General proof of a limit related to AR(k) model of Statistics

TL;DR

This paper provides a general proof for a limit related to AR(k) models, extending previous specific cases to a comprehensive formula, which is crucial for computing moments of estimators in autoregressive processes.

Contribution

It introduces a fully general proof of a limit formula for AR(k) models, building on prior work that handled only low-dimensional cases.

Findings

General formula for limits in AR(k) models

Extension of previous low-dimensional results

Foundation for computing moments of estimators

Abstract

Computing moments of various parameter estimators related to an autoregressive model of Statistics, one needs to evaluate several non-trivial limits. This was done by arXiv:1506.03131 for the case of two, three and four dimensions; in this article, we present a proof of a fully general formula, based on an ingenious solution of https://mathoverflow.net/users/4312/fedor-petrov.

Equations

$$X_i={\alpha }_1{X}_{i-1}+{\alpha }_2{X}_{i-2}+...+{\alpha }_k{X}_{i-k}+{\epsilon }_i$$

$${\lambda }^k={\alpha }_1{\lambda }^{k-1}+{\alpha }_2{\lambda }^{k-2}+...+{\alpha }_k$$

$${\rho }_j=A_1{\lambda }_1^{|j|}+A_2{\lambda }_2^{|j|}+...+A_k{\lambda }_k^{|j|}$$

$$\sum _{i=1}^n{X}_i$$

$$\sum _{i=1}^{n-j}{X}_i{X}_{i+j}$$

$$\sum _{i_1,i_2,...i_k=1}^{\tilde{n}}{\lambda }_1^{|i_1-i_2+s_1|}{\lambda }_2^{|i_2-i_3+s_2|}...{\lambda }_k^{|i_k-i_1+s_k|}$$

$$A\stackrel{\text{def}}{=}\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i_1,i_2,...i_k=1}^n{\lambda }_1^{|i_1-i_2-s_1|}{\lambda }_2^{|i_2-i_3-s_2|}...{\lambda }_k^{|i_k-i_1-s_k|}$$

$$B_S\stackrel{\text{def}}{=}\sum _{m_1+m_2+...+m_k=S}{\lambda }_1^{|m_1|}{\lambda }_2^{|m_2|}...{\lambda }_k^{|m_k|}$$

$$B_S<\sum _{m_2,m_3,...m_k=-\infty }^{\infty }|{\lambda }_2{|}^{|m_2|}|{\lambda }_3{|}^{|m_3|}...|{\lambda }_k{|}^{|m_k|}$$

$$i_p=i_1+\sum _{j=1}^{p-1}(m_j-s_j)\text{where }p=\mathrm{2...}k$$

$$1\le i_1+\underset{p=\mathrm{2...}k}{\min }\sum _{j=1}^{p-1}(m_j-s_j)$$

$$i_1+\underset{p=\mathrm{2...}k}{\max }\sum _{j=1}^{p-1}(m_j-s_j)\le n$$

$$1-\underset{p=\mathrm{2..}k}{\min }\sum _{j=1}^{p-1}(m_j-s_j)\le i_1\le n-\underset{p=\mathrm{2..}k}{\max }\sum _{j=1}^{p-1}(m_j-s_j)$$

$$n-\underset{p=\mathrm{2...}k}{\max }\sum _{j=1}^{p-1}(m_j-s_j)+\underset{p=\mathrm{2..}k}{\min }\sum _{j=1}^{p-1}(m_j-s_j)$$

$$F(t)\stackrel{\text{def}}{=}\sum _{S=-\infty }^{\infty }{t}^S\sum _{m_1+m_2+...+m_k=S}{\lambda }_1^{|m_1|}{\lambda }_2^{|m_2|}...{\lambda }_k^{|m_k|}=$$

$$\sum _{m_1,m_2,...,m_k=-\infty }^{\infty }{t}^{m_1}{\lambda }_1^{|m_1|}{t}^{m_2}{\lambda }_2^{|m_2|}...t^{m_k}{\lambda }_k^{|m_k|}=$$

$$\sum _{m_1,m_2,...,m_k=-\infty }^{\infty }\prod _{\ell =1}^k{t}^{m_{\ell }}{\lambda }_{\ell }{}^{|m_{\ell }|}=\prod _{\ell =1}^k\sum _{m=-\infty }^{\infty }{t}^m{\lambda }_{\ell }{}^{|m|}=$$

