Brother–sister mating. Consider the six state chain defined in Exercise 1.66. Show that the total…
Brother–sister mating. Consider the six state chain defined in Exercise 1.66. Show that the total number of A’s is a martingale and use this to compute the probability of getting absorbed into the 2,2 (i.e., all A’s state) starting from each initial state.
Brother–sister mating. In this genetics scheme two individuals (one male and one female) are retained from each generation and are mated to give the next. If the individuals involved are diploid and we are interested in a trait with two alleles, A and a, then each individual has three possible states AA, Aa, aa or more succinctly 2, 1, 0. If we keep track of the sexes of the two individuals the chain has nine states, but if we ignore the sex there are just six: 22, 21, 20, 11, 10, and 00. (a) Assuming that reproduction corresponds to picking one letter at random from each parent, compute the transition probability. (b) 22 and 00 are absorbing states for the chain. Show that the probability of absorption in 22 is equal to the fraction of A’s in the state. (c) Let