trend (version 0.2.0)

pettitt.test: Pettitt's test for change-point-detection

Description

Performes a nonparametric U-test after Pettitt in order to test for a shift in the central tendcency of a time series. The H0-hypothesis is: no change is tested against the HA-Hypothesis: change.

Usage

pettitt.test(x)

Arguments

x

a vector or a time series object of class "ts"

Value

An object of class "htest"

method

a character string indicating the chosen test

data.name

a character string giving the name(s) of the data

statistic

the maximum K of the absolute values of the U-statistics

estimate

probable change point at time step tau

nobs

number of total observations

alternative

a character string describing the alternative hypothesis

p.value

the p-value for the test

Warning

The current function is for complete observations only. The approximate probability is good for \(p \le 0.5\).

Details

The test is conducted in accordance to Pettitt's proposed method. First a vector that comprises the U-Statistics is calculated:

$$U_{i} = \sum_{k=1}^i{~\sum_{j=i+1}^n{ \textnormal{sgn}(x_k - x_j)}}$$

for \((1 \le i \le n )\). For a two-sided test, the maximum of the absolute obtained U-statistics is taken as the test statistic K:

$$K = \max \left| U_i \right|$$

Finally, the approximate probability for a two-sided test is calculated according to

$$2 \exp^{-6K^2 / (T^3 + T^2)}$$

References

CHR (ed., 2010): Das Abflussregime des Rheins und seiner Nebenfluesse im 20. Jahrhundert, Report no I-22 of the CHR, p. 172.

Pettitt, A. N., (1979). A non-parametric approach to the change point problem. Journal of the Royal Statistical Society Series C, Applied Statistics 28, 126-135.

See Also

efp sctest.efp

Examples

# NOT RUN {
data(maxau) ; plot(maxau[,"s"])
s.res <- pettitt.test(maxau[,"s"])
n <- s.res$nobs
i <- s.res$estimate
s.1 <- mean(maxau[1:i,"s"])
s.2 <- mean(maxau[(i+1):n,"s"])
s <- ts(c(rep(s.1,i), rep(s.2,(n-i))))
tsp(s) <- tsp(maxau[,"s"])
lines(s, lty=2)
print(s.res)


data(PagesData) ; pettitt.test(PagesData)
# }