Category Archives: MAT8181

Examen, Séries Chronologiques

22/04/2014 Arthur Charpentier Leave a comment

Après les exposés des dernières séances, l’examen du cours MAT8181, Séries Chronologiques avait lieu ce matin (et devrait finir dans quelques minutes, avec un peu de temps supplémentaire pour certain, compte tenu de la panne de métro qu’on a eu la joie de subir). L’énoncé est en ligne, et j’ai aussi écrit quelques éléments de correction. En cas de désaccord (mineur ou majeur) avec mes réponses, merci de me le faire savoir rapidement !

MAT8181, Time Series

Stationarity of ARCH processes

06/04/2014 Arthur Charpentier Leave a comment

In the context of AR(1) processes, we spent some time to explain what happens when $\phi$ is close to 1.

if $\phi<1$ the process is stationary,
if $\phi=1$ the process is a random walk
if $\phi>1$ the process will explode

Again, random walks are extremely interesting processes, with puzzling properties. For instance,

$\text{Var}(X_{t+h}\vert X_t)\sim \sigma^2 h\rightarrow\infty$

as $h\rightarrow\infty$ , and the process will cross the x-axis an infinite number of times…

Recently, in the MAT8181 course, we studied carefully properties of the ARCH(1) process, especially when $\alpha\sim 1$ . And again, what we get might be puzzling.

Consider some ARCH(1) process $(\varepsilon_t)$ , with a Gaussian noise, i.e.

$\varepsilon_t=\sigma_t\cdot \eta_t$

where

$\sigma^2_t=\omega+\alpha \varepsilon_{t-1}^2$

and $(\eta_t)$ is a sequence of i.i.d. $\mathcal{N}(0,1)$ variables. Here both $\omega$ and $\alpha$ have to be positive.

Recall that $\mathbb{E}(\varepsilon_t)=0$ since $\mathbb{E}(\eta_t)=0$ . Further

$\text{var}(\varepsilon_t)=\mathbb{E}(\varepsilon_t^2)=\omega+\alpha \mathbb{E}(\varepsilon_{t-1}^2)$

since $\mathbb{E}(\eta_t^2)=1$ , so the variance exists, and is constant only if $\alpha\in(0,1)$ , and in that case

$\sigma^2=\text{var}(\varepsilon_t)=\frac{\omega}{1-\alpha}\in[0,+\infty)$

Further, if $3\alpha^2<1$ , then the fourth moment can be obtained,

$\mathbb{E}(\varepsilon_t^4)=\frac{3\omega^2}{1-\alpha^2}\frac{1-\alpha^2}{1-3\alpha^2}$

since $\mathbb{E}(\eta_t^4)=3$ . Now, if we get back on the property obtained while studying the variance, what does that mean if $\alpha=1$ , or $\alpha>1$ ?

If we look at simulations, we can generate an ARCH(1) process with $\alpha=2$ for instance.

> n=600
> a=2
> w=0.2
> set.seed(1)
> eta=rnorm(n)
> epsilon=rnorm(n)
> sigma2=rep(w,n)
> for(t in 2:n){
+ sigma2[t]=w+a*epsilon[t-1]^2
+ epsilon[t]=eta[t]*sqrt(sigma2[t])
+ }
> plot(epsilon,type="l")

In order to understand what’s going on, we should keep in mind that, what we good is that $\alpha$ has to lie in $(0,1)$ to be able to compute the second moment of $(\varepsilon_t)$ . But it is possible to have a stationary process with infinite variance. And actually, this is what we have here.

Write

$\sigma^2_t=\omega+\alpha \varepsilon_{t-1}^2 = \omega+[\alpha \eta_{t-1}^2] \sigma_{t-1}^2$

and them, iterate

$\sigma^2_t= \omega+[\alpha \eta_{t-1}^2] \left( \omega+[\alpha \eta_{t-2}^2] \sigma_{t-2}^2\right)$

and iterate again, and again, and again…

$\sigma^2_t= \underbrace{\omega\left[ 1+\sum_{i=1}^h[\alpha \eta_{t-1}^2]\cdots [\alpha \eta_{t-i}^2] \right]}_{\Sigma_t(h)} +[\alpha \eta_{t-1}^2]\cdots[\alpha \eta_{t-h-1}^2]\sigma_{t-h-1}^2$

where

$\Sigma_t(h)=\sum_{i=1}^h\underbrace{[\alpha \eta_{t-1}^2]\cdots [\alpha \eta_{t-i}^2]}_{u_i}$

Here, we have a sum of positive terms, and we can use the so-called Cauchy rule: define

$\lambda=\text{limsup}\{ u_n^{1/n}\}$

then, if $\lambda<1$ , the series $\sum u_n$ converges. Here,

$u_n^{1/n}=\left[[\alpha \eta_{t-1}^2]\cdots [\alpha \eta_{t-n}^2]\right]^{1/n}$

which can also be written

$u_n^{1/n}=\exp\left[\frac{1}{n}\sum_{i=1}^n\log[\alpha \eta_{t-i}^2]\right]$

and from the law of large numbers, since we have here a sum of i.i.d. terms,

$u_n^{1/n}\rightarrow\exp\left[\mathbb{E}(\log[\alpha \eta^2])\right]$

So, if $\exp\left[\mathbb{E}(\log[\alpha \eta^2])\right]<1$ , then $\Sigma_t(h)$ will have a limit when $h$ goes to infinity.

The condition above can be written

$\gamma=\mathbb{E}(\log[\alpha \eta^2]<0$

which is called Lyapunov coefficient.

The equation

$\exp\left[\mathbb{E}(\log[\alpha \eta^2])\right]=\alpha\exp\left[-\mathbb{E}(\log[\eta^2])\right]<1$

is a condition on $\alpha$ .

In the case where $\eta\sim\mathcal{N}(0,1)$ , the numerical value of this upper bound is 3.56.

> 1/exp(mean(log(rnorm(1e7)^2)))
[1] 3.562517

In that case ( $\gamma<0$ ), the variance may be infinite, but the series is stationary. On the other hand, if $\gamma>0$ , then $\varepsilon_t^2$ will go to infinity almost surely, as $t$ goes to infinity.

But in order to observe this difference, we need a lot of observations. For instance, with $\alpha=0.8$ ,

and $\alpha=1.2$ ,

we can easily see a difference. I do not say that it’s easy to see that the distribution above has an infinite variance, but still. Actually, if we consider Hill’s plot on the series above, on the tails of positive $\varepsilon_t$ ‘s

> library(evir)
> hill(epsilon)

or on the tails of negative $\varepsilon_t$ ‘s

> hill(-epsilon)

we can see that the tail index is (strictly) smaller than 2 (meaning that the moment of order 2 does not exist).

Why is it puzzling? Maybe because here, $(\varepsilon_t)$ is not weakly stationary (in the $L^2$ sense), but it is strongly stationary. Which is not the usual way weak and strong are related. This might be why we will not call this strong stationarity, but strict.

MAT8181, Time Series

Inference for ARCH processes

02/04/2014 Arthur Charpentier 3 Comments

Consider some ARCH( $p$ ) process, say ARCH( $1$ ),

$\varepsilon_t=\sigma_t\cdot \eta_t$

where

$\sigma^2_t=\omega+\alpha \varepsilon_{t-1}^2$

with a Gaussian (strong) white noise $(\eta_t)$ .

> n=500
> a1=0.8
> a2=0.0
> w= 0.2
> set.seed(1)
> eta=rnorm(n)
> epsilon=rnorm(n)
> sigma2=rep(w,n)
> for(t in 3:n){
+ sigma2[t]=w+a1*epsilon[t-1]^2+a2*epsilon[t-2]^2
+ epsilon[t]=eta[t]*sqrt(sigma2[t])
+ }
> par(mfrow=c(1,1))
> plot(epsilon,type="l",ylim=c(min(epsilon)-.5,max(epsilon)))
> lines(min(epsilon)-1+sqrt(sigma2),col="red")

(the red line is the conditional variance process).

> par(mfrow=c(1,2))
> acf(epsilon,lag=50,lwd=2)
> acf(epsilon^2,lag=50,lwd=2)

We did mention in class that if $(\varepsilon_t)$ a ARCH( $1$ ), then $(\varepsilon_t^2)$ is an AR( $1$ ) process. So a first idea is to consider a regression, as we did for Gaussian AR( $1$ )

> db=data.frame(Y=epsilon[2:n]^2,X1=epsilon[1:(n-1)]^2)
> summary(lm(Y~X1,data=db))

Call:
lm(formula = Y ~ X1, data = db)

Residuals:
    Min      1Q  Median      3Q     Max 
-2.4538 -0.3618 -0.2626  0.0935  9.3667 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.34963    0.04342   8.052 6.08e-15 ***
X1           0.31123    0.04262   7.303 1.13e-12 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.8413 on 497 degrees of freedom
Multiple R-squared:  0.0969,	Adjusted R-squared:  0.09508 
F-statistic: 53.33 on 1 and 497 DF,  p-value: 1.129e-12

There is some significant autocorrelation here. But since our vectors cannot be considered as Gaussian, using least squares is perhaps not the best strategy. Actually, if our series is not Gaussian, it is still conditionally Gaussian, since we assumed that $(\eta_t)$ is a Gaussian (strong) white noise,

$\varepsilon_t\vert \underline{\boldsymbol{\epsilon}}_{t-1} \sim \mathcal{N}(0,\sigma^2_t)$

The likelihood is then

$\mathcal{L}=\prod_{t=1}^T \frac{1}{\sqrt{2\pi \sigma^2_t}}\exp\left(-\frac{\varepsilon_t^2}{2\sigma_t}\right)$

and the log-likelihood is

$\log \mathcal{L}=-\frac{1}{2} \sum_{t=1}^T \log (\sigma_t)-\frac{1}{2}\sum_{t=1}^T \frac{\varepsilon_t^2}{\sigma_t}$

And a natural idea is to define

$(\widehat{\omega},\widehat{\alpha})\in\text{argmax}\{\log \mathcal{L}(\omega,\alpha)\}$

The code is simply

> X=epsilon
> loglik=function(param){
+ w=exp(param[1])
+ a1=exp(param[2])
+ s2=rep(w,n)
+ for(t in 2:length(X)){s2[t]=w+a1*X[t-1]^2}
+ logL=-.5*sum(log(s2))-.5*sum(X^2/s2)
+ return(-logL)
+ }
> OPT=optim(par=
+ coefficients(lm(Y~X1,data=db)),fn=loglik)
> exp(OPT$par)
(Intercept)          X1 
  0.2482241   0.5858578

(since the parameters have to be positive, we assume here that they can be written as the exponential of some real values). Observe that those values are closer to the one used to generate our time series.

If we use R functions to estimate those parameters, we get

> library(tseries)
> summary(garch(epsilon,c(0,1)))
...

