How Truncation Alters Weighted Covariate Balance

Why Clipping Can Change Covariate Balance

Inverse-probability weights are designed to reweight treatment groups so that their measured covariate distributions resemble a common target distribution. With the correct, untruncated treatment probabilities, this balancing identity follows directly from conditional expectation. Clipping replaces some large weights by smaller ones, so strata that needed the largest corrections receive less weight. The result can be a new difference in covariate distributions, even when the untruncated weights would balance them exactly.

The aim here is to track that change in the covariate distribution itself. It is not enough to say that clipping makes weights smaller or less variable: the important question for balance is which observations have their weights changed, and how does that change their representation?

Definition: Let $A$ be a treatment taking values $0$ and $1$, and let $C$ be a measured covariate vector. Write $g_a(C)=P(A=a\mid C)$, with $g_a(C)>0$ almost surely for each treatment level being considered. The unstabilized inverse-probability weight for an individual receiving treatment $a$ is $W_a=1/g_a(C)$. For an upper clipping threshold $c>0$, the clipped weight is $W_{a,c}=\min\{W_a,c\}$. Weighted covariate balance means that the normalized covariate distributions within the treatment groups agree, exactly or approximately, on the covariates under consideration.

The positivity condition ensures that the inverse weights are defined. For the expectation identity below, it is also necessary that the expectations exist; for example, it is enough to take $h(C)$ bounded. Exact balance is a population statement when the treatment probabilities are known. In a finite sample, empirical weighted distributions need not match exactly, even if the treatment model is correct.

The Exact-Weight Balance Identity

For any bounded measurable function $h$ of the covariates, weighting the observations in treatment group $a$ by $1/g_a(C)$ recovers the population average of $h(C)$. Taking $h(C)$ to be an indicator gives the weighted prevalence of any covariate category; taking it to be a coordinate of $C$ gives a weighted mean.

Theorem: Suppose $g_a(C)=P(A=a\mid C)>0$ almost surely and $h(C)$ is bounded and measurable. Then $$ E\left[\frac{I(A=a)}{g_a(C)}h(C)\right]=E[h(C)] \quad\text{and}\quad E\left[\frac{I(A=a)}{g_a(C)}\right]=1. $$ Consequently, the population covariate distribution among treatment-$a$ observations normalized by their inverse-probability weights equals the marginal distribution of $C$.

Proof. By iterated expectation, condition first on $C$. Since $E[I(A=a)\mid C]=g_a(C)$,

E\left[\frac{I(A=a)}{g_a(C)}h(C)\right] =E\left[\frac{h(C)}{g_a(C)}E[I(A=a)\mid C]\right] =E[h(C)].

Setting $h(C)=1$ gives a weighted denominator with expectation one. Dividing the first identity by this denominator therefore gives the marginal expectation of every bounded measurable function $h$. In particular, indicator functions identify the whole covariate distribution. $\square$

Clipping changes the identity in a specific way. The next result shows that it is not simply a uniform reduction in weights: it tilts the represented covariate distribution according to the treatment probability.

Result: Under the same positivity condition, for bounded measurable $h$, $$ E[I(A=a)W_{a,c}h(C)] =E\left[\min\{1,cg_a(C)\}h(C)\right]. $$ Thus, after normalization, the clipped weighted distribution is proportional to the population covariate distribution multiplied by $r_a(C)=\min\{1,cg_a(C)\}$.

Derivation. Since $W_{a,c}=\min\{1/g_a(C),c\}$, conditioning on $C$ gives

\begin{aligned} E[I(A=a)W_{a,c}h(C)] &=E\left[g_a(C)\min\left\{\frac{1}{g_a(C)},c\right\}h(C)\right]\\ &=E[\min\{1,cg_a(C)\}h(C)]. \end{aligned}

The normalized clipped distribution therefore has density, relative to the marginal distribution of $C$, proportional to $\min\{1,cg_a(C)\}$. Where $1/g_a(C)\le c$, the factor is one and that stratum is not altered. Where the weight exceeds $c$, the factor is $cg_a(C)<1$, so that stratum is underrepresented relative to the exact-weighted target. Because the factor depends on $a$ and $C$, the two treatment groups can be tilted differently. $\square$

Worked Examples

Worked Example: A Binary Covariate and a Cap of 2.5

Suppose a population has 700 individuals. The binary covariate $X$ equals 1 for 100 individuals and 0 for 600. Treatment probabilities are $P(A=1\mid X=1)=0.8$ and $P(A=1\mid X=0)=0.2$. The expected counts in each covariate-treatment cell are integers, as shown below.

