One-Sided and Two-Sided Weight Truncation

Why the Side of the Cutoff Matters

A large weight can let one observation exert substantial influence on a weighted estimating equation. An upper cutoff caps that contribution. A lower cutoff does something different: it raises the contribution of observations whose weights are small. Applying both cutoffs therefore does not simply “clip more”; it changes the equation in two distinct ways.

Here, clipping means replacing weights while retaining the observations. It is not deletion. The cutoffs are applied to the weights used in the estimating equation, and the resulting equation may have a different solution. The calculations below describe that finite-sample change; they do not, by themselves, determine the causal interpretation of the new estimate.

Definition: Let $w_1,\ldots,w_n$ be nonnegative weights and choose finite cutoffs $0<L<U$. Upper-only clipping at $U$ replaces each weight by $w_i^{U}=\min(w_i,U)$. Two-sided clipping at $L$ and $U$ replaces it by $w_i^{L,U}=\max\{L,\min(w_i,U)\}$. Thus a weight below $L$ is raised to $L$, a weight between the bounds is unchanged, and a weight above $U$ is reduced to $U$.

The formulas distinguish the two modifications precisely. Upper-only clipping changes a weight if and only if $w_i>U$, and then reduces it by $w_i-U$. Two-sided clipping adds $L-w_i$ when $w_i<L$, leaves weights in $[L,U]$ unchanged, and subtracts $w_i-U$ when $w_i>U$. If $L=0$, the lower bound does not raise any nonnegative weight; a strictly positive lower cutoff is what gives small or zero weights positive weight in the equation.

How Clipping Enters a Weighted Score

Write a weighted estimating equation as $U_n(\theta;w)=n^{-1}\sum_{i=1}^n w_i\psi_i(\theta)=0$, where $\psi_i(\theta)$ is the observation-specific score contribution. The score may be scalar or vector-valued. If a solution exists, denote it by $\hat\theta(w)$. Replacing $w_i$ by clipped weights changes the score contribution by the weight change multiplied by $\psi_i(\theta)$; it does not change the contribution according to weight size alone.

Result: For any fixed $\theta$, the exact difference between the clipped and original scores is $$ U_n(\theta;w^*)-U_n(\theta;w)=\frac{1}{n}\sum_{i=1}^n (w_i^*-w_i)\psi_i(\theta). $$ For upper-only clipping this becomes $$ U_n(\theta;w^U)-U_n(\theta;w)=-\frac{1}{n}\sum_{i:w_i>U}(w_i-U)\psi_i(\theta). $$ For two-sided clipping it becomes $$ U_n(\theta;w^{L,U})-U_n(\theta;w)=\frac{1}{n}\sum_{i:w_i<L}(L-w_i)\psi_i(\theta)-\frac{1}{n}\sum_{i:w_i>U}(w_i-U)\psi_i(\theta). $$

To verify the result, subtract the two score definitions term by term: $$ U_n(\theta;w^*)-U_n(\theta;w) =\frac{1}{n}\sum_{i=1}^n\{w_i^*\psi_i(\theta)-w_i\psi_i(\theta)\} =\frac{1}{n}\sum_{i=1}^n(w_i^*-w_i)\psi_i(\theta). $$ For upper-only clipping, $w_i^U-w_i=-(w_i-U)$ when $w_i>U$, and is zero otherwise. For two-sided clipping, the difference is $L-w_i$ below $L$, zero between the bounds, and $-(w_i-U)$ above $U$. Substitution gives both formulas. They hold for every fixed $\theta$, whether or not either equation has a root. If one compares estimates, the existence and uniqueness of each root require additional conditions specific to the estimating equation. $\square$

The identity also cautions against assuming that an upper cutoff must move an estimate in a particular direction. The score change depends on the signs and sizes of the affected observations’ score contributions. Two-sided clipping adds a second set of terms, from the lower tail, which can oppose or reinforce the upper-tail change.

Worked Example: A Weighted Mean Under Two Clipping Rules

Suppose four outcomes are $y=(0,0,1,1)$, with weights $w=(0.2,1,4,12)$. Estimate their weighted mean by solving $\sum_i w_i(y_i-\theta)=0$. Because the sum of weights is positive, this equation has the unique root $\hat\theta=\sum_i w_i y_i/\sum_i w_i$.

With the original weights, the weighted numerator is $0.2(0)+1(0)+4(1)+12(1)=16$, and the denominator is $0.2+1+4+12=17.2$. Therefore $\hat\theta=16/17.2=40/43\approx0.9302$.

Now use upper-only clipping at $U=5$. The weights become $(0.2,1,4,5)$. The numerator is $9$, the denominator is $10.2$, and the clipped estimate is $9/10.2=15/17\approx0.8824$.

Instead use two-sided clipping at $L=0.5$ and $U=5$. The weights become $(0.5,1,4,5)$. The numerator remains $9$, because the weight raised from $0.2$ multiplies an outcome of zero. The denominator becomes $10.5$, so the estimate is $9/10.5=6/7\approx0.8571$. In this example, lower clipping decreases the estimate further by increasing the denominator attached to a zero outcome.

