Ideas in Forensic Mathematics

Likelihood ratios less than one

Paternity paradox: TPOX

Suppose, in a motherless paternity case, the child is pq at some locus – for example 8,11 in TPOX – and the alleged father is qr – say 8,9. He shares an allele with the child; people sometimes assume that that must be evidence that he is the father. But the formula for this situation is 1/4q, and the TPOX 8 allele is very common – q=0.55. Therefore the likelihood ratio is 1/2. Is that right? How can it be?

It is right. The given data is evidence of non-paternity. Real fathers typically share both alleles. (I thank Sandy Zabel for pointing out that this is an exaggeration. A little less pithy but more accurate would be: Real fathers often (probability p+q, where p, q are the probabilities of the child's alleles) are men both of whose alleles are found in the child. So my original statement is conditionally correct – depending on whether you accept that a qq man is properly described as "sharing both alleles" with a pq child – only when p+q>1/2. Therefore when q is the TPOX 8 allele I was right, but of course with the word "typically" I meant more than just for this particular case.)

Note that the situation is not even limited to motherless cases. If q is sufficiently common – the TPOX 8 allele is probably the only example among currently popular loci – then even if the mother is present and the 8 is known to be the paternal allele, still the likelihood ratio, 1/2q < 1. Real fathers share both alleles.

Mixture paradox

The same thing can happen with mixtures. Suppose a mixed DNA stain shows four alleles at some locus, two of which are very common. The mixture is alleged to be from the suspect and an accomplice who escaped. But if the two alleles that the suspect shares with the mixture are both common – frequencies over 30% for example – then evidence actually tends to exonerate the suspect. To see the point intuitively it may help to reflect that such a man could be a contributor to the mixture only if he has a buddy of the rare type that could supply both of the rare alleles. A more typical contributor would be a man who supplies at least one of the rare alleles himself; it would be much easier to imagine that his friend has the requisite complementary (common) alleles.

Example

Question (from Wiesbaden, Germany)

Case scenario
	Stain	Alleles 28/29/30/31,2/32.2
	Suspect	30 - homozygous
	Prosecution hypothesis: Suspect plus 2 unknown persons
	Defense hypothesis: 3 unknown persons.

In my opinion the evidence that the suspect contributed to the stain profile should be larger than 1, not smaller.

Can you explain what happened in this special case ?

Answer

The result that LR<1 is possible. It happens when the stain includes so many common alleles that few people are excluded, and the suspect has one or two quite common alleles. Hence, it can actually be harder to find two "accomplices" of him with the remaining (rare) alleles, than to find three random people who fit all the alleles. The evidence in such a situation is, in fact, in his favor. Remember, his being included is not surprising. Almost everyone is included. But he is in a sense "less included" than average because he explains none of the interesting – i.e. rare – alleles.

mixture exclusion

Note that the "exclusion" approach would be misleading in a case like this one, falsely suggesting that the evidence implicates the suspect. No wonder that the NRC II report commented that the exclusion method is "very hard to justify."