Slice Counts, Boundaries, and SIR Stability

Why SIR Depends on the Slices

Sliced inverse regression (SIR) summarizes how the predictor changes across groups formed from the response. That makes the groups more than a bookkeeping choice: changing their number or boundaries changes the conditional means being summarized, and therefore can change the estimated SIR matrix and its eigenstructure. In finite samples, the sizes of the groups matter too, because a mean based on only a few observations is particularly sensitive to those observations.

The practical question is not simply whether to use many or few slices. More slices can reveal changes that broad groups average away, but they also leave fewer observations per group and make the means less stable. Quantile slices balance group sizes when the response is continuous and ties are limited. Prespecified intervals instead keep boundaries fixed, which can be useful when cutpoints have a scientific interpretation or must match an analysis plan.

Definition: Let $X\in\mathbb{R}^p$ have mean $\mu$ and nonsingular covariance matrix $\Sigma_X$. Define the standardized predictor $Z=\Sigma_X^{-1/2}(X-\mu)$, so that $E[Z]=0$ and $\operatorname{Var}(Z)=I_p$. A slicing is a partition of the response range into disjoint sets $I_1,\ldots,I_H$. Write $p_h=P(Y\in I_h)$ and $m_h=E[Z\mid Y\in I_h]$ for slices with $p_h>0$. The population SIR matrix for this partition is $$ M_{\mathcal P}=\sum_{h=1}^H p_h m_hm_h^T. $$ Its leading eigenvectors describe directions with large between-slice variation in standardized inverse means.

The sample version uses a partition of the observed responses. If $S_h=\{i:Y_i\in I_h\}$, $n_h=|S_h|>0$, and $\bar Z_h=n_h^{-1}\sum_{i\in S_h}Z_i$, then

\widehat M_{\mathcal P}=\sum_{h=1}^H \frac{n_h}{n}\bar Z_h\bar Z_h^T.

Here the $Z_i$ are standardized using the chosen sample centering and covariance convention. This tutorial focuses on how the partition affects the means and matrix, not on alternative covariance estimators. A slice with no observations has no sample mean; in practice the partition must be changed or the empty slice omitted and the resulting partition stated clearly.

What Slice Count and Slice Size Change

With few slices, observations with different response values are combined. The resulting inverse mean is an average over a broad response region, so distinct patterns can cancel. Increasing the number of slices can separate those patterns. But it also reduces typical $n_h$, especially when slice sizes are unequal. The estimate of a slice mean then depends heavily on a small number of predictor observations.

There is also an exact rank constraint. Because the weighted slice means sum to zero, the rank of the sample SIR matrix cannot exceed $H-1$, regardless of predictor dimension. Thus a partition with very few slices cannot produce a high-rank sample SIR matrix. This is a limitation on the matrix constructed from that partition, not evidence that the underlying regression has at most $H-1$ relevant directions.

Result: If all $H$ slices are nonempty, then $\operatorname{rank}(\widehat M_{\mathcal P})\leq \min(p,H-1)$.

To see why, set $w_h=n_h/n$. The sample-wide standardized mean is zero, so $\sum_h w_h\bar Z_h=0$. Form the $p\times H$ matrix whose $h$th column is $\sqrt{w_h}\bar Z_h$. The sample SIR matrix is this matrix times its transpose, so its rank is at most $p$ and at most the rank of the column matrix. Its columns have the nontrivial linear relation $\sum_h\sqrt{w_h}(\sqrt{w_h}\bar Z_h)=0$; hence their rank is at most $H-1$. This proves the bound. $\square$

Worked Example: Two, Three, or Six Quantile Slices

Suppose the observed response and a single predictor have values $Y=X=-5,-4,\ldots,6$, one observation at each value. Use the empirical mean and covariance with divisor $n=12$ for standardization. The mean is $\bar X=0.5$, and the sum of squared deviations is $143$, so the empirical variance is $143/12$. For any slice, its standardized inverse mean is its raw $X$-mean minus $0.5$, divided by $\sqrt{143/12}$.

With two equal-size quantile slices, the groups are $-5,\ldots,0$ and $1,\ldots,6$. Their means are $-2.5$ and $3.5$, so their centered means are $-3$ and $3$, with weights $1/2$ each. The one-dimensional SIR matrix is the scalar

\widehat M_2 =\frac{\frac12(9)+\frac12(9)}{143/12} =\frac{9}{143/12} =\frac{108}{143} \approx 0.7552.

