Two Operations With Different Purposes
Inverse probability weights are often divided by their sample mean before an analysis. The resulting weights have mean one, which can make their scale easier to interpret. But a mean-one weight is not necessarily a stabilized weight. Stabilization changes the probability ratio used to construct each observation’s weight; normalization multiplies all weights by the same sample-specific constant. Those operations can produce different relative contributions to the analysis and can correspond to different weighted populations.
To make the distinction precise, consider one binary treatment \(A\in\{0,1\}\), measured with covariates \(L\). Write \(p(a\mid l)=P(A=a\mid L=l)\) for the treatment probability in the observed data. Let \(q(a\mid v)\) be a numerator treatment distribution, where \(V\) is a subset of \(L\), and suppose \(q(a\mid v)>0\) only where \(p(a\mid l)>0\). The earlier material on stabilized treatment weights as likelihood ratios defines the probability-ratio motivation for stabilization; here the focus is what these ratios are not equivalent to.
The numerator \(q\) is part of the definition of \(S\), not a consequence of dividing by \(\bar U\). In particular, normalization preserves all pairwise weight ratios: \(\widetilde U_i/\widetilde U_j=U_i/U_j\), whenever the weights are positive. Stabilization instead changes a weight by a factor \(q(A_i\mid V_i)\), which need not be the same for every observation.
What Distribution Does Each Weight Represent?
The distinction can be seen by deriving the weighted joint distribution of treatment and covariates. Assume the treatment probabilities are correctly specified, that the treatment is discrete with both levels possible in the population of interest, and that the required probabilities are positive. Let \(f(l)\) denote the distribution of \(L\). The observed joint probability of treatment \(a\) and covariates \(l\) is \(p(a\mid l)f(l)\).
Proof. Multiplying the observed joint probability by the stabilized weight cancels the denominator:
Summing over \(a\) gives \(f(l)\), because \(\sum_{a=0}^1q(a\mid v)=1\). Summing over \(l\) and \(a\) therefore gives total mass one. For the unstabilized weight, the same cancellation gives \(f(l)\) for each value of \(a\). The total mass over both treatment levels is \(2\), since \(\sum_l f(l)=1\). Dividing by this total yields \(f(l)/2\) for each \(a\). This proves both claims. \(\square\)
This result describes population weighting, not a guarantee that a finite weighted sample will exactly have the displayed distribution. In a finite sample, random variation and estimated treatment probabilities affect the observed weighted covariate distribution. The derivation nevertheless identifies the distinct targets of the two probability ratios. Mean-normalizing sample weights changes neither ratio nor its population weighting target.
Worked Examples
Worked Example: Normalizing Does Not Construct the Numerator
Suppose four observations have treatment-specific denominator probabilities \(p(A_i\mid L_i)=(0.20,0.80,0.40,0.60)\). Their treatments are \(1,0,0,0\), respectively. Choose numerator probabilities \(q(1)=0.25\) and \(q(0)=0.75\). Calculate both weight types and their mean-normalized values.
The unstabilized weights are the reciprocals of the denominator probabilities: \(U=(5,1.25,2.5,1.6667)\). Their sum is \(10.4167\), so \(\bar U=2.6042\). Dividing each by this mean gives \(\widetilde U\approx(1.92,0.48,0.96,0.64)\).
For the stabilized weights, multiply each reciprocal by the numerator probability for that observation’s treatment: \(S=(0.25\times5,\;0.75\times1.25,\;0.75\times2.5,\;0.75\times1.6667)\) \(=(1.25,0.9375,1.875,1.25)\). Their sum is \(5.3125\), so \(\bar S=1.328125\). Dividing by this mean gives \(\widetilde S\approx(0.9412,0.7059,1.4118,0.9412)\).
The two normalized vectors are not equal. In particular, their first-to-second weight ratios are \(1.92/0.48=4\) and \(0.9412/0.7059\approx1.3333\), respectively. Dividing by a common mean cannot change a within-vector ratio; the changed ratio comes from using the treatment-specific numerator in \(S\).
Worked Example: The Weighted Targets in a Binary-Treatment Population
Let \(L\) be binary with \(P(L=0)=P(L=1)=0.5\), and suppose \(p(1\mid0)=0.10\) and \(p(1\mid1)=0.40\). The marginal treatment probability is \(P(A=1)=0.5(0.10)+0.5(0.40)=0.25\); hence choose \(q(1)=0.25\) and \(q(0)=0.75\).
For each \(L\), the stabilized weighted joint mass at treatment \(a\) is \(q(a)P(L)\). Thus the four cells \((A,L)=(1,0),(1,1),(0,0),(0,1)\) have masses \(0.125,0.125,0.375,0.375\). These sum to \(1\); treatment has marginal probability \(0.25\) under the weighted distribution, and \(L\) remains equally likely to be 0 or 1 within each treatment group.
With the unstabilized weight, each of the four cells has weighted mass \(P(L)=0.5\). The total mass is \(2\), so after normalization each cell has mass \(0.25\). The treatment marginal is therefore \(0.5\), not \(0.25\). Both weighted distributions make treatment independent of \(L\) in this example, but they do not assign the same treatment proportions. Normalizing the unstabilized weights does not supply the marginal treatment probabilities in the stabilized numerator.
