About the QP transformation and Quaas (1988)

You might have heard about the QP transformation. This transformation is named after Quaas and Pollak (1981). These two scientists are among the pioneers of Quantitative Genetics and Animal Breeding. They have published numerous articles that are well ahead of their time. When I was in the University of Nebraska-Lincoln, I had the chance to meet Dr. John Pollak. However, I have not met Dr. Richard Louis Quaas. Later, I realised that he died in October 19, 2021, at the age of 77. The QP transformation enables genetic groups to be included in the pedigree, in contrast to previous approaches where genetic groups are included as a distinct fixed effect in the model.

What are genetic groups? In short, different groups of unknown parents are assigned to different genetic groups to enable them to represent different means. For example, an unknown sire with a progeny born in 1970 should have a different mean than an unknown dam with a progeny born in 1980. That also means the random vector has a non-zero mean.

After solving BLUP for solutions of animals ($\hat{\mathbf u}$) and genetic groups ($\hat{\mathbf g}$), genetic group contributions ($\mathbf Q \hat{\mathbf g}$) are added to $\hat{\mathbf u}$ to calculate animals’ breeding values, where Q, in simple words, is the matrix of coefficients between animals and genetic groups. This is an extra step, and Q is a huge matrix to read & write.

Later, Quaas (1988) reformulated the Animal Model with genetic groups, so that $\hat{\mathbf u} + \mathbf Q \hat{\mathbf g}$ is directly estimated via BLUP (Eq. 3 to Eq. 4 in his article):

$$ \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf X & \mathbf X' \mathbf R^{-1} \mathbf Z & \mathbf X' \mathbf R^{-1} \mathbf{ZQ} \\ \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Z' \mathbf R^{-1} \mathbf Z + \mathbf G^{-1} & \mathbf Z' \mathbf R^{-1} \mathbf{ZQ} \\ \mathbf Q' \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Q' \mathbf Z' \mathbf R^{-1} \mathbf Z & \mathbf Q' \mathbf Z' \mathbf R^{-1} \mathbf{ZQ} \end{bmatrix} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} \\ \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf Q' \mathbf Z' \mathbf R^{-1} \mathbf y \end{bmatrix} $$

$$ \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf X & \mathbf X' \mathbf R^{-1} \mathbf Z & \mathbf 0 \\ \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Z' \mathbf R^{-1} \mathbf Z + \mathbf G^{-1} & -\mathbf G^{-1} \mathbf Q \\ \mathbf 0 & -\mathbf Q' \mathbf G^{-1} & \mathbf Q' \mathbf G^{-1} \mathbf Q \end{bmatrix} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} + \mathbf Q \hat{\mathbf g} \\ \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf 0 \end{bmatrix}. $$

Quaas (1988) derived the mixed model equations (Eq. 4) by minimising the mean squared error through setting the partial derivative $\partial \text L / \partial \mathbf \Theta$ to zero, where $\text{L = log f}(\mathbf y, \mathbf u)$, and $\mathbf \Theta' = \begin{bmatrix} \mathbf b' & \mathbf u' & \mathbf g' \end{bmatrix}$ . Considering $\mathbf W = \begin{bmatrix} \mathbf X & \mathbf Z & \mathbf 0 \end{bmatrix}$ and $\mathbf U = \begin{bmatrix} \mathbf 0 & \mathbf I & -\mathbf Q \end{bmatrix}$ :

$$ \partial \text L / \partial \mathbf \Theta = \mathbf W’ \mathbf R^{-1} (\mathbf y - \mathbf{W \Theta}) - \mathbf U’ \mathbf G^{-1} \mathbf \Theta = \mathbf 0 $$

$$ \Rightarrow \begin{bmatrix} \mathbf W' \mathbf R^{-1} \mathbf W + \mathbf U' \mathbf G^{-1} \mathbf U \end{bmatrix} \mathbf \Theta = \mathbf W' \mathbf R^{-1} \mathbf y. $$

Substituting for W, U and $\mathbf \Theta$ yields Eq. 4. Also, note that

$$ \mathbf U' \mathbf G^{-1} \mathbf U = \begin{bmatrix} \mathbf 0 & \mathbf 0 & \mathbf 0 \\ \mathbf 0 & \mathbf G^{-1} & -\mathbf G^{-1} \mathbf Q \\ \mathbf 0 & -\mathbf Q' \mathbf G^{-1} & \mathbf Q' \mathbf G^{-1} \mathbf Q \end{bmatrix}, $$ $$ \begin{bmatrix} \mathbf I & \mathbf 0 & \mathbf 0 \\ \mathbf 0 & \mathbf I & \mathbf 0 \\ \mathbf 0 & -\mathbf Q' & \mathbf I \end{bmatrix} \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf Q' \mathbf Z' \mathbf R^{-1} \mathbf y \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf 0 \end{bmatrix} $$ $$ \Rightarrow \begin{bmatrix} \mathbf I & \mathbf 0 & \mathbf 0 \\ \mathbf 0 & \mathbf I & -\mathbf Q \\ \mathbf 0 & \mathbf 0 & \mathbf I \end{bmatrix}' \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf Q' \mathbf Z' \mathbf R^{-1} \mathbf y \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf 0 \end{bmatrix}, $$