Call:
garch(x = epsilon, order = c(0, 1))

Model:
GARCH(0,1)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.87023 -0.60836 -0.03426  0.66648  3.48443 

Coefficient(s):
    Estimate  Std. Error  t value Pr(>|t|)    
a0   0.24959     0.02470   10.104  < 2e-16 ***
a1   0.58306     0.09737    5.988 2.13e-09 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

so that the confidence interval for $\alpha$ is

> summary(garch(epsilon,c(0,1)))$coef[2,1]+
+ c(-1.96,1.96)*summary(garch(epsilon,c(0,1)))$coef[2,2]
[1] 0.3922088 0.7739088

Actually, since our main interest is this $\alpha$ parameter, it is possible to use profile likelihood techniques,

> proflik=function(a){
+ loglik=function(w){
+ s2=rep(w,n)
+ for(t in 2:length(X)){s2[t]=w+a*X[t-1]^2}
+ logL=-.5*sum(log(s2))-.5*sum(X^2/s2)
+ return(-logL)}
+ return(-optim(par=.3,fn=loglik)$value)}

> A=seq(0,2,by=.05)
> P=Vectorize(proflik)(A)
> par(mfrow=c(1,1))
> plot(A,P,type="l")
> OPT=optimize(function(x) -proflik(x), interval=c(0,2))
> t=-OPT$objective-qchisq(.95,df=1)
> abline(h=t,col="red")
> ainf=uniroot(function(x) proflik(x)-t,c(0,OPT$minimum))$root
> asup=uniroot(function(x) proflik(x)-t,c(OPT$minimum,2))$root
>  abline(v=ainf,lty=2)
>  abline(v=asup,lty=2)

Of course, all those techniques can be extended to higher order ARCH processes. For instance, if we assume that we have a ARCH( $2$ ) time series

$\varepsilon_t=\sigma_t\cdot \eta_t$

where now

$\sigma^2_t=\omega+\alpha_1 \varepsilon_{t-1}^2+\alpha_2 \varepsilon_{t-2}^2$

with a Gaussian (strong) white noise $(\eta_t)$ . The log-likelihood is still

$\log \mathcal{L}=-\frac{1}{2} \sum_{t=1}^T \log (\sigma_t)-\frac{1}{2}\sum_{t=1}^T \frac{\varepsilon_t^2}{\sigma_t}$

and we can define

$(\widehat{\omega},\widehat{\alpha}_1,\widehat{\alpha}_2)\in\text{argmax}\{\log \mathcal{L}(\omega,\alpha_1,\alpha_2)\}$

The code above can be changed, to take into account this additional component,

> db=data.frame(Y=epsilon[3:n]^2,
+ X1=epsilon[2:(n-1)]^2,
+ X2=epsilon[1:(n-2)]^2)
> X=epsilon
> loglik=function(param){
+ w=exp(param[1])
+ a1=exp(param[2])
+ a2=exp(param[3])
+ s2=rep(w,n)
+ for(t in 3:length(X)){s2[t]=w+a1*X[t-1]^2+a2*X[t-2]^2}
+ logL=-.5*sum(log(s2))-.5*sum(X^2/s2)
+ return(-logL)
+ }
> OPT=optim(par=
+ coefficients(lm(Y~X1+X2,data=db)),fn=loglik)
> exp(OPT$par)
(Intercept)          X1          X2 
 0.22710526  0.59475474  0.04741294

We can also consider some Generalized ARCH process, e.g. a GARCH( $1$ , $1$ ),

$\varepsilon_t=\sigma_t\cdot \eta_t$

where now

$\sigma^2_t=\omega+\alpha \varepsilon_{t-1}^2+\beta \sigma_{t-1}$

Again, maximum likelihood techniques can be used. Actually, we can also code Fisher-Scoring algorithm, since (in a very general context)

$\frac{\partial \log \mathcal{L}}{\partial \boldsymbol{\theta}}=-\frac{1}{2}\sum_{t=1}^T \frac{\partial \sigma_t^2}{\partial \boldsymbol{\theta}}\cdot \frac{1}{\sigma_t^2}\left(1-\frac{\varepsilon^2_t}{\sigma_t^2}\right)$

with here $\boldsymbol{\theta}=(\omega,\alpha,\beta)$ . Using a standard gradient descent algorithm, we get the following estimate for our GARCH process,

> X=epsilon
> theta=c(.2,.2,.2)
> G=rep(1,3)
> n=length(X)
> j=1
> while(sum(G^2)>1e-12){
+ s2=rep(theta[1],n)
+ for (i in 2:n){s2[i]=theta[1]+theta[2]*X[(i-1)]^2+theta[3]*s2[(i-1)]}
+ z=(X^2-s2)/s2^2
+ V=cbind(z[2:n],z[2:n]*X[1:(n-1)]^2,z[2:n]*s2[1:(n-1)])
+ H=(t(V)%*%V)
+ G=apply(V,2,sum)
+ theta=theta+solve(H)%*%G
+ j=j+1}
> as.numeric(theta)
[1] 0.20372918 0.59183911 0.08936159

The interesting point, here, is that we also derive the (asymptotic) variance

> (stdev=sqrt(diag(solve(H))))
[1] 0.01849067 0.04950477 0.02937233

MAT8181, MAT8595

Independence and correlation

29/03/2014 Arthur Charpentier Leave a comment

A short post to get back on a property I gave briefly in the MAT8595 class in January, and again in the MAT8181 class this week (to illustrate the distinction between weak and strong white noises). Recall that (real-valued) random variables $X$ and $Y$ are independent if for all $A, B\subset\mathbb{R}$ , $\mathbb{P}((X,Y)\in A\times B)=\mathbb{P}(X\in A)\cdot \mathbb{P}(Y\in B)$ Another characterization, for integrable variable is that $\mathbb{E}(f(X)\cdot g(Y)) = \mathbb{E}(f(X))\cdot \mathbb{E}( g(Y))$ for all $f,g:\mathbb{R}\mapsto \mathbb{R}$ , which can be written, if variables are square integrable $\text{corr}(f(X), g(Y))=0$ The idea to prove this characterization is to observe that if $X$ and $Y$ are independent $\mathbb{P}((X,Y)\in A\times B)=\mathbb{P}(X\in A)\cdot \mathbb{P}(Y\in B)$ can be written $\mathbb{E}(\boldsymbol{1}_A(X)\cdot \boldsymbol{1}_B(Y)) = \mathbb{E}(\boldsymbol{1}_A(X))\cdot \mathbb{E}( \boldsymbol{1}_B(Y))$ Using a standard argument in integration theory, equality $\mathbb{E}(f(X)\cdot g(Y)) = \mathbb{E}(f(X))\cdot \mathbb{E}( g(Y))$ is valid for step functions (not only indicators), and then to positive measurable functions, and finally to integrable functions. Proving this result is not that difficult. Observe that Rényi (1959) – inspired by Gebelein (1947) – followed by Sarmanov (1958) introduced the concept of maximal correlation, that can be related to this result, $\text{corr}^{\star}(X,Y)=\max\{\text{corr}(f(X),g(Y))\}$ where the maximum is taken over all functions $f$ and $g$ such that the correlation exist. Actually, it is possible to consider only transformations such that $\mathbb{E}(f(X))=0$ and $\text{Var}(f(X))=1$ (and similarly for $g$ , the idea is that we simple center and scale, which does not impact the correlation.Thus, $X$ and $Y$ are independent if and only if $\text{corr}^{\star}(X,Y)=0$ Algorithm to estimate that coefficient are interesting. The problem can be written, equivalently $(f^\star,g^\star)=\text{argmin}\{\mathbb{E}[f(X)-g(Y)]^2\}$ And if the minimization is considered over $g$ , assuming that $f$ is fixed, then the optimal transformation is $g(y)=\mathbb{E}[f(X)\vert Y=y]$ And similarly for $f$ . So using an iterative algorithm, it is possible to get $f^\star$ and $g^\star$ (see Breiman and Friedman (1985) for more details). Actually, those functions appear in nonlinear canonical analysis. As mentioned in Lancaster (1957), for a Gaussian random vector $(X,Y)$ $\text{corr}^{\star}(X,Y)=\vert \text{corr}(X,Y)\vert$ and in that case $f^\star$ and $g^\star$ are affine functions. This can be related to Hermite’s polynomial and to the expansion of the bivariate Gaussian density. I still hope that someone will go further for the project in the MAT8181 course.

MAT8181, Time Series

Seasonal Unit Roots

26/03/2014 Arthur Charpentier 2 Comments

As discussed in the MAT8181 course, there are – at least – two kinds of non-stationary time series: those with a trend, and those with a unit-root (they will be called integrated). Unit root tests cannot be used to assess whether a time series is stationary, or not. They can only detect integrated time series. And the same holds for seasonal unit root.

In a previous post, we’ve seen that it was difficult to model hourly temperature, since most test do not reject unit roots. Consider here the average monthly temperature, still in Montréal, QC.

> montreal=read.table("http://freakonometrics.free.fr/temp-montreal-monthly.txt")
> M=as.matrix(montreal[,2:13])
> X=as.numeric(t(M))
> tsm=ts(X,start=1948,freq=12)
> plot(tsm)

For those who don’t know Montréal, Winter and Summer are very different. We can visualize monthly differences using

> month=rep(1:12,length(tsm)/12)
> plot(month,as.numeric(tsm))
> lines(1:12,apply(M,2,mean),col="red",type="b",pch=19)

or, if install the uroot package, removed from the CRAN repository, we can use

> library(uroot)
> bbplot(tsm)

> bb3D(tsm)
Loading required package: tcltk

It looks like our time series is cyclic, because of the yearly seasonal pattern. The autocorrelation function is here

> acf(tsm,lag=120)

Again, this cycle can be visualized using

> persp(1948:2013,1:12,M,theta=-50,col="yellow",shade=TRUE,
+ xlab="Year",ylab="Month",zlab="Temperature",ticktype="detailed")

Now, the question is is there a seasonal unit root ? This would mean that our model should be

$(1-L^{12})\Phi(L)[X_t-\mu]=\Theta(L)\varepsilon_t$

If we forget about the autoregressive and the moving average component, we can estimate

$(1-{\color{red}\alpha}L^{12})[X_t-\mu]=\varepsilon_t$

If there is a seasonal unit root then ${\color{red}\alpha}$ should be close to 1. Somehow.

> arima(tsm,order=c(0,0,0),seasonal=list(order=c(1,0,0),period=12))

Call:
arima(x = tsm, order = c(0, 0, 0), seasonal = list(order = c(1, 0, 0), period = 12))

Coefficients:
        sar1  intercept
      0.9702     6.4566
s.e.  0.0071     2.1515

It is not far away from 1. Actually, it cannot be too close to 1. If it was, then we would get an error message…

To illustrate some interesting models, let us consider also quarterly temperatures,

> N=cbind(apply(montreal[,2:4],1,sum),apply(montreal[,5:7],1,sum),apply(montreal[,8:10],1,sum),apply(montreal[,11:13],1,sum))
> X=as.numeric(t(N))
> tsq=ts(X,start=1948,freq=4)
> persp(1948:2013,1:4,N,theta=-50,col="yellow",shade=TRUE,
+ xlab="Year",ylab="Quarter",zlab="Temperature",ticktype="detailed")

(again, the aim is just to be able to write down some equations, if necessary)

Why not consider a $VAR(1)$ model on the quarterly temperature? Something like

$\left(\begin{matrix}X_t^{(Q1)}\\ X_t^{(Q2)\\ X_t^{(Q3)\\ X_t^{(Q4)\end{matrix}\right)= A \left(\begin{matrix}X_{t-1}^{(Q1)}\\ X_{t-1}^{(Q2)\\ X_{t-1}^{(Q3)\\ X_{t-1}^{(Q4)\end{matrix}\right)+\left(\begin{matrix}\varepsilon_{t-1}^{(Q1)}\\ \varepsilon_{t-1}^{(Q2)\\ \varepsilon_{t-1}^{(Q3)\\ \varepsilon_{t-1}^{(Q4)\end{matrix}\right)$

i.e.