Worked Example: Mean Normalization Preserves Weight Concentration
Consider weights \(w=(1,1,8)\), whose mean is \(10/3\). Their normalized values are \((0.3,0.3,2.4)\). For an outcome vector \(y=(0,10,20)\), the weighted mean before normalization is \((1\times0+1\times10+8\times20)/10=170/10=17\). After normalization it is \((0.3\times0+0.3\times10+2.4\times20)/3=51/3=17\).
The usual effective sample size summary is \((\sum_i w_i)^2/\sum_i w_i^2\). Before normalization it is \(10^2/(1+1+64)=100/66\approx1.515\). Afterwards it is \(3^2/(0.09+0.09+5.76)=9/5.94\approx1.515\). The largest observation still carries most of the total weight. Setting the mean to one has changed the units, not the concentration.
What Normalization Can and Cannot Do
For a weighted mean, dividing all weights by their sample mean has no effect on the estimate. If \(y_i\) is an outcome and \(\bar w>0\), then
Both numerator and denominator acquire the same factor \(1/\bar w\), which cancels. More generally, if an estimating equation is \(\sum_i w_i\psi_i(\theta)=0\), replacing every \(w_i\) by \(w_i/c\) for a common positive \(c\) multiplies the whole equation by \(1/c\); its set of roots is unchanged. This is scale invariance, not a claim that every procedure using weights is unchanged: procedures with additional fixed penalties or scale-dependent thresholds need not have this property.
Normalization also leaves the ordering of positive weights unchanged and preserves their maximum-to-minimum ratio. The effective sample size in the example is another scale-invariant diagnostic. A mean of one is therefore not evidence that weights are moderate, that positivity holds, or that weighted covariates are balanced.
Proof. For any \(i,j\), \((w_i/c)/(w_j/c)=w_i/w_j\). For the effective sample size, the common factor cancels: \[ \frac{(\sum_i w_i/c)^2}{\sum_i(w_i/c)^2} =\frac{(\sum_i w_i)^2/c^2}{\sum_iw_i^2/c^2} =\frac{(\sum_iw_i)^2}{\sum_iw_i^2}. \] For the weighted mean, the same cancellation gives \[ \frac{\sum_i(w_i/c)y_i}{\sum_iw_i/c} =\frac{\sum_iw_i y_i}{\sum_iw_i}. \] Taking \(c=\bar w>0\) proves the stated claims for mean normalization. \(\square\)
Comparison, Diagnostics, and Limitations
| Operation | What changes | What does not change |
|---|---|---|
| Unstabilized weight \(1/p(A\mid L)\) | Weights observations according to the reciprocal treatment probability | No numerator treatment distribution is specified |
| Stabilization \(q(A\mid V)/p(A\mid L)\) | Relative contributions can change through the numerator; the weighted target retains the specified \(q\) | It does not guarantee bounded weights or eliminate positivity concerns |
| Divide any weights by their sample mean | The common numerical scale changes and the sample mean becomes one | Relative weights, ranking, and weighted means are preserved |
A useful diagnostic sequence is to identify the actual numerator and denominator used to create each weight, then calculate the sample mean, maximum, effective sample size, and weight distribution within treatment groups. Examine weighted covariate balance separately. If a mean-normalized weight has a mean of one by construction, that number is an arithmetic check—not a check of treatment-model fit, positivity, or balance. Extreme tails can remain after stabilization, as discussed in the earlier material on why stabilized weights can still have heavy tails.
A special case helps prevent an opposite overstatement. If \(q(A\mid V)\) is the same constant for every observation, then \(S\) is a constant multiple of \(U\). Their mean-normalized versions are identical. In a binary-treatment setting, this occurs with a uniform numerator \(q(0)=q(1)=0.5\). Outside that case, stabilization and normalization generally do not produce the same relative weights. Even when normalized weights coincide, that algebraic equality does not make mean normalization the operation that created the numerator probability ratio.
Neither operation repairs a treatment model that assigns very small probabilities to observed treatment, and normalization cannot remove the resulting extreme relative weights. Stabilization may alter their distribution, but it does not guarantee light tails. Mean normalization is a rescaling convention; stabilization encodes a numerator distribution in the probability ratio. Keeping these roles separate is essential when interpreting diagnostics and describing the weighted target.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- If every weight is divided by the same positive constant, can the weighted mean of an outcome change?
- For binary treatment, what treatment marginal does the population-normalized unstabilized weight induce under the assumptions in the theorem?
- When will mean-normalized stabilized and unstabilized weights coincide?
- Why does a sample mean of one fail to show that weights are not concentrated?
- Suppose \(q(1)=0.2\), \(q(0)=0.8\), and an observation has \(p(A\mid L)=0.4\). Find its unstabilized and stabilized weights for \(A=1\), and explain the difference.
Solutions. (1) No: the common factor cancels from the weighted-mean numerator and denominator. (2) It assigns probability \(1/2\) to each treatment level. (3) They coincide if the numerator is the same positive constant for every observation, so the stabilized weights are a common multiple of the unstabilized weights. (4) Normalization preserves weight ratios and effective sample size; one weight can still dominate the total. (5) The unstabilized weight is \(1/0.4=2.5\). The stabilized weight is \(0.2/0.4=0.5\). The numerator changes this observation’s relative contribution; dividing a collection of weights by its sample mean would only rescale that collection.