$$\prod _{\ell =1}^k\left(\sum _{m=0}^{\infty }(t{\lambda }_{\ell }{)}^m+\sum _{m=-\infty }^{-1}{t}^m{\lambda }_{\ell }{}^{-m}\right)=\prod _{\ell =1}^k\left(\sum _{m=0}^{\infty }(t{\lambda }_{\ell }{)}^m+\sum _{m=1}^{\infty }(\frac{{\lambda }_{\ell }}{t}{)}^m\right)=$$

$$=\prod _{\ell =1}^k\left(\frac{1}{1-t{\lambda }_{\ell }}+\frac{\frac{{\lambda }_{\ell }}{t}}{1-\frac{{\lambda }_{\ell }}{t}}\right)=\sum _{j=1}^k\left(\frac{C_j}{1-t{\lambda }_j}+\frac{D_j}{t-{\lambda }_j}\right)$$

$$C_j={F(t)(1-t{\lambda }_j)|}_{t={\lambda }_j^{-1}}=\prod _{\begin{array}{c}\ell =1\\ \ell \ne j\end{array}}^k\left(\frac{1}{1-\frac{{\lambda }_{\ell }}{{\lambda }_j}}+\frac{{\lambda }_j{\lambda }_{\ell }}{1-{\lambda }_j{\lambda }_{\ell }}\right)=$$

$$={\lambda }_j^{k-1}\prod _{\begin{array}{c}\ell =1\\ \ell \ne j\end{array}}^k\left(\frac{1}{{\lambda }_j-{\lambda }_{\ell }}+\frac{{\lambda }_{\ell }}{1-{\lambda }_j{\lambda }_{\ell }}\right)={\lambda }_j^{k-1}\prod _{\begin{array}{c}\ell =1\\ \ell \ne j\end{array}}^k\frac{1-{\lambda }_{\ell }^2}{({\lambda }_j-{\lambda }_{\ell })(1-{\lambda }_j{\lambda }_{\ell })}$$

$$\sum _{j=1}^k{\lambda }_j^S{C}_j$$

Full text

1 Introduction

The autoregressive model of Statistics generates a random sequence of observations by

[TABLE]

where $\varepsilon_{i}$ are independent, Normally distributed random variables with the mean of [math] and the same standard deviation, and $k$ is a fixed integer, usually quite small (e.g. $k=1$ defines the so called Markov model). The sufficient and necessary condition for the resulting sequence to be asymptotically stationary is that all $k$ solutions of the characteristic polynomial

[TABLE]

are, in absolute value, smaller than $1$ (this is then assumed from now on).

The $j^{th}$-order serial correlation coefficient $\rho_{j}$ (between $X_{i}$ and $X_{i+j}$) is then computed by

[TABLE]

where the $\lambda_{i}$’s are the $k$ roots of (2), and the $A_{i}$ coefficients are themselves simple functions of these roots. Note that the absolute value of each root must be smaller than $1$ if the resulting stochastic process is be stationary.

Computing the first few moments of various estimators (of the $\alpha_{i}$ parameters) boils down to computing moments of expressions of the

[TABLE]

and

[TABLE]

type, where $X_{1},$ $X_{2},...X_{n}$ is a collection of $n$ consecutive observations (assuming that the process has already reached its stationary phase).

This in turn requires evaluating various summations (see [4]), of which the most difficult has the form of

[TABLE]

where $\lambda_{1},$ $\lambda_{2},...\lambda_{k}$ are the $\lambda_{i}$ roots (some may be multiple), $s_{1},$ $s_{2},...s_{k}$ are (small) integers, and $\tilde{n}$ indicates that the upper limit equals to $n$, adjusted in the manner of (5).

For small $k$, it is possible (but rather messy - see [2]) to exactly evaluate (6) and realize that the answer will always consist of three parts:

• •

terms proportional to $\lambda_{i}^{n},$ which all tend to zero (as $n$ increases) ‘exponentially’,

• •

terms which stay constant as $n$ increases,

• •

terms proportional to $n.$

Luckily, to build an approximation which is usually deemed sufficient (see [4]), we need to find only the $n$ proportional terms. These can be extracted by dividing (6) by $n$ and taking the $n\rightarrow\infty$ limit. Incidentally, this results in the following (and most welcomed) simplification: the corresponding answer will be the same regardless of the $\tilde{n}$ adjustments (thus, we may as well use $n$ instead), and will similarly not depend on the individual $s_{i}$’s, but only on the absolute value of their sum, as the following statement indicates.

2 The main theorem

[TABLE]

where $S=|s_{1}+s_{2}+...+s_{k}|.$

Proof. Define

[TABLE]

where $S$ is the non-negative integer of the theorem.

