core25 min studyLesson 252 in Mathematics

Estimability and rank-deficient designs

Identify the parameter combinations a design can actually distinguish, including overparameterized group models.

On this page

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 E(Y)=XβE(Y)=X\beta, without yet specifying error variances or distributions. If vv is in the nullspace of XX, then β\beta and β+v\beta+v produce exactly the same mean vector. No information about that mean can distinguish the two parameter descriptions. A proposed quantity cTβc^{\mathsf T}\beta is therefore identifiable from the mean only if cTv=0c^{\mathsf T}v=0 for every such invisible change.

The orthogonal complement of the nullspace is the row space. Consequently the same condition says that cc must lie in the column space of XTX^{\mathsf T}. Equivalently, there is a vector aa with XTa=cX^{\mathsf T}a=c. Then aTYa^{\mathsf T}Y has expectation cTβc^{\mathsf T}\beta 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.

cTβ is linearly estimable  ⟺  c∈col⁡(XT)  ⟺  cTv=0∀v∈ker⁡Xc^{\mathsf T}\beta\text{ is linearly estimable}\iff c\in\operatorname{col}(X^{\mathsf T})\iff c^{\mathsf T}v=0\quad\forall v\in\ker X
This equivalence assumes the mean model E(Y)=Xβ and finite first moments. No Gaussian or equal-variance assumption is needed for the estimability test.↔ Scroll wide equations horizontally.

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.

Xβ^1=Xβ^2 ⟹ β^1−β^2∈ker⁡X,c⊥ker⁡X ⟹ cTβ^1=cTβ^2X\widehat\beta_1=X\widehat\beta_2\ \Longrightarrow\ \widehat\beta_1-\widehat\beta_2\in\ker X,\qquad c\perp\ker X\ \Longrightarrow\ c^{\mathsf T}\widehat\beta_1=c^{\mathsf T}\widehat\beta_2
Any two coefficient solutions representing the same fit agree on every estimable linear function, although individual coordinates may disagree.↔ Scroll wide equations horizontally.

Prediction also has an estimability condition

At a new predictor row x∗Tx_*^{\mathsf T}, the proposed mean is x∗Tβx_*^{\mathsf T}\beta. It is estimable only when x∗x_* 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 n−rn-r, where rr is the design rank; counting all printed coefficients would incorrectly count redundant directions as information consumed.

Two names for one constant mean

Three observations are y=(2,4,6)Ty=(2,4,6)^{\mathsf T} and both columns of XX equal (1,1,1)T(1,1,1)^{\mathsf T}. Determine the estimable combination, all least-squares solutions, and the minimum-norm solution.

  1. The fitted response is the constant β1+β2\beta_1+\beta_2. The design nullspace consists of multiples of (1,−1)T(1,-1)^{\mathsf T}, so c=(c1,c2)Tc=(c_1,c_2)^{\mathsf T} is orthogonal to it exactly when c1=c2c_1=c_2.

  2. The response mean is 44, so every least-squares solution satisfies β^1+β^2=4\widehat\beta_1+\widehat\beta_2=4. Write these as (t,4−t)T(t,4-t)^{\mathsf T} for any real tt; all produce fitted vector (4,4,4)T(4,4,4)^{\mathsf T}.

  3. Minimize t2+(4−t)2=2(t−2)2+8t^2+(4-t)^2=2(t-2)^2+8. The minimum-norm convention selects (2,2)T(2,2)^{\mathsf T}, but (0,4)T(0,4)^{\mathsf T} produces exactly the same fitted responses and residuals.

The result

The sum is estimable and estimated as 44. Neither coefficient is separately estimable, and the pseudoinverse does not turn the individual value 22 into a data-identified parameter.

Explore the model

Interactive lab

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.

Distinguish poor conditioning from missing rankThe 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. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

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.

Check your understanding

For a mean model with duplicate columns (1,1,1) and (1,1,1), which quantity is estimable?

Try it yourself

Let the columns of a design be 11, xx, and 2x2x, where x=(−1,0,1)Tx=(-1,0,1)^{\mathsf T}. Which of β0\beta_0, β1\beta_1, and β1+2β2\beta_1+2\beta_2 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 β0+θx\beta_0+\theta x with θ=β1+2β2\theta=\beta_1+2\beta_2. The constant and x columns are independent, so the intercept and θ are estimable. A change (0,−2t,t)(0,-2t,t) 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.

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.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.