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ISSN : 1598-7248 (Print)
ISSN : 2234-6473 (Online)
Industrial Engineering & Management Systems Vol.17 No.3 pp.434-443
DOI : https://doi.org/10.7232/iems.2018.17.3.434

# Analysis of the Preference Shift of Customer Brand Selection and its Matrix Structure: Expansion of its Block Matrix to the Third Order Lag and Compression of the Variables

Kazuhiro Takeyasu*
Corresponding Author, E-mail: takeyasu@sz.tokoha-u.ac.jp
May 13, 2017 June 26, 2017 August 14, 2017

## ABSTRACT

Focusing that consumers are apt to buy superior brand when they are accustomed or bored to use current brand, a new analysis method is introduced. It is often observed that consumers select the upper class brand when they buy the next time. Suppose that the former buying data and the current buying data are gathered. Also suppose that the upper brand is located upper in the variable array. Then the transition matrix becomes an upper triangle matrix under the supposition that the former buying variables are set input and the current buying variables are set output. Before buying data and after buying data is stated using linear model. When above stated events occur, transition matrix becomes an upper triangular matrix. The goods of the same brand group would compose the Block Matrix in the transition matrix. Condensing the variables of the same brand group into one, analysis becomes easier to handle and the transition of Brand Selection can be easily grasped. In this paper, equation using transition matrix stated by the Block Matrix is expanded to the third order lag and the method is newly re-built. Furthermore the method of condensing the variable stated above is also applied to this new model. These are confirmed by numerical examples. S-step forecasting model is also introduced. This approach makes it possible to identify brand position in the market and it can be utilized for building useful and effective marketing plan.

## 1. INTRODUCTION

It is often observed that consumers select upper class brand when they buy next time after they are bored to use current brand.

Suppose that former buying data and current buying data are gathered. Also suppose that upper brand is located upper in the variable array. Then transition matrix becomes upper triangular matrix under the supposition that former buying variables are set input and current buying variables are set output. If the top brand were selected from lower brand skipping intermediate brands, corresponding part in upper triangular matrix would be 0.

If transition matrix is identified, s-step forecasting can be executed. Generalized forecasting matrix components’ equations are introduced. Unless planners for products notice its brand position whether it is upper or lower than other products, matrix structure makes it possible to identify those by calculating consumers’ activities for brand selection. Thus, this proposed approach makes it effective to execute marketing plan and/or establish new brand.

Quantitative analysis concerning brand selection has been executed by Yamanaka (1982), Takahashi and Takahashi (2002). Yamanaka (1982) examined purchasing process by Markov Transition Probability with the input of advertising expense. Takahashi and Takahashi (2002) made analysis by the Brand Selection Probability model using logistics distribution.

In Takeyasu and Ishio (2011), Takeyasu and Higuchi (2008), Takeyasu (2014), matrix structure was analyzed for the case brand selection was executed toward upper class. The goods of the same brand group would compose the Block Matrix in the transition matrix. Condensing the variables of the same brand group into one, analysis becomes easier to handle and the transition of Brand Selection can be easily grasped.

In this paper, equation using transition matrix stated by the Block matrix is extended to the third order lag and the method is newly re-built and the method of condensing the variable stated above is also applied to this new model. Such research as this cannot be found as long as searched.

Hereinafter, matrix structure is clarified for the selection of brand in section 2. Block matrix structure is analyzed when brands are handled in group and s-step forecasting is formulated in section 3. Expansion of the model to the third order lag is executed in section 4. Numerical calculation is executed in section 5. Application of this method is extended in section 6.

In this paper, equation using transition matrix stated by the Block Matrix is expanded to the third order lag and the method of condensing the variable stated above is also applied to this new model. Such research cannot be found as long as searched.

The rest of the paper is organized as follows. Matrix structure is clarified for the selection of brand in section 2. A block matrix structure is analyzed when brands are handled in a group in section 3. Expansion of the block matrix structure to the second order lag is executed in section 4. A block matrix structure when condensing the variables of the same brand group is analyzed in section 5. Its expansion to the third order lag in stated in section 6. Numerical calculation is executed in section 7. Section 8 is a summary.

## 2. BRAND SELECTION AND ITS MATRIX STRUCTURE

### 2.1. Upper Shift of Brand Selection

Now, suppose that x is the most upper class brand, y is the second upper class brand, and z is the lowest class brand.

Consumer’s behavior of selecting brand might be $z → y , y → x , z → x$ etc. xz might be few.

Suppose that x is current buying variable, and xb is previous buying variable. Shift to x is executed from xb, yb, or zb.

Therefore, x is stated in the following equation. aij represents transition probability from j-th to t-th brand.

$x = a 11 x b + a 12 y b + a 13 z b$

Similarly,

$y = a 22 y b + a 23 z b$

and

$z = a 33 z b$

These are re-written as follows.(1)

$( x y z ) = ( a 11 a 12 a 13 0 a 22 a 23 0 0 a 33 ) ( x b y b z b )$
(1)

Set

$X = ( x y z ) , A = ( a 11 a 12 a 13 0 a 22 a 23 0 0 a 33 ) , X b = ( x b y b z b )$

then, X is represented as follows.(2)

$X = A X b$
(2)

Here,

$X ∈ R 3 , A ∈ R 3 × 3 , X b ∈ R 3$

A is an upper triangular matrix.