When $S\geq 0$, a term of $B_{S}$ and a term of the $A$ summation are considered identical (we also say that they match each other) only when $m_{1}=-i_{1}+i_{2}+s_{1},$ $m_{2}=-i_{2}+i_{3}+s_{2},$ … $m_{k-1}=-i_{k-1}+i_{k}+s_{k-1}$ (implying $m_{k}=-i_{k}+i_{1}+s_{k},$ since the $m$’s and $s$’s must add up to the same $S,$ and the $\acute{\imath}$’s cancel); note that this also implies (but not the reverse) that such matching terms have the same value. On the other hand, when $S<0$, we declare them identical when $m_{1}=i_{1}-i_{2}-s_{1},$ etc. instead. From now on, we assume that $S\geq 0$ to avoid a trivial duplication of all subsequent arguments.

Clearly, each term of the $A$ summation matches a term of $B_{S}$: just take $m_{p}=-i_{p}+i_{p+1}+s_{p}$ where $p=1,$ $2,...k$, with the understanding that $i_{k+1}=i_{1}$.

At the same time, no term of $B_{S}$ is matched by more than $n$ terms of the $A$ summation, since once you select $i_{1}$ (from any of its $n$ possible values), all the remaining $i$’s are uniquely determined by $i_{2}=m_{1}+i_{1}-s_{1,}$ $i_{3}=m_{2}+i_{2}-s_{2,}$ etc., resulting in a term of $A$ only when all of these turn out to be between $1$ and $n$ (inclusive).

This proves that $A\leq B_{S}.$

Since $|\lambda_{1}|^{|m_{1}|}\leq 1,$

[TABLE]

implying that the $B_{S}$ sum is (absolutely) convergent; let $B_{\infty}$ denote its actual value. This means that any number smaller that $B_{\infty}$ (say $B_{0}$) can be exceeded by a sum of finitely many terms of $B_{S}$ (this is true for any convergent series).

Now, let us go back to counting how many terms of the $A$ summation match a single, specific term of $B_{S}$; we have already seen that, starting with any one of the possible $n$ values of $i_{1},$ the subsequent $i$’s would be computed by

[TABLE]

matching a term of the $A$ summation only when they are all in the $1$ to $n$ range, i.e. when

[TABLE]

and

[TABLE]

This implies that, for each choice of $i_{1}$ which meet

[TABLE]

we get a legitimate term of the $A$ summation (matching and having the same value as the specific term of $B_{S}$); we thus have

[TABLE]

such terms in total. Dividing their sum by $n$ and taking the $n\rightarrow\infty$ limit thus yields the value of the specific $B_{S}$ term.

This can be repeated for any term of the finite sum of the previous paragraph; thus we get $A\geq B_{0}.$ And, since we can make $B_{0}$ as close to $B_{\infty}$ as we wish, this implies that $A\geq B_{\infty}.$

We have thus shown that (2) and (8) have the same value.

We now define the following Laurent series of the $B_{S}$ sequence (allowing $S$ to have any integer value, and assuming that $\max_{\ell=1...k}|\lambda_{\ell}|<|t|<\min_{\ell=1...k}|\lambda_{\ell}|^{-1}$), namely

[TABLE]

where the last expression is the partial-fraction expansion of the previous rational function of $t$ (the roots of the common denominator are the $\lambda$’s and their inverses). We can now get a formula for $B_{\infty}$ (and thus for our $A$ limit) as a coefficient of $t^{S}$ of the last expression. Since only the $C_{j}$ part contributes to non-negative powers of $t,$ and

[TABLE]

the final formula is therefore given by

[TABLE]

(note that the coefficient of $t^{S}$ in the expansion of $(1-t~{}\lambda_{j})^{-1}$ is $\lambda_{j}^{S}$).

This proves the original statement.

3 Conclusion

The formula of (2) then enables us to evaluate all the expected values needed to deal with any autoregressive model of type (1). Note that in some cases the set of $\lambda_{i}$ values may consist of only a subset of of roots of (2); this only reduces the value of $k$ and makes the result that much easier.

A modification of the formula is needed when some of the $\lambda_{i}$’s are identical; in that case all we have to do is to evaluate the formula’s corresponding limit, such as $\lambda_{5}\rightarrow\lambda_{2}$ when the two $\lambda$’s have the same value (in the case of triple roots, we would need to take two consecutive limits, etc.). This yields a multitude of new (and rather messy) formulas not worth quoting - suffices to say that they all result (as they must) in a finite expression.

A further challenge would be to find the constant part of (6).