$\boldsymbol{Y}_t=A \boldsymbol{Y}_{t-1}+\boldsymbol{\varepsilon}_t$

where $A$ is some matrix $4\times 4$ . This model can easily be estimated,

> library(vars)
> df=data.frame(N)
> names(df)=paste("y",1:4,sep="")
> model=VAR(df)
> model

VAR Estimation Results:
======================= 

Estimated coefficients for equation y1: 
======================================= 
Call:
y1 = y1.l1 + y2.l1 + y3.l1 + y4.l1 + const 

       y1.l1        y2.l1        y3.l1        y4.l1        const 
 -0.13943065   0.21451118   0.08921237   0.30362065 -34.74793931 

Estimated coefficients for equation y2: 
======================================= 
Call:
y2 = y1.l1 + y2.l1 + y3.l1 + y4.l1 + const 

      y1.l1       y2.l1       y3.l1       y4.l1       const 
 0.02520938  0.05288958 -0.13277377  0.05134148 40.68955266 

Estimated coefficients for equation y3: 
======================================= 
Call:
y3 = y1.l1 + y2.l1 + y3.l1 + y4.l1 + const 

      y1.l1       y2.l1       y3.l1       y4.l1       const 
 0.07740824 -0.21142726  0.11180518  0.12963931 56.81087283 

Estimated coefficients for equation y4: 
======================================= 
Call:
y4 = y1.l1 + y2.l1 + y3.l1 + y4.l1 + const 

      y1.l1       y2.l1       y3.l1       y4.l1       const 
 0.18842863 -0.31964579  0.25099508 -0.04452577  5.73228873

and matrix $A$ is here

> A=rbind(
+ coefficients(model$varresult$y1)[1:4],
+ coefficients(model$varresult$y2)[1:4],
+ coefficients(model$varresult$y3)[1:4],
+ coefficients(model$varresult$y4)[1:4])
> A
           y1.l1       y2.l1       y3.l1       y4.l1
[1,] -0.13943065  0.21451118  0.08921237  0.30362065
[2,]  0.02520938  0.05288958 -0.13277377  0.05134148
[3,]  0.07740824 -0.21142726  0.11180518  0.12963931
[4,]  0.18842863 -0.31964579  0.25099508 -0.04452577

Since stationary of this multiple time series is closely related to eigenvalues of this matrix, let us look at them,

> eigen(A)$values
[1]  0.35834830 -0.32824657 -0.14042175  0.09105836
> Mod(eigen(A)$values)
[1] 0.35834830 0.32824657 0.14042175 0.09105836

So it looks like there is no stationarity issue, here. A restricted model is the periodic autoregressive model, so called $PAR(1)$ model, discussed by Paap and Franses

$\boldsymbol{\Phi}_0 \boldsymbol{Y}_t=\boldsymbol{\Phi}_1 \boldsymbol{Y}_{t-1}+\boldsymbol{\varepsilon}_t$

where

$\boldsymbol{\Phi}_0=\left(\begin{array}{cccc}1&0&0&0\\ -\phi_{1,2}&1&0&0\\ 0&-\phi_{1,3}&1&0\\ 0&0&-\phi_{1,4}&1\end{array}\right)$

and

$\boldsymbol{\Phi}_1=\left(\begin{array}{cccc}0&0&0&\phi_{1,1}\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right)$

Keep in mind that this is a $VAR(1)$ model, since

$\boldsymbol{Y}_t=\underbrace{\boldsymbol{\Phi}_0^{-1} \boldsymbol{\Phi}_1}_A \boldsymbol{Y}_{t-1}+\boldsymbol{\varepsilon}_t$

This model can be estimated using a specific package (one can also look at the vignette, to get a better understanding of the syntax)

> library(partsm)
> detcomp <- list(regular=c(0,0,0), seasonal=c(1,0), regvar=0)
> model=fit.ar.par(wts=tsq, detcomp=detcomp, type="PAR", p=1)
> PAR.MVrepr(model)
----
    Multivariate representation of a PAR model.

  Phi0:

  1.000  0.000  0.000 0
 -0.242  1.000  0.000 0
  0.000 -0.261  1.000 0
  0.000  0.000 -0.492 1

  Phi1:

 0 0 0 0.314
 0 0 0 0.000
 0 0 0 0.000
 0 0 0 0.000

  Eigen values of Gamma = Phi0^{-1} %*% Phi1:
0.01 0 0 0 

  Time varing accumulation of shocks:

 0.010 0.040 0.155 0.314
 0.002 0.010 0.037 0.076
 0.001 0.003 0.010 0.020
 0.000 0.001 0.005 0.010

Here, the characteristic equation is

$|\boldsymbol{\Phi}_0-\boldsymbol{\Phi}_1 z |=(1-\phi_{1,1}\phi_{1,2}\phi_{1,3}\phi_{1,4}\cdot z)$

so there is a (seasonal) unit root if

$\phi_{1,1}\phi_{1,2}\phi_{1,3}\phi_{1,4}=1$

Which is clearly not the case, here. It is possible to perform Canova-Hansen test,

> CH.test(tsq)

  ------ - ------ ----
  Canova & Hansen test
  ------ - ------ ----

  Null hypothesis: Stationarity.
  Alternative hypothesis: Unit root.
  Frequency of the tested cycles: pi/2 , pi , 

  L-statistic: 1.122  
  Lag truncation parameter: 5 

  Critical values:

  0.10 0.05 0.025 0.01
 0.846 1.01  1.16 1.35

The idea is that polynomial $(1-L^4)$ has four root, in $\mathbb{C}$ ,

$\{+1,+i,-1,-i\}$

since

$(1-L^4)=(1-L)(1+L)(1+L^2)=(1-L)(1+L)(1-iL)(1+iL)$

If we get back to monthly data, $(1-L^{12})$ has twelve roots,

$\left\{\pm 1, \ \pm i, \ \pm\frac{1\pm \sqrt{3}i}{2},\pm\frac{\sqrt{3}\pm i}{2}\right\}$

each of them having different interpretations.

Here we can have 1 cycle per year (on 12 months), 2 cycles per year (on 6 months), 3 cycles per year (on 4 months), 4 cycles per year (on 3 months), even 6 cycles per year (on 2 months). This will depend on the argument of the root, with respectively

$\left\{\frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6},\pi\right\}$

The output of the test is here

> CH.test(tsm)

  ------ - ------ ----
  Canova & Hansen test
  ------ - ------ ----

  Null hypothesis: Stationarity.
  Alternative hypothesis: Unit root.
  Frequency of the tested cycles: pi/6 , pi/3 , pi/2 , 2pi/3 , 5pi/6 , pi , 

  L-statistic: 1.964  
  Lag truncation parameter: 20 

  Critical values:

 0.10 0.05 0.025 0.01
 2.49 2.75  2.99 3.27

And it looks like we reject the assumption of a seasonal unit root. I can even mention the following testing procedure

> library(forecast)
> nsdiffs(tsm, test="ch")
[1] 0

where the ouput: “1” means that there is a seasonal unit root and “0” that there is no seasonal unit root. Simple to read, isn’t it? If we consider the periodic autoregressive model on the monthly data, the output is

> model=fit.ar.par(wts=tsm, detcomp=detcomp, type="PAR", p=1)
> model
----
  PAR model of order 1 .

  y_t = alpha_{1,s}*y_{t-1} + alpha_{2,s}*y_{t-2} + ... + alpha_{p,s}*y_{t-p} + coeffs*detcomp + epsilon_t,  for s=1,2,...,12
----
  Autoregressive coefficients. 

          s=1  s=2  s=3  s=4  s=5 s=6 s=7  s=8  s=9 s=10 s=11 s=12
alpha_1s 0.15 0.05 0.07 0.33 0.11   0 0.3 0.38 0.31 0.19 0.15 0.37

So, whatever the test, we always reject the assumption that there is a seasonal unit root. Which does not mean that we can not have a strong cycle! Actually, the series is almost periodic. But there is no unit root! So all of this makes sense (I hardly believe that there might be unit root – seasonal, or not – in temperatures).

Just to make sure that we get it right, consider two times series, inspired from the previous one. The first one is a periodic sequence (with a very very small noise, just to avoid problem of non-definite matrices) and the second one is clearly integrated.

> Xp1=Xp2=as.numeric(t(M))
> for(t in 13:length(M)){
+ Xp1[t]=Xp1[t-12]
+ Xp2[t]=Xp2[t-12]+rnorm(1,0,2)
+ }
> Xp1=Xp1+rnorm(length(Xp1),0,.02)
> tsp1=ts(Xp1,start=1948,freq=12)
> tsp2=ts(Xp2,start=1948,freq=12)
> par(mfrow=c(2,1))
> plot(tsp1)
> plot(tsp2)

Linear ‘Prediction’ for AR Time Series

25/03/2014 Arthur Charpentier Leave a comment

In the exercises for the MAT8181 course, there are two Exercises (16 and 17) about prediction and extrapolation based on MA(1) and AR(1) time series. But before discussing those exercises (I had some request for hints), I wanted to recall the definition of the linear prediction,

$EL(Y\vert \boldsymbol{X})=\boldsymbol{\alpha}^{\star \text{\sffamily T}} \boldsymbol{X}$

where

$\boldsymbol{\alpha}^{\star}=\text{argmin}\left\{\mathbb{E}([Y-\boldsymbol{\alpha}^{\text{\sffamily T}} \boldsymbol{X}]^2)\right\}$

As discussed previously on this blog, we consider here on projection not on $\sigma\{\boldsymbol{X}\}$ (this would be the conditional expectancy) but on the linear subset.

The goal of Exercise 2 was to establish an important result, in the context of Gaussian random vectors. If $\boldsymbol{X}$ is a (multivariate) Gaussian vector, $\mathcal{N}(\boldsymbol{0},\boldsymbol{\Sigma})$ , then

$\boldsymbol{\alpha}^{\star}=\boldsymbol{\Sigma}^{-1}\boldsymbol{\gamma}$

where $\boldsymbol{\gamma}$ is the vector $\text{cov}(Y,\boldsymbol{X})$ .

Keep those results in mind, and let us look at Exercise 17, for instance. Here, $(X_t)$ is an AR(1) process, with innovation $(\varepsilon_t)$ ,

$X_{t}=\phi X_{t-1}+\varepsilon_t$

One observation (say $X_\tau$ ) is missing. We have here 3 questions:

what is the best linear prediction of $X_\tau$ given $X_{\tau-2}$ and $X_{\tau-1}$
what is the best linear prediction of $X_\tau$ given $X_{\tau+2}$ and $X_{\tau+1}$
what is the best linear prediction of $X_\tau$ given $X_{\tau-1}$ and $X_{\tau+1}$

Case 1. Here, we have to compute

$EL(X_{\tau}\vert X_{\tau-1}, X_{\tau-2})=\alpha_{-1}^\star X_{\tau-1}+\alpha_{-2}^\star X_{\tau-2}$

Since we have an AR(1) process, $\text{cov}(X_\tau,X_{\tau-1})=\phi$ , $\text{cov}(X_\tau,X_{\tau-2})=\phi^2$ and $\text{cov}(X_{\tau-1},X_{\tau-2})=\phi$ . Thus, from the relationship above

$\left(\begin{array}{ll}1 & \phi\\\phi & 1\end{array}\right)\left(\begin{array}{l}\alpha_{-2}^\star\\ \alpha_{-1}^\star\end{array}\right)=\left(\begin{array}{l}\phi^2\\\phi\end{array}\right)$

which can be written

$\left(\begin{array}{l}\alpha_{-2}^\star\\ \alpha_{-1}^\star\end{array}\right)=\left(\begin{array}{ll}1 & \phi\\\phi & 1\end{array}\right)^{-1}\left(\begin{array}{l}\phi^2\\\phi\end{array}\right)=\frac{1}{\phi^2-1}\left(\begin{array}{ll}-1 & \phi\\\phi & -1\end{array}\right)\left(\begin{array}{l}\phi^2\\\phi\end{array}\right)$

i.e. $(\alpha_{-2}^\star,\alpha_{-1}^\star)=(0,\phi)$ . Which makes sense actually: the AR(1) process is Markovian of order one, so

$EL(X_\tau\vert \underline{\boldsymbol{X}}_{\tau-1})=EL(X_\tau\vert X_{\tau-1})$

And we seen in class that for an AR(1) process

${}_{\tau}\widehat{X}_{\tau+1}=EL(X_\tau\vert \underline{\boldsymbol{X}}_{\tau-1})=\phi X_{\tau-1}$

So far, so good.

Case 2. Here, we have to compute

$EL(X_{\tau}\vert X_{\tau+1}, X_{\tau+2})=\alpha_{+1}^\star X_{\tau+1}+\alpha_{+2}^\star X_{\tau+2}$

Since we have an AR(1) process, $\text{cov}(X_\tau,X_{\tau+1})=\phi$ , $\text{cov}(X_\tau,X_{\tau+2})=\phi^2$ and $\text{cov}(X_{\tau+1},X_{\tau+2})=\phi$ . Thus, from the relationship above

$\left(\begin{array}{ll}1 & \phi\\\phi & 1\end{array}\right)\left(\begin{array}{l}\alpha_{+1}^\star\\ \alpha_{+2}^\star\end{array}\right)=\left(\begin{array}{l}\phi\\\phi^2\end{array}\right)$

i.e. $(\alpha_{+1}^\star,\alpha_{+2}^\star)=(\phi,0)$ .

