An iterative method for the computation of the correlation matrix implied by a recursive path model

Abstract

In Path Analysis, especially in social sciences studies, many researchers usually assume that errors in the model are uncorrelated with all exogenous variables as well as with each other. These assumptions, in most cases, are not valid in reality and were introduced to facilitate the model estimation. This article establishes a new algorithm for the computation of the correlation matrix implied by a recursive path model that overcomes these drawbacks. We compare our algorithm to two other methods used in the literature. The comparison was made mathematically through an illustrated example and numerically with a simulation study. The findings show that, unlike the classical methods, the proposed method gives more accurate results.

Appendices

Appendix 1: Proof of Theorem 2

Consider a recursive path model define by the equation (1) or equivalently (19) and let ${\varvec{\widehat{R}^{\star }}}$ be the implied correlation matrix of the whole model (including disturbances). Let ${\varvec{\widehat{R}^{new\star }}}$ be the matrix computed using Algorithm (2).

We will proceed by an induction to prove that each block ${\varvec{\widehat{R}^{\star }}}_{1:q+p+j,1:q+p+j}$ corresponds to the block ${\varvec{\widehat{R}^{new\star }}}_{1:q+p+j,1:q+p+j}$ for $j=0,\ldots ,p$.

For $j=0$, by the definition of the matrices:

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}_{1:q+p,1:q+p}={\varvec{\varPhi ^{\star }}}={\varvec{\widehat{R}^{\star }}}_{1:q+p,1:q+p} \end{aligned}$$

(28)

We suppose that, for j in $\{1,\ldots ,p-1\}$ :

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}_{1:q+p+j,1:q+p+j}={\varvec{\widehat{R}^{\star }}}_{1:q+p+j,1:q+p+j} \end{aligned}$$

(29)

Now we will prove it for $j+1$ (i.e. ${\varvec{\widehat{R}^{new\star }}}_{1:q+p+j+1,1:q+p+j+1}={\varvec{\widehat{R}^{\star }}}_{1:q+p+j+1,1:q+p+j+1}$).

Since the tow matrices are symmetric and with diagonal elements all equal to 1 (${\varvec{\widehat{R}^{new\star }}}$ according to Algorithm (2) and ${\varvec{\widehat{R}^{\star }}}$ because it is a correlation matrix), showing that :

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}_{1:q+p+j+1,1:q+p+j+1}={\varvec{\widehat{R}^{\star }}}_{1:q+p+j+1,1:q+p+j+1} \end{aligned}$$

(30)

is reduced to proving :

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}_{q+p+j+1,1:q+p+j}={\varvec{\widehat{R}^{\star }}}_{q+p+j+1,1:q+p+j} \end{aligned}$$

(31)

According to Algorithm (2):

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}_{1:q+p+j+1,1:q+p+j+1}={\varvec{A^{\star }}}_{j+1,1:q+p+j}{\varvec{\widehat{R}^{new\star }}}_{1:q+p+j,1:q+p+j+1} \end{aligned}$$

(32)

And by the hypothesis of induction (29):

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}_{1:q+p+j,1:q+p+j+1}={\varvec{\widehat{R}^{\star }}}_{1:q+p+j,1:q+p+j} \end{aligned}$$

(33)

Hence:

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}_{1:q+p+j+1,1:q+p+j+1}={\varvec{A^{\star }}}_{j+1,1:q+p+j}{\varvec{\widehat{R}^{\star }}}_{1:q+p+j,1:q+p+j} \end{aligned}$$

(34)

On the other hand, from (20):

$$\begin{aligned} {\varvec{\widehat{R}^{\star }}}_{q+p+j+1,1:q+p+j+1}={\mathbb {E}}[\eta _{j+1}({\varvec{\xi ^{\star t}}},\eta _1,\ldots ,\eta _j)] \end{aligned}$$

(35)

And from (19):

$$\begin{aligned} \eta _{j+1}={\varvec{A^{\star }}}_{j+1,1:q+2p}\begin{pmatrix} {\varvec{\xi ^{\star }}}\\ {\varvec{\eta }} \end{pmatrix} \end{aligned}$$

