# Inference for MA(q) Time Series

**Freakonometrics » R-english**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Yesterday, we’ve seen how inference for 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 time series

for some white noise .

> 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).

avec pour . We also have the following relationship on the variance of the process

With those three equations, for three unknown parameters, , and , 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 , we only observe our series . But we can try to rebuild that series (call it 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

and then, we can use least square techniques

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 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 time series.

**leave a comment**for the author, please follow the link and comment on their blog:

**Freakonometrics » R-english**.

R-bloggers.com offers

**daily e-mail updates**about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.