Case 3. Finally, we have to compute

$EL(X_{\tau}\vert X_{\tau-1}, X_{\tau+1})=\alpha_{-1}^\star X_{\tau-1}+\alpha_{+1}^\star X_{\tau+1}$

One more time, since we have an AR(1) process, $\text{cov}(X_\tau,X_{\tau-1})=\phi$ , $\text{cov}(X_{\tau-1},X_{\tau+1})=\phi^2$ and $\text{cov}(X_{\tau},X_{\tau+1})=\phi$ . So here, the relationship above becomes

$\left(\begin{array}{ll}1 & \phi^2\\\phi^2 & 1\end{array}\right)\left(\begin{array}{l}\alpha_{-1}^\star\\ \alpha_{+1}^\star\end{array}\right)=\left(\begin{array}{l}\phi\\\phi\end{array}\right)$ Here, we can write

$\left(\begin{array}{l}\alpha_{-1}^\star\\ \alpha_{+1}^\star\end{array}\right)=\left(\begin{array}{ll}1 & \phi^2\\\phi^2 & 1\end{array}\right)^{-1}\left(\begin{array}{l}\phi\\\phi\end{array}\right)=\frac{1}{\phi^4-1}\left(\begin{array}{ll}-1 & \phi^2\\\phi^2 & -1\end{array}\right)\left(\begin{array}{l}\phi\\\phi\end{array}\right)$

i.e.

$\alpha_{-1}^\star=\frac{\phi}{\phi^2+1}=\alpha_{+1}^\star$

So finally, what we got here is

$EL(X_{\tau}\vert X_{\tau-2}, X_{\tau-1})=\phi X_{\tau-1}$

$EL(X_{\tau}\vert X_{\tau+1}, X_{\tau+2})=\phi X_{\tau+1}$

and

$EL(X_{\tau}\vert X_{\tau-1}, X_{\tau+1})=\frac{\phi}{\phi^2+1}\left(X_{\tau-1}+X_{\tau+1}\right)$

The mean squared errors for each of those estimates are obtained computing

$\mathbb{E}\left([X_\tau-EL(X_{\tau}\vert\boldsymbol{X})]^2\right)$

I guess I should probably stop here… that’s a detailed hint actually.

MAT8181, Time Series

Seasonal, or periodic, time series

20/03/2014 Arthur Charpentier 2 Comments

Monday, in our MAT8181 class, we’ve discussed seasonal unit roots from a practical perspective (the theory will be briefly mentioned in a few weeks, once we’ve seen multivariate models). Consider some time series $(X_t)$ , for instance traffic on French roads,

> autoroute=read.table(
+ "http://freakonometrics.blog.free.fr/public/data/autoroute.csv",
+ header=TRUE,sep=";")
> X=autoroute$a100
> T=1:length(X)
> plot(T,X,type="l",xlim=c(0,120))
> reg=lm(X~T)
> abline(reg,col="red")

As discussed in a previous post, if there is a trend, we should remove it, and work on the residual $Y_t=X_t-[a+bt]$

> Y=residuals(reg)
> acf(Y,lag=36,lwd=3)

We can observe that there is some seasonality, here. A first strategy might be to assume that there is a seasonal unit root, so we consider $Z_t=(1-L^{12})Y_t$ , and we try to find some ARMA process. Consider the empirical autocorrelation function of that time series,

> Z=diff(Y,12)
> acf(Z,lag=36,lwd=3)

or the partial autocorrelation function

> pacf(Z,lag=36,lwd=3)

The first graph might suggest a MA(1) structure, while the second graph might suggest an AR(1) time series. Let us try both.

> model1=arima(Z,order=c(0,0,1))
> model1

Call:
arima(x = Z, order = c(0, 0, 1))

Coefficients:
          ma1  intercept
      -0.2367  -583.7761
s.e.   0.0916   254.8805

sigma^2 estimated as 8071255:  log likelihood = -684.1,  aic = 1374.2

> E1=residuals(model1)
> acf(E1,lag=36,lwd=3)

which can be considered as a white noise (if you’re not convinced, try either Box-Pierce or Ljung-Box test). Similarly,

> model2=arima(Z,order=c(1,0,0))
> model2

Call:
arima(x = Z, order = c(1, 0, 0))

Coefficients:
          ar1  intercept
      -0.3214  -583.0943
s.e.   0.1112   248.8735

sigma^2 estimated as 7842043:  log likelihood = -683.07,  aic = 1372.15

> E2=residuals(model2)
> acf(E2,lag=36,lwd=3)

which can be also considered as a white noise. So what we have, so far is

$(1-\phi L){Z_t}=(1-\phi L)\underbrace{(1-L^{12})Y_t}=\varepsilon_t$

for some white noise $(\varepsilon_t)$ . This suggest the following SARIMA structure on $(Y_t)$ ,

> model2b=arima(Y,order=c(1,0,0),
+               seasonal = list(order = c(0, 1, 0),
+               period=12)) 
> model2b

Call:
arima(x = Y, order = c(1, 0, 0), seasonal = list(order = c(0, 1, 0), period = 12))

Coefficients:
          ar1
      -0.2715
s.e.   0.1130

sigma^2 estimated as 8412999:  log likelihood = -685.62,  aic = 1375.25

So far, so good. Now, what if we consider that we do not have a seasonal unit root, but simply a large autoregressive coefficient in some AR structure. Let us try something like

$(1-\phi L){(1-{\color{red}\phi_{12}}L^{12})Y_t}=\varepsilon_t$

where a natural guess is that this coefficient should – probably – be close to one. Let us try this one,

> model3c=arima(Y,order=c(1,0,0),
+               seasonal = list(order = c(1, 0, 0), 
+               period = 12))
> model3c

Call:
arima(x = Y, order = c(1, 0, 0), seasonal = list(order = c(1, 0, 0), period = 12))

Coefficients:
          ar1    sar1  intercept
      -0.1629  0.9741  -684.9455
s.e.   0.1170  0.0115  3064.4040

sigma^2 estimated as 8406080:  log likelihood = -816.11,  aic = 1640.21

which is comparable with what we got previously (somehow), so we might assume that this model can be considered as an interesting one. We will discuss further the fact that the first coefficient might be considered as non-significant.

What is the difference from those two models? With a short term horizon, the two models are comparable. Clearly

> library(forecast)
> previ=function(model,h=36,b=40000){
+ prev=forecast(model,h)
+ T=1:85
+ Tfutur=86:(85+h)
+ plot(T,Y,type="l",xlim=c(0,85+h),ylim=c(-b,b))
+ polygon(c(Tfutur,rev(Tfutur)),c(prev$lower[,2],rev(prev$upper[,2])),col="orange",border=NA)
+ polygon(c(Tfutur,rev(Tfutur)),c(prev$lower[,1],rev(prev$upper[,1])),col="yellow",border=NA)
+ lines(prev$mean,col="blue")
+ lines(Tfutur,prev$lower[,2],col="red")
+ lines(Tfutur,prev$upper[,2],col="red")
+ }

Now, on a (very) long term perspective, the models are quite different: one is stationnary, so the forecast will tend to the average value (here 0, since the trend was removed), while the other one is (seasonaly) integrated, so the confidence interval will increase. For the non stationry, we get

> previ(model2b,600,b=60000)

and for the stationary one

> previ(model3c,600,b=60000)

But as mentioned in the introduction of this course, forecasts with those models are relevent only for short-term horizon (say not too large). And in that case, the prediction is almost the same here,

> previ(model2b,36,b=60000)

> previ(model3c,36,b=60000)

Now, if we come back on our second model, we did mention previously that the autoregressive coefficient might be considered as non-significant. What if we remove it?

> model3d=arima(Y,order=c(0,0,0),
+               seasonal = list(order = c(1, 0, 0), 
+               period = 12))
> (model3d)

Call:
arima(x = Y, order = c(0, 0, 0), seasonal = list(order = c(1, 0, 0), period = 12))

Coefficients:
        sar1  intercept
      0.9662  -696.5661
s.e.  0.0134  3182.3017

sigma^2 estimated as 8918630:  log likelihood = -817.03,  aic = 1640.07

If we look at a (short-term) forecast, we get

> previ(model3d,36,b=32000)

Do you see any difference? To be honest, I don’t… If we look at the figures, we get

> cbind(forecast(model2b,12)$mean,forecast(model3c,12)$mean,forecast(model3d,12)$mean)
Time Series:
Start = 86 
End = 97 
Frequency = 1 
1   -4908.4920  -5092.8999  -5520.8780
2  -10012.7837  -9640.8103  -9493.0339
3   -3880.2202  -3841.1960  -3828.2611
4  -18102.5211 -17638.4086 -17499.1828
5  -20602.7346 -20090.9117 -19934.1066
6  -10463.2212 -10209.0139 -10132.0439
7    2458.1538   2376.4897   2351.2377
8   -1680.3342  -1654.4844  -1647.0057
9     876.6837    836.2342    823.4934
10  18046.5642  17561.6520  17413.1463
11  21531.4820  20956.3451  20780.2836
12  -3217.6103  -3152.0446  -3132.4112

Figures are different, but not significantly (keep in mind the size of the confidence interval). This might explain why, in R, when we ask for an autoregressive process or order $https://latex.codecogs.com/gif.latex?p$ , then we get a model with $https://latex.codecogs.com/gif.latex?p$ parameters to estimate, and even if some are not significant, we usually keep them for the forecast. Most of the time, from forecasting point of view, it’s no big deal.

MAT8181, Time Series

Filtering a Stationary Time Series

13/03/2014 Arthur Charpentier 1 Comment

In the first part of the MAT8181 course, on linear (univariate) time series, I forgot to mention an important theorem. Let $(X_t)_{t\in\mathbb{Z}}$ be a stationary time series, and $(a_t)_{t\in\mathbb{Z}}$ a sequence of real numbers such that

$\sum_{t\in\mathbb{Z}} \vert a_t\vert < \infty$

then the time series $(Y_t)_{t\in\mathbb{Z}}$ defined as

$Y_t=\sum_{k\in\mathbb{Z}} a_k X_{t-k}= \left( \sum_{k\in\mathbb{Z}} a_k L^k\right) X_t$

is a stationary time series. Further, one can get easily that

$\gamma_Y(h)=\text{cov}(Y_t,Y_{t-k})=\sum_{r,s\in\mathbb{Z}} a_r a_s \underbrace{\gamma_X(h+r-s)}_{\text{cov}(X_{t-r},X_{t-s+h})}$

This result can be used, if necessary in the exercises (that might save some time actually). I did not include this property in my notes because it is a bit technical to establish that this sum exists, and that the time series is stationary. It is rather simple with the spectral density (since $f_Y(\omega)=\vert A(e^{i\omega})\vert^2f_X(\omega)$ where $A(\cdot)$ stands for the filter generating function), but I did not mention the spectral density since it requires some knowledge on Fourier analysis…

MAT8181, Tests, Time Series

Identification of ARMA processes

19/02/2014 Arthur Charpentier 3 Comments

Last week (in the MAT8181 course) in order to identify the orders of an ARMA process, we’ve seen the eacf method, and I mentioned the scan method, introduced in Tsay and Tiao (1985). The code below – to produce the output of the scan procedure – has been adapted from an old code by Steve Chen (where I included a visualization of the p-values, with the following colors)

