Statistics and the Evaluation of Evidence for Forensic Scientists. Franco Taroni

Statistics and the Evaluation of Evidence for Forensic Scientists - Franco Taroni


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images Relative frequency (%) 18.8 32.1 49.1 equation

      (see Section 3.5). He is still not too sure what this means but feels that it is more representative of the information available to him than the previous probability, since it takes account of the actual genotypes of the crime stain and the PoI.

      The genotype of the crime stain for locus LDLR is images. The genotype of the PoI is also images (if it were not they would not be a PoI). What is the value of this evidence? The discussion earlier suggests various possible answers.

      1 (1) The probability that two people chosen at random have the same genotype for locus LDLR. This is 0.379.

      2 (2) The probability that two people chosen at random have the same, pre‐specified, genotype. For genotype this is 0.103.

      3 (3) The probability that one person, chosen at random, has the same genotype as the crime stain. If the crime stain is of group , this probability is 0.321, from Table 1.1.

      The relative merits of these answers will be discussed in Section 3.5 for (1) and (2) and Section 2.4.5 for (3).

       It is important to keep in mind that in any crime investigation, random man is pure fiction: nobody was actually chosen at random in any population, and so probabilities calculated under an assumption of randomly sampled suspects have no direct bearing on evidential weight in actual cases. (p. 160)

      A comment on randomness is also made by Kingston and Kirk (1964) (see pp. 515–516). A discussion about the extension of the concept of ‘random man’ to the one of ‘unrelated person’ to the PoI, when dealing with DNA evidence evaluation, is discussed in Milot et al. (2020).

      1.3.3 Glass Fragments

      Section 1.3.2 discussed an example of the interpretation of the evidence of DNA profiling. Consider now an example concerning glass fragments and the measurement of the refractive index of these.

      Example 1.2 As before, consider the investigation of a crime. A window has been broken during the commission of the crime. A PoI is found with fragments of glass on their clothing, similar in refractive index to the broken window. Several fragments are taken for investigation and their refractive index measurements taken.

      Comparison in Example 1.2 has to be made between the two sets of fragments on the basis of their refractive index measurements. The evidential value of the outcome of this comparison has to be assessed. Notice that it is assumed that none of the fragments has any distinctive features and comparison is based only on the refractive index measurements.

Measurements from the window 1.518 44 1.518 48 1.518 44 1.518 50 1.518 40
1.518 48 1.518 46 1.518 46 1.518 44 1.518 48
Measurements from the PoI 1.518 48 1.518 50 1.518 48 1.518 44 1.518 46