To examine this, generating following data, which are all consisted by the data in which transition is made from lower brand to upper brand,

$X i = ( 1 0 0 ) ( 1 0 0 ) ⋯ ( 0 1 0 )$
(3)

$X b i = ( 0 1 0 ) ( 1 0 0 ) ⋯ ( 0 0 1 ) i = 1 , 2 , ⋯ , N$
(4)

parameter can be estimated by using least square method.

Suppose

$X i = A X b i + ε i$
(5)

where $ε i = ( ε 1 i ε 2 i ε 3 i ) i = 1 , 2 , ⋯ N$

and minimize following J(6)

$J = ∑ i = 1 N ε i T ε i → M i n$
(6)

$A ^$ which is an estimated value of A is obtained as follows.(7)

$A ^ = ( ∑ i = 1 N X i X b i T ) ( ∑ i = 1 N X b i X b i T ) − 1$
(7)

In the data group which are all consisted by the data in which transition is made from lower brand to upper brand, estimated value $A ^$ should be upper triangular matrix.

If following data which shift to lower brand are added only a few in equation (3) and (4),

$X i = ( 0 1 0 ) , X b i = ( 1 0 0 )$

$A ^$ would contain minute items in the lower part triangle.

### 2.2. Sorting Brand Ranking by Re-Arranging Row

In a general data, variables may not be in order as x, y, z. In that case, large and small value lie scattered in $A ^$ . But re-arranging this, we can set in order by shifting row. The large value parts are gathered in upper triangular matrix, and the small value parts are gathered in lower triangular matrix.(8)

(8)

### 2.3. Matrix Structure under the Case Skipping Intermediate Class Brand

It is often observed that some consumers select the most upper class brand from the most lower class brand and skip selecting the intermediate class brand.

We suppose v, w, x, y, z brands (suppose they are laid from upper position to lower position as v > w > x > y > z).

In the above case, selection shifts would be

$v ← z v ← y$

Suppose they do not shift to y, x, w from z, to x, w from y, and to w from x, then Matrix structure would be as follows.(9)

$( v w x y z ) = ( a 11 a 12 a 13 a 14 a 15 0 a 22 0 0 0 0 0 a 33 0 0 0 0 0 a 44 0 0 0 0 0 a 55 ) ( v b w b x b y b z b )$
(9)

We confirm this by numerical example in section 4.

## 3. BLOCK MATRIX STRUCTURE IN BRAND GOURPS AND S-STEP FORECASTING

Next, we examine the case in brand groups. Matrices are composed by Block Matrix.

### 3.1. Brand Shift Group-in the Case of Two Groups

Suppose brand selection shifts from Corolla class to Mark X class in car. In this case, it does not matter which company’s car they choose. Thus, selection of cars are executed in a group and brand shift is considered to be done from group to group. Suppose brand groups at time n are as follows.

X consists of p varieties of goods, and Y consists of q varieties of goods.

(10)

Here,

Make one more step of shift, then we obtain following equation.(12)

(11)

Make one more step of shift again, then we obtain following equation.

(12)

Similarly,(13)(14)

(13)

(14)

Finally, we get generalized equation for s -step shift as follows.

(15)

If we replace $n − s → n , n → n + s$ in equation (15), we can make s-step forecast.

### 3.2. Brand Shift Group－in the Case of Three Groups

Suppose brand selection is executed in the same group or to the upper group, and also suppose that brand position is x > y > z (x is upper position). Then brand selection transition matrix would be expressed as

(16)

where

$X n = ( x 1 n x 2 n ⋮ x p n ) , Y n = ( y 1 n y 2 n ⋮ y q n ) , Z n = ( z 1 n z 2 n ⋮ z r n )$

Here,

These are re-stated as(17)

$W n = A W n − 1$
(17)

where,

$W n = ( X n Y n Z n ) , A = ( A 11 , A 12 , A 13 0 , A 22 , A 23 0 , 0 , A 33 ) , W n − 1 = ( X n − 1 Y n − 1 Z n − 1 )$

Hereinafter, we shift steps as is done in the previous section.