With three equal-size slices, the groups are $-5,\ldots,-2$, $-1,\ldots,2$, and $3,\ldots,6$. Their means are $-3.5,0.5,4.5$, so the centered means are $-4,0,4$, with weights $1/3$. Therefore

\widehat M_3 =\frac{\frac13(16)+\frac13(0)+\frac13(16)}{143/12} =\frac{32/3}{143/12} =\frac{128}{143} \approx 0.8951.

With six equal-size slices, each group contains two consecutive values. The slice means are $-4.5,-2.5,-0.5,1.5,3.5,5.5$, and their centered means are $-5,-3,-1,1,3,5$. Each weight is $1/6$, giving

\widehat M_6 =\frac{(25+9+1+1+9+25)/6}{143/12} =\frac{70/6}{143/12} =\frac{140}{143} \approx 0.9790.

These values increase as the partition is refined, but this example alone does not establish that every increase in slice count must increase an estimated SIR matrix. Here the partitions are nested and the observations follow a simple ordered pattern. In general, partitions with different, nonnested boundaries can give different summaries, and sample noise can affect either one.

Boundaries, Quantiles, and Prespecified Intervals

Quantile slicing chooses boundaries from the observed response ranks to make slice counts approximately equal. This is attractive when there are enough distinct response values: it avoids having a few nearly empty groups simply because the response is skewed. Prespecified intervals choose boundaries in the response scale itself. They can preserve clinically or scientifically meaningful distinctions, but they may yield highly unequal slice sizes.

The difference can be seen even when the response is ordered exactly with a predictor. In the example below, the predictor and response take the same values as before. The quantile boundaries separate the observations into equal-size groups. The prespecified cutpoints instead retain the intervals $Y\leq -3$, $-3<Y\leq 3$, and $Y>3$. Since the observations are discrete, these intervals contain 3, 6, and 3 observations rather than 4 each.

Worked Example: Quantile Slices Versus Fixed Cutpoints

Use the 12 observations $X=Y=-5,-4,\ldots,6$, with mean $0.5$ and empirical variance $143/12$. For three quantile slices, the groups and means are $-5$ to $-2$ with mean $-3.5$, $-1$ to $2$ with mean $0.5$, and $3$ to $6$ with mean $4.5$. As calculated above, their centered means are $-4,0,4$, and the SIR matrix is $128/143\approx0.8951$.

For the prespecified intervals, the first group contains $-5,-4,-3$, with mean $-4$; the second contains $-2,-1,0,1,2,3$, with mean $0.5$; and the last contains $4,5,6$, with mean $5$. Subtracting the overall mean gives centered means $-4.5,0,4.5$, and the weights are $1/4,1/2,1/4$. Thus

\widehat M_{\mathrm{fixed}} =\frac{\frac14(4.5^2)+\frac12(0^2)+\frac14(4.5^2)}{143/12} =\frac{10.125}{143/12} =\frac{121.5}{143} \approx0.8497.

The two summaries differ because the cutpoints assign observations differently, changing both the inverse means and their weights. The comparison does not establish that quantile slicing is preferable: fixed intervals may be the appropriate choice when the cutpoints define meaningful response categories. It does show why a reported SIR analysis should specify how its slice boundaries were set.

Ties complicate quantile slicing. If many observations have the same response value at a proposed boundary, splitting those tied values across slices can be arbitrary and can make identical responses belong to different groups. Keeping tied values together avoids that problem, but may produce unequal slice sizes or fewer than the requested number of groups. For a discrete or heavily tied response, the actual cutpoints and resulting counts are more informative than simply reporting a requested number of quantile slices.

Why Nested Refinement Has a Population Guarantee

At the population level, a refined partition cannot reduce the SIR matrix in positive-semidefinite order when it is genuinely nested and all conditional means exist. This statement is about the population matrices for the two partitions, not about arbitrary sample estimates from different cutpoints.

Theorem: Let $\mathcal P_c$ be a coarse response partition and $\mathcal P_f$ a finer partition such that every fine slice is contained in a coarse slice. Assume $E[\|Z\|^2]<\infty$, and omit any slices of probability zero. Then $$ M_{\mathcal P_f}-M_{\mathcal P_c} $$ is positive semidefinite.

