Simple and fast convergent procedure to estimate recursive path analysis model

Abstract

BFGS procedure is classically used for the estimation of the parameters of a recursive Path Analysis model. In practice, BFGS does not present any problem of convergence. However, to date, no proof of its convergence is available. The present paper introduces an alternative procedure and establishes its convergence properties. Numerical experiments will be presented, concluding that the proposed alternative seems to converge faster than BFGS procedure.

Appendices

Appendix 1

Proof of Lemma 1

To prove Lemma 1, it is sufficient to prove that $\widehat{\mathbf{R }}({\varvec{\theta }})$ is affine with respect to each parameter. Let $1\le t\le T$.

• 1. If $t\le n_{{\varvec{\Phi }}}$, then $\theta _t$ is an element of ${\varvec{\Phi }}$.

  • Initialization: FIM algorithm gives $\widehat{\mathbf{R }}({\varvec{\theta }})_{1:p,1:p}={\varvec{\Phi }}$. However, ${\varvec{\Phi }}$ is obviously affine with respect to each of its elements; in particular, ${\varvec{\Phi }}$ is affine with respect to $\theta _t$. Consequently, $\widehat{\mathbf{R }}({\varvec{\theta }})_{1:p,1:p}$ is affine with respect to $\theta _t$.

  • Step 1: Let i be an integer, such that $1\le i\le q$ and suppose that the block $\widehat{\mathbf{R }}({\varvec{\theta }})_{1:p+i-1,1:p+i-1}$ is affine with respect to $\theta _t$. Since $\theta _t$ is an element of ${\varvec{\Phi }}$, then $\mathbf{A }$ is constant with respect to $\theta _t$. As a consequence, $\widehat{\mathbf{R }}({\varvec{\theta }})_{p+i,1:p+i-1}=\mathbf{A }_{i,1:p+i-1}\widehat{\mathbf{R }}({\varvec{\theta }})_{1:p+i-1,1:p+i-1}$ is affine with respect to $\theta _t$.

  • Step 2: Since $\widehat{\mathbf{R }}({\varvec{\theta }})_{1:p+i-1,p+i}=\widehat{\mathbf{R }'}({\varvec{\theta }})_{p+i,1:p+i-1}$, then $\widehat{\mathbf{R }}({\varvec{\theta }})_{1:p+i-1,p+i}$ is affine with respect to $\theta _t$.

  • Step 3: Since $\widehat{\mathbf{R }}({\varvec{\theta }})_{p+i,p+i}=1$, then $\widehat{\mathbf{R }}({\varvec{\theta }})_{p+i,p+i}$ is affine with respect to $\theta _t$. As a result, the block $\widehat{\mathbf{R }}({\varvec{\theta }})_{1:p+i,1:p+i}$ is affine with respect to $\theta _t$.

• 2. If $t> n_{{\varvec{\Phi }}}$, then $\theta _t$ is an element of $\mathbf{A }$. And the proof of this part is given in (El Hadri et al. 2020).

As a consequence, $\widehat{\mathbf{R }}({\varvec{\theta }})$ can be decomposed as given in (1). $\square$

Appendix 2

Proof of Lemma 2

Let $t\in \{1,\ldots ,T\}$,

• 1. If $t\le n_{{\varvec{\Phi }}}$, then $\theta _t$ is an element of ${\varvec{\Phi }}$ and $\exists (k,j) \in \{1,\ldots ,p\}$, such that $j\ne k$ and $\theta _t={\varvec{\Phi }}_{kj}$. Using the initialization step of FIM, it comes

  $$\begin{aligned} \widehat{\mathbf{R }}_{k,j}({\varvec{\theta }})={\varvec{\Phi }}_{kj}=\theta _t. \end{aligned}$$

  (26)

  However, from Theorem 1, we get

  $$\begin{aligned} \widehat{\mathbf{R }}_{k,j}({\varvec{\theta }})=\left[ \mathbf{M }({\varvec{\theta }}_{(-t)})\right] _{k,j}+\theta _t\left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] _{k,j}. \end{aligned}$$

  (27)

  Identifying (26) and (27) gives, $\left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] _{k,j}=1$. And y symmetry $\left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] _{j,k}=1$. As a consequence, $\parallel \mathbf{N }({\varvec{\theta }}_{(-t)})\parallel _F^2\ge \left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] _{k,j}^2+\left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] _{j,k}^2=2.$

• 2. If $t> n_{{\varvec{\Phi }}}$, then $\theta _t$ is an element of $\mathbf{A }$ and $\exists k\in \{1,\ldots ,q\}$ and $\exists j\in \{1,\ldots ,p+k-1\}$. such that $\theta _t=\mathbf{A }_{kj}$. Step 1 of FIM gives $\widehat{\mathbf{R }}_{p+k,j}({\varvec{\theta }})=\mathbf{A }_{k,1:p+k-1}\widehat{\mathbf{R }}_{1:p+k-1,j}({\varvec{\theta }})$. Thus, $\widehat{\mathbf{R }}_{p+k,j}({\varvec{\theta }})=\sum _{l=1}^{p+k-1}\mathbf{A }_{k,l}\widehat{\mathbf{R }}_{l,j}({\varvec{\theta }})$. As a consequence

  $$\begin{aligned} \widehat{\mathbf{R }}_{p+k,j}({\varvec{\theta }})=\theta _t+\sum _{l=1,l\ne j}^{p+k-1}\mathbf{A }_{k,l}\widehat{\mathbf{R }}_{l,j} ({\varvec{\theta }}). \end{aligned}$$

  (28)

  However, from Theorem 1, we get

  $$\begin{aligned} \widehat{\mathbf{R }}_{p+k,j}({\varvec{\theta }})=\left[ \mathbf{M }({\varvec{\theta }}_{(-t)})\right] _{p+k,j}+\theta _t\left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] _{p+k,j}. \end{aligned}$$

  (29)

  Thus, by identification of (28) and (29), we get $\left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] _{p+k,j}=1$. And by symmetry, $\left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] _{j,p+k}=1$. As a consequence, $\parallel \mathbf{N }({\varvec{\theta }}_{(-t)}) \parallel _F^2\ge \left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] ^2_{p+k,j}+\left[ \mathbf{N }({\varvec{\theta }}_{(-t)})\right] ^2_{j,p+k}=2.$

$\square$