In the general description, we state as

$W n = A ( s ) W n − s$
(18)

Here,

$A ( s ) = ( A 11 ( s ) , A 12 ( s ) , A 13 ( s ) 0 , A 22 ( s ) , A 23 ( s ) 0 , 0 , A 33 ( s ) ) , W n − s = ( X n − s Y n − s Z n − s )$

From definition,(19)

$A ( 1 ) = A$
(19)

In the case s = 2 , we obtain(20)

$A ( 2 ) = ( A 11 , A 12 , A 13 0 , A 22 , A 23 0 , 0 , A 33 ) ( A 11 , A 12 , A 13 0 , A 22 , A 23 0 , 0 , A 33 ) = ( A 11 2 , A 11 A 12 + A 12 A 22 , A 11 A 13 + A 12 A 23 + A 13 A 33 0 , A 22 2 , A 22 A 23 + A 23 A 33 0 , 0 , A 33 2 )$
(20)

Next, in the case s = 3 , we obtain(21)

$A ( 3 ) = ( Α 11 3 , Α 11 2 Α 12 + Α 11 Α 12 Α 22 + Α 12 Α 22 2 , Α 11 2 Α 13 + Α 11 Α 12 Α 23 + Α 11 Α 13 Α 33 + Α 12 Α 22 Α 23 + Α 12 Α 23 Α 33 + Α 13 Α 33 2 0 , Α 22 3 , Α 22 2 Α 23 + Α 22 Α 23 Α 33 + Α 23 Α 33 2 0 , 0 , Α 33 3 )$
(21)

In the case s = 4 , equations become wide-spread, so we express each Block Matrix as follows.(22)

$A 11 ( 4 ) = A 11 4 A 12 ( 4 ) = A 11 3 A 12 + A 11 2 A 12 A 22 + A 11 A 12 A 22 2 + A 12 A 22 3 A 13 ( 4 ) = A 11 3 A 13 + A 11 2 A 12 A 23 + A 11 2 A 13 A 33 + A 11 A 12 A 22 A 23 + A 11 A 12 A 23 A 33 + A 11 A 13 A 33 2 + A 12 A 22 2 A 23 + A 12 A 22 A 23 A 33 + A 12 A 23 A 33 2 + A 13 A 33 3 A 22 ( 4 ) = A 22 4 A 23 ( 4 ) = A 22 3 A 23 + A 22 2 A 23 A 33 + A 22 A 23 A 33 2 + A 23 A 33 3 A 33 ( 4 ) = A 33 4 }$
(22)

In the case s = 5, we obtain the following equations similarly.(23)

$A 11 ( 5 ) = A 11 5 A 12 ( 5 ) = A 11 4 A 12 + A 11 3 A 12 A 22 + A 11 2 A 12 A 22 2 + A 11 A 12 A 22 3 + A 12 A 22 4 A 13 ( 5 ) = A 11 4 A 13 + A 11 3 A 12 A 23 + A 11 3 A 13 A 33 + A 11 2 A 12 A 22 A 23 + A 11 2 A 12 A 23 A 33 + A 11 2 A 13 A 33 2 + A 11 A 12 A 22 2 A 23 + A 11 A 12 A 22 A 23 A 33 + A 11 A 12 A 23 A 33 2 + A 11 A 13 A 33 3 + A 12 A 22 3 A 23 + A 12 A 22 2 A 23 A 33 + A 12 A 22 A 23 A 33 2 + A 12 A 23 A 33 3 + A 13 A 33 4 A 22 ( 5 ) = A 22 5 A 23 ( 5 ) = A 22 4 A 23 + A 22 3 A 23 A 33 + A 22 2 A 23 A 33 2 + A 22 A 23 A 33 3 + A 23 A 33 4 A 33 ( 5 ) = A 33 5 }$
(23)

In the case s = 6 , we obtain(25)

$A 11 ( 6 ) = A 11 6 A 12 ( 6 ) = A 11 5 A 12 + A 11 4 A 12 A 22 + A 11 3 A 12 A 22 2 + A 11 2 A 12 A 22 3 + A 11 A 12 A 22 4 + A 12 A 22 5 A 13 ( 6 ) = A 11 5 A 13 + A 11 4 A 12 A 23 + A 11 4 A 13 A 33 + A 11 3 A 12 A 22 A 23 + A 11 3 A 12 A 23 A 33 + A 11 3 A 13 A 33 2 + A 11 2 A 12 A 22 2 A 23 + A 11 2 A 12 A 22 A 23 A 33 + A 11 2 A 12 A 23 A 33 2 + A 11 2 A 13 A 33 3 + A 11 A 12 A 22 3 A 23 + A 11 A 12 A 22 2 A 23 A 33 + A 11 A 12 A 22 A 23 A 33 2 + A 11 A 12 A 23 A 33 3 + A 11 A 13 A 33 4 + A 12 A 22 4 A 23 + A 12 A 22 3 A 23 A 33 + A 12 A 22 2 A 23 A 33 2 + A 12 A 22 A 23 A 33 3 + A 12 A 23 A 33 4 + A 13 A 33 5 }$
(24)

We get generalized equations for s-step shift as follows.

