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I'm analyzing the time series DAX (I call Z) from the dataset EuStockMarket. Could you please verify if my analysis makes sense? Thank you so much for your help!

  • First, I plot Z.

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It's from the plot that Z does not have constant mean and thus is not stationary.

  • Then, I plot its ACF

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This plot suggests strong correlation at even big lags. This means no pattern of moving average term.

  • Then I plot its PACF

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This pattern indicates an autoregressive term of order 1.

  • I differentiate Z to get diff1Z

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The plot suggests diff1Z has constant mean of $0$.

  • Then I plot its ACF and PACF.

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It seems that diff1Z's PACF vanishes after $23$ lags and its ACF vanishes after $20$ lags. As such, I propose $\texttt{Z} \sim \operatorname{ARMA}(23, 1, 20)$.

Update: I have taken the log-differences of the data and plot the series as well as its ACF and PACF.

T <- diff(log(Z))
plot(T)
acf(T)
pacf(T)

enter image description here enter image description here enter image description here

Both ACF and PACF vanish after $11$ lags.

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    $\begingroup$ The AR and MA orders you have selected are ridiculously high given the context. For a time series of financial returns (stock returns, index returns), ARMA(0,0) is the standard and ARMA(1,1) would be the largest reasonable model. In most cases, including this one, I would go with ARMA(0,0). But do what Ben suggests and work with log-returns instead. $\endgroup$
    – Richard Hardy
    Commented Dec 24, 2020 at 7:26
  • $\begingroup$ @RichardHardy I have taken the log-differences of the data. The resulted ACF and PACF vanished after $11$ lags. From this, I would propose ARMA(11, 11), but this is still faraway from your largest reasonable model ARMA(1,1). Could you please elaborate more on how to proceed? I'm still open with the possibility to use ARCH/GARCH models. $\endgroup$
    – Akira
    Commented Dec 24, 2020 at 10:04
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    $\begingroup$ Try ARMA(11,11) vs. ARMA(1,1) vs. ARMA(1,0) vs. ARMA(0,0) and see which one does best at out-of-sample forecasting. Use a rolling window approach. I expect you will find ARMA(11,11) distinctly worst and ARMA(0,0) best, with ARMA(1,0) a close competitor. ARMA(1,1) will either have nearly cancelling AR and MA roots and be a close competitor to ARMA(0,0) or have distinct roots and be a less close competitor. Financial theory simply persuades me more than ACF and PACF plots when it comes to stock index returns. $\endgroup$
    – Richard Hardy
    Commented Dec 24, 2020 at 10:49
  • $\begingroup$ Thank you so much for your help @RichardHardy. I wish you a happy Christmas with your family ^^ $\endgroup$
    – Akira
    Commented Dec 24, 2020 at 10:50
  • $\begingroup$ Happy modeling while you are at it* and then a Merry Christmas to you as well! *(I am running some simulations myself but will quit starting from lunch time which is coming soon.) $\endgroup$
    – Richard Hardy
    Commented Dec 24, 2020 at 10:51

1 Answer 1

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When analysing stock market data, or other similar kinds of time-series variables, one usually examines the log-differences of the data. This is done in order to first filter out the effect of exponential growth, and instead analyse the changes in growth rates over time. To do this, you first apply the transformation:

$$X_t = \log Z_t - \log Z_{t-1}.$$

In your analysis you have taken a straight difference, but you have not taken the logarithm, so you are not correcting for the exponential growth in the original variable. This is likely to mean that your variable will display non-stationarity that will be hard to model on that scale. I recommend you use the standard log-difference transformation and then examine the ACF and PACF of the resulting variable.

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  • $\begingroup$ I have updated my question with the ACF and PACF of the log-differences of the data. Both function vanish after $11$ lags. Hence I would propose ARMA(11, 11) for this transformed data. However, @Richard Hardy said that ARMA(1,1) would be the largest reasonable model for these kind of data. I'm quite confused. Could you please elaborate on which model I should choose? $\endgroup$
    – Akira
    Commented Dec 24, 2020 at 9:58
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    $\begingroup$ I would add that you can see very clearly that this is necessary from the plot of the difference (the 4th chart above), where the variance clearly increases with the level of the original series (i.e. follows the peaks and valleys in the early part, then grows much higher when the level increases more dramatically towards the end). $\endgroup$
    – Chris Haug
    Commented Dec 24, 2020 at 13:56
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    $\begingroup$ I agree roughly with Richard. ARMA(11, 11) has at least 23 free parameters, which is likely to cause serious overfitting on a single time series. $\endgroup$
    – Ben
    Commented Dec 24, 2020 at 21:44

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