(36)

Since the matrix ${\varvec{B}}$ is strictly lower triangular: ${\varvec{B}}_{j+1,k}=\beta _{j+1,k}=0$ for $j+1 \leqslant k \leqslant p$. This also means that ${\varvec{A^{\star }}}_{j+1,q+p+k}=0$ for $j+1 \leqslant k \leqslant p$ or equivalently :

$$\begin{aligned}{\varvec{A^{\star }}}_{j+1,q+p+j+1:q+2p}={\varvec{0}}\end{aligned}$$

Hence, from (36):

$$\begin{aligned} \eta _{j+1}={\varvec{A^{\star }}}_{j+1,1:q+2p}\begin{pmatrix} {\varvec{\xi ^{\star }}}\\ {\varvec{\eta }} \end{pmatrix}={\varvec{A^{\star }}}_{j+1,1:q+p+j}\begin{pmatrix} {\varvec{\xi ^{\star t}}}&\eta _1&\ldots&\eta _j \end{pmatrix}^{t} \end{aligned}$$

(37)

Using (37), we can write (35) as :

$$\begin{aligned} {\varvec{\widehat{R}^{\star }}}_{q+p+j+1,1:q+p+j+1}= & {} {\mathbb {E}}[\eta _{j+1}({\varvec{\xi ^{\star t}}},\eta _1,\ldots ,\eta _j)]\nonumber \\= & {} {\mathbb {E}}[{\varvec{A^{\star }}}_{j+1,1:q+p+j}( {\varvec{\xi ^{\star t}}}, \eta _1 ,\ldots , \eta _j )^{t}({\varvec{\xi ^{\star t}}},\eta _1,\ldots ,\eta _j)] \end{aligned}$$

(38)

Therefore,

$$\begin{aligned} {\varvec{\widehat{R}^{\star }}}_{q+p+j+1,1:q+p+j+1}= & {} {\varvec{A^{\star }}}_{j+1,1:q+p+j}{\mathbb {E}}[( {\varvec{\xi ^{\star t}}}, \eta _1 ,\ldots , \eta _j )^{t}({\varvec{\xi ^{\star t}}},\eta _1,\ldots ,\eta _j)]\nonumber \\= & {} {\varvec{A^{\star }}}_{j+1,1:q+p+j}{\varvec{\widehat{R}^{\star }}}_{1:q+p+j,1:q+p+j} \end{aligned}$$

(39)

From (34) and (39), we deduce that:

$$\begin{aligned}{\varvec{\widehat{R}^{new\star }}}_{1:q+p+j+1,1:q+p+j+1}={\varvec{\widehat{R}^{\star }}}_{1:q+p+j+1,1:q+p+j+1}\end{aligned}$$

Finally:

$$\begin{aligned}{\varvec{\widehat{R}^{\star }}}={\varvec{\widehat{R}^{new\star }}}\end{aligned}$$

Appendix 2: Proof of Corollary 1

Let :

• ${\varvec{\widehat{R}^{\star }}}$ (respectively ${\varvec{\widehat{R}}}$) be the correlation matrix of a path model including error terms (respectivlly excluding error terms).

• ${\varvec{\widehat{R}^{new \star }}}$ (respectively ${\varvec{\widehat{R}^{new}}}$) be the matrix computed using Algorithm (2)(respectivlly defined by (22)).

According to Theorem (2):

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}= {\varvec{\widehat{R}^{\star }}} \end{aligned}$$

(40)

Then from the definition of ${\varvec{\widehat{R}^{new}}}$:

$$\begin{aligned} {\varvec{\widehat{R}^{new\star }}}={\varvec{G\widehat{R}^{new\star }G^t}}={\varvec{G\widehat{R}^{\star }G^t}} \end{aligned}$$

(41)

