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I computed an ARMA model for conditional mean on a financial time series of log returns. The model that minimise the AIC was an ARMA(5,5) and all but 2 of the coefficients ( AR $\phi_4$ and MA $\theta_4$) resulted statistical significant. The ACF looked cleaned out. The ACF of squared residuals seems autocorrelated.

However, after implementing an ARMA(5,5)-GARCH(1,1) to model also conditional variance and get rid of heteroskedasticity, the estimated $\omega$, $\alpha_1$ and $\beta_1$ were statistical significant, but now all the arma coefficient (but $\phi_5$ at 10% level) turned out to be non-significant now.

Questions:

  • What's going on? How would you comment the results?
  • I was refusing the the weak form of the Efficient Market Hypothesis (EMH) on the basis of the significativity of the ARMA coefficient estimates. Indeed past returns were found to have an effect on future returns. But now the results from GARCH say the opposite. Who's right? Could I still reject EMH or should I revise my hypothesis?
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  • $\begingroup$ Why would you expect the conditional mean parameters not to change when you change the conditional variance parameters? You should not. Now, at least one of the models is misspecified. You should not trust a misspecified model if you are doing hypothesis testing. Significance of individual parameters in an ARMA model is not that imporant, in my opinion, and there are many representations that are close (AR models approximate MA models and vice versa). I would pay more attention to residual diagnostics. But you can always estimate a model that restricts the insignificant parameters to zero. $\endgroup$ – Richard Hardy May 5 '18 at 19:45
  • $\begingroup$ @RichardHardy thank you for your answer! You say "You should not trust a misspecified model if you are doing hypothesis testing". That sounds extremely right. Financial time series often exhibit volatility clustering, fat tail and leverage: does it mean that if I notice them I should only test hypothesis on, for example, ARMA-EGARCH with t-distributed innovation (being the one that account for many of the series peculiarities)? Shouldn't I rely on AIC? $\endgroup$ – toyo10 May 5 '18 at 20:03
  • $\begingroup$ @RichardHardy with "rely on AIC" I mean test just the model with the lowest AIC. (in my case ARMA has a lower AIC than ARMA-GARCH) $\endgroup$ – toyo10 May 5 '18 at 20:16
  • $\begingroup$ This is a complicated matter. AIC will select a model that will be best at forecasting, even if that model will not pass specification tests and have somewhat poor residual diagnostics. So if you are interested in forecasting and have restricted yourself to prespecified a pool of models, then AIC is the best you can do (short of combining forecasts from different models). But if you want to test some hypotheses, then the validity of the result depends on the assumptions to be met, e.g. that errors are i.i.d. (But in the end I still feel this is a complicated matter...) $\endgroup$ – Richard Hardy May 6 '18 at 7:39
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What's going on? How would you comment the results?

Why would you expect the conditional mean parameters not to change when you change the conditional variance parameters? Generally, you should not.

I was refusing the the weak form of the Efficient Market Hypothesis (EMH) on the basis of the significance of the ARMA coefficient estimates. Indeed past returns were found to have an effect on future returns. But now the results from GARCH say the opposite. Who's right?

The validity of hypothesis testing depends on the assumptions being satisfied, e.g. that model errors are i.i.d. Your ARMA model had autocorrelation in squared residuals, which is an indication the errors were not i.i.d. What about the ARMA-GARCH model? If it passes residual diagnostic test, you can keep the model and use it as a basis for your hypothesis test. I would not worry too much about significance of model coefficients. Since AR models approximate MA models and vice versa, you may have a representation that is suboptimal (as there are so many similar ones to choose from); but as long as it does a good job modelling the patterns in the data, you should be fine. Also, you may estimate a model that restricts the insignificant parameters to zero and see if the residual diagnostic remain favourable. If they do, keep the restrictions, if not, drop them.

Could I still reject EMH or should I revise my hypothesis?

Regarding the null hypothesis itself, if you are interested in testing the EMH, then I do not see why you should revise it. Regarding the result of the test, you should rely on one that is achieved when the assumptions are satisfied, as discussed above.

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  • $\begingroup$ Thank you for the answer! From that I got that, for example, in my case I should rely for testing hypothesis basically on the ARMA-GARCH because of the lack of autocorrelation in the residuals. Just for clarification, let's consider my residuals still present fat-tail after GARCH application and that I managed to "get rid of it" by specifying a t-distributed error. If in this case the shape parameter results significant, should I better rely on this model for testing hypothesis on the parameters? $\endgroup$ – toyo10 May 6 '18 at 11:46
  • $\begingroup$ In general, the estimated error distribution should match the assumed one. So if you assumed a normal distribution but got fat tails, that is a problem. (Recall that in GARCH models, we make assumptions on standardized innovations and thus look at standardized residuals, not raw ones.) Is there any natural reason to expect the shape parameter to have a certain value? If so, then you can test whether it is significantly different from that value. Otherwise you cannot test that, so it does not make sense to discuss the significance of it. $\endgroup$ – Richard Hardy May 6 '18 at 12:04
  • $\begingroup$ Great explanation! So in this case I would use the model that minimise the AIC for forecasting and the model with "better specification" for testing hypothesis. The two can not coincide, and it's perfectly fine right? $\endgroup$ – toyo10 May 6 '18 at 12:42
  • $\begingroup$ The two may coincide (that is possible), but they need not to. $\endgroup$ – Richard Hardy May 6 '18 at 14:47

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