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Scaling and Normalizing Stabilized Weights

You will derive how common and arm-specific rescaling affect stabilized-weight estimates, then work through their implications for treatment-group means and contrasts.

Intermediate 10 min read

What You'll Learn

  • Define marginal and within-treatment-arm normalization of stabilized weights
  • Prove why a common positive rescaling leaves a weighted mean unchanged
  • Show why treatment-arm means and their difference survive separate rescaling of each arm
  • Distinguish invariant weighted means from pooled weighted summaries that can change
  • Compare normalized Hájek estimates with weighted totals that depend on scale
  • Diagnose normalization errors and interpret the limits of rescaling

Why Scaling Weights Needs Care

Stabilized inverse probability weights can make a weighted analysis easier to work with, but their numerical scale is not unique. A single common multiplier changes the weights’ totals without changing a ratio-based weighted mean. Multipliers chosen separately for treated and control units also preserve each arm’s weighted mean, but they can change a pooled weighted summary. Knowing which quantities are invariant prevents a harmless rescaling from being mistaken for a change in the treatment effect.

Consider a binary treatment \(A\), baseline covariates \(X\), and outcome \(Y\). Let \(e(X)=P(A=1\mid X)\), and let \(p=P(A=1)\). For an average treatment effect (ATE) analysis, the stabilized weight for unit \(i\) is

$$ w_i=A_i\frac{p}{e(X_i)}+(1-A_i)\frac{1-p}{1-e(X_i)}. $$

Thus a treated unit receives \(p/e(X_i)\), while a control unit receives \((1-p)/(1-e(X_i))\). The marginal treatment probability \(p\) may be known or estimated; the same algebra below applies either way, provided the same numerator is used for all units in the corresponding arm.

Definition: Marginal normalization multiplies every stabilized weight by one common positive constant, often chosen so the overall mean weight is 1. Within-treatment-arm normalization multiplies treated weights by one positive constant and control weights by another, often choosing each arm’s mean weight to be 1.

The causal interpretation of an inverse probability weighted comparison still requires the usual conditions: consistency, conditional exchangeability of treatment given \(X\), and positivity, so that both treatment probabilities are positive for covariate values in the target population. If \(e(X)\) is estimated, adequate estimation of the propensity score is also needed. Normalization changes none of these conditions and cannot repair their failure.

Which Weighted Quantities Are Invariant?

For arm \(a\in\{0,1\}\), write the weighted outcome mean as

$$ \hat\mu_a(w)= \frac{\sum_{i=1}^{n} I(A_i=a)w_iY_i} {\sum_{i=1}^{n} I(A_i=a)w_i}, \qquad \hat\tau(w)=\hat\mu_1(w)-\hat\mu_0(w). $$

Here \(I(A_i=a)\) equals 1 when unit \(i\) is in arm \(a\) and 0 otherwise. The arm-specific means estimate the corresponding potential-outcome means under the causal assumptions; their difference estimates the ATE. These are ratio-based, or Hájek, estimates.

Suppose all weights are multiplied by the same constant \(c>0\). The constant appears in both the numerator and denominator of each arm mean and cancels. The same cancellation holds if each arm has its own positive constant \(c_a\).

$$ \hat\mu_a(c_aw)= \frac{\sum_i I(A_i=a)c_aw_iY_i} {\sum_i I(A_i=a)c_aw_i} = \frac{c_a\sum_i I(A_i=a)w_iY_i} {c_a\sum_i I(A_i=a)w_i} =\hat\mu_a(w). $$

Consequently, both arm means and their difference are unchanged by either marginal normalization or separate within-arm normalization. This result depends on using a ratio within each arm. It does not say that every statistic calculated from the weights is invariant.

A common marginal normalization is to divide each weight by the overall mean weight, \(\bar w=n^{-1}\sum_i w_i\). The normalized weights then have mean 1. Dividing by \(\sum_i w_i\) instead makes their total 1; this also applies one common multiplier, so the arm means and contrast are unchanged. Within-arm mean-one normalization divides each weight in arm \(a\) by \(\bar w_a=n_a^{-1}\sum_{i:A_i=a}w_i\), where \(n_a\) is the number of observed units in that arm. Each arm then has normalized weight total \(n_a\).

Worked Examples

Worked Example: Marginal and Arm-Specific Normalization

Suppose four hypothetical units have the following stabilized weights and outcomes. The weights are positive, and the two treatment arms have different weight totals.

ArmWeightOutcome
Treated48
Treated212
Control0.55
Control1.57

The treated weighted outcome total is \(4(8)+2(12)=32+24=56\), and the treated weight total is \(4+2=6\). Thus

$$ \hat\mu_1=\frac{56}{6}=\frac{28}{3}\approx9.333. $$

The control weighted outcome total is \(0.5(5)+1.5(7)=2.5+10.5=13\), and the control weight total is \(0.5+1.5=2\). Therefore

$$ \hat\mu_0=\frac{13}{2}=6.5, \qquad \hat\tau=\frac{28}{3}-\frac{13}{2} =\frac{56-39}{6}=\frac{17}{6}\approx2.833. $$

The estimated contrast is about 2.833 outcome units. The overall weight total is \(6+2=8\), so the overall mean weight is \(8/4=2\). Dividing every weight by 2 gives overall mean-one weights. The treated numerator and denominator become \(56/2=28\) and \(6/2=3\); their ratio is \(28/3\). The control numerator and denominator become \(13/2=6.5\) and \(2/2=1\); their ratio is 6.5. The contrast remains \(17/6\).

