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I have a question about the Diebold-Mariano test. I have different forecasting horizons (n-ahead = 1 to 40) and different forecasting origins (26). I.e. I employ a rolling origin evaluation approach. I want to compare the forecasting results of two empirical models. Now, should I:

  1. Compare all n-ahead=1 forecasting errors for the two models, set h=1, and perform the test. Then, I would do the same for n-ahead=2 forecasts, but set h=2, and so on.
  2. Alternatively, is it more common to compare the average errors over all forecasting origins? However, this would result in information loss, wouldn't it?
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The answer depends on what forecasts you actually want to compare.

  • If you are interested in each of the 40 horizons $h=1,\dots,40$, you could do 40 separate Diebold-Mariano tests. E.g. each horizon's forecast leads to a separate decision, and consequences of each decision can be identified using a loss function defined on the forecast error. (The loss functions could be different for each horizon.) Thus you would get a specific answer for each horizon.
    You may want to adjust the significance level to account for multiple testing.
    You may also consider treating this as a test of a joint hypothesis.
  • If you treat the 40 horizons as one multivariate forecast, you could do a single test. E.g. there is a single decision based on the 40 forecasts and you can define a loss function on the multivariate forecast error such as $L(e_1,\dots,e_{40})=\sum_{h=1}^{40} w_h|e_h|$ with nonnegative weights $w_h$ for $h=1,\dots,40$; here $e_h$ for $h=1,\dots,40$ are forecast errors. Thus you would get a single answer covering all the horizons.

The choice between these approaches should be guided by what you really are interested in from the subject-matter perspective.

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  • $\begingroup$ Thanks a lot! I am interested in the former. However, I am still not sure how to deal with the different forecasting origins. Can I just pool them over the forecasting horizons? And what would be the correct denominator then? Should I still include the forecasting horizon in the denominator (set h=1, h=2, etc.), or can I discard this and set h=1 for all forecasting horizons, as I am not comparing them over all forecasting horizons? $\endgroup$ Commented Apr 21, 2023 at 7:36
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    $\begingroup$ @Aneconomist, if you want to do 40 tests, you would specify the horizon separately for each test. If you want to do a single test, I think you would have to specify h=40, since this is the overlap of at least some forecast errors that you will have when rolling the window one period at a time. $\endgroup$ Commented Apr 21, 2023 at 8:32
  • $\begingroup$ Thanks. Regarding the former statement "if you want to do 40 tests, you would specify the horizon separately for each test," this turns out to be a problem since the number of interactions is 26 and h goes up to 40, so we have h > n, which results in issues (k <- ((n + 1 - 2 * h + (h/n) * (h - 1))/n)^(1/2)). Is there a standard solution for this problem? $\endgroup$ Commented Apr 21, 2023 at 8:45
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    $\begingroup$ @Aneconomist, the argument h=... is used for estimating the long-run variance of the loss differential. Instead of using the dm.test in R, you could take the loss differential series, estimate the long-run variance manually and then do a $t$-test using this estimate. There are probably many functions that estimate long-run variance based on Newey-West, Andrews or other methods automatically, without relying on h. (They select the lag length according to some sensible rules.) $\endgroup$ Commented Apr 21, 2023 at 9:14
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    $\begingroup$ @Aneconomist, yes. The only function of h is for determining the lag length. However, when h is large, the standard way o using it for that may no longer be optimal. (E.g. trying to obtain autocorrelation at lag 40 from a time series of length 26 breaks down.) $\endgroup$ Commented Apr 21, 2023 at 9:50

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