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I would appreciate if you could let me know to conclude about the modified Diebold and Mariano test when our alternative hypothesis is less? How about greater?

library(forecast)
forecast <- ts(c( 96, 99,102, 96,105, 99, 99,103, 98,106))
observed <- ts(c( 96,101,107,108, 93,103, 99,105,103, 98))
forecast2 <- ts(c(105, 94,107,101,111,115,104,111,111,116))
print(dm.test((forecast-observed), (forecast2-observed),alternative = "two.sided"))  
print(dm.test((forecast-observed), (forecast2-observed),alternative = "less"))  
print(dm.test((forecast-observed), (forecast2-observed),alternative = "greater"))  
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  • $\begingroup$ Do you have any trouble interpreting the unmodified DM test? The interpretation of the modified test is the same. Modification simply allows for nonzero autocorrelation in forecast errors. $\endgroup$ Commented Mar 2, 2023 at 17:01
  • $\begingroup$ @RichardHardy. Yes, could you please explain it? $\endgroup$
    – ebrahimi
    Commented Mar 2, 2023 at 17:07
  • $\begingroup$ Have you read a textbook entry on the Diebold-Mariano test? What exactly did you find confusing? It is just a $t$-test that can be interpreted in a pretty regular way. Also, I can warmly recommend Diebold's (2015) paper on the use and abuse of the test. $\endgroup$ Commented Mar 2, 2023 at 17:17

1 Answer 1

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If you happen to reject the null hypothesis of equal expected predictive loss, $H_0\colon E(L(e_1))=E(L(e_2))$, then

  • under $H_{1a}\colon E(L(e_1))\neq E(L(e_2))$, you favor the view that the expected losses are unequal (alternative="two.sided");
  • under $H_{1b}\colon E(L(e_1))> E(L(e_2))$, you favor the view that the expected loss from the first forecast is larger (alternative="less");
  • under $H_{1c}\colon E(L(e_1))< E(L(e_2))$, you favor the view that the expected loss from the second forecast is larger (alternative="greater").

You can find the same interpretation phrased another way in the functions' help file.

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