The procedure was described in the course, last Thursday,

arma.scan=function(z,ar.max=15,ma.max=15,alpha=0.01)
{
  ym=function(z,t,m){return(z[t:(t-m)])}
  n=length(z)
  z=z - mean(z)
  cmax=ma.max + 1
  rmax=ar.max + 1
  corref=matrix(0,nrow=rmax,ncol=cmax)
  cmj.table=matrix(0,nrow=rmax,ncol=cmax)
  pv=matrix(0,nrow=rmax,ncol=cmax)
  mark=matrix(rep("X",(rmax)*(cmax)),nrow=rmax,ncol=cmax)
  Rnames=paste("AR",0:(ar.max),sep="-")
  Cnames=paste("MA",0:(ma.max),sep="-")
  rownames(corref)=Rnames
  colnames(corref)=Cnames
  rownames(cmj.table)=Rnames
  colnames(cmj.table)=Cnames
  rownames(pv)=Rnames
  colnames(pv)=Cnames
  rownames(mark)=Rnames
  colnames(mark)=Cnames
  for (m in 0:ar.max)
  {
   m1=m+1
   for (j in 0:ma.max)
   {
   j1=j+1 
   if (m == 0 && j != 0)  
   {
      racf=acf(z,plot=FALSE)$acf[1:(j+1)]    
      lamb=racf[j+1]^2    
      corref[m1,j]=round(lamb,4)
      dmj=1 + 2*sum(racf[1:j]^2)
      cmj=-1*(n-m-j)*log(1.0 - lamb/dmj)
      pvalue =pchisq(cmj,1,lower.tail=FALSE)
      pv[m1,j]=round(pvalue,4)
      cmj.table[m1,j]=round(cmj,4)
      mark[m1,j]=ifelse(pvalue > alpha,"O","X")    
    } 
    else if (m != 0 && j == 0) 
    {
      racf=pacf(z,plot=FALSE)$acf[1:(m+1)]
      lamb=racf[m+1]^2
      corref[m1,j1]=round(lamb,4)
      dmj = 1
      cmj=-1*(n-m-j)*log(1.0 - lamb/dmj)    
      pvalue =pchisq(cmj,1,lower.tail=FALSE)
      pv[m1,j1]=round(pvalue,4)
      cmj.table[m1,j1]=round(cmj,4)    
      mark[m1,j1]=ifelse(pvalue > alpha,"O","X")
    } 
    else
    {        
      mat1=matrix(0,nrow=m1,ncol=m1)
      mat2=matrix(0,nrow=m1,ncol=m1) 
      mat3=matrix(0,nrow=m1,ncol=m1)
      mat4=matrix(0,nrow=m1,ncol=m1)     
      for (t in (j+m+2):n)
      {
         tj1=t-j-1
         ym1=ym(z,tj1,m)
         ym2=ym(z,t,m)    

         mat1=mat1 + as.matrix(ym1)%*%ym1    
         mat2=mat2 + as.matrix(ym1)%*%ym2    
         mat3=mat3 + as.matrix(ym2)%*%ym2    
         mat4=mat4 + as.matrix(ym2)%*%ym1                
      }  
      b1=solve(mat1)%*%mat2
      b2=solve(mat3)%*%mat4
      A=b2%*%b1
      eig <-eigen(A)
      eig.val <-eig$values
      eig.val=Re(eig.val)
      eig.len=length(eig.val)
      eig.vector=eig$vectors
      lamb=min(eig.val)
      eig.vector0=eig.vector[,which.min(eig.val)]
      eig.vector0 = eig.vector0/eig.vector0[1]
      resid=(1:n)*0 
      for (t in (j+m+1):n)
      {
        z0=z[seq(t,t-m,-1)]      
        resid[t]=sum(z0 * eig.vector0)
      } 
      jm1=j + m + 1
      rx=Re(resid[jm1:n])
      racf=acf(rx,plot=FALSE)$acf[1:j]
      dmj=1 + 2*sum(racf^2)
      cmj=-1*(n-m-j)*log(1.0 - lamb/dmj)     
      pvalue =pchisq(cmj,df=1,lower.tail=FALSE)
      corref[m1,j1]=round(lamb,4)     
      pv[m1,j1]=round(pvalue,4)
      cmj.table[m1,j1]=round(cmj,4)    
      mark[m1,j1]=ifelse(pvalue > alpha,"O","X")
    }
   } 
  } 

  cat("\n\nSCAN: Smallest CANonical Correlation Method for ARIMA(p,d,q)\n\n")
  cat("Estimates of Squared Canonical Correlation \n\n")
  print(corref)
  cat("\n\nC(m,j)\n\n")
  print(cmj.table)
  cat("\n\nChi-Square(1) Test p-value\n\n")
  print(pv)
  cat("\nSCAN Matrix \n\n")
  print(mark)

plot(0:1,0:1,col="white",xlim=c(0,nrow(pv)-1),ylim=c(0,ncol(pv)-1),axes=FALSE,xlab="AR",ylab="MA")
axis(1); axis(2)
library(RColorBrewer)
CL=brewer.pal(6, "RdBu")[c(1,2,3,5)]
cpv=matrix(as.numeric(cut(as.vector(pv),c(-1,.01,.05,.1,2))),nrow(pv),ncol(pv))
for(i in 1:nrow(pv)){
for(j in 1:ncol(pv)){
 polygon(c(i-1,i-1,i,i)-.5,c(j-1,j,j,j-1)-.5,
 col=CL[cpv[i,j]])
}}
}

Consider the following simulated time series,

> s=arima.sim(n=200,model=list(ar=c(0,0,0,.4,0,0,0,.5),ma=c(0,0,1))) 
> plot(s,type="l")

The output is here

> arma.scan(s,6,6)

SCAN: Smallest CANonical Correlation Method for ARIMA(p,d,q)

Estimates of Squared Canonical Correlation 

       MA-0   MA-1   MA-2   MA-3   MA-4   MA-5   MA-6
AR-0 0.0614 0.0104 0.1862 0.3516 0.0971 0.0128 0.0000
AR-1 0.0302 0.0294 0.1501 0.0943 0.0855 0.0127 0.0385
AR-2 0.3070 0.2781 0.2140 0.0006 0.1589 0.1884 0.2243
AR-3 0.1627 0.0037 0.1927 0.2311 0.1379 0.0207 0.0376
AR-4 0.2087 0.3947 0.3653 0.3075 0.1502 0.1364 0.1013
AR-5 0.1677 0.1219 0.0110 0.0263 0.0332 0.0350 0.0044
AR-6 0.0114 0.0485 0.0561 0.0427 0.0009 0.0089 0.0308

C(m,j)

        MA-0    MA-1    MA-2    MA-3   MA-4   MA-5    MA-6
AR-0  4.1161  0.6585 12.0315 20.6512 4.5388 0.5620  0.0000
AR-1  6.1127  1.9499  9.9356  4.9145 4.7219 0.4642  1.9015
AR-2 72.6011 19.1679 14.3512  0.0337 7.9668 9.6479 11.4573
AR-3 34.9724  0.2386 10.1620 13.4082 6.7875 0.8725  1.4071
AR-4 45.8691 27.5070 19.1422 20.2835 7.3339 5.5374  3.5874
AR-5 35.7981  8.0498  0.6280  1.3543 1.8470 1.7930  0.2338
AR-6  2.2147  3.1466  3.5990  1.9904 0.0511 0.4816  1.6440

Chi-Square(1) Test p-value

       MA-0   MA-1   MA-2   MA-3   MA-4   MA-5   MA-6
AR-0 0.0425 0.4171 0.0005 0.0000 0.0331 0.4534 0.0000
AR-1 0.0134 0.1626 0.0016 0.0266 0.0298 0.4957 0.1679
AR-2 0.0000 0.0000 0.0002 0.8543 0.0048 0.0019 0.0007
AR-3 0.0000 0.6252 0.0014 0.0003 0.0092 0.3503 0.2355
AR-4 0.0000 0.0000 0.0000 0.0000 0.0068 0.0186 0.0582
AR-5 0.0000 0.0046 0.4281 0.2445 0.1741 0.1806 0.6287
AR-6 0.1367 0.0761 0.0578 0.1583 0.8212 0.4877 0.1998

SCAN Matrix 

     MA-0 MA-1 MA-2 MA-3 MA-4 MA-5 MA-6
AR-0 "O"  "O"  "X"  "X"  "O"  "O"  "X" 
AR-1 "O"  "O"  "X"  "O"  "O"  "O"  "O" 
AR-2 "X"  "X"  "X"  "O"  "X"  "X"  "X" 
AR-3 "X"  "O"  "X"  "X"  "X"  "O"  "O" 
AR-4 "X"  "X"  "X"  "X"  "X"  "O"  "O" 
AR-5 "X"  "X"  "O"  "O"  "O"  "O"  "O" 
AR-6 "O"  "O"  "O"  "O"  "O"  "O"  "O"

with the following graph

Of course, it is possible to ask for larger values,

> arma.scan(s,12,12)

The graph is now

MAT8181, Time Series

Temperatures Series as Random Walks

12/02/2014 Arthur Charpentier 3 Comments

Last year, I did mention in a post that unit-root tests are dangerous, because they might lead us to strange models. For instance, in a post, I did obtain that the temperature observed in January 2013, in Montréal, might be considered as a random walk process (or at leat an integrated process). The code to extract the data has changed (since the website has been updated), so here, we use

library(RCurl)
library(XML)
options(RCurlOptions = list(useragent = "R"))
HEURE=0:23
extracttemp=function(Y,M,D){
url=paste(
"http://climate.weather.gc.ca/climateData/hourlydata_e.html?timeframe=1&Prov=QC&StationID=5415&Year=",Y,"&Month=",
M,"&Day=",D,sep="")
wp <- getURLContent(url)
doc <- htmlParse(wp, asText = TRUE) 
docName(doc) <- url
tmp <- readHTMLTable(doc)
basejour=data.frame(Year=Y,Month=M,Day=D,
Hour=HEURE,Temp=as.numeric(as.character(data.frame(tmp[2])[,2]))[2:25])
return(basejour)}
B=NULL
for(y in 1955:2013){
for(d in 1:31){
B=rbind(B,extracttemp(y,1,d))}}

Here are all the temperatures observed, and 2013,

plot(B$X,B$Temp,cex=.5,col="light blue",xlab="January, in Montreal",ylab="Temperature (Celsius)")
I=which(B$Year==2013)
lines(B$X[I],B$Temp[I],col="red")

In the previous post, one test only was used, and one year was considered. I was wondering if this behavior was observed only with temperature of 2013 (or not), and how the other tests (mentioned in a previous post too) were performing.

I might need a function, because those tests cannot be used if there is a missing value, even only one. So I did use the value observed one hour before (just to make sure that the tests can be done)

correcty=function(Y){
I=which(is.na(Y))
	if(length(I)==0){Yc=Y}
	if(length(I)>0){Yc=Y;for(i in I) Yc[i]=Yc[i-1]}	
return(Yc)
}

Now, we can compute the p-values, for all the years, and the three different three (keeping in mind that two test if the series is non-stationary, and one if the series is stationary)

DF=matrix(NA,2013-1954,3)
library(urca)
for(y in 1955:2013){
Z=B$Temp[which(B$Year==y)]
	Zc=correcty(Z)
	DF[y-1954,2]=as.numeric(pp.test(Zc)$p.value)	
	DF[y-1954,1]=as.numeric(kpss.test(Zc)$p.value)	
	DF[y-1954,3]=as.numeric(adf.test(Zc)$p.value)
}

Visually, if red means stationary, and blue means non-stationary, we get

DFP=DF
DFP[,1]=DF[,1]<.05
DFP[,2:3]=DF[,2:3]>.05
library(RColorBrewer)
CL=brewer.pal(6, "RdBu")
plot(0:1,0:1,xlim=c(1950,2015),ylim=c(0,3),axes=FALSE,xlab="",ylab="")
axis(1)
text(1952,.5,"KPSS")
text(1952,1.5,"PP")
text(1952,2.5,"ADF")
for(y in 1955:2013){
for(i in 1:3){
polygon(y+c(-1,-1,1,1)/2.2,i-.5+c(-1,1,1,-1)/2.2,col=CL[1+(DFP[y-1954,i]==1)*5],border=NA)}}

Quite frequently, we conclude that the temperature is a random walk. Which does not make sense (from a physical point of view). But again, it might come from the fact that temperature are stationary, but with some fractional behavior (as suggested in the previous post).