$A 11 ( s ) = A 11 s A 12 ( s ) = A 11 s − 1 A 12 + ∑ k = 2 s − 1 A 11 s − k A 12 A 22 k − 1 + A 12 A 22 s − 1 A 13 ( s ) = A 11 s − 1 A 13 + A 11 s − 2 ( ∑ k = 1 2 A 1 ( k + 1 ) A ( k + 1 ) 3 ) + ∑ j = 1 s − 3 [ A 11 S − 2 − j { A 12 ( ∑ k = 1 j + 1 A 22 j + 1 − k A 23 A 33 k − 1 ) + A 13 A 33 j + 1 } ] A 22 ( s ) = A 22 s A 23 ( s ) = ∑ K − 1 s Α 22 s − k Α 23 Α 33 k − 1 A 33 ( s ) = A 33 s }$
(25)

Expressing them in matrix, it follows.(26)

$A ( S ) = ( A 11 s , A 11 s − 1 A 12 + ∑ k = 2 s − 1 Α 11 s − k Α 12 Α 22 k − 1 + Α 12 Α 22 s − 1 , A 11 s − 1 A 13 + A 11 s − 2 ( ∑ k = 1 2 A 1 ( k + 1 ) A ( K + 1 ) 3 ) + ∑ j = 1 s − 3 [ A 11 s − 2 − j { A 12 ( ∑ k = 1 j + 1 A 22 j + 1 − k A 23 A 33 k − 1 ) + A 13 A 33 j + 1 } ] 0 , A 22 s , ∑ k = 1 s A 22 s − k A 23 A 33 k − 1 0 , 0 , A 33 s )$
(26)

Generalizing them to m groups, they are expressed as(27)

$( X n ( 1 ) X n ( 2 ) ⋮ X n ( m ) ) = ( A 11 A 12 ⋯ A 1 m A 21 A 22 ⋯ A 2 m ⋮ ⋮ ⋮ A m 1 A m 2 ⋯ A m m ) ( X n − 1 ( 1 ) X n − 1 ( 2 ) ⋮ X n − 1 ( m ) ) X n ( 1 ) ∈ R k 1 , X n ( 2 ) ∈ R k 2 , ... , X n ( m ) ∈ R k m , A i j ∈ R k i × k j ( i = 1 , ⋯ , m ) ( j = 1 , ⋯ , m )$
(27)

## 4. EXPANSION TO THE THIRD ORDER LAG

Expansion of the above stated Block Matrix model to the third order lag is executed in the following method.

Here we take three groups case.

Generating Eq. (16) and Eq. (18),we state the model as follows. Here we set P = 3.(28)

$( X n Y n Z n ) = ( Α , B , C D , E , F G , H , J ) ( X n − 1 Y n − 1 Z n − 1 )$
(28)

where(29)

$X n = ( x 1 n x 2 n x 3 n ) , Y n = ( y 1 n y 2 n y 3 n ) , Z n = ( z 1 n z 2 n z 3 n )$
(29)

Here,

$X n ∈ R 3 ( n = 1 , 2 , ⋯ ) , Y n ∈ R 3 ( n = 1 , 2 , ⋯ ) , Z n ∈ R 3 ( n = 1 , 2 , ⋯ ) , { A , B , C , D , E , F , G , H , J } ∈ R 3 × 3$

These are re-stated as:(30)(31)(32)(33)

$W n = P W n − 1$
(30)

$W n = ( X n Y n Z n )$
(31)

$P = ( A , B , C D , E , F G , H , J )$
(32)

$W n − 1 = ( X n − 1 Y n − 1 Z n − 1 )$
(33)

If N amount of data exist, we can derive the following equation similarly as Eq. (5).

$W n i = P W n − 1 i + ε n i ( i = 1 , 2 , ⋯ , N )$
(34)

and(35)

$J n = ∑ i = 1 N ε n i T ε n i → M i n$
(35)

$P ∧$ which is an estimated value of P is obtained as follows.(36)

$P ∧ = ( ∑ i = 1 N W n i W n − 1 i T ) ( ∑ i = 1 N W n − 1 i W n − 1 i T ) − 1$
(36)

Now, we expand Eq.(34) to the third order lag model as follows.

$W n i = P 1 W n − 1 i + P 2 W n − 2 i + P 3 W n − 3 i + ε n i$
(37)

Here

$P 1 = ( A 1 , B 1 , C 1 D 1 , E 1 , F 1 G 1 , H 1 , J 1 ) , P 2 = ( A 2 , B 2 , C 2 D 2 , E 2 , F 2 G 2 , H 2 , J 2 ) , P 3 = ( A 3 , B 3 , C 3 D 3 , E 3 , F 3 G 3 , H 3 , J 3 )$
(38)

It we set

$P = ( P 1 , P 2 , P 3 )$
(39)

then $P ∧$ can be estimated as follows.

