0
$\begingroup$

I have a dataset of 400 time series. I want to evaluate if there is some relationship between consecutive data points. Thus, I have calculated the auto-correlation (AC) of the time series with different lags. For the best lag value, I obtain an auto-correlation of about 0.59 on average.

Now my problem is how can I know if this is a good AC value? Is there some way to evaluate if an auto-correlation value like 0.59 is good?

I have tried the following approaches to evaluate if this is a good value.

  1. I have calculated the AC of random sequences to compare it with the AC of my time series. I have found that the AC of random sequences is always close to 0. This is interesting as 0.59 is far from 0 but still does not tell me clearly if 0.59 is a good value.

  2. I have thus decided to do another experiment to compare the AC of my time series with the auto-corrrelation of an ascending time series (a function that is linearly increasing) and a random time series. In that experiment, I randomly tamper each sequence according to a tampering ratio that is varied from 0 to 100%. The result is like this:

Comparison of auto-correlation

It can be observed that the AC of my time series is close to that of the ascending time series and decrease quickly when the tampering ratio is increased. This seems to indicate that there is a good AC in my time series. However, it does not really tell me what is a good value for the AC.

Thus, is there other ways to determine what is a good AC value? Or do you have any other suggestions about how to determine what is a good AC value?

$\endgroup$
1
$\begingroup$

Comment: According to data collected in the late 1970s on eruptions of Old Faithful geyser in Yellowstone National Park, lengths of eruptions varied between short $0$ (less than 2 min.) and long $1$ (more than 2 min.) approximately according to a 2-state Markov chain in which there are never two consecutive short eruptions, and short eruptions follow long ones with probability $0.44.$ Consequently, one can show that over the long run about 70% of eruptions are long.

However, short and long eruptions are not independent Bernoulli trials, as for a coin with Heads probability 0.7, but form an autocorrelated series according to a 2-state Markov Chain.

Two thousand successive steps of such a chain can be simulated in R as shown below.

set.seed(2020)
n = 2000; x = numeric(n); x[1]=0
for (i in 2:n) {
 if (x[i-1]==0) x[i] = 1
 else x[i] = rbinom(1, 1, .56) }
mean(x)
[1] 0.7005

In R, one can make an autocorrelation plot for several lags. Of course the autocorrelation for lag $0$ is $1.000.$ Autocorrelations that fall outside the horizontal blue dotted lines are considered significantly different from $0.$ So, it seems for 2000 observations from the Old Faithful process, that autocorrelations greater in absolute value than about $0.035$ or $0.04$ are considered significantly different from $0.$

acf(x)

enter image description here

Specific lags can be obtained by using acf with the parameter plot=F.

acf(x, plot=F)

Autocorrelations of series ‘x’, by lag

     0      1      2      3      4      5      6      7 
 1.000 -0.426  0.203 -0.085  0.018 -0.009  0.016 -0.025 
     8      9     10     11     12     13     14     15 
 0.002 -0.030 -0.004  0.004 -0.025  0.033 -0.043  0.032 
    16     17     18     19     20     21     22     23 
-0.006  0.006  0.009 -0.001  0.005  0.014 -0.028  0.002 
    24     25     26     27     28     29     30     31 
-0.002  0.008 -0.018 -0.020  0.039 -0.009  0.013  0.010 
    32     33 
 0.005 -0.037 

In an ergodic (convergent) Markov chain, the Markov dependence 'wears off' after a few lags so that observations taken far apart along the sequence are nearly independent.

In your application, in order to say whether an autocorrelation is 'good', you need to specify the relevant lag, and to have a specific test criterion for 'significant' autocorrelation. From your question, I don't know enough about your process or application to give a specific answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.