binseg.meanvar.norm: Multiple Changes in Mean and Variance using Binary Segmentation method - Normal Data

Description

Calculates the optimal positioning and number of changepoints for Normal data using Binary Segmentation method. Note that this is an approximate method.

Usage

binseg.meanvar.norm(data, Q=5, pen=0)

Arguments

data

A vector containing the data within which you wish to find changepoints.

Numeric value of the maximum number of changepoints you wish to search for, default is 5.

pen

Numeric value of the linear penalty function. This value is used in the decision as to the optimal number of changepoints.

Value

A list is returned containing the following items
cps2xQ Matrix containing the changepoint positions on the first row and the test statistic on the second row.
op.cptsThe optimal changepoint locations for the penalty supplied.
penPenalty used to find the optimal number of changepoints.

Details

This function is used to find a multiple changes in mean and variance for data that is assumed to be normally distributed. The value returned is the result of finding the optimal location of up to Q changepoints using the log of the likelihood ratio statistic. Once all changepoint locations have been calculated, the optimal number of changepoints is decided using pen as the penalty function.

References

Binary Segmentation: Scott, A. J. and Knott, M. (1974) A Cluster Analysis Method for Grouping Means in the Analysis of Variance, Biometrics 30(3), 507--512

Change in mean and variance: Chen, J. and Gupta, A. K. (2000) Parametric statistical change point analysis, Birkhauser

Examples

Run this code

# Example of multiple changes in mean and variance at 50,100,150 in simulated normal data
set.seed(1)
x=c(rnorm(50,0,1),rnorm(50,5,3),rnorm(50,10,1),rnorm(50,3,10))
binseg.meanvar.norm(x,Q=5, pen=4*log(200)) # returns optimal number as 4 and the locations as c(50,100,150,152)
binseg.meanvar.norm(x,Q=3, pen=4*log(200)) # returns optimal number as 2 as this is the maximum number of changepoints it can find.  If you get the maximum number, you need to increase Q until this is not the case.

# Example no change in mean or variance
set.seed(1)
x=rnorm(200,0,1)
binseg.meanvar.norm(x,pen=4*log(200)) # returns optimal number as 0

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