$P = ( ∑ i = 1 N W t i ( W t − 1 i W t − 2 i W t − 3 i ) T ) ( ∑ i = 1 N ( W t − 1 i W t − 2 i W t − 3 i ) ( W t − 1 i W t − 2 i W t − 3 i ) T ) − 1$
(40)

We further develop this equation as follows.(41)

$P = ( P 1 , P 2 , P 3 ) = ( A 1 , B 1 , C 1 , D 1 , E 1 , F 1 , G 1 , H 1 , J 1 , A 2 , B 2 , C 2 , D 2 , E 2 , F 2 , G 2 , H 2 , J 2 , A 3 , B 3 , C 3 D 3 , E 3 , F 3 G 3 , H 3 , J 3 ) = ( ∑ i = 1 N W t i W t − 1 i T , ∑ i = 1 N W t i W t − 2 i T , ∑ i = 1 N W t i W t − 3 i T ) × ( ∑ i = 1 N W t − 1 i W t − 1 i T , ∑ i = 1 N W t − 1 i W t − 2 i T , ∑ i = 1 N W t − 1 i W t − 3 i T ∑ i = 1 N W t − 2 i W t − 1 i T , ∑ i = 1 N W t − 2 i W t − 2 i T , ∑ i = 1 N W t − 2 i W t − 3 i T ∑ i = 1 N W t − 3 i W t − 1 i T , ∑ i = 1 N W t − 3 i W t − 2 i T , ∑ i = 1 N W t − 3 i W t − 3 i T ) − 1 = ( ∑ i = 1 N ( x t i y t i z t i ) ( x t − 1 i T , y t − 1 i T , z t − 1 i T ) , ∑ i = 1 N ( x t i y t i z t i ) ( x t − 2 i T , y t − 2 i T , z t − 2 i T ) , ∑ i = 1 N ( x t i y t i z t i ) ( x t − 3 i T , y t − 3 i T , z t − 3 i T ) ) ( ∑ i = 1 N ( x t − 1 i y t − 1 i z t − 1 i ) ( x t − 1 i T , y t − 1 i T , z t − 1 i T ) , ∑ i = 1 N ( x t − 1 i y t − 1 i z t − 1 i ) ( x t − 2 i T , y t − 2 i T , z t − 2 i T ) , ∑ i = 1 N ( x t − 1 i y t − 1 i z t − 1 i ) ( x t − 3 i T , y t − 3 i T , z t − 3 i T ) , ∑ i = 1 N ( x t − 2 i y t − 2 i z t − 2 i ) ( x t − 1 i T , y t − 1 i T , z t − 1 i T ) , ∑ i = 1 N ( x t − 2 i y t − 2 i z t − 2 i ) ( x t − 2 i T , y t − 2 i T , z t − 2 i T ) , ∑ i = 1 N ( x t − 2 i y t − 2 i z t − 2 i ) ( x t − 3 i T , y t − 3 i T , z t − 3 i T ) ∑ i = 1 N ( x t − 3 i y t − 3 i z t − 3 i ) ( x t − 1 i T , y t − 1 i T , z t − 1 i T ) , ∑ i = 1 N ( x t − 3 i y t − 3 i z t − 3 i ) ( x t − 2 i T , y t − 2 i T , z t − 2 i T ) , ∑ i = 1 N ( x t − 3 i y t − 3 i z t − 3 i ) ( x t − 3 i T , y t − 3 i T , z t − 3 i T ) ) − 1 = ( ∑ i = 1 N x t i x t − 1 i T , ∑ i = 1 N x t i y t − 1 i T , ∑ i = 1 N x t i z t − 1 i T , ∑ i = 1 N y t i x t − 1 i T , ∑ i = 1 N y t i y t − 1 i T , ∑ i = 1 N y t i z t − 1 i T , ∑ i = 1 N z t i x t − 1 i T , ∑ i = 1 N z t i y t − 1 i T , ∑ i = 1 N z t i z t − 1 i T , ∑ i = 1 N x t i x t − 2 i T , ∑ i = 1 N x t i y t − 2 i T , ∑ i = 1 N x t i z t − 2 i T , ∑ i = 1 N y t i x t − 2 i T , ∑ i = 1 N y t i y t − 2 i T , ∑ i = 1 N y t i z t − 2 i T , ∑ i = 1 N z t i x t − 2 i T , ∑ i = 1 N z t i y t − 2 i T , ∑ i = 1 N z t i z t − 2 i T , ∑ i = 1 N x t i x t − 3 i T , ∑ i = 1 N x t i y t − 3 i T , ∑ i = 1 N x t i z t − 3 i T ∑ i = 1 N y t i x t − 3 i T , ∑ i = 1 N y t i y t − 3 i T , ∑ i = 1 N y t i z t − 3 i T ∑ i = 1 N z t i x t − 3 i T , ∑ i = 1 N z t i y t − 3 i T , ∑ i = 1 N z t i z t − 3 i T ) × ( ∑ i = 1 N x t − 1 i x t − 1 i T , ∑ i = 1 