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Sufficient Dimension Reduction · Tutorial 461 of 975

The Population SIR Operator

Learn how the population SIR operator is constructed and how its nonzero eigenspace and rank are determined by conditional predictor means.

Advanced 11 min read

What You'll Learn

  • Define the standardized predictor and the population SIR matrix.
  • Prove that the operator is positive semidefinite and bounded above by the identity.
  • Characterize its range through the conditional mean vectors.
  • Relate rank to the span of those means and to the number of response categories.
  • Calculate the operator and its eigenspace in explicit models.

From Inverse Means to a Population Matrix

Inverse regression studies the predictor distribution after conditioning on the response. For a multivariate predictor, the conditional mean \(E[X\mid Y]\) is a vector, and its variation across response values can be summarized by a matrix. That matrix is the population SIR operator: it records which standardized predictor directions have conditional means that vary with the response.

Standardization matters because the coordinates of a predictor can have different units and scales. It gives the operator a common reference geometry and makes its eigenvectors interpretable as directions in standardized predictor coordinates. The matrix describes variation in conditional means; by itself, its rank is the dimension of that inverse-mean signal.

Definition: Let \(X\) be a \(p\)-dimensional random vector with finite second moments, mean \(\mu=E[X]\), and positive-definite covariance matrix \(\Sigma=\operatorname{Var}(X)\). Define the standardized predictor \(Z=\Sigma^{-1/2}(X-\mu)\), where \(\Sigma^{-1/2}\) is the symmetric inverse square root. The population SIR operator is \(M_{\mathrm{SIR}}=\operatorname{Var}(E[Z\mid Y])\).

Write \(m(Y)=E[Z\mid Y]\). Conditional expectation is understood componentwise, up to almost-sure equality. Since \(E[Z]=0\), the tower property gives \(E[m(Y)]=0\), and consequently the operator also has the useful expression \(M_{\mathrm{SIR}}=E[m(Y)m(Y)^T]\). The conditional mean has finite second moments: conditional Jensen's inequality gives \(E[\|m(Y)\|^2]\leq E[\|Z\|^2]=p\).

Thus the operator is a \(p\)-by-\(p\) matrix even when the response is continuous. It summarizes the full conditional-mean function \(m(y)\) through its second moment, rather than listing one mean vector for each possible value of \(Y\).

Basic Properties and the Nonzero Eigenspace

The first property explains why eigenanalysis is appropriate: the SIR operator is symmetric and positive semidefinite. Its positive-eigenvalue directions are exactly the linear span of the conditional mean vectors, with null directions corresponding to contrasts that have zero conditional mean at almost every response value.

Theorem: Under the conditions in the definition, \(M_{\mathrm{SIR}}\) is symmetric and positive semidefinite, with \(0\preceq M_{\mathrm{SIR}}\preceq I_p\). Let \(S\) be the smallest linear subspace of \(\mathbb{R}^p\) such that \(m(Y)\in S\) almost surely. Then \(\operatorname{Range}(M_{\mathrm{SIR}})=S\), and the span of the eigenvectors with positive eigenvalues is also \(S\). In particular, \(\operatorname{rank}(M_{\mathrm{SIR}})=\dim(S)\).

Proof. For any \(a\in\mathbb{R}^p\),

$$ a^T M_{\mathrm{SIR}}a =E[(a^T m(Y))^2]\geq 0. $$

The matrix is symmetric because it is an expectation of symmetric outer products, and the displayed quadratic form proves positive semidefiniteness. To obtain the upper bound, apply the law of total covariance:

$$ I_p=\operatorname{Var}(Z) =\operatorname{Var}(E[Z\mid Y])+E[\operatorname{Var}(Z\mid Y)] =M_{\mathrm{SIR}}+E[\operatorname{Var}(Z\mid Y)]. $$

Conditional covariance matrices are positive semidefinite, so their expectation is positive semidefinite. Therefore \(I_p-M_{\mathrm{SIR}}\) is positive semidefinite, proving the upper bound.

