Coherent forecasting for stationary time series of discrete data

Abstract

Coherent forecasting for discrete-valued stationary time series is considered in this article. In the context of count time series, different methods of coherent forecasting such as median forecasting and mode forecasting are used to obtain $h$-step ahead coherent forecasting. However, there are not many existing works in the context of categorical time series. Here, we consider the case of a finite number of categories with different possible models, such as the Pegram’s operator-based ARMA($p$,$q$) model, the mixture transition distribution model and the logistic regression model, and study their $h$-step ahead coherent forecasting. Some theoretical results are derived along with some numerical examples. To facilitate comparison among the three models, we use some forecasting measures. The procedure is illustrated using one real-life categorical data, namely the infant sleep status data.

Appendix

1.1 Appendix A : Proof of Theorem 2

From the model (3.2), the 1-step ahead conditional distribution is given by

$$\begin{aligned} \begin{array}{lcl} p_1(i|i_1,\ldots ,i_p)&{}=&{}P(Y_{n+1}=C_{i}|Y_{n}=C_{i_1},\ldots ,Y_{n-p+1}=C_{i_p})\\ &{}=&{}\eta _{11}I(i_1=i)+\cdots +\eta _{1p}I(i_p=i)+(1-\eta _{11}-\cdots -\eta _{1p})p_i, \end{array} \end{aligned}$$

with $\eta _{1l}=\phi _{l}$, $l=1,\ldots ,p$. Then the two-step ahead conditional distribution is given by

$$\begin{aligned} p_2(i|i_1,\ldots ,i_p)&= P(Y_{n+2}=C_{i}|Y_{n}=C_{i_1},\ldots ,Y_{n-p+1}=C_{i_p})\\&= \displaystyle \sum _{j=0}^{k}P(Y_{n+2}=C_{i}|Y_{n+1}=C_{j},Y_{n}=C_{i_1},\ldots ,Y_{n-p+2}=C_{i_{p-1}})\\&\times P(Y_{n+1}=C_{j}|Y_{n}=C_{i_1},\ldots ,Y_{n-p+1}=C_{i_{p}})\\&= \displaystyle \sum _{j=0}^{k}\left\{ \eta _{11}I(j=i)+\cdots +\eta _{1p}I(i_{p-1}=i)+(1-\eta _{11}-\cdots -\eta _{1p})p_i\right\} \\&\times \left\{ \phi _1I(i_1=j)+\cdots +\phi _pI(i_p=j)+(1-\phi _1-\cdots -\phi _p)p_j\right\} \\&= \eta _{21}I(i_1=i)+\cdots +\eta _{2p}I(i_p=i)+(1-\eta _{21}-\cdots -\eta _{2p})p_i \end{aligned}$$

where $\varvec{\eta _2}=\varvec{\Phi }\varvec{\phi }$. So the result is true for $h=2$. Let it be true for $(h-1)$, that is $\varvec{\eta _{h-1}}=\varvec{\Phi }^{h-2}\varvec{\phi }$. Then by induction it is straightforward to show that the $h$-step ahead conditional distribution is given by (3.5).

1.2 Appendix B : Proof of Theorem 3

To prove the Theorem 3, it is enough to show that $\displaystyle \lim _{h\rightarrow \infty }\eta _{hi}=0$ for all $i$. To show this we use the result that for any $n \times n$ matrix $A$ with its eigenvalues $\lambda _1, \lambda _2,\ldots ,\lambda _s$, $\displaystyle \lim _{k \rightarrow \infty }A^k=0$ if the spectral radius of $A$, $\rho (A)<1$ where $\rho (A)=\max \{\left| \lambda _1\right| , \left| \lambda _2\right| ,\ldots ,\left| \lambda _s\right| \}$ (See Atkinson 2008). Outline of the proof is given follows.

From the Jordan normal theorem, for any $n \times n$ matrix $A$, there exist a non-singular matrix $V$ and a block diagonal matrix $J$ such that

$$\begin{aligned} A=VJV^{-1} \end{aligned}$$

for

where the $m_{i}\times m_{i}$ matrix $J_{m_{i}}(\lambda _{i})$ being

Now

$$\begin{aligned} A^k=VJ^kV^{-1} \end{aligned}$$

and, since $J$ is block diagonal,

Now a standard result on the $k$th power of an $m \times m$ Jordan block states that, for $k \ge m$,

$$\begin{aligned} J_{m}^k(\lambda )=\begin{pmatrix} \lambda ^k &{}\quad {{k} \atopwithdelims (){1}} \lambda ^{k-1} &{}\quad {{k}\atopwithdelims (){2}} \lambda ^{k-2} &{}\quad \cdots &{}\quad {{k}\atopwithdelims (){m-1}} \lambda ^{k-m+1} \\ 0 &{}\quad \lambda ^k &{}\quad {{k}\atopwithdelims (){1}} \lambda ^{k-1} &{}\quad \cdots &{}\quad {{k}\atopwithdelims (){m-2}} \lambda ^{k-m+2}\\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \lambda ^k \end{pmatrix}. \end{aligned}$$