MAT8181, Time Series

Unit Root Tests

12/02/2014 Arthur Charpentier 5 Comments

This week, in the MAT8181 Time Series course, we’ve discussed unit root tests. According to Wold’s theorem, if $https://latex.codecogs.com/gif.latex?(Y_t)$ is (weakly) stationnary then

$https://latex.codecogs.com/gif.latex?%20%20%20%20%20Y_{t}=\sum%20_{{j=0}}^{\infty%20}\psi_{j}\varepsilon%20_{{t-j}}+\xi%20_{t}$

where $https://latex.codecogs.com/gif.latex?%20(\varepsilon%20_{{t}})$ is the innovation process, and where $https://latex.codecogs.com/gif.latex?%20(\xi%20_{{t}})$ is some deterministic series (just to get a result as general as possible). Observe that

$https://latex.codecogs.com/gif.latex?\sum%20_{{j=0}}^{{\infty%20}}|\psi_{{j}}|^{2}%20%3C%20\infty$

as discussed in a previous post. To go one step further, there is also the Beveridge-Nelson decomposition : an integrated of order one process, defined as

$https://latex.codecogs.com/gif.latex?%20%20%20%20%20\Delta%20Y_{t}=(1-L)%20Y_t=\sum%20_{{j=0}}^{\infty%20}\psi_{j}\varepsilon%20_{{t-j}}+\xi=\Psi(L)\varepsilon%20_{{t}}+\xi$ can be represented as

a linear trend $https://latex.codecogs.com/gif.latex?+$ a random walk $https://latex.codecogs.com/gif.latex?+$ a stationary remaining term

i.e.

$https://latex.codecogs.com/gif.latex?%20Y_{t}=\underbrace{Y_0%20+%20\xi%20t%20%C2%A0}+\underbrace{\Psi(1)\sum_{j=1}^t\varepsilon%20_{{i}}}+\underbrace{\tilde\Psi(L)\varepilon_0-\tilde\Psi(L)\varepsilon_t}$

where $https://latex.codecogs.com/gif.latex?%20\tilde\Psi(\cdot)$ is the polynomial with terms $https://latex.codecogs.com/gif.latex?%20\tilde\psi_j$ , where

$https://latex.codecogs.com/gif.latex?%20\tilde\psi_j%20=\sum_{i=j+1}^\infty\psi_i$

For unit-root tests, we will use various representation of the process. In order to illustrate the implementation of those tests, consider the following series

> E=rnorm(240)
> X=cumsum(E)
> plot(X,type="l")

Dickey Fuller (standard)

Here, for the simple version of the Dickey-Fuller test, we assume that

$https://latex.codecogs.com/gif.latex?%20Y_t=\alpha+\beta%20t+\varphi%20Y_{t-1}+\varepsilon_t$

and we would like to test if $https://latex.codecogs.com/gif.latex?%20\varphi=1$ (or not). We can write the previous representation as

$https://latex.codecogs.com/gif.latex?%20\Delta%20Y_t=\alpha+\beta%20t+[\varphi-1]%20Y_{t-1}+\varepsilon_t$

so we simply have to test if the regression coefficient in the linear regression is – or not – null. Which can be done with Student’s test. If we consider the previous model without the linear drift, we have to consider the following regression

> lags=0
> z=diff(X)
> n=length(z)
> z.diff=embed(z, lags+1)[,1]
> z.lag.1=X[(lags+1):n]
> summary(lm(z.diff~0+z.lag.1 ))

Call:
lm(formula = z.diff ~ 0 + z.lag.1)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.84466 -0.55723 -0.00494  0.63816  2.54352 

Coefficients:
         Estimate Std. Error t value Pr(>|t|)
z.lag.1 -0.005609   0.007319  -0.766    0.444

Residual standard error: 0.963 on 238 degrees of freedom
Multiple R-squared:  0.002461,	Adjusted R-squared:  -0.00173 
F-statistic: 0.5873 on 1 and 238 DF,  p-value: 0.4442

Our testing procedure will be based on the Student’s t value,

> summary(lm(z.diff~0+z.lag.1 ))$coefficients[1,3]
[1] -0.7663308

which is exactly the value computed using

> library(urca)
> df=ur.df(X,type="none",lags=0)
> df

############################################################### 
# Augmented Dickey-Fuller Test Unit Root / Cointegration Test # 
############################################################### 

The value of the test statistic is: -0.7663

The interpretation of this value can be done using critical values (99%, 95%, 90%)

> qnorm(c(.01,.05,.1)/2)
[1] -2.575829 -1.959964 -1.644854

If the statistics exceeds those values, then the series is not stationnary, since we cannot reject the assumption that $https://latex.codecogs.com/gif.latex?%20\varphi-1=0$ . So we might conclude that there is a unit root. Actually, those critical values are obtained using

> summary(df)

############################################### 
# Augmented Dickey-Fuller Test Unit Root Test # 
############################################### 

Test regression none 

Call:
lm(formula = z.diff ~ z.lag.1 - 1)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.84466 -0.55723 -0.00494  0.63816  2.54352 

Coefficients:
         Estimate Std. Error t value Pr(>|t|)
z.lag.1 -0.005609   0.007319  -0.766    0.444

Residual standard error: 0.963 on 238 degrees of freedom
Multiple R-squared:  0.002461,	Adjusted R-squared:  -0.00173 
F-statistic: 0.5873 on 1 and 238 DF,  p-value: 0.4442

Value of test-statistic is: -0.7663 

Critical values for test statistics: 
      1pct  5pct 10pct
tau1 -2.58 -1.95 -1.62

The problem with R is that there are several packages that can be used for unit root tests. Just to mention another one,

> library(tseries)
> adf.test(X,k=0)

	Augmented Dickey-Fuller Test

data:  X
Dickey-Fuller = -2.0433, Lag order = 0, p-value = 0.5576
alternative hypothesis: stationary

We do have here also a test where the null hypothesis is that there is a unit root. But the p-value is quite different. What is odd is that we have

> 1-adf.test(X,k=0)$p.value
[1] 0.4423705
> df@testreg$coefficients[4]
[1] 0.4442389

(but I think it is a coincidence).

Augmented Dickey Fuller

It is possible to had some lags in the regression. For instance, we can consider

$https://latex.codecogs.com/gif.latex?%20\Delta%20Y_t=\alpha+\beta%20t+[\varphi-1]%20Y_{t-1}+\psi%20\Delta%20Y_{t-1}+\varepsilon_t$

Again, we have to check if one coefficient is null, or not. And this can be done using Student’s t test.

> lags=1
> z=diff(X)
> n=length(z)
> z.diff=embed(z, lags+1)[,1]
> z.lag.1=X[(lags+1):n]
> k=lags+1
> z.diff.lag = embed(z, lags+1)[, 2:k]
> summary(lm(z.diff~0+z.lag.1+z.diff.lag ))

Call:
lm(formula = z.diff ~ 0 + z.lag.1 + z.diff.lag)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.87492 -0.53977 -0.00688  0.64481  2.47556 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
z.lag.1    -0.005394   0.007361  -0.733    0.464
z.diff.lag -0.028972   0.065113  -0.445    0.657

Residual standard error: 0.9666 on 236 degrees of freedom
Multiple R-squared:  0.003292,	Adjusted R-squared:  -0.005155 
F-statistic: 0.3898 on 2 and 236 DF,  p-value: 0.6777

> summary(lm(z.diff~0+z.lag.1+z.diff.lag ))$coefficients[1,3]
[1] -0.7328138

This value is the one obtained using

> df=ur.df(X,type="none",lags=1)
> summary(df)

############################################### 
# Augmented Dickey-Fuller Test Unit Root Test # 
############################################### 

Test regression none 

Call:
lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.87492 -0.53977 -0.00688  0.64481  2.47556 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
z.lag.1    -0.005394   0.007361  -0.733    0.464
z.diff.lag -0.028972   0.065113  -0.445    0.657

Residual standard error: 0.9666 on 236 degrees of freedom
Multiple R-squared:  0.003292,	Adjusted R-squared:  -0.005155 
F-statistic: 0.3898 on 2 and 236 DF,  p-value: 0.6777

Value of test-statistic is: -0.7328 

Critical values for test statistics: 
      1pct  5pct 10pct
tau1 -2.58 -1.95 -1.62

And again, other pckages can be used:

> adf.test(X,k=1)

	Augmented Dickey-Fuller Test

data:  X
Dickey-Fuller = -1.9828, Lag order = 1, p-value = 0.5831
alternative hypothesis: stationary

Hopefully, the conclusion is the same (we should reject the assumption that the series is stationary, but I am not sure about the computation of the p-value).

Augmented Dickey Fuller with trend and drift

So far, we have not included the drift in our model. But this is simple to do (this will be called the augmented version of the previous procedure): we just have to include a constant in the regression,

> summary(lm(z.diff~1+z.lag.1+z.diff.lag ))

Call:
lm(formula = z.diff ~ 1 + z.lag.1 + z.diff.lag)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.91930 -0.56731 -0.00548  0.62932  2.45178 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  0.29175    0.13153   2.218   0.0275 *
z.lag.1     -0.03559    0.01545  -2.304   0.0221 *
z.diff.lag  -0.01976    0.06471  -0.305   0.7603  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9586 on 235 degrees of freedom
Multiple R-squared:  0.02313,	Adjusted R-squared:  0.01482 
F-statistic: 2.782 on 2 and 235 DF,  p-value: 0.06393

The statistics of interest are obtained here considering some analysis of variance outputs, where this model is compared with the one without the integrated part, and the drift,

> summary(lm(z.diff~1+z.lag.1+z.diff.lag ))$coefficients[2,3]
[1] -2.303948
> anova(lm(z.diff ~ z.lag.1 + 1 + z.diff.lag),lm(z.diff ~ 0 + z.diff.lag))$F[2]
[1] 2.732912

Those two values are the ones obtained also with

> df=ur.df(X,type="drift",lags=1)
> summary(df)

############################################### 
# Augmented Dickey-Fuller Test Unit Root Test # 
############################################### 

Test regression drift 

Call:
lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.91930 -0.56731 -0.00548  0.62932  2.45178 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  0.29175    0.13153   2.218   0.0275 *
z.lag.1     -0.03559    0.01545  -2.304   0.0221 *
z.diff.lag  -0.01976    0.06471  -0.305   0.7603  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9586 on 235 degrees of freedom
Multiple R-squared:  0.02313,	Adjusted R-squared:  0.01482 
F-statistic: 2.782 on 2 and 235 DF,  p-value: 0.06393

Value of test-statistic is: -2.3039 2.7329 

Critical values for test statistics: 
      1pct  5pct 10pct
tau2 -3.46 -2.88 -2.57
phi1  6.52  4.63  3.81

And we can also include a linear trend,

> temps=(lags+1):n
> summary(lm(z.diff~1+temps+z.lag.1+z.diff.lag ))

Call:
lm(formula = z.diff ~ 1 + temps + z.lag.1 + z.diff.lag)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.87727 -0.58802 -0.00175  0.60359  2.47789 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)  
(Intercept)  0.3227245  0.1502083   2.149   0.0327 *
temps       -0.0004194  0.0009767  -0.429   0.6680  
z.lag.1     -0.0329780  0.0166319  -1.983   0.0486 *
z.diff.lag  -0.0230547  0.0652767  -0.353   0.7243  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9603 on 234 degrees of freedom
Multiple R-squared:  0.0239,	Adjusted R-squared:  0.01139 
F-statistic:  1.91 on 3 and 234 DF,  p-value: 0.1287

> summary(lm(z.diff~1+temps+z.lag.1+z.diff.lag ))$coefficients[3,3]
[1] -1.98282
> anova(lm(z.diff ~ z.lag.1 + 1 + temps+ z.diff.lag),lm(z.diff ~ 1+ z.diff.lag))$F[2]
[1] 2.737086

while R function returns

> df=ur.df(X,type="trend",lags=1)
> summary(df)

############################################### 
# Augmented Dickey-Fuller Test Unit Root Test # 
############################################### 

Test regression trend 

Call:
lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.87727 -0.58802 -0.00175  0.60359  2.47789 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)  
(Intercept)  0.3227245  0.1502083   2.149   0.0327 *
z.lag.1     -0.0329780  0.0166319  -1.983   0.0486 *
tt          -0.0004194  0.0009767  -0.429   0.6680  
z.diff.lag  -0.0230547  0.0652767  -0.353   0.7243  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.9603 on 234 degrees of freedom
Multiple R-squared:  0.0239,	Adjusted R-squared:  0.01139 
F-statistic:  1.91 on 3 and 234 DF,  p-value: 0.1287

Value of test-statistic is: -1.9828 1.8771 2.7371 

Critical values for test statistics: 
      1pct  5pct 10pct
tau3 -3.99 -3.43 -3.13
phi2  6.22  4.75  4.07
phi3  8.43  6.49  5.47

KPSS test

Here, in the KPSS testing procedure, two models can be considerd : with a drift, or with a linear trend. Here, the null hypothesis is that the series is stationnary.

