Estimability and rank-deficient designs
Identify the parameter combinations a design can actually distinguish, including overparameterized group models.
On this page
The question for today
What you’ll understand
- Use the nullspace to test whether a linear parameter function is estimable.
- Explain why identifying constraints do not create information.
- Separate unique fitted means from nonunique coefficient vectors.
A software package can print one coefficient vector even when the data cannot distinguish it from infinitely many others. Estimability asks a sharper question than whether a matrix routine returned numbers: which linear combinations of coefficients are determined by the observable mean? Redundant intercepts, categorical indicators, and perfectly aliased experimental factors make this question practical. Rather than treating a singular design as a complete failure, identify the combinations that remain meaningful and report those combinations explicitly.
Invisible parameter changes live in the nullspace
Work in the fixed-design mean model , without yet specifying error variances or distributions. If is in the nullspace of , then and produce exactly the same mean vector. No information about that mean can distinguish the two parameter descriptions. A proposed quantity is therefore identifiable from the mean only if for every such invisible change.
The orthogonal complement of the nullspace is the row space. Consequently the same condition says that must lie in the column space of . Equivalently, there is a vector with . Then has expectation for every coefficient vector. This is linear estimability: it describes existence of an unbiased linear estimator, not whether its variance is small or whether a particular data realization is convincing.
Constraints select a representation
Consider a group-means parameterization with an intercept and an indicator for every group. Adding a constant to the intercept while subtracting it from every group coefficient leaves all fitted means unchanged. Dropping one indicator, setting one group coefficient to zero, or imposing a sum-to-zero constraint can choose coordinates. These conventions do not supply extra observations. A group mean or a difference of group means can be estimable even though an isolated coefficient in the redundant parameterization is not.
A Moore–Penrose pseudoinverse chooses the coefficient vector of smallest Euclidean norm among least-squares solutions. That is a mathematical convention that depends on parameter scaling; it is not evidence that the true coefficients have that norm. All least-squares solutions differ by nullspace vectors, so an estimable combination has the same numerical value for every solution. A nonestimable combination can change while the fitted responses and the residual sum of squares remain exactly fixed.
Prediction also has an estimability condition
At a new predictor row , the proposed mean is . It is estimable only when belongs to the row space of the observed design. This is a stronger issue than ordinary extrapolation uncertainty: a new combination of perfectly aliased factors can be unidentified even if each factor level appeared in the data. Printing a prediction after silently choosing one coefficient convention can hide that ambiguity rather than resolve it.
Exact deficiency and near deficiency deserve different responses. Exact redundancy makes some combinations unidentified. Nearly dependent columns can leave every coefficient identifiable but make certain estimates highly variable and numerically sensitive. Singular values, scientific design knowledge, and the precise prediction target help distinguish the cases. With spherical error covariance, residual degrees of freedom are , where is the design rank; counting all printed coefficients would incorrectly count redundant directions as information consumed.
Work it through
Two names for one constant mean
Three observations are and both columns of equal . Determine the estimable combination, all least-squares solutions, and the minimum-norm solution.
The fitted response is the constant . The design nullspace consists of multiples of , so is orthogonal to it exactly when .
The response mean is , so every least-squares solution satisfies . Write these as for any real ; all produce fitted vector .
Minimize . The minimum-norm convention selects , but produces exactly the same fitted responses and residuals.
The sum is estimable and estimated as . Neither coefficient is separately estimable, and the pseudoinverse does not turn the individual value into a data-identified parameter.
The living notebook
Explore the model
Interactive lab · matrix
Distinguish poor conditioning from missing rank
Pause and Reset; set Diagonal scale a to 1 and Shear b to 0. Compare with a fresh Reset followed by a=0.25 and b=2. Predict a smaller minimum singular value and inspect the condition number. Every supported a is positive, so this family remains full rank. It illustrates sensitivity near compression, not an exactly rank-deficient regression or an estimability test.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forSmall singular directions can be fragile while still identifiable; an exact null direction is a different condition.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let the columns of a design be , , and , where . Which of , , and are estimable? Describe an invisible parameter change.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Collect the terms multiplying x, and then find a nonzero coefficient change that leaves that total unchanged.
Compare with the solution
The mean is with . The constant and x columns are independent, so the intercept and θ are estimable. A change leaves the mean unchanged for every t. It changes β₁, so β₁ alone is not estimable. Adding a constraint can name one representation but cannot distinguish these originally equivalent coefficients.
Keep the idea
Take it with you
- Estimable linear functions annihilate every design null vector.
- Identifying constraints and pseudoinverses choose coordinates without adding information.
- New-row mean predictions require their own row-space check.
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