Tests the null hypothesis that a benchmark forecasting method has unconditional superior predictive ability over all competing alternatives, using Hansen's (2005) stationary bootstrap approach.
Arguments
- Y
An
n x Jmatrix of loss differentials with respect to the benchmark. Positive values indicate the benchmark outperforms competitorjin periodt.- level
Significance level (e.g., 0.05).
- B
Integer; number of bootstrap replications. Default
5000.- q
Numeric in \((0, 1)\); parameter for the geometric distribution in the stationary bootstrap. Controls the average block length (\(1/q\)). Default
0.25(average block length of 4).
Value
A list with class "spa_test" containing:
- statistic
The SPA test statistic (consistent).
- pvalue
Bootstrap p-value.
- reject
Logical; whether the null is rejected.
- level
Significance level used.
- d_bar
Sample mean of loss differentials for each competitor.
- n
Number of observations.
- J
Number of competitors.
- B
Number of bootstrap replications.
Details
The SPA test statistic is:
$$T^{SPA} = \max_{1 \le j \le J} \frac{\sqrt{n}\, \bar{d}_j}{\hat{\sigma}_j}$$
where \(\bar{d}_j\) is the mean loss differential for competitor j
and \(\hat{\sigma}_j\) is its estimated standard deviation.
The null hypothesis is rejected when the benchmark is not superior to all competitors, i.e., some competitor significantly outperforms it.
Hansen's (2005) consistent test uses a pre-selection step to remove clearly inferior competitors from the critical value computation.
References
Hansen, P.R. (2005). A Test for Superior Predictive Ability. Journal of Business & Economic Statistics, 23(4), 365-380.
White, H. (2000). A Reality Check for Data Snooping. Econometrica, 68(5), 1097-1126.
Examples
if (FALSE) { # \dontrun{
sim <- do_sim(J = 3, n = 250, a = 1, c = 0, rho_u = 0.4)
spa_test(sim$Y, level = 0.05)
} # }