In openpharma/mmrm: Mixed Models for Repeated Measures

Weighted Least Square (WLS) Empirical Covariance

Following the notation we have without weights, @bell2002bias and @pustejovsky2018small suggest

[ v = s C^\top(X^\top W X)^{-1}\sum_{i}{X_i^\top W_i A_i \epsilon_i \epsilon_i^\top A_i W_i X_i} (X^\top W X)^{-1} C ]

where $A_i$ takes $I_i$, $(I_i - H_{ii})^{-\frac{1}{2}}$, or $(I_i - H_{ii})^{-1}$ is unchanged, but $H$ is changed

[ H = X (X^\top W X)^{-1} X^\top W ]

For the degrees of freedom, we have

[ G_{ij} = g_i^\top \Phi g_j ]

where

[ g_i = s^{\frac{1}{2}} (I - H)_i^\top A_i W_i X_i (X^\top X)^{-1} C ]

Difference of Implementations

Comparing the previous section with our implementation, we can find out the differences. Since they have nearly the same symbols, to differentiate the different part, we use subscript $1$ to denote the implementation suggested by @bell2002bias and @pustejovsky2018small, and use $2$ to denote the our implementation of covariance estimator in mmrm, we have

[ v_{1} = s C^\top(X^\top W X)^{-1}\sum_{i}{X_i^\top W_i A_{1, i} \epsilon_i \epsilon_i^\top A_{1, i} W_i X_i} (X^\top W X)^{-1} C ]

[ v_{2} = s C^\top(X^\top W X)^{-1}\sum_{i}{X_i^\top L_i A_{2, i} L_i^\top \epsilon_i \epsilon_i^\top L_i A_{2, i} L_i^\top X_i} (X^\top W X)^{-1} C ]

Here we will prove that they are identical.

Proof of Identity

Proof for Covariance Estimator

First of all, we assume that all $A_i$ matrix, in any form, are positive-definite. Comparing $v_{1}$ and $v_{2}$, we see that the different part is

[ M_{1, d, i} = W_i A_{1, i} ] and [ M_{2, d, i} = L_i A_{2, i} L_i^\top ]

Substitute $H_{1}$ and $H_{2}$ with its expression, we have

[ M_{1, d, i} = W_i (I_i - X_i (X^\top W X)^{-1} X_i^\top W_i)^d ]

[ M_{2, d, i} = L_i (I_i - L_i^\top X_i (X^\top W X)^{-1} X_i^\top L_i)^d L_i^\top ]

Where $d$ takes $0$, $-1/2$ and $-1$ respectively.

Apparently, if $d=0$, these two are identical because $W_i = L_i L_i^\top$.

When $d = -1$, we have

[ M_{2, -1, i} = L_i (I_i - L_i^\top X_i (X^\top W X)^{-1} X_i^\top L_i)^{-1} L_i^\top \ = (L_i^{-1})^{-1} (I_i - L_i^\top X_i (X^\top W X)^{-1} X_i^\top L_i)^{-1} ((L_i^\top)^{-1})^{-1} \ = [((L_i^\top)^{-1})(I_i - L_i^\top X_i (X^\top W X)^{-1} X_i^\top L_i)(L_i^{-1})]^{-1} \ = [(L_i^\top)^{-1}L_i^{-1} - X_i (X^\top W X)^{-1} X_i^\top]^{-1} \ = (W_i^{-1} - X_i (X^\top W X)^{-1} X_i^\top)^{-1} ]

[ M_{1, -1, i} = W_i (I_i - X_i (X^\top W X)^{-1} X_i^\top W_i)^{-1} \ = (W_i^{-1})^{-1} (I_i - X_i (X^\top W X)^{-1} X_i^\top W_i)^{-1} \ = [(I_i - X_i (X^\top W X)^{-1} X_i^\top W_i)((W_i^{-1}))]^{-1} \ = (W_i^{-1} - X_i (X^\top W X)^{-1} X_i^\top)^{-1} ]

Obviously, $M_{2, -1, i} = M_{1, -1, i}$, and use the following notation

[ M_{2, -1, i} = L_i B_{2, i} L_i^\top ]

[ M_{1, -1, i} = W_i B_{1, i} ]

we have

[ B_{1, i} = W_i^{-1} L_i B_{2, i} L_i^\top \ = (L_i^\top)^{-1} B_{2, i} L_i^\top ]

When $d = -1/2$, we have the following

[ M_{2, -1/2, i} = L_i (I_i - L_i^\top X_i (X^\top W X)^{-1} X_i^\top L_i)^{-1/2} L_i^\top \ = L_i B_{2, i}^{1/2} L_i^\top ]

[ M_{1, -1/2, i} = W_i (I_i - X_i (X^\top W X)^{-1} X_i^\top W_i)^{-1/2} \ = W_i B_{1, i}^{1/2} ]

Apparently if $B_{1, i}^{1/2} \ne (L_i^\top)^{-1} B_{2, i}^{1/2} L_i^\top$, we should also have [ B_{1, i}^{1/2} B_{1, i}^{1/2} \ne (L_i^\top)^{-1} B_{2, i}^{1/2} L_i^\top (L_i^\top)^{-1} B_{2, i}^{1/2} L_i^\top ]

leading to

[ B_{1, i} \ne (L_i^\top)^{-1} B_{2, i} L_i^\top ]

which is contradictory with our previous result. Thus, these covariance estimator are identical.

