In this section, we give the proofs of the theorems.

*Proof of Theorem 4.1* By (3.1) we have

Note that for all |\alpha |<1,

Let {u}_{t}={\epsilon}_{t}+({\theta}_{t}^{r}-{\theta}_{t}^{l})/3, then

\begin{array}{rl}{W}_{t}& ={x}_{t}+({\xi}_{t}^{r}-{\xi}_{t}^{l})/3\\ =\alpha {x}_{t-1}+{\epsilon}_{t}+[\alpha ({\xi}_{t-1}^{r}-{\xi}_{t-1}^{l})+({\theta}_{t}^{r}-{\theta}_{t}^{l})]/3\\ =\alpha {W}_{t-1}+{u}_{t}.\end{array}

This means that

\begin{array}{rcl}{M}_{T}& =& \sum _{t=1}^{T}{W}_{t}{W}_{t-1}-\alpha \sum _{t=1}^{T}{W}_{t-1}^{2}\\ =& \sum _{t=2}^{T}{u}_{t}{u}_{t-1}+\alpha \sum _{t=3}^{T}{u}_{t}{u}_{t-2}+\cdots +{\alpha}^{T-2}{u}_{T}{u}_{1}+{W}_{0}\sum _{t=1}^{T}{\alpha}^{t-1}{u}_{t}.\end{array}

(5.1)

Because the last term has mean 0 and variance {W}_{0}^{2}{\sigma}^{2}(1-{\alpha}^{2T})/(1-{\alpha}^{2}), using Chebyshev’s inequality, we have

\left({W}_{0}\sum _{t=1}^{T}{\alpha}^{t-1}{u}_{t}\right)/\sqrt{T}\stackrel{p}{\to}0,\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

(5.2)

Let

Since {H}_{T} is a linear combination of terms {u}_{t}{u}_{s} (t\ne s), and each term is uncorrelated with other terms, we have

In what follows, we prove that

{Z}_{T,S}\stackrel{L}{\to}N(0,{\sigma}^{4}(1-{\alpha}^{2(S+1)})/(1-{\alpha}^{2})),\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

(5.5)

Further, let

{Z}_{T,S}^{\ast}=\left(\sum _{t=S+2}^{T}[{u}_{t}{u}_{t-1}+\alpha {u}_{t}{u}_{t-2}+\cdots +{\alpha}^{S}{u}_{t}{u}_{t-S-1}]\right)/\sqrt{T}.

Note that {Z}_{T,S}^{\ast} and {Z}_{T,S} have the same limiting distribution as T\to \mathrm{\infty}. Next, we show that

{Z}_{T,S}^{\ast}\stackrel{L}{\to}N(0,{\sigma}^{4}(1-{\alpha}^{2(S+1)})/(1-{\alpha}^{2})),\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

Now, let

{y}_{t}={u}_{t}{u}_{t-1}+\alpha {u}_{t}{u}_{t-2}+\cdots +{\alpha}^{S}{u}_{t}{u}_{t-S-1}.

Then

Thus, combining Lemma 4.1, we prove (5.5). The theorem follows from (5.1)-(5.5) and Lemma 4.2. □

*Proof of Theorem 4.2*

Note that

\begin{array}{rcl}{B}_{T}& =& \sum _{t=1}^{T}{W}_{t-1}^{2}\\ =& {W}_{0}^{2}+{({u}_{1}+\alpha {W}_{0})}^{2}+\cdots +{({u}_{T-1}+\alpha {u}_{T-2}+\cdots +{\alpha}^{T-1}{W}_{0})}^{2}\\ =& [{u}_{1}^{2}(1+{\alpha}^{2}+\cdots +{\alpha}^{2(T-2)})+\cdots +{u}_{T-1}^{2}]\\ +2[\alpha ({u}_{2}{u}_{1}+\cdots +{u}_{T-1}{u}_{T-2})+\cdots +{\alpha}^{T-2}{u}_{T-1}{u}_{1}]\\ +2{W}_{0}[{u}_{1}(\alpha +{\alpha}^{3}+\cdots +{\alpha}^{2T-3})+\cdots +{\alpha}^{T-1}{u}_{T-1}]\\ +{W}_{0}^{2}[1+{\alpha}^{2}+\cdots +{\alpha}^{2(T-1)}]\\ =& {B}_{{T}_{1}}+{B}_{{T}_{2}}+{B}_{{T}_{3}}+{B}_{{T}_{4}}.\end{array}

It is easy to see that

{B}_{{T}_{4}}/T\to 0,\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

(5.6)

Because E({B}_{{T}_{3}}/T)=0 and E{({B}_{{T}_{3}}/T)}^{2}\le 4{W}_{0}^{2}{\sigma}^{2}{\alpha}^{2}/(T{(1-{\alpha}^{2})}^{2}), we have

{B}_{{T}_{3}}/T\stackrel{p}{\to}0,\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

(5.7)

Further, from

similarly, we get

{B}_{{T}_{2}}/T\stackrel{p}{\to}0,\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

(5.8)

Lastly, we will show that {B}_{{T}_{1}}/T\stackrel{p}{\to}{\sigma}^{2}/(1-{\alpha}^{2}). Observe that

\begin{array}{rl}\frac{1}{1-{\alpha}^{2}}\sum _{t=1}^{T-1}{u}_{t}^{2}-{B}_{{T}_{1}}& ={u}_{1}^{2}({\alpha}^{2(T-1)}+{\alpha}^{2T}+\cdots )+\cdots +{u}_{T-1}^{2}({\alpha}^{2}+{\alpha}^{4}+\cdots )\\ ={u}_{1}^{2}{\alpha}^{2(T-1)}/(1-{\alpha}^{2})+\cdots +{\alpha}^{2}{u}_{T-1}^{2}/(1-{\alpha}^{2}).\end{array}

Since this is a nonnegative random variable with expected value

\left({\alpha}^{2}{\sigma}^{2}(1-{\alpha}^{2(T-1)})\right)/{(1-{\alpha}^{2})}^{2},

using Markov’s inequality, we have

(\frac{1}{1-{\alpha}^{2}}\sum _{t=1}^{T-1}{u}_{t}^{2}-{B}_{{T}_{1}})/T\stackrel{p}{\to}0,\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

(5.9)

Further, by the law of large numbers, we get

\left(\frac{1}{1-{\alpha}^{2}}\sum _{t=1}^{T-1}{u}_{t}^{2}\right)/T\stackrel{p}{\to}{\sigma}^{2}/(1-{\alpha}^{2}),\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

Then it holds that

{B}_{{T}_{1}}/T\stackrel{p}{\to}{\sigma}^{2}/(1-{\alpha}^{2})\phantom{\rule{1em}{0ex}}\text{as}T\to \mathrm{\infty}.

(5.10)

This, together with (5.6)-(5.9), completes the proof. □

*Proof of Theorem 4.3*

Note that

\sqrt{T}(\stackrel{\u02c6}{\alpha}-\alpha )=({M}_{T}/\sqrt{T})/({B}_{T}/T).

With the application of Slusky’s theorem and Theorems 4.1 and 4.2, we prove Theorem 4.3. □