Now we investigate the Yule–Walker estimators for the RrRCINAR(1) model. Since the marginal distribution of the RrRCINAR(1) model varies at different circumstances, we can not use the Yule–Walker estimation method directly. Therefore, in order to use the Yule–Walker estimation methods, we set a sample belonging to the same environment state.
Let us assume that the data that are in the same cluster are observed under the same state. Select a sample of size N in model (2), \({X_{1}}({z_{1}}),{X_{2}}({z_{2}}), \ldots,{X_{N}}({z_{N}})\). When the sample corresponds to the environment \(k \in{E_{r}}\), then \(\exists i,n \in\mathbb{N}\), \({z_{i}} \ne k\), \({z_{i + 1}} = {z_{i + 2}} = \cdots = {z_{n}} = k\), \({z_{n + 1}} \ne k\), we call \({X_{i + 1}}(k),{X_{i + 2}}(k), \ldots,{X_{n}}(k)\) a subsample if \({X_{i}}(j),{X_{i + 1}}(k),{X_{i + 2}}(k), \ldots,{X_{n}}(k), {X_{n + 1}}(l)\), \(j \ne k\), \(l \ne k\), \(j, k, l \in{E_{r}}\).
The sample \({X_{1}}({z_{1}}),{X_{2}}({z_{2}}), \ldots,{X_{N}}({z_{N}})\) can be partitioned into subsample with different states, for \(k \in\{ 1,2, \ldots,r\} \), let
$${I_{k}} = \bigl\{ i \in\{ 1,2, \ldots,N\}| {z_{i}} = k \bigr\} $$
be the subscript of the sample \({X_{1}}({z_{1}}),{X_{2}}({z_{2}}),\ldots ,{X_{N}}({z_{N}})\) corresponding to the environment k.
Let \({n_{k}}\) be the number of the sample \({X_{1}}({z_{1}}),{X_{2}}({z_{2}}), \ldots,{X_{N}}({z_{N}})\) in the circumstance k, then we have
$$\bigcup_{k = 1}^{r} {{I_{k}}} = \{ 1,2 ,\ldots,N\},\quad | {I_{k}} | = {n_{k}}, {n_{1}} + {n_{2}} + \cdots+{n_{r}} = N. $$
Let
$${U_{k}} = \bigl({X_{{k_{1}}}}(k),{X_{{k_{2}}}}(k), \ldots ,{X_{{k_{{n_{k}}}}}}(k) \bigr),\quad {k_{i}} \in{I_{k}}, {k_{i}} < {k_{i + 1}}, \forall i \in\{ 1,2, \ldots,{n_{k}} - 1\}, $$
it represents a set of elements in the state k. Subsamples are denoted as \({U_{k,1}},{U_{k,2}}, \ldots,{U_{k,{i_{k}}}}\), where \(\{ {1,2,}\ldots,{i_{k}}\}\) represents the order of subsamples in the state k.
Similar to Zhang [12], the Yule–Walker estimation of the sample mean \({\hat{\mu}_{k,l}}\), the sample variance \(\hat{\gamma}_{0,l}^{(k)}\), and the first-order sample covariance \(\hat{\gamma}_{{1,}l}^{(k)}\) of the set \({U_{k,l}}\) are given by
$$\begin{gathered} {\hat{\mu}_{k,l}} = \frac{1}{{{n_{k,l}}}}\sum_{i \in {J_{k,l}}} {{X_{i}}(k)}, \\ \hat{\gamma}_{0,l}^{ ( k )} = \frac{1}{{{n_{k,l}}}}\sum _{i \in{J_{k,l}}} {{{ \bigl({X_{i}} ( k ) - {{ \hat{\mu}}_{k,l}} \bigr)}^{2}}} , \\ \hat{\gamma}_{{1},l}^{(k)} = \frac{1}{{{n_{k,l}}}}\sum _{\{ i,i + 1\} \subseteq{J_{k,l}}} { \bigl({X_{i + 1}}(k) - {{\hat{\mu}}_{k,l}} \bigr) \bigl({X_{i}}(k) - {{\hat{\mu}}_{k,l}} \bigr)} , \end{gathered}$$
where \({J_{k,l}} = \{ i \in\{ 1,2, \ldots,N\} |{X_{i}}({z_{i}}) \in {U_{k,l}}\} \), \(|{J_{k,l}}| = {n_{k,l}}\), \({n_{k,1}} + {n_{k,2}} + \cdots+{n_{k,{i_{k}}}} = {n_{k}}\). By inequality of \(0 < {\alpha^{2}} + \sigma_{\alpha}^{2} < 1\), we can get that estimators \({\hat{\mu}_{k,l}}\), \(\hat{\gamma}_{0,l}^{(k)}\), and \(\hat{\gamma}_{{1,}l}^{(k)}\) are strongly consistent.