On the other hand, from (19 ):

$$\begin{aligned}{\varvec{\widehat{R}^{\star }}}=\begin{pmatrix} {\mathbb {E}}[{\varvec{\xi ^{\star }}}{\varvec{\xi ^{\star t}}}] &{} {\mathbb {E}}[{\varvec{\xi ^{\star }}}{\varvec{\eta ^{t}}}]\\ {\mathbb {E}}[{\varvec{\eta \xi ^{\star t}}}] &{} {\mathbb {E}}[{\varvec{\eta \eta ^{t}}}] \end{pmatrix}= \begin{pmatrix} {\mathbb {E}}[{\varvec{\xi \xi ^{t}}}]&{} {\mathbb {E}}[{\varvec{\xi \zeta ^{t}}}] &{} {\mathbb {E}}[{\varvec{\xi \eta ^{t}}}] \\ {\mathbb {E}}[{\varvec{\zeta \xi ^{t}}}] &{} {\mathbb {E}}[{\varvec{\zeta \zeta ^{t}}}] &{} {\mathbb {E}}[{\varvec{\zeta \eta ^{t}}}] \\ {\mathbb {E}}[{\varvec{\eta \xi ^{t}}}] &{} {\mathbb {E}}[{\varvec{\eta \zeta ^{t}}}] &{} {\mathbb {E}}[{\varvec{\eta \eta ^{t}}}] \end{pmatrix}\end{aligned}$$

Hence, from (22):

$$\begin{aligned} G\widehat{R}^{\star }G^{t}&=\begin{pmatrix} {\varvec{I_{q}}}&{}{\varvec{0}}&{}{\varvec{0}}\\ {\varvec{0}}&{}{\varvec{0}}&{}{\varvec{I_{p}}} \end{pmatrix}\begin{pmatrix} {\mathbb {E}}[{\varvec{\xi \xi ^{t}}}]&{} {\mathbb {E}}[{\varvec{\xi \zeta ^{t}}}] &{} {\mathbb {E}}[{\varvec{\xi \eta ^{t}}}] \\ {\mathbb {E}}[{\varvec{\zeta \xi ^{t}}}] &{} {\mathbb {E}}[{\varvec{\zeta \zeta ^{t}}}] &{} {\mathbb {E}}[{\varvec{\zeta \eta ^{t}}}] \\ {\mathbb {E}}[{\varvec{\eta \xi ^{t}}}] &{} {\mathbb {E}}[{\varvec{\eta \zeta ^{t}}}] &{} {\mathbb {E}}[{\varvec{\eta \eta ^{t}}}] \end{pmatrix} \begin{pmatrix} {\varvec{I_{q}}}&{}{\varvec{0}}&{}{\varvec{0}}\\ {\varvec{0}}&{}{\varvec{0}}&{}{\varvec{I_{p}}} \end{pmatrix}^{t}\\&=\begin{pmatrix} {\mathbb {E}}[{\varvec{\xi \xi ^{t}}}]&{} {\mathbb {E}}[{\varvec{\xi \zeta ^{t}}}] &{} {\mathbb {E}}[{\varvec{\xi \eta ^{t}}}] \\ {\mathbb {E}}[{\varvec{\eta \xi ^{t}}}] &{} {\mathbb {E}}[{\varvec{\eta \zeta ^{t}}}] &{} {\mathbb {E}}[{\varvec{\eta \eta ^{t}}}] \end{pmatrix}\begin{pmatrix} {\varvec{I_{q}}}&{}{\varvec{0}}&{}{\varvec{0}}\\ {\varvec{0}}&{}{\varvec{0}}&{}{\varvec{I_{p}}} \end{pmatrix}^t \\&=\begin{pmatrix} {\mathbb {E}}[{\varvec{\xi \xi ^{t}}}]&{} {\mathbb {E}}[{\varvec{\xi \eta ^{t}}}] \\ {\mathbb {E}}[{\varvec{\eta \xi ^{t}}}] &{} {\mathbb {E}}[{\varvec{\eta \eta ^{t}}}] \end{pmatrix}\\&={\varvec{\widehat{R}}} \end{aligned}$$

Therefore:

$$\begin{aligned}{\varvec{\widehat{R}^{new}}}={\varvec{\widehat{R}}}\end{aligned}$$