Question 18
Each member of a population has two genes (possibly repeated) among the genes $G_1, G_2, \dots, G_n$. Suppose that, in generation 0 of this population, $p_{ij}(0)$ ($1 \le i, j \le n$) is the proportion of members of this population with genotype $G_i G_j$ (or $G_j G_i$, which is biologically identical to $G_i G_j$).
To generate each member of generation $k+1$, two members of generation $k$ are chosen independently and at random, and one gene of each member is chosen independently and at random: these two genes will be the genotype of their breed. Let $p_{ij}(k)$ be the probability that a member of generation $k$ has genotype $G_i G_j$ (or $G_j G_i$). Suppose the population is big enough in a way that at least two members are generated.
Prove that $p_{ij}(2012) = p_{ij}(2013)$ for all $i, j$ ($1 \le i, j \le n$).
To generate each member of generation $k+1$, two members of generation $k$ are chosen independently and at random, and one gene of each member is chosen independently and at random: these two genes will be the genotype of their breed. Let $p_{ij}(k)$ be the probability that a member of generation $k$ has genotype $G_i G_j$ (or $G_j G_i$). Suppose the population is big enough in a way that at least two members are generated.
Prove that $p_{ij}(2012) = p_{ij}(2013)$ for all $i, j$ ($1 \le i, j \le n$).