Exposure biomarker conversions from data
In table ExposureBiomarkerConversions, conversion factors are specified to convert measured biomarkers to the biomarkers of interest. Conversion factors may be dependent on individual properties like age and gender. Specify in table ‘ExposureBiomarkerConversionSGs’ (exposure biomarker conversion subgroups) age and/or gender specific conversion factors.
For age, specify the lower bound of the age interval (in years) of the exposure biomarker conversion subgroup. Individuals belong to a subgroup when the age of the individual is equal or greater than the specified lower bound and smaller than the specified lower age of the next subgroup.
Check option Use exposure biomarker conversion factors subgroup to use age and/or gender specific conversion factors in your assessment.
Uniform distribution
The uniform distribution is defined on the interval [\(a\), \(b\)] where \(a\) and \(b\) are the minimum and maximum value.
The conversion factor is the nominal value of the uniform distribution, here equal to the mean \(1/2 (a + b)\). The variability upper value is defined as parameter \(b\). Parameter \(a\) is calculated as \(factor - range\) where \(range\) = \(b - factor\).
Lognormal distribution
The lognormal distribution is characterised by parameters \(\mu\) and \(\sigma\), which are the mean and standard deviation on log-scale.
The conversion factor is the nominal value of the lognormal distribution, \(\mu = ln(conversion factor)\). The variability upper value specifies the p95 of the standard lognormal. Parameter \(\sigma\) is calculated as:
Beta distribution
The standard beta distribution is defined on the interval (0, 1) and is usually characterised by two parameters \(a\) and \(b\), with \(a>0\), \(b>0\) (see e.g. Mood et al. (1974)). Alternatively, it can be parameterised by the mean
and the variance
Note that:
The conversion factor is the nominal value of the beta distribution, here the mean. The variability upper value is the variance of the beta distribution. The equations solved for \(a\) and \(b\) show that:
and