Distribution of “T” statistic for Residual CUSUM test

I believe I understand how to calculate the test statistic of the (Lee et al. 2004, Journal of the Japan Statistical Society, doi:10.14490/jjss.34.173) CUSUM test (p. 175).

$$T_n := \frac{1}{\sqrt{n} \tau} \max_{1 \le k \le n} \left| \sum_{t = 1}^k \xi_t^2 - \left( \frac{k}{n} \right) \sum_{t = 1}^n \xi_t^2 \right|.$$

I now wonder about the distribution of this T statistic? I've read in several places that it is to be compared to a Brownian bridge but I do not understand how to create a test with a specified significance out of this.

$T_n$ is the absolute maximum of a CUSUM of squares process. The definition you cite above pertains to the theoretical process based on the true disturbances. When replacing the disturbances with the empirical residuals $\hat \xi_t$ Lee et al. obtain the analogous empirical statistic $\hat T_n$. And in their Theorem 1 they show that $\hat T_n$ converges in distribution to the supremum of a Brownian bridge on the unit interval:

$$\hat T_n \stackrel{d}{\rightarrow} \sup_{0 \le u \le 1} \left| B^0(u) \right|,$$

where $B^0$ is a standard Brownian bridge. The corresponding critical values can easily be obtained by simulation. However, for the absolute maximum there is also a series expansion for the $p$-values available. For example, Equation (20) from Ploberger & Krämer (1992, Econometrica, http://www.jstor.org/stable/2951597) provides:

$$\mathrm{Prob}\left( \sup_{0 \le u \le 1} \left| B^0(u) \right| ~>~ T \right) ~=~ 2 \sum_{j = 1}^\infty (-1)^{j + 1} \exp(-2 j^2 T^2)$$

Thus, you can easily evaluate the first few (say 100) terms of this series and sum them up. Unless you want probabilities close to 1, 100 terms are more than sufficient. (The first term computes the probability for a single boundary crossing, the second for a crossing of two opposite boundaries, the third for three crossings etc. Except for small $T$ values these terms become extremely small.)

In R:

bbpvalue <- function(x, nterms = 100) {
j <- 1L:nterms
2 * sum((-1)^(j + 1) * exp(-2 * j^2 * x^2))
}
bbpvalue(1.23)
## [1] 0.0970269


Note that the same functionality is also used in the R package strucchange which provides a variety of CUSUM and related tests (albeit not a CUSUM of squares test for GARCH models).

strucchange::pvalue.efp(1.23, lim.process = "Brownian bridge",
functional = "max", k = 1, alt.boundary = FALSE)
## [1] 0.0970269

• Thank you very much for this most helpful and detailed reply. – robinsa Oct 5 '16 at 7:35