Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 39 / 4 / v

Let $A=R_{11}+2R_{12}+R_{22}$. Among the ten complete trios, six have two heterozygous parents: three have child $11$, two have child $12$ and one has child $22$. Their conditional likelihood contribution isTwo more trios have a heterozygous parent and a $22$ parent, with child $12$ in each; they contribute $[R_{12}/(R_{12}+R_{22}){]}^2$. The remaining two parent pairs can produce only child $12$, so their conditional probabilities are one. Multiplying givesThe factor $4$ does not depend on the genotype relative risks, so the conventional proportional likelihood function isThis is the expression in the original PDF, up to the parameter-independent constant. Its last denominator is $R_{12}+R_{22}$; the converted TeX's $R_{11}+R_{22}$ is a transcription error. The factor $4$ in the exact probability comes from the two heterozygous children of two heterozygous parents, each of which has two labelled transmission routes.