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Overlap, Trimming and Positivity · Tutorial 253 of 699

Positivity, Extreme Weights, and Information

You will derive how extreme treatment probabilities inflate inverse-probability weights and learn to assess the resulting instability with numerical diagnostics.

Intermediate 10 min read

What You'll Learn

  • Define positivity for the average treatment effect and distinguish it from a stronger bounded-positivity condition.
  • Derive how inverse-probability weights grow as treatment probabilities approach zero or one.
  • Connect large weights to second moments, estimator instability, and information concentration.
  • Calculate arm-specific effective sample size and interpret it as a diagnostic rather than a literal sample count.
  • Identify weight patterns that warrant investigation without treating estimated scores as proof of a positivity violation.
  • Compare weighting diagnostics with matching, outcome modeling, and trimming.

Why Positivity Matters for Weighting

Inverse-probability weighting uses each observation to represent people with similar measured covariates who received the other treatment. When a person’s treatment is highly predictable from those covariates, that representation can require a large weight. A handful of such observations can then dominate an estimate, making it sensitive to individual outcomes and reducing the amount of useful information in the weighted sample.

Let \(A\) be a binary treatment, with \(A=1\) for treated and \(A=0\) for control, and let \(X\) denote measured pretreatment covariates. The propensity score is \(e(X)=P(A=1\mid X)\). Let \(Y(1)\) and \(Y(0)\) be the potential outcomes under treatment and control, and define the average treatment effect (ATE) as \(\tau=E[Y(1)-Y(0)]\). In the examples below, all data are hypothetical.

Definition: Positivity for the ATE requires \(0<e(x)<1\) for covariate values \(x\) occurring in the target population. Thus, at each such covariate profile, both treatment assignments must have a positive probability. A stronger condition, often called bounded positivity, requires some \(\epsilon>0\) such that \(\epsilon\le e(x)\le 1-\epsilon\) throughout the target population. This stronger condition prevents probabilities from getting arbitrarily close to either boundary.

Positivity works alongside consistency and conditional exchangeability, \(Y(1),Y(0)\perp A\mid X\), for identifying the ATE from observed data. Positivity alone does not remove confounding. Conversely, even if positivity holds strictly at every covariate value, probabilities can be so close to zero or one that weighting is practically unstable in a finite sample.

From Small Probabilities to Large Weights

For the ATE, the inverse-probability weight is \(1/e(X)\) for a treated observation and \(1/[1-e(X)]\) for a control observation. The Horvitz–Thompson form of the IPW estimator is

$$ \hat{\tau}_{\mathrm{IPW}}=\frac{1}{n}\sum_{i=1}^{n}\left\{\frac{A_iY_i}{e(X_i)}-\frac{(1-A_i)Y_i}{1-e(X_i)}\right\}. $$

When \(e(X_i)\) is small, a treated person receives a large weight \(1/e(X_i)\). When \(e(X_i)\) is near one, a control person receives a large weight \(1/[1-e(X_i)]\). The key is the probability of the treatment actually received: a treated person with \(e=.02\) has weight \(50\), while a control person with \(e=.98\) also has weight \(50\). In either case, that person’s outcome can exert substantial influence on the estimate.

Worked Example: Translating Propensity Scores Into Weights

Consider four hypothetical observations. The first two are treated, with propensity scores .05 and .40. The other two are controls, with scores .95 and .60. Calculate each observation’s ATE weight.

For a treated observation the weight is \(1/e(X)\). The treated weights are \(1/.05=20\) and \(1/.40=2.5\).

For a control observation the weight is \(1/[1-e(X)]\). The control weights are \(1/(1-.95)=20\) and \(1/(1-.60)=2.5\).

The two observations with scores .05 and .95 each receive weight 20, while the other two receive weight 2.5. The large weights indicate that those first two observations represent relatively uncommon treatment assignments for their covariate profiles. They are diagnostic signals of potential instability, not by themselves proof that the corresponding population probabilities are zero.

A direct justification for why large weights matter comes from their second moments. Suppose the propensity score is the true \(e(X)\), conditional exchangeability holds, and potential outcomes have finite second moments. Write the summand in the estimator as \(\psi=AY/e(X)-(1-A)Y/[1-e(X)]\). Conditional on \(X\), the treated and control terms cannot both be nonzero. By consistency and exchangeability,

$$ E[\psi^2\mid X] =\frac{E[Y(1)^2\mid X]}{e(X)} +\frac{E[Y(0)^2\mid X]}{1-e(X)}. $$

For example, the treated contribution follows because \(A^2=A\) and \(E[A\mid X]=e(X)\), so \(E[A Y^2/e(X)^2\mid X]=E[Y(1)^2\mid X]/e(X)\). The control contribution follows in the same way using \(1-A\), whose conditional probability is \(1-e(X)\). As \(e(X)\) approaches zero, the first term can grow sharply; as it approaches one, the second can. These second moments help explain why extreme weights can make estimates variable even when all probabilities are technically positive. The outcome magnitudes also matter, so a large weight does not determine the variance by itself.

Weight Concentration and Effective Sample Size

A useful summary of how weights are distributed within an arm is the Kish effective sample size. For nonnegative weights \(w_1,\ldots,w_m\), define \(n_{\mathrm{eff}}=(\sum_j w_j)^2/\sum_j w_j^2\). Equal weights give \(n_{\mathrm{eff}}=m\); concentration in a few weights makes it smaller. The measure is unchanged if all weights are multiplied by the same positive constant.

Interpretation: Effective sample size is a weight-concentration diagnostic. It describes how many equally weighted observations would produce the same ratio of squared total weight to sum of squared weights. It is not generally the literal number of independent observations or a complete formula for an estimator’s variance.