With a drift, the code is

> summary(ur.kpss(X,type="mu"))

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: mu with 4 lags. 

Value of test-statistic is: 0.972 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.347 0.463  0.574 0.73

while it will be, in the case there is a trend

> summary(ur.kpss(X,type="tau"))

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 4 lags. 

Value of test-statistic is: 0.5057 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216

One more time, it is possible to use another package to get the same test (but again, a different output)

> kpss.test(X,"Level")

	KPSS Test for Level Stationarity

data:  X
KPSS Level = 1.1997, Truncation lag parameter = 3, p-value = 0.01

> kpss.test(X,"Trend")

	KPSS Test for Trend Stationarity

data:  X
KPSS Trend = 0.6234, Truncation lag parameter = 3, p-value = 0.01

At least, there is some kind of consistency, since we keep rejecting the stationnary assumption, for that series.

Philipps-Perron test

The Philipps-Perron test is based on the ADF procedure. The code is here

> PP.test(X)

	Phillips-Perron Unit Root Test

data:  X
Dickey-Fuller = -2.0116, Truncation lag parameter = 4, p-value = 0.571

with again, a possible alternative with the other package

> pp.test(X)

	Phillips-Perron Unit Root Test

data:  X
Dickey-Fuller Z(alpha) = -7.7345, Truncation lag parameter = 4, p-value
= 0.6757
alternative hypothesis: stationary

Comparison

I will not spend more time comparing the different codes, in R, to run those tests. Let us spend some additional time on a quick comparison of those three procedure. Let us generate some autoregressive processes, with more or less autocorrelation, as well as some random walk, and let us see how those tests perform :

> n=100
> AR=seq(1,.7,by=-.01)
> P=matrix(NA,3,31)
> M1=matrix(NA,1000,length(AR))
> M2=matrix(NA,1000,length(AR))
> M3=matrix(NA,1000,length(AR))

> for(i in 1:(length(AR)+1)){
+ for(s in 1:1000){
+ if(i==1) X=cumsum(rnorm(n))
+ if(i!=1) X=arima.sim(n=n,list(ar=AR[i]))
+ library(urca)
+ M2[s,i]=as.numeric(pp.test(X)$p.value)
+ M1[s,i]=as.numeric(kpss.test(X)$p.value)
+ M3[s,i]=as.numeric(adf.test(X)$p.value)
+ }}

Here, we would like to count how many times the p-value of our tests exceed 5%,

> prop05=function(x) mean(x>.05)
+ P[1,]=1-apply(M1,2,prop05)
+ P[2,]=apply(M2,2,prop05)
+ P[3,]=apply(M3,2,prop05)
+ }
> plot(AR,P[1,],type="l",col="red",ylim=c(0,1),ylab="proportion of non-stationnary 
+ series",xlab="autocorrelation coefficient")
> lines(AR,P[2,],type="l",col="blue")
> lines(AR,P[3,],type="l",col="green")
> legend(.7,1,c("ADF","KPSS","PP"),col=c("green","red","blue"),lty=1,lwd=1)

We can see here how poorly Dickey-Fuller test behave, since a 50% (at least) of our autoregressive processes are considered as non-stationnary.

MAT8181, Time Series

Inference for ARMA(p,q) Time Series

30/01/2014 Arthur Charpentier 1 Comment

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

MAT8181, Time Series

Inference for MA(q) Time Series

29/01/2014 Arthur Charpentier 2 Comments

Yesterday, we’ve seen how inference for $https://latex.codecogs.com/gif.latex?AR(p)$ time series was possible. I started with that one because it is actually the simple case. For instance, we can use ordinary least squares. There might be some possible bias (see e.g. White (1961)), but asymptotically, estimators are fine (consistent, with asymptotic normality). But when the noise is (auto)correlated, then it is more complex. So, consider here some $MA(2)$ time series

$X_t=\varepsilon_t +\theta_1 \varepsilon_{t-1}+\theta_2\varepsilon_{t-2}$

for some white noise $https://latex.codecogs.com/gif.latex?(\varepsilon_t)$ .

> theta1=.25
> theta2=.7
> n=1000
> set.seed(1)
> e=rnorm(n)
> Z=rep(0,n)
> for(t in 3:n) Z[t]=e[t]+theta1*e[t-1]+theta2*e[t-2]
> Z=Z[800:1000]
> plot(Z,type="l")

Using the empirical autocorrelations

The first idea might be to use the first two (empirical) autocorrelations (the two that are supposed to be – theoretically – non null).

$\rho(1)=\frac{\theta_1+\theta_1\theta_2}{1+\theta_1^2+\theta_2^2}$

$\rho(2)=\frac{\theta_2}{1+\theta_1^2+\theta_2^2}$

with $\rho(h)=0$ when $h\geq0$ . We also have the following relationship on the variance of the process

$\text{Var}(X_t)=\sigma^2\cdot(1+\theta_1^2+\theta_2^2)$

With those three equations, for three unknown parameters, $https://latex.codecogs.com/gif.latex?\theta_1$ , $https://latex.codecogs.com/gif.latex?\theta_2$ and $https://latex.codecogs.com/gif.latex?\sigma$ , we simply have to solve (numerically) that system of equations,

> v=c(as.numeric(acf(Z)$acf[2:3]),var(Z))
> v
[1] 0.1658760 0.3823053 1.6379498
> library(rootSolve)
> seteq=function(x){
+ F1=v[1]-(x[1]+x[1]*x[2])/(1+x[1]^2+x[2]^2)
+ F2=v[2]-(x[2])/(1+x[1]^2+x[2]^2)
+ F3=v[3]-(1+x[1]^2+x[2]^2)*x[3]^2
+ return(c(F1,F2,F3))}
> multiroot(f=seteq,start=c(.1,.1,1))
$root
[1] 0.1400579 0.4766699 1.1461636

$f.root
[1]  7.876355e-10  4.188458e-09 -2.839977e-09

$iter
[1] 5

$estim.precis
[1] 2.605357e-09

We are a bit far away from the true values, used to generate our sample. And if we consider 1,000 sample (instead of only one), we still have the bias, and a large variance for our three estimators,

Using least square techniques

We can try something quite different here. The problem we have is that we do not observe the noise $https://latex.codecogs.com/gif.latex?(\varepsilon_t)$ , we only observe our series $https://latex.codecogs.com/gif.latex?(X_t)$ . But we can try to rebuild that series (call it $(u_t)$ since we’re not sure it will be a reconstruction of the noise). As suggested in Box & Jenkins (1967), assume that the first two values are null. And then, use

$u_t =X_t-\left(\theta_1u_{t-1}+\theta_2un_{t-2}\right)$

and then, we can use least square techniques

$\min_{(\theta_1,\theta_2)}\left\{\sum_{i=1}^tu_i^2\right\}$

The code will be

> V=function(p){
+ theta1=p[1]
+ theta2=p[2]
+ u=rep(0,length(Z))
+ for(t in 3:length(Z)) u[t]=Z[t]-theta1*u[t-1]-theta2*u[t-2]
+ return(sum(u^2))
+ }

If we try to minimize the sum of the squares of the residuals, we get

> optim(par=c(.1,.1),V)
$par
[1] 0.2751667 0.6723909

$value
[1] 225.8104

$counts
function gradient 
      77       NA 

$convergence
[1] 0

$message
NULL

which is close to the true value. Another good thing is that, if we compare that rebuilt noise with the true one (since we actually have it), then we have the same vector,

> plot(e[800:1000],col="blue",type="l")
> theta1=0.2751667
> theta2=0.6723909
> u=rep(0,length(Z))
> for(t in 3:length(Z)) u[t]=Z[t]-theta1*u[t-1]-theta2*u[t-2]
> lines(1:201,u,col="red")

So far, so good. And if we look at 1,000 samples, we get

It looks like we have some bias here. And since the two estimators should be negatively correlated, one over-estimates, while the other one under-estimates.

Using the (global) maximum likelihood technique

And a final method might be to use the maximum likelihood technique (globally). Again, if we assume that we have a Gaussian i.i.d noise, then the vector $https://latex.codecogs.com/gif.latex?\boldsymbol{Y}=(Y_1,\cdots,Y_t)$ is Gaussian, with a simple variance matrix (since a lot of elements will be null),

> library(mnormt)
> GlobalLogLik=function(A,TS){
+ n=length(TS)
+ theta1=A[1];  theta2=A[2]
+ sigma=A[3]
+ SIG=matrix(0,n,n)
+ rho=rep(0,n)
+ rho[1]=1
+ rho[2]=(theta1+theta1*theta2)/(1+theta1^2+theta2^2)
+ rho[3]=(theta2)/(1+theta1^2+theta2^2)
+ for(i in 1:n){for(j in 1:n){
+ SIG[i,j]=rho[abs(i-j)+1]}}
+ gamma0=(1+theta1^2+theta2^2)*sigma^2
+ SIG=gamma0*SIG
+ return(dmnorm(TS,rep(0,n),SIG,log=TRUE))}
> LogL=function(A) -GlobalLogLik(A,TS=Z)
> optim(c(.1,.1,1),LogL)
$par
[1] 0.2584144 0.6826530 1.0669820

$value
[1] 298.8699

$counts
function gradient 
      86       NA 

$convergence
[1] 0

$message
NULL

Here, the values that minimize the likelihood are rather close to the ones used to generate our sample. And if we run this algorithm on 1,000 samples, we can see that those estimates are fine,

I could not find other ideas, to estimate those parameters. I guess we can use the partial autocorrelation function, since we have relationships that can be related to Yule-Walker equations for $https://latex.codecogs.com/gif.latex?AR(p)$ time series.

Inference, MAT8181, Time Series

Inference for AR(p) Time Series

28/01/2014 Arthur Charpentier 2 Comments

Consider a (stationary) autoregressive process, say of order 2,

$https://latex.codecogs.com/gif.latex?Y_t%20=\varphi_1%20Y_{t-1}+\varphi_2%20Y_{t-2}+\varepsilon_t$

for some white noise with variance . Here is a code to generate such a process,

> phi1=.25
> phi2=.7
> n=1000
> set.seed(1)
> e=rnorm(n)
> Z=rep(0,n)
> for(t in 3:n) Z[t]=phi1*Z[t-1]+phi2*Z[t-2]+e[t]
> Z=Z[800:1000]
> n=length(Z)
> plot(Z,type="l")

Here, we have to estimate two sets of parameters: the autoregressive coefficients, and the variance of the innovation process . Several techniques can be used to estimate those parameters.

using least square regression

A natural idea is to see here a regression model, since (if we consider a matrix formulation)

Here we can run (conditional) ordinary least squares estimation,

> base=data.frame(Y=Z[3:n],X1=Z[2:(n-1)],X2=Z[1:(n-2)])
> regression=lm(Y~0+X1+X2,data=base)
> summary(regression)

Call:
lm(formula = Y ~ 0 + X1 + X2, data = base)

Residuals:
    Min      1Q  Median      3Q     Max 
-3.0268 -0.7063  0.1065  0.6925  3.2566 

Coefficients:
   Estimate Std. Error t value Pr(>|t|)    
X1  0.23400    0.05463   4.283 2.88e-05 ***
X2  0.62863    0.05476  11.479  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.062 on 197 degrees of freedom
Multiple R-squared:  0.6349,	Adjusted R-squared:  0.6312 
F-statistic: 171.3 on 2 and 197 DF,  p-value: < 2.2e-16

so we get the following estimators, for the autocorrelation coefficients, and the volatility of the noise

> regression$coefficients
       X1        X2 
0.2339959 0.6286321 
> summary(regression)$sigma
[1] 1.061839

using Yule-Walker equations

As we’ve seen in class, we can easily get the following equations for the autocovariance functions,

which can also be written (again, using a matrix expression)

So we just have to solve a simple linear system of equations. Note that if we divide by the variance, those equations can be written in terms of the autocorrelation functions

The code is the following

> rho1=cor(Z[1:(n-1)],Z[2:n])
> rho2=cor(Z[1:(n-2)],Z[3:n])
> A=matrix(c(1,rho1,rho1,1),2,2)
> b=matrix(c(rho1,rho2),2,1)
> (PHI=solve(A,b))
          [,1]
[1,] 0.2256270
[2,] 0.6315329

Now, we need to extract the estimated innovation process, from this set of parameters

> estWN=base$Y-(PHI[1]*base$X1+PHI[2]*base$X2)
> sd(estWN)
[1] 1.058558

This estimator is probably not the best one (we can take into account that we’ve lost two degrees of freedom), but as a starting point, let us consider this one.

