In charlotte-ngs/lbgfs2021: Livestock Breeding and Genomics (ETHZ) - Fall Semester 2021

knitr::opts_chunk$set(echo = TRUE)
# rmdhelp::show_knit_hook_call()
knitr::knit_hooks$set(hook_convert_odg = rmdhelp::hook_convert_odg)

Problem 1: Inverse Numerator Relationship Matrix

During the lecture the method of computing the inverse numerator relationship matrix $A^{-1}$ directly was introduced. The computation is based on the LDL-decomposition. As a result, we can write

$$A^{-1} = (L^T)^{-1} \cdot D^{-1} \cdot L^{-1}$$ where $L^{-1} = I-P$, and $D^{-1}$ is a diagonal matrix with $(D^{-1})_{ii} * \sigma_u^{-2} = var(m_i)^{-1}$.

Tasks

Use the example pedigree given below and compute the matrices $L^{-1}$ and $D^{-1}$ to compute $A^{-1}$
Verify your result using the function getAinv() from package pedigreemm.

Pedigree

nr_animal <- 6
tbl_pedigree <- tibble::tibble(Calf = c(1:nr_animal),
                               Sire = c(NA, NA, NA, 1 ,3, 4),
                               Dam = c(NA, NA, NA, 2, 2, 5))
tbl_pedigree

Solution

The matrix $P$ comes from the simple decomposition and can be constructed using the pedigree.

P = matrix(0, nrow = nr_animal, ncol = nr_animal)
for (i in 1:nr_animal){
  s <- tbl_pedigree$Sire[i]
  d <- tbl_pedigree$Dam[i]
  if (!is.na(s)){
    P[i,s] <- 0.5
  }
  if(!is.na(d)){
    P[i,d] <- 0.5
  }
}
P

With that the matrix $L^{-1}$ is

I <- diag(1, nrow = nr_animal, ncol = nr_animal)
Linv <- I - P
Linv

The matrix $D$ is obtained from package pedigreemm

ped <- pedigreemm::pedigree(sire = tbl_pedigree$Sire,
                            dam  = tbl_pedigree$Dam,
                            label = as.character(1:nr_animal))
D <- pedigreemm::Dmat(ped = ped)
Dinv <- diag(1/D, nrow = nr_animal, ncol = nr_animal)
Dinv

The inverse numerator relationship matrix is

Ainv <- t(Linv) %*% Dinv %*% Linv
Ainv

Verification

pedigreemm::getAInv(ped = ped)

Problem 2: Rules

The following diagram helps to illustrate the rules for constructing $A^{-1}$

#rmdhelp::use_odg_graphic(ps_path = "odg/inv-num-mat.odg")
knitr::include_graphics(path = "odg/inv-num-mat.png")

Tasks

Go through the list of animals in the pedigree and write down the contributions that are made to the different elements of matrix $A^{-1}$
Based on the different contributions, try to come up with some general rules

Solution

In what follows, we use the following convention $\delta_i = (D^{-1})_{ii}$.

Animal 1

We start with animal $1$.

#rmdhelp::use_odg_graphic(ps_path = "odg/inm-animal1.odg")
knitr::include_graphics(path = "odg/inm-animal1.png")

Animal $1$ has no parents and therefore the diagonal element $\delta_1 = (D^{-1}){11}$ of matrix $D^{-1}$ is $\delta{1} = 1$. By looking at the red boxes, we can see that $\delta_1$ is added as a contribution to $(A^{-1}){11}$. So far we are still missing a contribution of $0.5$ to the element $(A^{-1}){11}$. Again by inspecting the red boxes in the above diagram, we can see that this contribution corresponds to $\delta_4 /4$ which comes from offspring $4$ of parent $1$. Hence diagonal elements of $(A^{-1})_{ss}$ corresponding to parents $s$ of offsprint $i$ receive $\delta_i/4$ as contribution. More details on that is obtained when inspecting the contributions of animal $4$. Animals $2$ and $3$ do not have parents and are therefore analogous to animal $1$.

Animal 4

Animal $4$ has parents $1$ and $2$.

#rmdhelp::use_odg_graphic(ps_path = "odg/inm-animal4.odg")
knitr::include_graphics(path = "odg/inm-animal4.png")

The red boxes in the above diagram show that for animal $4$ there is a contribution of $\delta_4$ to the diagonal. Then there are contributions of $\delta_4/4$ for the elements $(A^{-1}){11}$, $(A^{-1}){22}$, $(A^{-1}){12}$ and $(A^{-1}){21}$. Furthermore there are negative contributions of $\delta_4/2$ to $(A^{-1}){41}$, $(A^{-1}){14}$, $(A^{-1}){24}$ and $(A^{-1}){42}$.

General Rules

From this the general rules which were first published by Henderson can be deduced as

Both Parents Known
- add $\delta_i$ to the diagonal-element $(i,i)$
- add $-\delta_i/2$ to off-diagonal elements $(s,i)$, $(i,s)$, $(d,i)$ and $(i,d)$
- add $\delta_i/4$ to elements $(s,s)$, $(d,d)$, $(s,d)$, $(d,s)$
Only One Parent Known
- add $\delta_i$ to diagonal-element $(i,i)$
- add $-\delta_i/2$ to off-diagonal elements $(s,i)$, $(i,s)$
- add $\delta_i/4$ to element $(s,s)$
Both Parents Unknown
- add $\delta_i$ to diagonal-element $(i,i)$