t * u t − 1 = O p ( T 3 2 ) , η = O p ( T − 3 / 2 ) , u ¯ 1 = O p ( T − 1 2 ) and η = O p ( T − 1 ) ,

I T = ∑ t = 1 T u t − 1 2 + O P ( T 1 2 ) . (A.33)

Since ∑ t = 1 T u t − 1 2 / T → P m 0 ,

I T T → P m 0 . (A.34)

From (A.28), (A.31), and (A.34), we get

ρ ˜ → P m 1 m 0 = r 1 . (A.35)