Inference for ARMA(p,q) Time Series

As we mentioned in our previous post, as soon as we have a moving average part, inference becomes more complicated. Again, to illustrate, we do not need a two general model. Consider, here, some https://latex.codecogs.com/gif.latex?ARMA(1,1) process,

https://latex.codecogs.com/gif.latex?X_t=\phi%20X_{t-1}+\varepsilon_t+\theta%20\varepsilon_{t-1}

where https://latex.codecogs.com/gif.latex?(\varepsilon_t) is some white noise, and assume further that https://latex.codecogs.com/gif.latex?\theta+\phi\neq0.

> theta=.7
> phi=.5
> n=1000
> Z=rep(0,n)
> set.seed(1)
> e=rnorm(n)
> for(t in 2:n) Z[t]=phi*Z[t-1]+e[t]+theta*e[t-1]
> Z=Z[800:1000]
> plot(Z,type="l")

  • A two step procedure

To start with something simple, assume that we did miss the moving average component,

https://latex.codecogs.com/gif.latex?X_t=\phi%20X_{t-1}+u_t

The estimator of https://latex.codecogs.com/gif.latex?\phi – by least squares – is not longer consistent. But still. We can still compute it

> base=data.frame(Y=Z[2:n],X=Z[1:(n-1)])
> regression=lm(Y~0+X,data=base)
> summary(regression)

Call:
lm(formula = Y ~ 0 + X, data = base)

Residuals:
    Min      1Q  Median      3Q     Max 
-3.2445 -0.7909  0.0626  0.9707  3.0685 

Coefficients:
  Estimate Std. Error t value Pr(>|t|)    
X  0.69571    0.05101   13.64   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.225 on 199 degrees of freedom
  (799 observations deleted due to missingness)
Multiple R-squared:  0.4832,	Adjusted R-squared:  0.4806 
F-statistic:   186 on 1 and 199 DF,  p-value: < 2.2e-16

and then, we cancompute the autocorrelation of the noise,

> n=200
> cor(residuals(regression)[2:n],residuals(regression)[1:(n-1)])
[1] 0.2663076

or more formally, use Durbin-Watson estimator, to get autocorrelation of the noise (and some significance test)

> library(car)
> durbinWatsonTest(regression)
 lag Autocorrelation D-W Statistic p-value
   1       0.2656555       1.46323       0
 Alternative hypothesis: rho != 0

The point, here, is that we would like to assume that

https://latex.codecogs.com/gif.latex?u_t=\varepsilon_t+\theta\varepsilon_{t-1}

meaning that https://latex.codecogs.com/gif.latex?(u_t) should be some https://latex.codecogs.com/gif.latex?MA(1) process. And

https://latex.codecogs.com/gif.latex?\rho(1)=\frac{\theta}{1+\theta^2}

i.e. https://latex.codecogs.com/gif.latex?\theta is a root of this quadratic problem,

> polyroot(c(1,-1/cor(residuals(regression)[2:n],residuals(regression)[1:(n-1)]),1))
[1] 0.2884681+0i 3.4665883+0i

Here, we do have two positive roots. I would go for the one smaller than one, in order to be able to invert the polynomial, if necessary…

  •  Use of the empirical autocorrelation function

An alternative might be to use properties of the autocorrelation function,

https://latex.codecogs.com/gif.latex?\gamma(0)=\frac{1+\theta^2+2\phi\theta}{1-\phi^2}\cdot\sigma^2

https://latex.codecogs.com/gif.latex?\gamma(1)=\phi\gamma(0)+\theta\sigma^2

and

https://latex.codecogs.com/gif.latex?\gamma(2)=\phi\gamma(1)

Again, we have a set of three equations, with three unknown parameters. Numerically, it is possible to find some roots. If we run the code, we get

> v=c(as.numeric(acf(Z)$acf[2:3]),1)*var(Z)
> library(rootSolve)
> seteq=function(x){
+ F1=v[1]-x[3]^2*(x[2]^2+2*x[1]*x[2]+1)/(1-x[1]^2)
+ F2=v[2]-(x[1]*v[1]+x[2]*x[3]^2)
+ F3=v[3]-x[1]*v[2]
+ return(c(F1,F2,F3))}
> multiroot(f=seteq,start=c(.1,.1,1))
$root
[1]  3.643734 -3.188145  1.427759

$f.root
[1]  1.371170e-11 -3.714573e-11  0.000000e+00

$iter
[1] 8

$estim.precis
[1] 1.695248e-11

Here, we have a situation…

  • Use of least square techniques

We can use, here, the algorithm described in the context of https://latex.codecogs.com/gif.latex?MA(q) processes.

> V=function(p){
+ phi=p[1]
+ theta=p[2]
+ u=rep(0,length(Z))
+ for(t in 2:length(Z)) u[t]=Z[t]-phi*Z[t-1]-theta*u[t-1]
+ return(sum(u^2))
+ }
> p=optim(par=c(.1,.1),V)$par
[1] 0.3637783 0.7773845
> coef=c(p,sqrt(V(p)/(length(Z))))

which is not so bad. Actually, if we run that procecure on 1,000 samples, we get the following output

  • Use of maximum likelihood techniques

Last, but not least, one more time, we can use (global) maximum likelihood techniques, since the process is a Gaussian process (all finite dimensional vector will have a joint Gaussian distribution) if we assume that the noise is Gaussian.

> library(mnormt)
> GlobalLogLik=function(A,TS){
+ n=length(TS)
+ phi=A[1];  theta=A[2]
+ sigma=A[3]
+ SIG=matrix(0,n,n)
+ rho=rep(0,n)
+ rho[1]=sigma^2*(theta^2+2*phi*theta+1)/(1-phi^2)
+ rho[2]=phi*rho[1]+theta*sigma^2
+ for(h in 3:n) rho[h]=phi*rho[h-1]
+ for(i in 1:n){for(j in 1:n){
+ SIG[i,j]=rho[abs(i-j)+1]}}
+ return(dmnorm(TS,rep(0,n),SIG,log=TRUE))}
> LogL=function(A) -GlobalLogLik(A,TS=Z)
> optim(c(.1,.1,1),LogL)$par
[1] 0.3890991 0.7672036 1.0731340

It works fine, one more time. But maybe we got lucky here. We’ve seen in the post on autoregressive time series that the algorithm might fell if the time series is not stationary. In order to avoid such problems, we can consider a constraint optimization problem, where we simply recall that https://latex.codecogs.com/gif.latex?\phi\in(-1,1),

> U=matrix(c(1,-1,0,0,0,0),2,3)
> C=-c(.999,.999)
> constrOptim(c(.1,.1,1),LogL,grad=NULL,ui=U,ci=C)
$par
[1] 0.3890991 0.7672036 1.0731340

$value
[1] 300.1956

$counts
function gradient 
     118       NA 

$convergence
[1] 0

$message
NULL

$outer.iterations
[1] 2

$barrier.value
[1] -1.536358e-05

If we run that algorithm 1,000 times, on simulated time series (with the same parameters), we get


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