For the subspace characterization, the quadratic-form identity shows

$$ a\in\operatorname{Null}(M_{\mathrm{SIR}}) \quad\Longleftrightarrow\quad E[(a^T m(Y))^2]=0 \quad\Longleftrightarrow\quad a^T m(Y)=0\ \text{almost surely}. $$

The last equivalence holds because a nonnegative random variable has expectation zero if and only if it is zero almost surely. The vectors orthogonal to \(S\) are precisely those \(a\) for which \(a^Tm(Y)=0\) almost surely. Hence \(\operatorname{Null}(M_{\mathrm{SIR}})=S^\perp\). For a symmetric matrix, its range is the orthogonal complement of its null space, so \(\operatorname{Range}(M_{\mathrm{SIR}})=S\). By the spectral theorem, the range of a positive semidefinite matrix is the span of its eigenvectors with positive eigenvalues. Its rank is the dimension of its range, completing the result. \(\square\)

This theorem identifies the operator's nonzero eigenspace without requiring a separate guess about which directions matter. It is the smallest fixed subspace that contains the conditional mean vectors, ignoring response values that occur only on a probability-zero set. Eigenvectors within that subspace give an orthonormal basis, but if eigenvalues are repeated, individual eigenvectors need not be unique.

Worked Example: A Binary Response Gives a Rank-One Operator

Let \(P(Y=1)=1/4\) and \(P(Y=0)=3/4\). Suppose the standardized predictor is two-dimensional and its conditional distributions are \(Z\mid Y=1\sim N((1,0)^T,\operatorname{diag}(2/3,1))\) and \(Z\mid Y=0\sim N((-1/3,0)^T,\operatorname{diag}(2/3,1))\).

The conditional means are \(m_1=(1,0)^T\) and \(m_0=(-1/3,0)^T\). Their weighted mean is \((1/4)m_1+(3/4)m_0=(0,0)^T\), as required for a centered standardized predictor. The operator is their probability-weighted outer-product sum:

$$ M_{\mathrm{SIR}} =\frac14 \begin{pmatrix}1&0\\0&0\end{pmatrix} +\frac34 \begin{pmatrix}1/9&0\\0&0\end{pmatrix} = \begin{pmatrix}1/3&0\\0&0\end{pmatrix}. $$

Its eigenvalues are \(1/3\) and \(0\), with corresponding directions \(e_1=(1,0)^T\) and \(e_2=(0,1)^T\). Thus its rank is one, and its positive-eigenvalue subspace is the span of \(e_1\). The stated conditional covariance contributes \(\operatorname{diag}(2/3,1)\); adding the between-mean matrix above gives \(\operatorname{Var}(Z)=I_2\), verifying that the model is standardized.

Finite Response Categories and Rank

When \(Y\) takes finitely many values, the population operator has a direct sum representation. Let its categories be \(1,\ldots,K\), with \(p_k=P(Y=k)>0\), and let \(m_k=E[Z\mid Y=k]\). Since \(E[Z]=0\), the category means satisfy \(\sum_{k=1}^K p_km_k=0\).

Result: For a response with \(K\) categories of positive probability, \(M_{\mathrm{SIR}}=\sum_{k=1}^K p_km_km_k^T\), and \(\operatorname{rank}(M_{\mathrm{SIR}})\leq\min(p,K-1)\).

Derivation. On the event \(Y=k\), the random vector \(m(Y)\) equals \(m_k\), so taking its outer-product expectation gives the stated sum. Its range is the span of the \(K\) vectors \(m_1,\ldots,m_K\), by the theorem. That span has dimension at most \(p\). Also, the weighted-zero-mean relation expresses one mean as a linear combination of the others:

$$ m_K=-\frac{1}{p_K}\sum_{k=1}^{K-1}p_km_k. $$

Therefore their span has dimension at most \(K-1\), proving the bound. The bound is about the number of response categories and the predictor dimension; it does not guarantee that the rank reaches either limit.

