two.level.lindley.LR {comparison}

Likelihood ratio calculation using Lindley's approach

Description

Takes a compitem object which represents some control item, and a compitem object which represents a recovered item, then uses information from a compcovar object, which represents the information from the population, to calculate a likelihood ratio as a measure of the evidence given by the observations for the same/different source propositions.

Usage

two.level.lindley.LR(control, recovered, background)

Arguments

control

a compitem object with the control item information

recovered

a compitem object with the recovered item information

background

a compcovar object with the population information

Details

Does the likelihood ratio calculations for a two-level model assuming that the between item distribution is univariate normal. This function is taken from the approach devised by Denis Lindley in his 1977 paper (details below) and represents the progenitor of all the functions in this package.

Value

an estimate of the likelihood ratio

Author(s)

David Lucy

References

Lindley, D. (1977) A problem in forensic Science. Biometrika: 64; 207-213.

Examples

# load Greg Zadora's glass data
data(glass)

# calculate a compcovar object based upon dat
# using K
Z = two.level.components(glass, 7, 1)

# calculate a compitem object representing the control item
control = two.level.comparison.items(glass[1:6,], 7)

# calculate a compitem object representing the recovered item
# known to be from the same item (item 1)
recovered.1 = two.level.comparison.items(glass[7:12,], 7)

# calculate a compitem object representing the recovered item
# known to be from a different item (item 2)
recovered.2 = two.level.comparison.items(glass[19:24,], 7)


# calculate the likelihood ratio for a known
# same source comparison - should be 6.323941
# This value is 6.323327 in this version and in the last version written by David (1.0-4)
lr.1 = two.level.lindley.LR(control, recovered.1, Z)
lr.1

# calculate the likelihood ratio for a known
# different source comparison - should be 0.004422907
# This value is 0.004421978 in this version and the last version written by David (1.0-4)
lr.2 = two.level.lindley.LR(control, recovered.2, Z)
lr.2