For a proof, fix a coarse slice $C$. Let its fine sub-slices be $F_j$, with probabilities $q_j=P(Y\in F_j\mid Y\in C)$ and standardized inverse means $m_j=E[Z\mid Y\in F_j]$. Write $m_C=E[Z\mid Y\in C]$. The law of total expectation gives $m_C=\sum_j q_jm_j$. The coarse contribution from $C$ is $P(Y\in C)m_Cm_C^T$; the fine contributions within $C$ are $P(Y\in C)\sum_jq_jm_jm_j^T$. Their difference is

P(Y\in C)\left\{\sum_jq_jm_jm_j^T-m_Cm_C^T\right\} =P(Y\in C)\sum_jq_j(m_j-m_C)(m_j-m_C)^T.

The equality follows by expanding the right-hand side and using $\sum_jq_jm_j=m_C$ and $\sum_jq_j=1$. Each outer product is positive semidefinite: for any $a\in\mathbb R^p$, its quadratic form is $(a^T(m_j-m_C))^2\geq0$. Summing over coarse slices proves that the difference of the two population matrices is positive semidefinite. $\square$

The theorem explains why a nested refinement can add population between-slice information. It does not say that every extra slice is useful in a finite sample, that eigenvectors must change smoothly, or that the estimated matrix must increase under arbitrary changes to boundaries. Those claims would require additional assumptions and, for sample comparisons, consideration of estimation variability.

Stability and a Common Pitfall

A direct way to see the role of slice size is to consider the mean itself. If one predictor observation in a slice of size $n_h$ changes by $\Delta$, while the other observations in that slice stay fixed, its slice mean changes by $\Delta/n_h$. The same change therefore moves a mean based on two observations by $\Delta/2$, but moves one based on twenty observations by only $\Delta/20$. Since the SIR matrix uses outer products of these means, changes in small-slice means can have an appreciable effect on the matrix and its eigenvectors.

More generally, for fixed weights $p_h$, suppose slice means $m_h$ are perturbed to $m_h+e_h$. The matrix change is

\sum_h p_h\{(m_h+e_h)(m_h+e_h)^T-m_hm_h^T\}.

Using $\|uv^T\|_{\mathrm{op}}=\|u\|\|v\|$ and the triangle inequality, its operator norm is at most

\sum_h p_h\left(2\|m_h\|\|e_h\|+\|e_h\|^2\right).

Thus changes in slice means matter most when their errors are large, especially in slices with substantial weight or large inverse means. This bound holds for fixed weights and does not account for changes in weights or standardization; those changes can also affect the sample matrix.

Common pitfall: “More slices always give a better SIR estimate because the population matrix gains information under refinement.” The first part confuses a population result for nested partitions with finite-sample performance. The theorem guarantees a positive-semidefinite difference only for nested population partitions. In a sample, additional slices can have small or empty counts, making their means unstable or undefined; changing to nonnested boundaries also does not satisfy the theorem’s condition. Choose slice count and boundaries with both resolution and sample size in view.

A useful stability check is to repeat the analysis over a small range of reasonable slice counts or cutpoints, while reporting the slice sizes and the resulting directions or subspaces. Agreement across nearby choices is reassuring but is not a proof of validity. Large changes signal that the inferred directions depend strongly on the discretization and should be interpreted cautiously. If the response has ties, report how they were handled; if intervals were prespecified, report their endpoints and counts.

Worked Example: A Requested Quantile Count Meets Ties

Suppose ten responses consist of six zeros and four ones, and an analyst requests four quantile slices. If tied response values are kept together, there are only two distinct response groups: $Y=0$ with six observations and $Y=1$ with four. Trying to form four nonempty slices without splitting ties is impossible, because each response value must remain in one group.

If the analyst instead divides the six zero responses among multiple rank-based slices, observations with identical response values are assigned to different slices. The resulting inverse means then compare predictor observations across groups that do not represent different response levels. A defensible analysis would keep the ties together and use two groups, or explain a different scientifically justified partition; it should not report four slices as though four distinct response strata had been observed.

Key takeaway: SIR is built from weighted inverse means, so slice boundaries determine what is averaged and slice sizes determine how much data support each mean. Quantile slicing often balances counts, while prespecified intervals preserve chosen response distinctions. Neither choice removes the need to inspect counts, ties, and sensitivity to reasonable alternatives.

Check Your Understanding

Use the definitions and results above to answer the following.

Why can a sample SIR matrix based on three nonempty slices have rank at most two?
What condition on two response partitions is needed for the population positive-semidefinite refinement result?
In the numerical comparison, why did the quantile and fixed-interval matrices differ even though both used three slices?
How does changing one predictor value by the same amount affect slice means of sizes two and twenty?
What difficulty arises when quantile boundaries fall among many tied response values?