Worked Examples With Score Equations

Worked Example: Clipping Changes a Weighted Regression Slope

Consider weighted least squares with an intercept and a slope for the three points $(x_i,y_i)=(0,0),(1,1),(2,5)$, using weights $(0.1,1,9)$. The weighted slope is $$ \hat\beta_1=\frac{S_{xy}-S_xS_y/S_w}{S_{xx}-S_x^2/S_w}, $$ where $S_w=\sum_iw_i$, $S_x=\sum_iw_ix_i$, $S_y=\sum_iw_iy_i$, $S_{xx}=\sum_iw_ix_i^2$, and $S_{xy}=\sum_iw_ix_iy_i$. In each calculation below, the denominator is positive, so the weighted design has enough variation in $x$ to identify the slope.

For the original weights, the five sums are $S_w=10.1$, $S_x=19$, $S_y=46$, $S_{xx}=37$, and $S_{xy}=91$. Thus $$ \hat\beta_1 =\frac{91-(19)(46)/10.1}{37-19^2/10.1} =\frac{4.4653}{1.2574} \approx3.551. $$

With upper-only clipping at $U=3$, the weights are $(0.1,1,3)$. The sums become $4.1,7,16,13,31$, respectively. Hence $$ \hat\beta_1^U =\frac{31-(7)(16)/4.1}{13-7^2/4.1} =\frac{3.6829}{1.0488} \approx3.512. $$

With two-sided clipping at $L=0.5$ and $U=3$, the weights are $(0.5,1,3)$. The sums are $4.5,7,16,13,31$, giving $$ \hat\beta_1^{L,U} =\frac{31-(7)(16)/4.5}{13-7^2/4.5} =\frac{6.1111}{2.1111} \approx2.895. $$

The lower cutoff raises the weight on the point $(0,0)$, while leaving the other two weights unchanged from the upper-clipped case. That added contribution changes both the weighted centering of $x$ and the weighted relationship between $x$ and $y$; it is not merely an additional cap on influential observations.

Worked Example: An Unchanged Weight Can Have a Different Relative Contribution

Take two weights $w=(1,10)$ and apply upper-only clipping at $U=5$. The first weight remains numerically equal to $1$, while the second changes from $10$ to $5$. Before clipping, the normalized shares are $1/11$ and $10/11$. Afterwards they are $1/6$ and $5/6$. The first observation’s relative share has therefore increased, even though its own weight did not change.

For example, if the outcomes are $y=(1,0)$, the weighted mean changes from $1/11$ to $1/6$. This follows directly from the altered denominator: the unchanged weight of the first observation is now a larger fraction of the total weight. The safe statement is that upper-only clipping leaves each weight below $U$ numerically unchanged; it does not guarantee that its relative contribution to the whole weighted equation is unchanged.

Choosing and Checking the Rule

Upper-only clipping targets the high tail while preserving the numerical values of weights at or below $U$. Two-sided clipping also raises weights below $L$. In a weighted score, those increases matter whenever the corresponding $\psi_i(\theta)$ is nonzero. In a weighted mean, for instance, the effect depends on the outcomes attached to the raised weights, as the first example showed.

These rules differ from several related operations. Weight normalization rescales weights but does not cap high values or raise low ones. Deletion removes observations rather than retaining them with modified weights. Quantile cutoffs specify bounds using the sample distribution of weights; once bounds are selected, the clipping formulas are still the ones given here. Each choice changes what enters the equation, and none is automatically a substitute for another.

Useful diagnostics make the operation visible. Report the lower and upper cutoffs, the number and proportion of weights changed at each end, and how much total weight was removed from the upper tail or added to the lower tail. Compare the resulting estimate across defensible cutoffs and against the estimate without clipping. Also check that the weighted equation remains identifiable: for example, clipping must not leave a weighted regression design without variation in a predictor.

Common pitfall: “Upper clipping only changes the high weights, so all other observations keep the same relative contribution.” This confuses a weight’s numerical value with its share of the total weighted equation. An unchanged weight can have a larger or smaller relative share when other weights change. With two-sided clipping, low weights are changed directly as well. Inspect both the modified weights and the resulting estimating equation.

Clipping is a sensitivity choice, not a guarantee of improved causal estimation. A fixed cutoff can reduce the influence of extreme weights but can also move the solution of the estimating equation away from the solution based on the original weights. A lower bound can further amplify observations that originally had little weight. The direction and size of these changes depend on the score contributions; they cannot be inferred from the cutoff values alone. For that reason, state the rule clearly and interpret comparisons across cutoffs as informative about sensitivity, not as proof that one cutoff is universally correct.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For weights $(0.4,2,8)$, what are the upper-only clipped weights at $U=5$? What are the two-sided clipped weights at $L=1$ and $U=5$?
In the two-weight example $(1,10)$, why does the first weight’s normalized share change after upper clipping at $5$, although its numerical value stays at $1$?
For the weighted mean in the first worked example, what is the two-sided clipped estimate at $L=0.5$, $U=5$, and why does lower clipping change it?
Can the upper-clipped score change at a fixed $\theta$ be determined from the weights and cutoff alone? Explain using the score identity.

Solutions. (1) Upper-only clipping gives $(0.4,2,5)$; two-sided clipping gives $(1,2,5)$. (2) Its normalized share changes from $1/11$ to $1/6$, because the second weight falls from $10$ to $5$, reducing the total weight. (3) The estimate is $9/10.5=6/7\approx0.8571$. The lower cutoff raises the weight on an observation with outcome zero, increasing the denominator without changing the numerator. (4) No. The score change is $-n^{-1}\sum_{i:w_i>U}(w_i-U)\psi_i(\theta)$, so it also depends on the affected observations’ score contributions.

One-Sided and Two-Sided Weight Truncation

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