Covariate	Treated, $A=1$	Untreated, $A=0$	Total
$X=1$	80	20	100
$X=0$	120	480	600
Total	200	500	700

The untreated group has $20/500=0.04$ with $X=1$, while the treated group has $80/200=0.40$. For treated individuals the exact weights are $1/0.8=1.25$ when $X=1$, and $1/0.2=5$ when $X=0$. Their weighted counts are therefore $80(1.25)=100$ and $120(5)=600$, giving weighted prevalence $100/(100+600)=1/7$, or about $0.1429$.

For untreated individuals, the exact weights are $1/(1-0.8)=5$ when $X=1$, and $1/(1-0.2)=1.25$ when $X=0$. Their weighted counts are $20(5)=100$ and $480(1.25)=600$, also giving prevalence $100/700=1/7$, or about $0.1429$. Both groups now match the population prevalence $100/700=1/7$.

Now clip every weight above $2.5$ to $2.5$. In the treated group, the $X=1$ weight stays at $1.25$, but the $X=0$ weight falls from 5 to 2.5. The clipped weighted counts are $80(1.25)=100$ and $120(2.5)=300$, so the treated weighted prevalence is $100/400=0.25$. In the untreated group, the $X=1$ weight falls from 5 to 2.5, while the $X=0$ weight stays at 1.25. The counts become $20(2.5)=50$ and $480(1.25)=600$, so the untreated weighted prevalence is $50/650\approx0.0769$. The clipped groups differ by about $0.25-0.0769=0.1731$, despite exact balance before clipping.

The calculation also checks the direction of the change: clipping reduces the representation of the low-probability covariate-treatment cells. For treated observations, that is the $X=0$ stratum; for untreated observations, it is the $X=1$ stratum. Those are different strata, so the two clipped distributions move apart.

Worked Example: A Less Severe Cap

Keep the same cell counts, but use a cap of $c=4$. In the treated group, only the $X=0$ weight changes, from 5 to 4. The weighted counts are $100$ for $X=1$ and $120(4)=480$ for $X=0$. Thus the treated weighted prevalence is $100/(100+480)=100/580\approx0.1724$.

In the untreated group, only the $X=1$ weight changes, from 5 to 4. The weighted counts are $20(4)=80$ for $X=1$ and $480(1.25)=600$ for $X=0$. The untreated weighted prevalence is $80/(80+600)=80/680\approx0.1176$. The remaining difference is $0.1724-0.1176\approx0.0548$.

The cap of 4 leaves less imbalance than the cap of 2.5 in this example, but it does not restore exact balance. A higher cap reduces the alteration here because fewer observations are moved as far from their exact weights; it does not guarantee that every balance measure improves monotonically in more complicated data.

Worked Example: The Same Cap on Stabilized Weights

In the same population, $P(A=1)=200/700=2/7$ and $P(A=0)=5/7$. Consider stabilized weights that multiply each inverse-probability weight by the marginal probability of the treatment actually received. Within a treatment group this is a constant multiplier, so without clipping it cancels from the normalized covariate distribution.

For treated individuals, the stabilized weights are $(2/7)/0.8=5/14$ for $X=1$ and $(2/7)/0.2=10/7$ for $X=0$. The weighted counts are $80(5/14)=200/7$ and $120(10/7)=1200/7$. Their normalized $X=1$ prevalence is $(200/7)/(1400/7)=1/7$.

For untreated individuals, the stabilized weights are $(5/7)/0.2=25/7$ for $X=1$ and $(5/7)/0.8=25/28$ for $X=0$. The weighted counts are $20(25/7)=500/7$ and $480(25/28)=3000/7$. The normalized prevalence is $(500/7)/(3500/7)=1/7$.

Apply the same numerical cap, 2.5, to these stabilized weights. No treated weight reaches the cap, so the treated prevalence remains $1/7$. The untreated $X=1$ weight is clipped to 2.5, giving a weighted count of $20(2.5)=50$; the $X=0$ weighted count remains $3000/7$. The clipped untreated prevalence is $50/(50+3000/7)=350/3700\approx0.0946$. Thus the clipped groups differ by about $0.1429-0.0946=0.0483$. The numerical cap has a different effect on stabilized and unstabilized weights because their scales differ.