Proof for Degrees of Freedom

To prove [ G_{1, ij} = g_{1, i}^\top \Phi g_{1, j} ] and [ G_{2, ij} = g_{2, i}^\top g_{2, j} ] are identical, we only need to prove that

[ L^{-1} g_{1, i} = g_{mmrm_i} ]

where $\Phi = W^{-1}$ according to our previous expression.

We first expand $L^{-1} g_{1, i}$ and $g_{mmrm_i}$

[ L^{-1} g_{1, i} = L^{-1} (I - X(X^\top W X)^{-1}X^\top W) S_i^\top A_{1, i}^d W_i X_i (X^\top W X)^{-1} C ]

[ g_{2, i} = (I - L_i^\top X(X^\top W X)^{-1}X^\top L_i) S_i^\top A_{2, i}^d L_i^\top X_i (X^\top W X)^{-1} C ]

where $S_i$ is the row selection matrix.

We will prove the inner part equal [ L^{-1} (I - X(X^\top W X)^{-1}X^\top W) S_i^\top A_{1, i}^d W_i = (I - L^\top X(X^\top W X)^{-1}X^\top L) S_i^\top A_{2, i}^d L_i^\top ]

With the previous proof of covariance estimators, we already have

[ M_{1, d, i} = W_i A_{1, i}^d = L_i A_{2, i}^d L_i^\top = M_{2, d, i} ] we then need to prove [ L^{-1} (I - X(X^\top W X)^{-1}X^\top W) S_i^\top = (I - L^\top X(X^\top W X)^{-1}X^\top L) S_i^\top L_i^{-1} ]

and note the relationship between $(I - X(X^\top W X)^{-1}X^\top W)$ and $(I - L^\top X(X^\top W X)^{-1}X^\top L)$ has already been proved in covariance estimator section, we only need to prove

[ L^{-1} (I - X(X^\top W X)^{-1}X^\top W) S_i^\top = (I - L^\top X(X^\top W X)^{-1}X^\top L) S_i^\top L_i^{-1} ]

Apparently

[ L^{-1} (I - X(X^\top W X)^{-1}X^\top W) S_i^\top = L^{-1} S_i^\top - L^{-1} X(X^\top W X)^{-1}X_i^\top W_i ]

[ (I - L^\top X(X^\top W X)^{-1}X^\top L) S_i^\top L_i^{-1} = S_i^\top L_i^{-1} - L^\top X(X^\top W X)^{-1}X_i^\top ]

And obviously [ L^{-1} S_i^\top = S_i^\top L_i^{-1} ]

[ L^{-1} X(X^\top W X)^{-1}X_i W_i = L^\top X(X^\top W X)^{-1}X_i^\top ]

because of the following [ (X(X^\top W X)^{-1}X_i W_i){i} = X_i(X^\top W X)^{-1}X_i W_i \ = W_i X_i(X^\top W X)^{-1}X_i^\top \ = (W X(X^\top W X)^{-1}X_i^\top){i} ]

Special Considerations in Implementations

Pseudo Inverse of a Matrix

Empirical covariance matrix is involved with the inverse of a matrix, or symmetric square root of a matrix. To calculate this, we usually requires that the matrix is positive-definite. However, @young2016improved suggest that this is not always assured in practice.

Thus, following @pustejovsky2018small, we use the pseudo inverse to avoid this. We follow the following logic (see the corresponding C++ function pseudoInverseSqrt) to obtain the pseudo inverse:

Conduct singular value decomposition.
Use cpow to obtain the square root of the reciprocals of singular values, if the value is larger than a computational threshold; otherwise replace the value with 0.
Reconstruct the pseudo inverse matrix from modified singular values and U/V matrix.

In Eigen package, the pseudo inverse method is already implemented in Eigen::CompleteOrthogonalDecomposition< MatrixType_ >::pseudoInverse, but it is not used for the following reason:

The pseudo inverse method is not stable and can lead to NAN in calculations.
To find out the symmetric square root, singular value decomposition is still needed, so not using the method but instead calculating directly the square root of the pseudo inverse can be simpler.