N x t − 1 i y t − 1 i T , ∑ i = 1 N x t − 1 i z t − 1 i T , ∑ i = 1 N y t − 1 i x t − 1 i T , ∑ i = 1 N y t − 1 i y t − 1 i T , ∑ i = 1 N y t − 1 i z t − 1 i T , ∑ i = 1 N z t − 1 i x t − 1 i T , ∑ i = 1 N z t − 1 i y t − 1 i T , ∑ i = 1 N z t − 1 i z t − 1 i T , ∑ i = 1 N x t − 1 i x t − 2 i T , ∑ i = 1 N x t − 1 i y t − 2 i T , ∑ i = 1 N x t − 1 i z t − 2 i T , ∑ i = 1 N y t − 1 i x t − 2 i T , ∑ i = 1 N y t − 1 i y t − 2 i T , ∑ i = 1 N y t − 1 i z t − 2 i T , ∑ i = 1 N z t − 1 i x t − 2 i T , ∑ i = 1 N z t − 1 i y t − 2 i T , ∑ i = 1 N z t − 1 i z t − 2 i T , ∑ i = 1 N x t − 1 i x t − 3 i T , ∑ i = 1 N x t − 1 i y t − 3 i T , ∑ i = 1 N x t − 1 i z t − 3 i T ∑ i = 1 N y t − 1 i x t − 3 i T , ∑ i = 1 N y t − 1 i y t − 3 i T , ∑ i = 1 N y t − 1 i z t − 3 i T ∑ i = 1 N z t − 1 i x t − 3 i T , ∑ i = 1 N z t − 1 i y t − 3 i T , ∑ i = 1 N z t − 1 i z t − 3 i T ∑ i = 1 N x t − 2 i x t − 1 i T , ∑ i = 1 N x t − 2 i y t − 1 i T , ∑ i = 1 N x t − 2 i z t − 1 i T , ∑ i = 1 N y t − 2 i x t − 1 i T , ∑ i = 1 N y t − 2 i y t − 1 i T , ∑ i = 1 N y t − 2 i z t − 1 i T , ∑ i = 1 N z t − 2 i x t − 1 i T , ∑ i = 1 N z t − 2 i y t − 1 i T , ∑ i = 1 N z t − 2 i z t − 1 i T , ∑ i = 1 N x t − 2 i x t − 2 i T , ∑ i = 1 N x t − 2 i y t − 2 i T , ∑ i = 1 N x t − 2 i z t − 2 i T , ∑ i = 1 N y t − 2 i x t − 2 i T , ∑ i = 1 N y t − 2 i y t − 2 i T , ∑ i = 1 N y t − 2 i z t − 2 i T , ∑ i = 1 N z t − 2 i x t − 2 i T , ∑ i = 1 N z t − 2 i y t − 2 i T , ∑ i = 1 N z t − 2 i z t − 2 i T , ∑ i = 1 N x t − 2 i x t − 3 i T , ∑ i = 1 N x t − 2 i y t − 3 i T , ∑ i = 1 N x t − 2 i z t − 3 i T ∑ i = 1 N y t − 2 i x t − 3 i T , ∑ i = 1 N y t − 2 i y t − 3 i T , ∑ i = 1 N y t − 2 i z t − 3 i T ∑ i = 1 N z t − 2 i x t − 3 i T , ∑ i = 1 N z t − 2 i y t − 3 i T , ∑ i = 1 N z t − 2 i z t − 3 i T ∑ i = 1 N x t − 3 i x t − 1 i T , ∑ i = 1 N x t − 3 i y t − 1 i T , ∑ i = 1 N x t − 3 i z t − 1 i T , ∑ i = 1 N y t − 3 i x t − 1 i T , ∑ i = 1 N y t − 3 i y t − 1 i T , ∑ i = 1 N y t − 3 i z t − 1 i T , ∑ i = 1 N z t − 3 i x t − 1 i T , ∑ i = 1 N z t − 3 i y t − 1 i T , ∑ i = 1 N z t − 3 i z t − 1 i T , ∑ i = 1 N x t − 3 i x t − 2 i T , ∑ i = 1 N x t − 3 i y t − 2 i T , ∑ i = 1 N x t − 3 i z t − 2 i T , ∑ i = 1 N y t − 3 i x t − 2 i T , ∑ i = 1 N y t − 3 i y t − 2 i T , ∑ i = 1 N y t − 3 i z t − 2 i T , ∑ i = 1 N z t − 3 i x t − 2 i T , ∑ i = 1 N z t − 3 i y t − 2 i T , ∑ i = 1 N z t − 3 i z t − 2 i T , ∑ i = 1 N x t − 3 i x t − 3 i T , ∑ i = 1 N x t − 3 i y t − 3 i T , ∑ i = 1 N x t − 3 i z t − 3 i T ∑ i = 1 N y t − 3 i x t − 3 i T , ∑ i = 1 N y t − 3 i y t − 3 i T , ∑ i = 1 N y t − 3 i z t − 3 i T ∑ i = 1 N z t − 3 i x t − 3 i T , ∑ i = 1 N z t − 3 i y t − 3 i T , ∑ i = 1 N z t − 3 i z t − 3 i T ) − 1$
(41)

