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How to statistically compare two time series?

After reading the above responses, I realized that what I should be doing for my data is exactly what https://stats.stackexchange.com/users/3382/irishstat recommended. However, I am having trouble determining the correct parameters for the arima(p,d,q) function in r. Could someone provide a detailed tutorial for how to correctly choose these values? I know that the time series need to be stationary, which is simple enough to test. I also know the acf/pacf graphs are used for this, but it's still a bit foggy and I want my statistics to be sound. I have been unable to find a successful guide- most of the walkthroughs I have seen skim over the part of choosing the p,d,q values.

EDIT: I've added the 2 time series that I have been working on. The data is sampled bi-monthly for 9 months (excepting Dec, Jan, and Feb). alpha, in this case, is the TN data aggregated by month and year, which allowed me to coerce it to a time series. For our purposes, monthly averages are good enough to display what we are trying to look at. Another important note, these 2 time series have the proper time window, etc., but they are merely examples. They are not the final time series that I will use in the analysis. Final note, the NAs in the first time series caused gaps in the plots, but it is not difficult to apply a gap-filling function, which would generate a plot much like the second.

These time series are clearly not models, but simply points stored in the time series format. I believe that using a model is the best way to compare them, but I am not sure. I was also able to decompose the model to remove the seasonal data, and I was curious if it would be better to compare the series with the seasonal component removed.

alpha <- with(alpha, 
 aggregate(TN~Month+Year, FUN = mean, na.rm=T)
 )
> alpha
   Month Year       TN
1      3 2011 2.550675
2      4 2011 2.166793
3      5 2011 1.666279
4      6 2011 1.235067
5      7 2011 1.306130
6      8 2011 1.380530
7      9 2011 1.434623
8     10 2011 1.599755
9     11 2011 2.261617
10     3 2012 2.529887
11     4 2012 1.938779
12     5 2012 1.700785
13     6 2012 1.160013
14     7 2012 1.099877
15     8 2012 1.244322
16     9 2012 1.471384
17    10 2012 1.695036
18    11 2012 1.959646
19     3 2013 2.808547
20     4 2013 2.546227
21     5 2013 2.112756
22     6 2013 1.875753
23     7 2013 1.882885
24     8 2013 2.010292
25     9 2013 1.771419
26    10 2013 1.820127
27    11 2013 2.351775
28     3 2014 3.002976
29     4 2014 2.286398
30     5 2014 2.177926
31     6 2014 1.772718
32     7 2014 1.245376
33     8 2014 1.274671
34     9 2014 1.378657
35    10 2014 1.554602
36    11 2014 1.702360
37     3 2015 2.771875
38     4 2015 2.373219
39     5 2015 2.025162
40     6 2015 1.607793
41     7 2015 1.656044
42     8 2015 1.525059
43     9 2015 1.494547
44    10 2015 1.746673
45    11 2015 1.942351

V1 <- as.numeric(rep(c(12,1,2),4))
V2 <- as.numeric(c(2011, rep(2012,3), rep(2013,3), rep(2014,3),rep(2015,2)))
V3 <- rep(NA, 12)
df1 <- data.frame(V1,V2,V3)
names(df1) <- c('Month', 'Year', 'TN')

TNMonthsFilled <- rbind(alpha, df1) #adds in dates for Dec-Feb as NAs

TNMonthsFilled$Date <- as.Date(with(TNMonthsFilled, paste(Year, as.numeric(Month), as.numeric(15),sep="-")), "%Y-%m-%d") #converts values back to dates


TNOrdered <- TNMonthsFilled[order(TNMonthsFilled$Date),] #orders dates for coercion to ts

TNts <- with(TNOrdered, ts(TN, start = c(2011,3), end = c(2015,11), frequency = 12)) #TS with NAs

enter image description here

plot(TNts)

Time Series for First River

Nan_Up_Ag <- with(Nan_Up, 
     aggregate(TN~Date, FUN = mean, na.rm=T)
     )

> Nan_Up_Ag
         Date        TN
1  2011-04-15 3.3625000
2  2011-05-15 2.8972222
3  2011-06-15 3.6680000
4  2011-07-15 1.8125000
5  2011-08-15 1.5816667
6  2011-09-15 2.5475000
7  2011-10-15 3.4550000
8  2011-11-15 3.9600000
9  2012-04-15 3.5683333
10 2012-05-15 2.6991667
11 2012-06-15 2.0630000
12 2012-07-15 1.6233333
13 2012-08-15 1.4250000
14 2012-09-15 2.3961111
15 2012-10-15 2.9616667
16 2012-11-15 3.8216667
17 2013-03-15 4.6820000
18 2013-04-15 5.6225000
19 2013-05-15 3.5658333
20 2013-06-15 0.4916667
21 2013-07-15 2.8266667
22 2013-08-15 2.9216667
23 2013-09-15 2.7991667
24 2013-10-15 2.5875000
25 2013-11-15 3.4583333
26 2014-03-15 4.6200000
27 2014-04-15 3.9941667
28 2014-05-15 3.4116667
29 2014-06-15 2.8661111
30 2014-07-15 2.2483333
31 2014-08-15 2.0416667
32 2014-09-15 2.1316667
33 2014-10-15 2.6465000
34 2014-11-15 3.2133333
35 2015-03-15 4.1460000
36 2015-04-15 4.1650000
37 2015-05-15 3.6616667
38 2015-06-15 3.0483333
39 2015-07-15 2.8133333
40 2015-08-15 2.4466667
41 2015-09-15 2.1983333
42 2015-10-15 3.0233333
43 2015-11-15 3.3060000

Nan_Up_ts <- with(Nan_Up_Ag, ts(TN, start = c(2011,3), end = c(2015,11), frequency = 12))

enter image description here

plot.ts(Nan_Up_ts)

River 2 Time Series

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  • $\begingroup$ why don't you [post the two tome series and I will try and help you (and others ) that have similar problems. $\endgroup$ – IrishStat Aug 25 '17 at 21:34
  • $\begingroup$ Choosing p,d, and q is essentially an optimization problem. Have you tried auto.arima? $\endgroup$ – Nate Diamond Aug 26 '17 at 0:34
  • $\begingroup$ Choosing an ARIMA order has been asked dozens if not hundreds of times here. Your question must be a duplicate, you just have to choose which of the older threads you prefer (because there have been other duplicates before). See e.g. these questions. Besides, time series textbooks covering ARIMA usually have a section on order selection. Have you tried those? What did you not understand there? $\endgroup$ – Richard Hardy Aug 26 '17 at 8:04

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