We can obtain Yule–Walker estimators \({\hat{\mu}_{k}}\), \(\hat{\gamma}_{0}^{(k)}\), and \(\hat{\gamma}_{1}^{(k)}\) of the set \({U_{k}}\) in the same way, these estimators are defined as follows:
$$\begin{gathered} {\hat{\mu}_{k}} = \frac{1}{{{n_{k}}}}\sum_{i \in{I_{k}}} {{X_{i}}(k)} , \\ \hat{\gamma}_{0}^{(k)} = \frac{1}{{{n_{k}}}}\sum _{i \in{I_{k}}} {{{ \bigl({X_{i}}(k) - {{\hat{\mu}}_{k}} \bigr)}^{2}}}, \\ \hat{\gamma}_{1}^{(k)} = \frac{1}{{{n_{k}}}}\sum _{\{ i,i + 1\} \subseteq{I_{k}}} { \bigl({X_{i + 1}}(k) - {{\hat{\mu}}_{k}} \bigr) \bigl({X_{i}}(k) - {{\hat{\mu}}_{k}} \bigr)} .\end{gathered} $$
Next, we prove the strong consistency of estimators \({\hat{\mu}_{k}}\), \(\hat{\gamma}_{0}^{(k)}\), and \(\hat{\gamma}_{1}^{(k)}\).
Theorem 2
The estimators
\({\hat{\mu}_{k}}\), \(\hat{\gamma}_{0}^{(k)}\), and
\(\hat{\gamma}_{1}^{(k)}\)under the circumstancekare strongly consistent.
Proof
First, let us prove that the estimator \({\hat{\mu}_{k}}\) is strongly consistent, that is, \(P({\hat{\mu}_{k}} \to{\mu_{k}}, {n_{k}} \to \infty) = 1\). We assume that
$$\begin{gathered} {n_{k,l}} \to\infty,\quad l \in \{ {1,2, \ldots,d} \}, \\ {n_{k,j}} \to{c_{j}} < \infty,\quad j \in \{ {d + 1,d + 2, \ldots ,{i_{k}}} \},\end{gathered} $$
where \({n_{k}} = {n_{k,1}} + {n_{k,2}} + \cdots + {n_{k,{i_{k}}}}\), when \({n_{k}} \to\infty\).
It holds that
$$\begin{aligned} {{\hat{\mu}}_{k}} &= \frac{1}{{{n_{k}}}}\sum _{i \in{I_{k}}} {{X_{i}}(k)} = \frac{1}{{{n_{k}}}} \sum_{l = 1}^{{i_{k}}} {\sum _{i \in {J_{k,l}}} {{X_{i}}(k)} } = \sum _{l = 1}^{{i_{k}}} {\frac {{{n_{k,l}}}}{{{n_{k}}}}\frac{{1}}{{{n_{k.l}}}}\sum _{i \in {J_{k,l}}} {{X_{i}}(k)} } \\ &= \sum_{l = 1}^{{i_{k}}} {\frac{{{n_{k,l}}}}{{{n_{k}}}}{{\hat{\mu}}_{k,l}}}= \sum_{l = 1}^{d} { \frac{{{n_{k,l}}}}{{{n_{k}}}}} {{\hat{\mu}}_{k,l}} + \sum _{l = d + 1}^{{i_{k}}} {\frac {{{n_{k,l}}}}{{{n_{k}}}}} {{\hat{\mu}}_{k,l}} , \end{aligned} $$
because \(\lim _{{n_{k}} \to\infty} \frac {{{n_{k,j}}}}{{{n_{k}}}} = 0\), \(j \in \{ {d + 1,d + 2, \ldots,{i_{k}}} \}\), we can rewrite
$${{\hat{\mu}}_{k}}=\sum_{l = 1}^{d} { \frac{{{n_{k,l}}}}{{{n_{k}}}}} {{\hat{\mu}}_{k,l}}. $$
Estimator \({\hat{\mu}_{k,l}}\) is strongly consistent, that is,
$$P \bigl({\hat{\mu}_{k,l}} \to{\mu_{k}},{n_{k,l}} \to \infty,\forall l \in\{ 1,2, \ldots,d\} \bigr)= 1. $$
The sum \(\sum_{j = d + 1}^{{i_{k}}} {{n_{k,j}}} \to\sum_{j = d + 1}^{{i_{k}}} {{c_{j}}} \) is negligible compared with \({n_{k,l}} \to \infty\), so we can rewrite \({n_{k}}\) as \({n_{k}} = {n_{k,1}} + {n_{k,2}} + \cdots + {n_{k,d}}\). This implies
$$\begin{aligned} {{\hat{\mu}}_{k}} &= \sum _{l = 1}^{d} {\frac{{{n_{k,l}}}}{{{n_{k}}}}} {{\hat{\mu}}_{k,l}} = \sum_{l = 1}^{d} { \frac {{{n_{k,l}}}}{{{n_{k}}}}} \bigl({\mu_{k}} + o({n_{k,l}})\bigr) = \sum _{l = 1}^{d} {\frac{{{n_{k,l}}}}{{{n_{k}}}}{ \mu_{k}} + } \sum_{l = 1}^{d} { \frac{{{n_{k,l}}}}{{{n_{k}}}}o({n_{k,l}})} \\ &= {\mu_{k}}\frac{{\sum_{l = 1}^{d} {{n_{k,l}}} }}{{{n_{k}}}} + \sum_{l = 1}^{d} {\frac{{{n_{k,l}}}}{{{n_{k}}}}o({n_{k,l}})} = {\mu _{k}} + \sum _{l = 1}^{d} {\frac{{{n_{k,l}}}}{{{n_{k}}}}o({n_{k,l}})} \\ & = {\mu_{k}} + \sum_{l = 1}^{d} { \frac {{{n_{k,l}}}}{{{n_{k}}}}o({n_{k,l}})} , \end{aligned} $$
where \(\lim_{{n_{k,l \to\infty,\forall l \in\{ 1,2, \ldots,d\} }}}\frac{{{n_{k,l}}}}{{{n_{k}}}} < \infty\). Therefore, \({{\hat{\mu}}_{k}} \to{\mu_{k}}\), \({n_{k,l}} \to\infty\), \(\forall l \in\{ 1,2, \ldots,d\}\).