Worked Example: Three Categories and Two Positive Eigenvalues

Suppose \(p=2\), the response has three equally likely categories, and the conditional standardized means are \(m_1=(1,0)^T\), \(m_2=(0,1)^T\), and \(m_3=(-1,-1)^T\). They have weighted mean zero. The operator is

$$ M_{\mathrm{SIR}} =\frac13\left[ \begin{pmatrix}1&0\\0&0\end{pmatrix} +\begin{pmatrix}0&0\\0&1\end{pmatrix} +\begin{pmatrix}1&1\\1&1\end{pmatrix} \right] = \begin{pmatrix}2/3&1/3\\1/3&2/3\end{pmatrix}. $$

For direction \((1,1)^T\), multiplication by the matrix gives \((1,1)^T\), so the eigenvalue is \(1\). For \((1,-1)^T\), multiplication gives \((1/3,-1/3)^T\), so the eigenvalue is \(1/3\). Both are positive; the rank is two, and the positive-eigenvalue subspace is all of \(\mathbb{R}^2\). This reaches the bound \(\min(2,3-1)=2\). The matrix \(I_2-M_{\mathrm{SIR}}\) has eigenvalues \(0\) and \(2/3\), so it is positive semidefinite, consistent with the covariance bound in the theorem.

Continuous Responses and the Span of a Curve

For a continuous response, the conditional mean \(m(y)\) can trace a curve or surface in predictor space. The operator still depends on its second moment, and its rank depends on the linear span of that curve, not on how many distinct response values are possible. The following model shows how a nonlinear conditional mean can occupy a low-dimensional subspace.

Worked Example: A Continuous Response with Rank Two

Let \(Y\sim N(0,1)\), let \(\varepsilon\) be independent of \(Y\) with mean zero and covariance \(\operatorname{diag}(3/4,7/9,1)\), and define \(Z=((1/2)Y,\ (1/3)(Y^2-1),\ 0)^T+\varepsilon\). Then \(E[Z\mid Y]=((1/2)Y,\ (1/3)(Y^2-1),\ 0)^T\).

For a standard normal variable, \(E[Y]=0\), \(E[Y^2]=1\), \(E[Y^3]=0\), and \(E[Y^4]=3\). Thus both nonzero conditional-mean coordinates have mean zero, their covariance is \((1/6)E[Y(Y^2-1)]=0\), and their variances are \((1/4)E[Y^2]=1/4\) and \((1/9)E[(Y^2-1)^2]=(1/9)(3-2+1)=2/9\). Therefore

$$ M_{\mathrm{SIR}}= \begin{pmatrix} 1/4&0&0\\ 0&2/9&0\\ 0&0&0 \end{pmatrix}. $$

The eigenvalues are \(1/4,2/9,0\); the positive-eigenvalue subspace is the span of the first two coordinate directions, and the rank is two. The residual covariance was chosen so that \(\operatorname{Var}(Z)=\operatorname{diag}(1/4,2/9,0)+\operatorname{diag}(3/4,7/9,1)=I_3\), verifying standardization.

Interpretation and a Common Pitfall

A positive eigenvalue means that conditional means vary along the associated direction in standardized predictor coordinates. A zero eigenvalue means that the projection of the conditional mean onto that direction is zero almost surely. The rank therefore counts dimensions of variation in the conditional mean, not dimensions of variation in the predictors themselves: a standardized predictor can vary in a direction even when its conditional mean does not vary there.

A frequent mistake is to treat the rank as automatically equal to the dimension of the full regression structure. The matrix is constructed from conditional means only. Its rank precisely describes the span of those means, as proved above; any further interpretation as a regression-relevant subspace requires additional assumptions and arguments. Another error is to count response categories and declare the rank equal to \(K-1\). The rank is only bounded above by \(K-1\); the category mean vectors may span a smaller space or even all be zero.

The population operator is useful because it turns a function of the response, \(E[Z\mid Y]\), into a finite-dimensional matrix whose range and rank can be characterized exactly. Those properties describe what the SIR matrix itself contains, without conflating its inverse-mean signal with other features of the conditional distribution.

Check Your Understanding

Use the definitions and results above to answer these questions.

  1. Why does \(E[Z]=0\) imply that \(M_{\mathrm{SIR}}=E[m(Y)m(Y)^T]\)?
  2. What does \(a^TM_{\mathrm{SIR}}a=0\) imply about \(a^Tm(Y)\) almost surely?
  3. If a response has four categories and the predictor has dimension three, what is the largest possible rank of the population SIR operator?
  4. Can a standardized predictor direction have positive variance while its corresponding eigenvalue in \(M_{\mathrm{SIR}}\) is zero? Explain.
  5. In the continuous-response example, why does the operator have rank two even though \(Y\) has infinitely many possible values?
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