The definition of LD coefficient for multiple loci can be found in Slakin 1972 Genetics.
Hello Lei
You may find another definition of multilocus LD based on the approach of Geiringer and Bennett in Gorelick and Laubichler 2004, Genetics. There is another notion based on "cumulants" discussed in Neher and Shraiman, 2011, Reviews of Modern Physics.
With three loci A, B and C, each with two alleles, 1 and 2, there are three two locus(first order) LD coefficients D(AB), D(AC), and A(BC) following the definition given in the question in terms of the haplotype frequencies(represented by y with indices 1 and 2 for the alleles) and one three locus(second order) LD coefficient given by
D(ABC) = (y111y222 - y112y221) - (y121y212 - y122y211)
@Victor Garcia, Hi, Victor, I have a quick look at Gorelick and Laubichler 2004. As far as I can see, It gives a clear math difinition about higher order linkage disequilibrium. However, it is not helpful for this question. Since the definition in Slakin 1972 for 3 loci is the same as the definition in Gorelick and Laubichler 2004. This definition for me is not "symmetrical", For 3 loci case, one can always write down 8 different formula to define D(ABC)...
The expression for D(ABC) given in my answer is entirely in terms of the eight possible three-locus gametic frequencies. An alternative expression, following Slatkin (1972), would be, in terms of the three-locus gametic frequency, the three gene frequencies and the three first-order LD coefficients, as
D(ABC) = p(ABC) - p(A) D(BC) - p(B) D(AC) - p(C) D(AB) - p(A) p(B) p(C).
As per Bennett (1954), the effects of first-order LD coefficients, are removed by subtraction.
Acording to my recent calculation, three first-order LD coefficients based on Slakin 1972 is
D(ABC)=[(y111+y122+y212+y221)-(y112+y121+y211+y222)]/8-D(AB)(2P(C)-1)/2-D(AC)(2P(B)-1)/2-D(BC)(2P(A)-1)/2-(2P(A)-1)(2P(B)-1)(2P(C)-1)/8.
It is a symetric result though very ugly (conplex).
Please check whether the right-hand side of the expression reduces to zero when all LD coefficients are zero.
@Prem Narain, As there is no LD, y111=P(A)P(B)P(C), y122=P(A)Q(B)Q(C), y212=Q(A)P(B)Q(C), y221=Q(A)Q(B)P(C), y112=P(A)P(B)Q(C), y121=P(A)Q(B)P(C), y211=Q(A)P(B)P(C), y222=Q(A)Q(B)Q(C), thus, [(y111+y122+y212+y221)-(y112+y121+y211+y222)]/8=(2P(A)-1)(2P(B)-1)(2P(C)-1)/8. And No LD, D(AB)=D(AC)=D(BC)=0, thus RHS=0.
what if I exchange teh first two terms or the last two terms, it will be correct?
D(ABC) = (y112y221- y111y222 -) - (y121y212 - y122y211)
or
D(ABC) = (y111y222 - y112y221) - (y122y211 - y121y212 )
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