I was reading the book "Genetics and Analysis of Quantitative Traits", by Lynch and Walsh. I how the covariance between two individuals with IBD $\Theta$ gets divided into just the additive variance and dominance variance component, even in the simple $1$ locus case.
Here my understanding of the modelling (for the simple one allele case):
Given a genotypic value $G_{i,j}$ of mean $0$, $i,j \in \{0,1\}$ we find numbers $\alpha_0$ and $\alpha_1$ minimising the least squares of the following form $\mathbb{E}(G_{i,j}-\alpha_i-\alpha_j)^2$, where the expectation is over the population.
We next define the error terms in each case as $\delta_{i,j}=G_{i,j}-\alpha_i-\alpha_j$. From the properties, viewed as functions of the population $\alpha_i$ is independent of $\delta_{i,j}$, and both have mean $0$.
The claim made in the book is that given two individuals, with IBD $\Theta$ and probability that the genotype is equal $\Delta$, the covariance of the genotypes $G_{i,j}$ and $G_{k,l}$ is given by,
$$\text{cov}(G_{i,j},G_{k,l}) = 2\Theta \sigma_A^2 +\Delta \sigma_D^2,$$
where $\sigma_A^2= \text{Var}(\alpha_i)$, and $\sigma_D^2 = \text{Var}(\delta_{i,j})$.
Expanding the LHS of the expression, showing that $\mathbb{E}[(\alpha_i +\alpha_j)(\alpha_k+\alpha_l)] =\Theta \sigma_A^2$ is quite easy. It also seems to follow from the that the the terms $\mathbb{E}[(\alpha_i +\alpha_j)\delta_{k,l}]=0$ from independence of errors from the $\alpha$.
On analysing $\mathbb{E}[\delta_{i,j}\delta_{k,l}]$, we see that if both genotypes are equal, which occurs with probability $\Delta$, then this reduces to $\sigma_D^2$. This gives us a term $\Theta \sigma_D^2$. Further, if both $i,j$ and $k,l$ are not IBD then the covariance is $0$. However when one of the two alleles are IBD, then it is not clear to me that this the covariance will still be $0$.
The book seems to claim that unless both alleles are IBD, $\delta_{i,j}$ and $\delta_{k,l}$ are independent. I do not see why this is the case. Am I missing anything here? I'd appreciate any help wrt this.