According to the assumptions, we can get
$${{n_{k}} \to\infty}\quad \Leftrightarrow\quad {{n_{k,l}} \to\infty,\quad\forall l \in \{ {1,2, \ldots,d} \}}, $$
and then
$$\lim _{{n_{k,l}} \to\infty,\forall l \in \{ {1,2, \ldots,d} \}} {\hat{\mu}_{k}}= \lim _{{n_{k}} \to\infty} {\hat{\mu}_{k}} ={\mu_{k}}. $$
Here, we complete the proof that the estimator \({\hat{\mu}_{k}}\) is strongly consistent. The proofs of estimators \(\hat{\gamma}_{0}^{(k)}\) and \(\hat{\gamma}_{{1}}^{(k)}\) are similar to the proof of estimator \({\hat{\mu}_{k}}\). □
The parameter estimator α̂ of α can be expressed as
$$\hat{\alpha}= \sum_{k = 1}^{r} { \frac{{{n_{k}}}}{N}} { \hat{\alpha}_{k}}, $$
where
$${\hat{\alpha}_{k}} = \frac{{\hat{\gamma}_{1}^{(k)}}}{{\hat{\gamma}_{0}^{(k)}}}. $$
Theorem 3
The estimatorsα̂and
\({\hat{\alpha}_{k}}\)are strongly consistent.
Proof
It is easy to get the strong consistency of estimator \({\hat{\alpha}_{k}}\) by the strong consistency of estimators \(\hat{\gamma}_{0}^{(k)}\) and \(\hat{\gamma}_{{1}}^{(k)}\).
Next, we prove the strong consistency of estimator \(\hat{\alpha}= \sum_{k = 1}^{r} {\frac{{{n_{k}}}}{N}} {\hat{\alpha}_{k}}\), when \({{n_{k}} \to\infty}\). Because \({\hat{\alpha}_{k}}\) is strongly consistent and \(( {N \to\infty} ) \Leftrightarrow ( {{n_{k}} \to \infty,\forall k \in\{ 1,2, \ldots,r\} } )\), it is obvious to prove that α̂ is strongly consistent by using the subadditivity of the probability. □
The remaining estimator \({\hat{\mu}_{{\varepsilon_{n}}(i,j)}}\) of \({ \mu _{{\varepsilon_{n}}(i,j)}}\) is given by
$${{\hat{\mu}}_{{\varepsilon_{n}}(i,j)}} = {{\hat{\mu}}_{j}} - \hat{\alpha}\cdot{{ \hat{\mu}}_{i}}. $$
Theorem 4
The estimator
\({\hat{\mu}_{{\varepsilon_{n}}(i,j)}}\)is strongly consistent.
Proof
Using strong consistency of estimators \({\hat{\mu}_{j}}\), \({\hat{\mu}_{i}}\), and α̂, it is obvious to prove that estimator \({\hat{\mu}_{{\varepsilon_{n}}(i,j)}}\) is strongly consistent. □
Let
$${P_{M}} = \left ( { \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} {{p_{11}}} & {{p_{12}}} & \cdots & {{p_{1r}}} \\ {{p_{21}}} & {{p_{22}}} & \cdots & {{p_{2r}}} \\ \vdots & \vdots & \ddots & \vdots \\ {{p_{r1}}} & {{p_{r2}}} & \cdots & {{p_{rr}}} \end{array}\displaystyle } \right ) $$
be a transition probability matrix, where r is the number of states, and \({p_{kj}}\), \(k,j \in\{ 1,2, \ldots,r\}\) is the transition probability from state k to state j. The estimator of \({p_{kj}}\) is defined as
$${{\hat{p}}_{kj}} = \frac{{{n_{kj}}}}{{{n_{k}}}}, $$
where \({n_{k}}\) represents the number of elements in the state k, \({n_{kj}}\) represents the number of elements transferring from state k to state j. We can easily come to the following conclusion.
Theorem 5
The estimator
\({{\hat{p}}_{kj}}\)is strongly consistent.