Worked Example: Calculating Arm-Specific Effective Sample Size

In a hypothetical weighted sample, two treated observations have propensity scores .10 and .80. Calculate the treated weights and their effective sample size.

The treated weights are \(1/.10=10\) and \(1/.80=1.25\). Their sum is \(10+1.25=11.25\), and the sum of their squares is \(10^2+1.25^2=100+1.5625=101.5625\).

Therefore, \(n_{\mathrm{eff}}=(11.25)^2/(10^2+1.25^2)=126.5625/101.5625=1.246\), to three decimal places. Although there are two treated observations, their unequal weights make the weighted sample’s concentration comparable, under this diagnostic, to about 1.246 equally weighted observations. This does not mean that the study literally contains a fraction of a person or that its variance can be read off from this number alone.

Calculate this diagnostic separately for treated and control weights: each arm has a different weight formula and may have a different concentration pattern. It can also be informative to inspect the largest weights, upper quantiles, and the share of total weight carried by a small number of observations. A single summary can conceal whether concentration is driven by one extreme case or by a broader tail.

How Extreme Weights Can Move an Estimate

The effect of a large weight depends on both the weight and the observation’s outcome. A normalized, or Hájek, weighted mean in the treated arm is \(\sum_i A_iY_i/e(X_i)\) divided by \(\sum_i A_i/e(X_i)\). The next example shows how one observation can have substantial leverage on that mean.

Worked Example: Sensitivity to One High-Weight Outcome

Suppose two hypothetical treated observations have outcomes 40 and 60. Their propensity scores are .05 and .50, respectively. Calculate the normalized weighted mean, then determine how much it changes if the first outcome increases by 10 units.

The weights are \(1/.05=20\) and \(1/.50=2\). Thus the weighted mean is \((20(40)+2(60))/(20+2)=(800+120)/22=41.82\) outcome units, to two decimal places.

If the first outcome increases from 40 to 50, the weighted mean becomes \((20(50)+2(60))/22=1120/22=50.91\). The estimate changes by \(50.91-41.82=9.09\) units (using unrounded values, the change is \(20(10)/22=9.091\)). The high-weight observation accounts for \(20/22\) of the total weight, so its outcome strongly influences this weighted mean. This calculation illustrates sensitivity; it does not establish that the estimate is biased or that this observation should automatically be removed.

Diagnostics, Alternatives, and Limitations

Weight diagnostics are most useful when interpreted together. Inspect the propensity-score distributions by treatment arm, the largest arm-specific weights, effective sample sizes, and how much total weight is carried by the most heavily weighted cases. Then check whether the extreme scores correspond to substantively unusual covariate profiles or sparse treatment cells. The earlier material on support and tail diagnostics motivates these checks; here the emphasis is on the information consequences of the resulting weights.

Distinguish a structural positivity failure from practical instability. If a covariate profile in the target population can only receive one treatment, strict positivity fails there, and the ATE is not identified from those data without additional assumptions. If probabilities are positive but very small, positivity may hold mathematically while the finite-sample estimate remains highly variable. Estimated scores close to zero or one flag a concern, but do not prove the true probabilities are exactly zero or one. This tutorial concerns the consequences of weight patterns, not sampling variability in estimated support.

Different approaches address different parts of the problem. Matching seeks comparable treated and control observations, while outcome regression models outcomes conditional on covariates; neither makes unsupported treatment comparisons automatically credible. Trimming or restricting to a supported region can reduce extreme weights, but changes the population to which the estimate applies. No method can recover information about a treatment contrast that the observed design does not support without extra assumptions. The choice should therefore be guided by the target population and the scientific question, not by a preference for smaller weights alone.

  • Do not equate a large weight with an error. It reflects the weighting scheme for the target estimand; investigate its implications rather than deleting it automatically.
  • Do not treat effective sample size as a variance estimate. It summarizes weight concentration but does not incorporate every feature of outcomes, dependence, or the estimator.
  • Do not assume that clipping weights is cost-free. Capping weights can reduce the influence of extreme observations but generally changes the weighting and may introduce bias for the original estimand.
  • Do not infer adequate overlap from a moderate average weight. Averages can hide a small number of very large weights or a poorly supported subgroup.
Key takeaway: Near-zero treatment probabilities create large treated weights; near-one probabilities create large control weights. Their concentration can magnify outcome sensitivity and reduce effective information. Use the pattern of scores, weights, and arm-specific effective sample sizes to diagnose risk, while keeping the target population and identification assumptions explicit.

Check Your Understanding

Answer the questions, then compare your reasoning with the solutions.

  1. A control observation has propensity score .90. What is its ATE weight, and why is it large?
  2. For weights 2, 2, and 2, calculate the Kish effective sample size. What does the result tell you?
  3. For weights 8 and 2, calculate the Kish effective sample size and compare it with the number of observations.
  4. Does an estimated propensity score of .01 prove a structural positivity violation? Explain.
  5. Why can trimming extreme weights change the interpretation of an estimate?

Solutions. (1) The control weight is \(1/(1-.90)=10\); control assignment is relatively unlikely at that score. (2) \(n_{\mathrm{eff}}=6^2/(2^2+2^2+2^2)=36/12=3\), matching the three equally weighted observations. (3) \(n_{\mathrm{eff}}=10^2/(8^2+2^2)=100/68=1.471\), below the two observations because weight is concentrated on the first. (4) No. It flags potential practical instability, but an estimated value does not prove that the true treatment probability is zero. (5) Trimming changes which observations contribute and can redefine the retained target population or the weighting estimand.

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