An alternative could be to include the variance term in Yule-Walker equations, to get a three dimensional linear equation,

$https://latex.codecogs.com/gif.latex?\left\{\begin{array}{l}%20\gamma_0%20=%20\varphi_1%20\gamma_1+\varphi_2%20\gamma_2+\sigma^2\\%20\gamma_1=\varphi_1%20\gamma_0+\varphi_2%20\gamma_1%20\\%20\gamma_2=\varphi_1%20\gamma_1+\varphi_2%20\gamma_0\end{array}\right.$

It is not much more complicated to solve, actually,

> gamma0=var(Z[1:n])
> gamma1=var(Z[1:(n-1)],Z[2:n])
> gamma2=var(Z[1:(n-2)],Z[3:n])
> A=matrix(c(gamma1,gamma0,gamma1,gamma2,gamma1,gamma0,1,0,0),3,3)
> b=matrix(c(gamma0,gamma1,gamma2),3,1)
> (PHISIGMA=solve(A,b))
          [,1]
[1,] 0.2283151
[2,] 0.6283431
[3,] 1.1335501

using (conditional) likelihood estimators

Finally, we can assume some distribution for the innovation process. The standard model is a Gaussian model, i.e.

$https://latex.codecogs.com/gif.latex?Y_t\vert%20Y_{t-1}=y_{t-1},Y_{t-2}=y_{t-2}$

has a Gaussian distribution

$https://latex.codecogs.com/gif.latex?\mathcal{N}(\varphi_1y_{t-1}+\varphi_2y_{t-2},\sigma^2)$

In that case, the conditional log likelihood (conditional since we set the first two observations here) is

> CondLogLik=function(A,TS){
+ phi1=A[1];  phi2=A[2]
+ sigma=A[3]; L=0
+ for(t in 3:length(TS)){
+ L=L+dnorm(TS[t],mean=phi1*TS[t-1]+
+ phi2*TS[t-2],sd=sigma,log=TRUE)}
+ return(-L)}

Now, we can run standard optimization procedures,

> LogL=function(A) CondLogLik(A,TS=Z)

> optim(c(0,0,1),LogL)
$par
[1] 0.2339589 0.6285002 1.0565613

$value
[1] 293.3042

$counts
function gradient 
     106       NA 

$convergence
[1] 0

$message
NULL

It is also possible to consider a global maximum likelihood optimisation problem, since the variance matrix of vector $https://latex.codecogs.com/gif.latex?\boldsymbol{Y}=(Y_1,\cdots,Y_t)$ has a know form.

using (unconditional) likelihood estimators

The variance matrix of $https://latex.codecogs.com/gif.latex?\boldsymbol{Y}=(Y_1,\cdots,Y_t)$ is $https://latex.codecogs.com/gif.latex?\boldsymbol{\Gamma}=[\gamma(\vert%20i-j\vert)]$ , where autocovariances are not not know, be can easily be computed using a recursive relationship.

> library(mnormt)
> GlobalLogLik=function(A,TS){
+ n=length(TS)
+ phi1=A[1];  phi2=A[2]
+ sigma=A[3]
+ SIG=matrix(0,n,n)
+ rho=rep(0,n)
+ rho[1]=1
+ rho[2]=phi1/(1-phi2)
+ for(h in 3:n) rho[h]=phi1*rho[h-1]+phi2*rho[h-2]
+ for(i in 1:n){for(j in 1:n){
+ SIG[i,j]=rho[abs(i-j)+1]}}
+ gamma0=(1-phi2)*sigma^2/((1+phi2)*((1-phi2)^2-phi1^2))
+ SIG=gamma0*SIG
+ return(dmnorm(TS,rep(0,n),SIG,log=TRUE))}
> LogL=function(A) -GlobalLogLik(A,TS=Z)
> optim(c(.1,.1,1),LogL)
Error in chol.default(x, pivot = FALSE) : 
Error in pd.solve(varcov, log.det = TRUE) : 
  x appears to be not positive definite

The problem is that there is a strong constraint on the pair $https://latex.codecogs.com/gif.latex?(\varphi_1,\varphi_2)$ to get a stationary process (we are not far away, here, from the border of the triangle, where the process become non stationary). To be more specific (this was mentioned in a previous post), we should have

$https://latex.codecogs.com/gif.latex?\left\{\begin{array}{l}%20\phi_2-\phi_1%3C1%20\\\phi_2+\phi_1%3C1\\%20\vert\phi_2\vert%3C1\end{array}\right.$

i.e. in a standard matrix form

$https://latex.codecogs.com/gif.latex?\left[\begin{array}{cc}%20+1%20&%20-1%20\\%20-1%20&%20-1%20\\%200%20&%20+1\end{array}\right]\left[\begin{array}{c}%20\varphi_1%20\\%20\varphi_2\end{array}\right]%20%3E%20\left[\begin{array}{c}%20-1%20\\%20-1%20\\%20-1\end{array}\right]$

(we can add an additional constraint on the variance parameter, to insure that it will be positive). To run a contrained optimization routine, consider

> U=matrix(c(1,0,0,-1,0,1,0,-1,0,0,1,0),4,3)
> C=c(0,0,0,-.99999)
> constrOptim(c(.1,.1,1),LogL,grad=NULL,ui=U,ci=C)
$par
[1] 0.2238892 0.6342850 1.0613388

$value
[1] 297.9202

$counts
function gradient 
     108       NA 

$convergence
[1] 0

$message
NULL

$outer.iterations
[1] 2

$barrier.value
[1] 0.000189892

(here, to faster, we restrain the parameters so that they will be positive).

comparing those estimates

Here, our five estimators are rather close. Let us run more samples to see more precisely how they behave. For the first parameter $https://latex.codecogs.com/gif.latex?\widehat{\varphi_1}$ , we get

and for the second one, $https://latex.codecogs.com/gif.latex?\widehat{\varphi_2}$ , we have

The bias we observe is probably coming from the fact that, with this numerical example, we are not far away from the non-stationary case (the sum of the true parameters should be less than 1, and it is 0.95). When we estimate the parameters, we force them to be inside the triangle, since those parameters can be estimated only if the process is stationary.

Observe that the standard-deviation of the innovation process $https://latex.codecogs.com/gif.latex?\widehat{\sigma}$ is here, well estimated,

(with clearly some estimators that perform better than others).

MAT8181, Time Series

Defining Properly MA(∞) Time Series

28/01/2014 Arthur Charpentier Leave a comment

In order to properly define $https://latex.codecogs.com/gif.latex?MA(\infty)$ series, we need to get back on some properties of infinite sequences, as briefly mentioned yesterday in the MAT8181 course. Consider some sequence $https://latex.codecogs.com/gif.latex?(a_i)_{i\in\mathbb{N}}$ . The sequence is said to be summable if

$https://latex.codecogs.com/gif.latex?S_n=\sum_{i=0}^n%20a_i$

is convergent, i.e. if the limit of $https://latex.codecogs.com/gif.latex?S_n$ exists when $https://latex.codecogs.com/gif.latex?n\rightarrow\infty$ .

From Cauchy criterion, $https://latex.codecogs.com/gif.latex?\sum%20a_i$ converges if and only if for each $https://latex.codecogs.com/gif.latex?\eta%3E0$ , there is $https://latex.codecogs.com/gif.latex?m\in\mathbb{N}$ for which

$https://latex.codecogs.com/gif.latex?\vert%20a_i+a_{i+1}+\cdots+a_{j-1}+a_j\vert%3C\eta$

when $https://latex.codecogs.com/gif.latex?%20i,j%3Em$ . The sequence $https://latex.codecogs.com/gif.latex?(a_i)_{i\in\mathbb{N}}$ is said to be absolutely summable if

$https://latex.codecogs.com/gif.latex?%20\sum_{i=0}^\infty%20\vert%20a_i\vert%20%3C\infty$

and square-summable if

$https://latex.codecogs.com/gif.latex?%20\sum_{i=0}^\infty%20a_i^2%20%3C\infty$

Observe that absolute summability will imply square summability (since for $https://latex.codecogs.com/gif.latex?j$ ‘s large enough $https://latex.codecogs.com/gif.latex?%20\vert%20a_j\vert%20\leq1$ , and then $https://latex.codecogs.com/gif.latex?%20a_j^2\leq\vert%20a_j\vert$ )

Consider now some $https://latex.codecogs.com/gif.latex?MA(\infty)$ time series

$https://latex.codecogs.com/gif.latex?%20X_t=\sum_{h=0}^\infty%20\theta_h%20\varepsilon_{t-h}$

If the sequence of coefficients $https://latex.codecogs.com/gif.latex?%20(\theta_i)$ is square-summable, then

$https://latex.codecogs.com/gif.latex?%20S_T%20=%20\sum_{h=0}^T%20\theta_h%20\varepsilon_{t-h}$

converges in $https://latex.codecogs.com/gif.latex?%20L_2$ to some random varible as $https://latex.codecogs.com/gif.latex?%20T\rightarrow\infty$ . This can be proved easily using Cauchy criteria, in the sense that for any $https://latex.codecogs.com/gif.latex?\eta%3E0$ , there is a $https://latex.codecogs.com/gif.latex?%20T$ large enough such that, for any $https://latex.codecogs.com/gif.latex?%20h$ ,

$\underbrace{\mathbb{E}\left(\sum_{i=T}^h\theta_i \varepsilon_{t-i}\right)^2}_{\sigma^2 \cdot \sum_{i=T}^h \theta_i^2} \leq \eta$

In that case, if the sequence of coefficients $https://latex.codecogs.com/gif.latex?%20(\theta_i)$ is square-summable, then $https://latex.codecogs.com/gif.latex?%20(X_t)$ is stationary (in the $https://latex.codecogs.com/gif.latex?%20L_2$ sense) since the process is centered, and

$https://latex.codecogs.com/gif.latex?\gamma(h)=\sigma^2%20\cdot%20\sum_{i=0}^\infty%20\theta_i%20\theta_{i+h}$

for all $https://latex.codecogs.com/gif.latex?h\in\mathbb{N}$ .

Further, ergodicity of the time series, define as the absolute summability of the autocovariance sequence, is obtained when the sequence of coefficients $https://latex.codecogs.com/gif.latex?%20(\theta_i)$ is absolutely summable.

Freakonometrics

Category Archives: MAT8181

Examen, Séries Chronologiques

Stationarity of ARCH processes

Inference for ARCH processes

Independence and correlation

Seasonal Unit Roots

Linear ‘Prediction’ for AR Time Series

Seasonal, or periodic, time series

Filtering a Stationary Time Series

Identification of ARMA processes

Temperatures Series as Random Walks

Unit Root Tests

Inference for ARMA(p,q) Time Series

Inference for MA(q) Time Series

Inference for AR(p) Time Series

Defining Properly MA(∞) Time Series

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