We set this as:(42)

$P = ( P 1 , P 2 , P 3 ) P = ( K 1 K 2 K 3 K 4 K 5 K 6 K 7 K 8 K 9 L 1 L 2 L 3 L 4 L 5 L 6 L 7 L 8 L 9 M 1 M 2 M 3 M 4 M 5 M 6 M 7 M 8 M 9 ) × ( N 1 N 2 N 3 N 4 N 5 N 6 N 7 N 8 N 9 Q 1 Q 2 Q 3 Q 4 Q 5 Q 6 Q 7 Q 8 Q 9 R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 S 1 S 2 S 3 S 4 S 5 S 6 S 7 S 8 S 9 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 T 9 U 1 U 2 U 3 U 4 U 5 U 6 U 7 U 8 U 9 V 1 V 2 V 3 V 4 V 5 V 6 V 7 V 8 V 9 α 1 α 2 α 3 α 4 α 5 α 6 α 7 α 8 α 9 β 1 β 2 β 3 β 4 β 5 β 6 β 7 β 8 β 9 ) − 1$
(42)

Then when all consist of the same level shifts or the upper level shifts (suppose X > Y > Z),

$K 4 , K 7 , K 8 , L 4 , L 7 , L 8 , M 2 , M 7 , M 8 , N 2 , N 3 , N 6 , T 2 , T 3 , T 6 , β 2 , β 3 , β 6 , Q 4 , Q 7 , Q 8 , R 4 , R 7 , R 8 , U 4 , U 7 , U 8$

are all 0.

therefore they are all 0.

$N 1 , N 5 , N 9 , T 1 , T 5 , T 9 , β 1 , β 5 , β 9$ become diagonal Matrices.

Matrices

Using a symbol “∗” as a diagonal matrix, P becomes as follows by using the relation stated above.(43)

$P = ( K 1 , K 2 , K 3 , 0 , K 5 , K 6 , 0 , 0 , K 9 , L 1 , L 2 , L 3 , 0 , L 5 , L 6 , 0 , 0 , L 9 , M 1 , M 2 , M 3 0 , M 5 , M 6 0 , 0 , M 9 ) × ( ∗ , 0 , 0 , N 1 , N 2 , N 3 , 0 , ∗ , 0 , 0 , N 5 , N 6 , 0 , 0 , ∗ , 0 , 0 , N 9 , S 1 , 0 0 , ∗ , 0 , 0 , S 4 , S 5 , 0 , 0 , ∗ , 0 , S 7 , S 8 , S 9 , 0 , 0 , ∗ , R 1 , R 2 , R 3 0 , R 5 , R 6 0 , 0 , R 9 U 1 U 2 U 3 0 , U 5 U 6 0 , 0 , U 9 V 1 , 0 , 0 , α 1 , 0 , 0 , * , 0 , 0 , V 4 , V 5 , 0 , α 4 , α 5 , 0 , 0 , * , 0 , V 7 , V 8 , V 9 , α 7 , α 8 , α 9 , 0 , 0 , * ) − 1$
(43)

## 5. MATRIX STRUCTURE WHEN CONDENSING THE VARIABLES OF THE SAME CLASS

Suppose the customer selects Bister from Corolla (Bister is an upper class brand automobile than Corolla.) when he/she buy next time. In that case, there are such brand automobiles as bluebird, Gallant sigma and 117 coupes for the corresponding same brand class group with Bister in other companies.

Someone may select another automobile from the same brand class group. There is also the case that the consumer select another company’s automobile of the same brand class group when he/she buy next time.

Matrix structure would be, then, as follows.

Suppose w, x, y in the same brand class group in the example of 2.3. If there exist following shifts:

$y , x , w , v ← z b x , w ← y b y , w ← x b x , y ← w b v ← y b v ← x b v ← w b y ← y b y ← x b y ← w b$

then, transition equation is expressed as follows.(44)

$( v w x y z ) = ( a 11 a 12 a 13 a 14 a 15 0 a 22 a 23 a 24 a 25 0 a 32 a 33 a 34 a 35 0 a 42 a 43 a 44 a 45 0 0 0 0 a 55 ) ( v b w b x b y b z b )$
(44)

Expressing these in Block Matrix form, it becomes as follows.(45)

$X = ( a 11 A 12 a 15 0 A 22 Α 23 0 0 a 55 ) X b$
(45)

where,

$X ∈ R 5 , A 12 ∈ R 1 × 3 , A 22 ∈ R 3 × 3 , A 23 ∈ R 3 × 1 , X b ∈ R 5$

As w, x, y are in the same class brand group, condensing these variables into one, and expressing it as w, then Eq. (10) becomes as follows.(46)

$( v w ¯ z ) = ( a 11 a ¯ 12 a 15 0 a ¯ 22 a ¯ 23 0 0 a 55 ) ( v b w ¯ b z )$
(46)

As aij satisfies the following equation :(48)(49)(50)

$∑ i = 1 5 a i j = 1 ( ∀ j )$
(47)

Condensed version of Block Matrix A22 are as follows, where sum of each item of Block matrix is taken and they are divided by the number of variables.