and

$$ \begin{bmatrix} \mathbf I & \mathbf 0 & \mathbf 0 \\ \mathbf 0 & \mathbf I & \mathbf Q \\ \mathbf 0 & \mathbf 0 & \mathbf I \end{bmatrix} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} \\ \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} + \mathbf Q \hat{\mathbf g} \\ \hat{\mathbf g} \end{bmatrix} $$ $$ \Rightarrow \begin{bmatrix} \mathbf I & \mathbf 0 & \mathbf 0 \\ \mathbf 0 & \mathbf I & -\mathbf Q \\ \mathbf 0 & \mathbf 0 & \mathbf I \end{bmatrix}^{-1} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} \\ \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} + \mathbf Q \hat{\mathbf g} \\ \hat{\mathbf g} \end{bmatrix}. $$

Since Eq. 3 and Eq. 4 are equivalent models, it was not clear to me how to find my way from Eq. 3 to Eq. 4. So, I started exploring. I began with changing $\hat{\mathbf g}$ to $\mathbf Q \hat{\mathbf g}$ in Eq. 3:

$$ \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf X & \mathbf X' \mathbf R^{-1} \mathbf Z & \mathbf X' \mathbf R^{-1} \mathbf Z \\ \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Z' \mathbf R^{-1} \mathbf Z + \mathbf G^{-1} & \mathbf Z' \mathbf R^{-1} \mathbf Z \\ \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Z' \mathbf R^{-1} \mathbf Z & \mathbf Z' \mathbf R^{-1} \mathbf Z \end{bmatrix} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} \\ \mathbf Q \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \end{bmatrix}. $$

We then absorb $\mathbf Q \hat{\mathbf g}$ into $\hat{\mathbf u}$ by

$$ \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf X & \mathbf X' \mathbf R^{-1} \mathbf Z \\ \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Z' \mathbf R^{-1} \mathbf Z + \mathbf G^{-1} \end{bmatrix} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} + \mathbf Q \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y - \mathbf G^{-1} \mathbf Q \hat{\mathbf g} \end{bmatrix}. $$

However, the problem is the $\hat{\mathbf g}$ at the right-hand-side, where we can only have known matrices and vectors, not a vector of predictions (unknowns). To get away from this problem, we write the above equation as

$$ \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf X & \mathbf X' \mathbf R^{-1} \mathbf Z & \mathbf 0 \\ \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Z' \mathbf R^{-1} \mathbf Z + \mathbf G^{-1} & \mathbf G^{-1} \end{bmatrix} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} + \mathbf Q \hat{\mathbf g} \\ -\mathbf Q \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \end{bmatrix} $$ $$ \Rightarrow \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf X & \mathbf X' \mathbf R^{-1} \mathbf Z & \mathbf 0 \\ \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Z' \mathbf R^{-1} \mathbf Z + \mathbf G^{-1} & \mathbf G^{-1} \\ \mathbf 0 & \mathbf 0 & \mathbf 0 \end{bmatrix} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} + \mathbf Q \hat{\mathbf g} \\ -\mathbf Q \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf 0 \end{bmatrix}. $$

The left-hand-side should be symmetric. Therefore

The last (third) block row indicates that $\mathbf G^{-1} \hat{\mathbf u} = \mathbf 0$. Finally, changing $-\mathbf Q \hat{\mathbf g}$ predictions to $\hat{\mathbf g}$, we arrive to Eq. 4:

$$ \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf X & \mathbf X' \mathbf R^{-1} \mathbf Z & \mathbf 0 \\ \mathbf Z' \mathbf R^{-1} \mathbf X & \mathbf Z' \mathbf R^{-1} \mathbf Z + \mathbf G^{-1} & -\mathbf G^{-1} \mathbf Q \\ \mathbf 0 & -\mathbf Q' \mathbf G^{-1} & \mathbf Q' \mathbf G^{-1} \mathbf Q \end{bmatrix} \begin{bmatrix} \hat{\mathbf b} \\ \hat{\mathbf u} + \mathbf Q \hat{\mathbf g} \\ \hat{\mathbf g} \end{bmatrix} = \begin{bmatrix} \mathbf X' \mathbf R^{-1} \mathbf y \\ \mathbf Z' \mathbf R^{-1} \mathbf y \\ \mathbf 0 \end{bmatrix}. $$

At last, I could make sense of the equivalence between the two models (Eq. 3 and Eq. 4) in Quaas (1988). I hope you found this exercise useful.