0
$\begingroup$

As I asked in my answer to this question: does anyone know if the DM test (in R in this case) is supposed to be made with h=h-1?

If not, am I supposed to make several prediction sets (with h observations each) and perform h-step predictions on them in order to have more degrees of freedom? That is, am I supposed to make n h-step predictions for the DM test?

When I attempt to perform the test with h=18 (there are 18 obs.) it says that DM variance is 0 (it consumes the degrees of freedom?).

Error in dm.test(error.311arima, error.013arima, : Variance of DM statistic is zero

The residuals are different for the time series predictions.


EDIT

I have "checked under the hood" of the dm.test function in R, and apparently what happens is that when it executes the following operation:

m.test
function (e1, e2, alternative = c("two.sided", "less", "greater"), 
    h = 1, power = 2) 
{
    alternative <- match.arg(alternative)
    d <- c(abs(e1))^power - c(abs(e2))^power
    d.cov <- acf(d, na.action = na.omit, lag.max = h - 1, type = "covariance", 
        plot = FALSE)$acf[, , 1]
    d.var <- sum(c(d.cov[1], 2 * d.cov[-1]))/length(d)
    dv <- d.var
    if (dv > 0) 
        STATISTIC <- mean(d, na.rm = TRUE)/sqrt(dv)
    else stop("Variance of DM statistic is zero")

the d.var part yields a negative number, i.e. the variance estimator is negative???

I presume it is due to a lack of covariance stationarity on the errors, but then what should I do? I can only think of manually differencing the series (or the error of the series) and then use those forecasts, which should be in turn, covariance stationary.

error 311arima  error 013arima
-0.8758334  -0.8623334
-0.571964484    -0.614464484
-1.5118295  -1.3041295
-2.034418946    -1.918818946
-1.422745494    -1.805045494
-1.777216545    -2.240316545
-1.671314551    -2.260914551
-2.61727251 -3.40957251
-4.29605616 -5.10285616
-4.87524124 -5.78274124
-6.29766468 -7.27736468
-7.458823814    -8.455323814
-0.983687537    -1.110287537
-0.49127652 -0.92037652
0.652971123 0.048671123
2.52594615  1.73144615
0.651429228 -0.409670772
3.416551133 2.252951133
3.302610752 2.023910752
0.929007023 -0.460192977
1.959733687 0.522733687
-0.710917583    -2.217517583
-2.09341202 -3.64921202
-2.320514415    -3.910114415

The answer to this question can be found in a later question here: Diebold-Mariano test for multiple prediction horizons

We cannot compare forecasts for VECTORS of differing (in this case increasing) points in time, we must use a rolling window approach with a fixed h to compare these two models' h-point forecasts along time

| cite | improve this question | | | | |
$\endgroup$
3
$\begingroup$

The answer to this question can be found in a later question here: Diebold-Mariano test for multiple prediction horizons

We cannot compare forecasts for VECTORS of differing (in this case increasing) points in time, we must use a rolling window approach with a fixed h to compare these two models' h-point forecasts along time

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.