$a ¯ 22 = 1 3 ∑ i = 2 4 ∑ j = 2 4 a i j$
(48)

$a ¯ 12 = 1 3 ∑ j = 2 4 a 1 j$
(49)

$a ¯ 23 = ∑ i = 2 4 a i 5$
(50)

Generalizing this, it becomes as follows.(51)

$( X 1 X 2 ⋮ X r ) = ( A 11 A 12 ⋯ A 1 r 0 A 21 ⋯ A 2 r ⋮ ⋮ ⋮ 0 0 ⋯ A r r ) ( X 1 , b X 2 , b ⋮ X r , b )$
(51)

where

$( X 1 X 2 ⋮ X r ) = ( A 11 A 12 ⋯ A 1 r 0 A 21 ⋯ A 2 r ⋮ ⋮ ⋮ 0 0 ⋯ A r r ) ( X 1 , b X 2 , b ⋮ X r , b )$

When the variables of each Block Matrix are concondensed into one, transition matrix is expressed as follows.(52)

$( x ¯ 1 x ¯ 2 ⋮ x ¯ r ) = ( a ¯ 11 , a ¯ 12 , ⋯ a ¯ 1 r 0 , a ¯ 22 , ⋯ a ¯ 2 r ⋮ ⋮ ⋮ 0 , 0 , ⋯ a ¯ r r ) ( x ¯ 1 , b x ¯ 2 , b ⋮ x ¯ r , b )$
(52)

where(53)

$a ¯ i j = 1 p j r ∑ k = 1 p i ∑ l = 1 p j a k l i j ( i = 1 , ⋯ , r ) , ( j = i , ⋯ , r )$
(53)

Here, $a k l i j ( k = 1 , ⋯ , P i ) , ( l = 1 , ⋯ , P i )$ means the item of Aij.

Taking these operations, the variables of the same brand class group are condensed into one and the transition condition among brand class can be grasped easily and clearly. Judgment when and where to put the new brand becomes easy and may be executed properly.

## 6. EXPANSION TO THE THIRD ORDER LAG

Here we take up Eq. (37), (38), (39) and expand the method stated in section 5.

When the variables of each matrix are condensed into one, transition matrix is expressed as follows in the same way stated in section 5.(54)

$( x ¯ n y ¯ n z ¯ n ) = ( a ¯ 1 , b ¯ 1 , c ¯ 1 d ¯ 1 , e ¯ 1 , f ¯ 1 g ¯ 1 , h ¯ 1 , j ¯ 1 ) ( x ¯ n − 1 y ¯ n − 1 z ¯ n − 1 ) + ( a ¯ 2 , b ¯ 2 , c ¯ 2 d ¯ 2 , e ¯ 2 , f ¯ 2 g ¯ 2 , h ¯ 2 , j ¯ 2 ) ( x ¯ n − 2 y ¯ n − 2 z ¯ n − 2 ) + ( a ¯ 3 , b ¯ 3 , c ¯ 3 d ¯ 3 , e ¯ 3 , f ¯ 3 g ¯ 3 , h ¯ 3 , j ¯ 3 ) ( x ¯ n − 3 y ¯ n − 3 z ¯ n − 3 )$
(54)

where

$a ¯ 1 = 1 3 ∑ i = 1 3 ∑ j = 1 3 a i j 1 b ¯ 1 = 1 3 ∑ i = 1 3 ∑ j = 1 3 b i j 1 ⋮ a ¯ 2 = 1 3 ∑ i = 1 3 ∑ j = 1 3 a i j 2 b ¯ 2 = 1 3 ∑ i = 1 3 ∑ j = 1 3 b i j 2$

Here,

$a i j 1 , b i j 1 , ⋯ , a i j 2 , b i j 2 , ⋯$ are items of each Block Matrix $A 1 , B 1 , ⋯ , A 2 , B 2 , ⋯ ,$ respectively and $a ¯ 1 , b ¯ 1 , ⋯ a ¯ 2 , b ¯ 2 , ⋯$ are the condensed variables of each Block Matrix $A 1 , B 1 , ⋯ , A 2 , B 2 ⋯$.

## 7. NUMERICAL EXAMPLE

We consider the case that brand selection shifts to the same class or upper classes. As above-referenced, transition matrix must be an upper triangular matrix.

Suppose following events occur.

Vector in these cases are expressed as follows. We show some of them as an example.

• $X t − 3 X t − 2 X t − 1 X t ( X t − 3 Y t − 3 Z t − 3 ) = ( 0 0 0 0 0 0 0 0 1 ) , ( X t − 2 Y t − 2 Z t − 2 ) = ( 0 0 0 0 0 0 0 0 1 ) , ( X t − 1 Y t − 1 Z t − 1 ) = ( 0 0 0 0 0 0 0 0 1 ) , ( X t Y t Z t ) = ( 0 0 0 0 0 0 0 0 1 )$

• $( X t − 3 Y t − 3 Z t − 3 ) = ( 0 0 0 0 0 0 0 0 1 ) , ( X t − 2 Y t − 2 Z t − 2 ) = ( 0 0 0 0 0 0 0 0 1 ) , ( X t − 1 Y t − 1 Z t − 1 ) = ( 0 0 0 0 0 0 0 0 1 ) , ( X t Y t Z t ) = ( 0 0 0 0 0 0 0 1 0 )$

• $( X t − 3 Y t − 3$