Above results lead to(since below can be written as ):
Theorem 1 is the proj matrix onto
Thus
The solutions for the NEs are solutions for
can be decomposed uniquely as where and . Hence
All the solutions of NEs can be given by:
Theorem 2 If , then and is the projection matrix onto
Proof is clear from the definition and . Then
If we have , which often comes from we include a intercept in the model, then we have is idempotent.
thus we have the corrected SSR is
and we also have (holds even ):
hence we have .
Reparameterization
Linear models and are equivalent or reparameterizations iff . Recall that if , then there exist and , s.t. and .We now show some important results:
If , then , which implies and .
If solves the NEs in , then solves the NEs in since
Gram-Schmidt Orthonormalization
For any matrix , it can be decomposed as a unitary matrix and a upper triangle matrix, i.e.
Such QR factorization also related to the Cholesky factorization:
So is the transpose of .
Estimability and Least Squares EstimatorsAssumptions for the Linear Mean Model
For the linear model
We now assume that the error have zero mean:
Such Assumption is also equivalent to assume
Estimator and Estimability
Definition:
An estimator is unbiased for iff for all .An estimator is linear iff for some and . is linearly estimable iff there exists a linear unbiased estimator for it, otherwise the function is called nonestimabel.
The following statements are equivalent:
is linearly estimable.There exists a vector s.t.
We can prove this follow: and .
Thus when , is estimable. That is, if has full column rank, all of components of are estimable.
is estimable are free of the choice of is a linear unbiased estimator of Reparameterization Revisited
Suppose and are equivalent with and .
If is estimable in and solves the NEs in , then is the least squares estimator of since .If is estimable in and solves the NEs in , then is estimable in and is its least squares estimator.
We often choose to be full-column rank so that the solution to the NEs is unique and any components of is estimable. Another important advantage is interpretation of the new parameters will be more clear or make diagonal or block diagonal.
Conditions for a Unique Solution
Suppose we impose a condition on the solution to NEs:
now the NEs become:
where is , and is , , (so that we may have a unique solution). To achieve this purpose, below equation should be hold:
Thus have
And the system of equations above is equivalent to
since iff both side is zero.
Then we have the following important results:
is the unique solution of above system. is a generalized inverse of .
Proof We only prove 5, consider , then
Note that for idempotent matrix , holds:
The proof is finished since invertible idempotent matrix is identity.
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