Projection Sums of Squares for Nested Model Spaces

Why a Nested-Model Increment Is a Projection

Suppose a smaller linear model can fit vectors in a space \(S_0\), and a larger model can fit vectors in a space \(S_1\), where \(S_0\subseteq S_1\subseteq\mathbb{R}^n\). Least squares projects the observed response vector \(y\) onto the model space. The larger model can therefore improve the fit only through the part of its space that is not already available to the smaller model.

That new part is the subspace of \(S_1\) orthogonal to \(S_0\). The sum of squares added by the larger model is the squared length of the projection of \(y\) onto this subspace. This formulation applies whether or not the columns used to describe the larger model are orthogonal to the smaller model.

An orthogonal projection matrix is symmetric and idempotent: \(P'=P\) and \(P^2=P\). Its projected vector is the least-squares fit in its subspace. The fitted values are \(P_0y\) and \(P_1y\), and their difference is the portion of the larger fit that the smaller fit could not provide.

The Projection Difference Theorem

Proof. Since \(S_0\subseteq S_1\), projecting a vector first onto \(S_0\) and then onto \(S_1\) leaves it unchanged after the first projection. Thus \(P_1P_0=P_0\). Taking transposes gives \(P_0P_1=P_0\). Therefore \(Q\) is symmetric, and \[ Q^2=(P_1-P_0)^2=P_1-P_1P_0-P_0P_1+P_0=P_1-P_0=Q. \] So \(Q\) is an orthogonal projection matrix.

For any \(x\), \(Qx\) belongs to \(S_1\), since both \(P_1x\) and \(P_0x\) belong to \(S_1\). Also \(P_0Q=P_0(P_1-P_0)=0\), so \(Qx\) is orthogonal to \(S_0\). Hence the range of \(Q\) is contained in \(S_1\cap S_0^\perp\). Conversely, if \(w\in S_1\cap S_0^\perp\), then \(P_1w=w\) and \(P_0w=0\), so \(Qw=w\). This proves that the range is exactly the orthogonal increment space. Its dimension is \(\dim(S_1)-\dim(S_0)\), by the orthogonal decomposition of \(S_1\) into \(S_0\) and \(S_1\cap S_0^\perp\). Finally, symmetry and idempotence give \(y'Qy=(Qy)'(Qy)=\|Qy\|^2\). Since \(Qy=P_1y-P_0y\), the stated identity follows. \(\square\)

Connection to Residual Sum of Squares

For a model space \(S_j\), define its residual sum of squares by \(\operatorname{RSS}_j=\|(I-P_j)y\|^2\). In nested spaces, the improvement in fit and the reduction in residual sum of squares are the same quantity.

Derivation. The residual from the smaller space splits into two orthogonal vectors: \[ (I-P_0)y=(I-P_1)y+(P_1-P_0)y. \] The first term is orthogonal to \(S_1\), while the second belongs to \(S_1\). Their inner product is therefore zero. By the Pythagorean identity, \[ \|(I-P_0)y\|^2=\|(I-P_1)y\|^2+\|(P_1-P_0)y\|^2. \] Subtracting the first term on the right and applying the projection difference theorem yields the result. This equality concerns sums of squares for the same response vector and nested spaces; it does not require the observed vector to lie in either model.

Exact Normal-Theory Distribution

The projection representation also gives the exact distribution of the added sum of squares under spherical normal errors. Assume \(y\sim N(\mu,\sigma^2 I)\), where \(\sigma^2>0\), and let \(Q=P_1-P_0\) have rank \(r\). The result below describes one increment; it does not require a decomposition into multiple model increments.

Proof. Choose an orthonormal basis for the range of \(Q\), and place its \(r\) basis vectors in the columns of an \(n\times r\) matrix \(U\). Then \(U'U=I_r\) and \(Q=UU'\). Consequently \(y'Qy=(U'y)'(U'y)\). A linear transformation of a multivariate normal vector is normal, so \(U'y\) has mean \(U'\mu\) and covariance \(U'(\sigma^2 I)U=\sigma^2 I_r\). Its coordinates, divided by \(\sigma\), are independent normal random variables with unit variance and means given by the coordinates of \(U'\mu/\sigma\). The sum of their squares has a noncentral chi-square distribution with \(r\) degrees of freedom and noncentrality \[ \left\|U'\mu/\sigma\right\|^2=\frac{\mu'UU'\mu}{\sigma^2}=\frac{\mu'Q\mu}{\sigma^2}. \] If \(\mu\in S_0\), then \(Q\mu=0\), so the noncentrality parameter is zero. \(\square\)

Using the Result Carefully

The geometry depends on model spaces, not on a particular choice of coefficient coordinates. If a design matrix has redundant columns, coefficients may not be uniquely determined, but its column space still has a unique orthogonal projection. For a full-column-rank design matrix \(X\), the projection is \(P_X=X(X'X)^{-1}X'\); in general, it is safest to define the projection directly as the projection onto \(\operatorname{col}(X)\).

A common mistake is to treat an added predictor as though its raw column were automatically the new orthogonal direction. In the three-dimensional example, projecting \(y\) directly onto \(z\) gives squared length \((y'z)^2/(z'z)=25/2\), not the correct increment \(32/3\). The raw column overlaps the smaller model space. The correct direction is its residual after projection onto that space, \(w=(I-P_0)z\).

The rank \(r=\dim(S_1)-\dim(S_0)\) is the number of independent directions added, not necessarily the number of columns appended to a design matrix. This distinction matters when columns are redundant or when an added column already belongs to the smaller space. If \(r=0\), the spaces are equal, \(P_1-P_0=0\), and the added sum of squares is zero for every \(y\).