Balance Diagnostics and Interpretation

Balance should be checked on the actual weights used in the analysis, after clipping and any subsequent normalization. For a binary covariate, compare its normalized weighted prevalence across treatment groups, as in the examples. For a continuous covariate, compare weighted means and, where relevant, other features of its distribution. A standardized mean difference can summarize a mean difference relative to a scale, but it does not establish equality of the full distributions.

A practical diagnostic sequence is:

Compute balance before and after clipping.
Use the same covariates and balance summaries for both sets of weights. This isolates how much the clipping rule changed balance.

Identify the altered observations.
Tabulate which treatment-covariate strata contain weights above the threshold. These strata are the ones whose representation is directly reduced.

Compare across thresholds and scales.
Check whether conclusions depend strongly on the cap and whether the rule applies to stabilized or unstabilized weights. The same numerical threshold need not have the same effect on both scales.

Balance diagnostics concern measured covariates and the weighted sample. They do not show that all confounding has been removed, that the treatment model is correct, or that an outcome model is adequate. Conversely, an imbalance after clipping does not by itself quantify the resulting change in an outcome estimate; that requires examining the estimating procedure and outcome information.

Common Pitfalls and Alternatives

Assuming clipping preserves exact balance. The exact-weight identity uses $1/g_a(C)$. After clipping, the factor becomes $\min\{1,cg_a(C)\}$, which generally varies across covariate strata. Recompute balance rather than assuming it remains.
Using the wrong group prevalence in a weighted calculation. The weighted prevalence is a ratio of weighted counts within the treatment group. In the first example, treated $X=1$ has prevalence $0.40$, but its exact weighted prevalence is $0.1429$, not $0.40$: $80(1.25)/[80(1.25)+120(5)]=100/700$.
Thinking normalization repairs the balance change. Dividing all weights in a treatment group by their sum changes the common scale, not the relative weights of covariate strata. It cannot undo clipping’s stratum-specific reductions.
Confusing clipping with deletion. Clipping retains an observation with a reduced weight; deletion removes it from the weighted representation. Neither operation generally preserves the exact-weight balance identity.

Leaving weights untruncated preserves the population identity when the probabilities are correct, but large weights can make finite-sample estimates unstable. Clipping trades some of that influence for a changed weighted covariate distribution. Deletion has a different effect because it discards observations rather than retaining them with capped influence. Alternatives that deliberately target a different covariate distribution, such as overlap weighting, should be interpreted according to their own target; they are not a way to claim that clipped inverse-probability weights still balance to the original population.

The central limitation is therefore precise: a clipped analysis can be useful as a stability choice, but its weighted sample is not automatically the same pseudo-population as the one defined by exact inverse-probability weights. Report the clipping rule and assess balance after applying it, on the scale actually used.

Key takeaway: Exact inverse-probability weights reproduce the marginal covariate distribution because their treatment-probability factor cancels. Clipping replaces that cancellation with a covariate-dependent tilt, so balance must be recalculated rather than presumed.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

In the 2.5-cap example, what are the clipped weighted prevalences of $X=1$ in the treated and untreated groups, and what is their difference?
For what covariate-treatment observations does upper clipping have no effect?
Why does multiplying all weights in one treatment group by a positive constant preserve that group’s normalized weighted covariate distribution, while clipping need not?
In the stabilized-weight example, which group has a weight clipped at 2.5, and what is its resulting weighted $X=1$ prevalence?
Name two balance summaries that could be checked after clipping, and state one limitation of those summaries.

Solutions. (1) The prevalences are $100/400=0.25$ and $50/650\approx0.0769$; their difference is about $0.1731$. (2) Clipping has no effect when $1/g_a(C)\le c$. (3) A common positive multiplier cancels between numerator and denominator of a normalized weighted distribution; clipping changes some weights relative to others. (4) The untreated $X=1$ weight is clipped; its prevalence becomes $50/(50+3000/7)\approx0.0946$. (5) One could check weighted prevalence for binary covariates and weighted means or standardized mean differences for continuous covariates. Such summaries concern measured features and do not establish balance of every feature of the distributions or absence of unmeasured confounding.