Since $\rho (A)<1$, i.e., $|\lambda _i|<1$ for all $i$ and $\displaystyle \lim _{k \rightarrow \infty }{k \atopwithdelims ()i}\lambda ^{k-i}=0$, and hence $\displaystyle \lim _{k \rightarrow \infty } J_m^k(\lambda )=0$, . This implies that $\displaystyle \lim _{k \rightarrow \infty }J^k=0$. Therefore,

$$\begin{aligned} \displaystyle \lim _{k \rightarrow \infty }A^k=\displaystyle \lim _{k \rightarrow \infty }VJ^kV^{-1}=V(\displaystyle \lim _{k \rightarrow \infty }J^k)V^{-1}=0. \end{aligned}$$

Note that the eigenvalues of $\varvec{\Phi }$ are $\phi _1, \ldots ,\phi _p$ all of which lie between 0 and 1, and hence $\displaystyle \lim _{h \rightarrow \infty }\varvec{\Phi }^h=0$. Consequently $\displaystyle \lim _{h \rightarrow \infty } \varvec{\eta }_h=\displaystyle \lim _{h \rightarrow \infty }\varvec{\Phi }^{h-1}\varvec{\phi } =(\displaystyle \lim _{h \rightarrow \infty }\varvec{\Phi }^{h-1})\varvec{\phi }=0$.

1.3 Appendix C: Pegram’s MA(2) model

Here for $h=1$,

$$\begin{aligned}&P(Y_{n+1}=C_i|Y_{n}=C_j,Y_{n-1}=C_k)\\&\quad =\frac{\displaystyle \sum \nolimits _{r=0}^2\displaystyle \sum \nolimits _{s=0}^2\displaystyle \sum \nolimits _{t=0}^2\theta _r\theta _s\theta _tP(\epsilon _{n+1-r}=C_i,\epsilon _{n-s}=C_j,\epsilon _{n-1-t}=C_k)}{\displaystyle \sum \nolimits _{s=0}^2\displaystyle \sum \nolimits _{t=0}^2\theta _s\theta _tP(\epsilon _{n-s}=C_j,\epsilon _{n-1-t}=C_k)}, \end{aligned}$$

where

$$\begin{aligned} P(\epsilon _{n+1-r}=C_{i},\epsilon _{n-s}=C_{j},\epsilon _{n-1-t}=C_{k})&= p_ip_jp_kI(r-1\ne s\ne t+1) \\&\quad + p_ip_jI(j = k)I(r-1\ne s=t+1)\\&\quad + p_ip_jI(i = k)I(r-1 =t+1 \ne s)\\&\quad + p_ip_kI(i = j)I(r-1 =s\ne t+1)\\&\quad + p_iI(i=j=k)I(r-1=s=t\!+\!1) \end{aligned}$$

and

$$\begin{aligned} P(\epsilon _{n-s}=C_{j},\epsilon _{n-1-t}=C_{k})=p_{j}p_{k} I(s\ne t+1) + p_jI(j = k)I(s=t+1). \end{aligned}$$

Similarly for $h=2$,

$$\begin{aligned}&P(Y_{n+2}=C_{i}|Y_{n},\ldots ,Y_{1})\\&\quad = P(Y_{n+2}=C_{i}|Y_{n}=C_{j},Y_{n-1}=C_{k})\\&\quad =\frac{P(Y_{n+2}=C_{i},Y_{n}=C_{j},Y_{n-1}=C_{k})}{P(Y_{n}=C_{j},Y_{n-1}=C_{k})}\\&\quad =\frac{\displaystyle \sum \nolimits _{r=0}^2\displaystyle \sum \nolimits _{s=0}^2\displaystyle \sum \nolimits _{t=0}^2\theta _r\theta _s\theta _tP(\epsilon _{n+2-r}=C_{i},\epsilon _{n-s}=C_{j},\epsilon _{n-1-t}=C_{k})}{\displaystyle \sum \nolimits _{s=0}^2\displaystyle \sum \nolimits _{t=0}^2\theta _s\theta _tP(\epsilon _{n-s}=C_{j},\epsilon _{n-1-t}=C_{k})} \end{aligned}$$

where

$$\begin{aligned} P(\epsilon _{n+2-r}=C_{i},\epsilon _{n-s}=C_{j},\epsilon _{n-1-t}=C_{k})&= p_ip_jp_kI(r-2\ne s\ne t+1) \\&\quad + p_ip_jI(j = k)I(r-2\ne s=t+1)\\&\quad + p_ip_jI(i = k)I(r-2 =t+1 \ne s)\\&\quad + p_ip_kI(i = j)I(r-2 =s\ne t+1) \\&\quad + p_iI(i\!=\!j\!=\!k)I(r\!-\!2=s=t\!+\!1), \end{aligned}$$

and

$$\begin{aligned} P(\epsilon _{n-s}=C_{j},\epsilon _{n-1-t}=C_{k})=p_jp_kI(s\ne t+1) + p_jI(j = k)I(s=t+1). \end{aligned}$$

And for $h>2,\; P(Y_{n+h}=C_{i}| Y_{n}, Y_{n-1}, \ldots )=p_{i}$.