For within-arm mean-one normalization, the treated weights are divided by their arm mean \(6/2=3\), while the control weights are divided by their arm mean \(2/2=1\). The resulting arm weight totals are both 2. The treated weighted outcome total is \(56/3\), giving mean \((56/3)/2=28/3\). The control total remains 13, giving mean \(13/2=6.5\). The treatment contrast is again \(17/6\).

The pooled weighted outcome mean does change under arm-specific normalization. Before that normalization it is \((56+13)/(6+2)=69/8=8.625\). After normalization it is

$$ \frac{56/3+13}{2+2} =\frac{95/3}{4} =\frac{95}{12}\approx7.917. $$

This pooled mean is not the difference in arm means and is not preserved by separate arm scaling. The corrected control total, 13, is essential to both pooled calculations.

Worked Example: Stabilized Versus Unstabilized Arm Means

Consider a hypothetical sample with two treated and two control units, so the sample treatment proportion is \(p=0.5\). Let the treated units have \(e(X)\) values 0.2 and 0.5, with outcomes 10 and 14. Let the controls have \(e(X)\) values 0.25 and 0.5, with outcomes 4 and 8.

The treated unstabilized weights \(1/e(X)\) are 5 and 2. The corresponding stabilized weights \(p/e(X)\) are 2.5 and 1, exactly one-half of the unstabilized weights. The treated mean using stabilized weights is

$$ \frac{2.5(10)+1(14)}{2.5+1} =\frac{39}{3.5} \approx11.14. $$

Using the unstabilized weights gives \(\{5(10)+2(14)\}/(5+2)=78/7\approx11.14\), the same mean. For controls, the unstabilized weights are \(1/(1-0.25)=4/3\) and \(1/(1-0.5)=2\). The stabilized weights \((1-p)/(1-e(X))\) are \(2/3\) and 1, again one-half of the unstabilized weights. The stabilized control mean is

$$ \frac{(2/3)(4)+1(8)}{2/3+1} =\frac{32/3}{5/3} =6.4. $$

With the unstabilized weights it is \(\{(4/3)(4)+2(8)\}/(4/3+2)=(64/3)/(10/3)=6.4\). Both weighting schemes therefore give the same contrast, \(11.14-6.4\approx4.743\) outcome units. In each arm stabilization has multiplied all weights by the same numerator, which cancels from that arm’s ratio.

Worked Example: Separate Rescaling Changes a Pooled Mean

Suppose treated units have weights 2 and 1 and outcomes 6 and 10; control units have weights 1 and 3 and outcomes 2 and 4. The arm means are

$$ \hat\mu_1=\frac{2(6)+1(10)}{3}=\frac{22}{3}, \qquad \hat\mu_0=\frac{1(2)+3(4)}{4}=\frac{14}{4}=3.5. $$

The contrast is \(22/3-7/2=23/6\approx3.833\) outcome units. Now multiply treated weights by 5 and control weights by 0.25. The treated mean remains \((5)(22)/(5)(3)=22/3\); the control mean remains \((0.25)(14)/(0.25)(4)=3.5\). Thus the contrast remains \(23/6\).

But the pooled weighted outcome mean changes. Originally it is \((22+14)/(3+4)=36/7\approx5.143\). After rescaling, the treated weighted outcome total is 110 and its weight total is 15; the control weighted outcome total is 3.5 and its weight total is 1. Therefore the new pooled mean is \((110+3.5)/(15+1)=113.5/16=7.094\). The shift reflects different relative contributions of the two arms, not a change in either arm’s weighted mean.

Why the Cancellation Works—and Where It Stops

The cancellation is algebraic rather than an approximation: a positive constant multiplying every term of a ratio’s numerator and denominator cancels exactly. For stabilized ATE weights, the treated numerator \(p\) is common to all treated observations and the control numerator \(1-p\) is common to all controls. Consequently, the arm-specific Hájek means from stabilized weights equal those from the corresponding unstabilized weights, assuming the same propensity scores and observed units. This statement concerns these weighted ratios; it does not claim that every weighting estimator is identical.

In particular, a weighted total is not invariant. If \(T_a=\sum_i I(A_i=a)w_iY_i\), then multiplying arm \(a\)’s weights by \(c_a\) changes that total to \(c_aT_a\). A Horvitz–Thompson-style total or mean formed directly from weighted sums therefore depends on the chosen scale unless its target-specific divisor is adjusted consistently. A Hájek mean divides by the corresponding sum of weights and is scale-invariant. This is the key distinction between normalized ratios and totals.

Likewise, a pooled weighted mean,

$$ \frac{\sum_i w_iY_i}{\sum_i w_i}, $$ depends on the relative scale assigned to the treatment arms. It is a weighted summary of observed outcomes, not the ATE. Multiplying all weights by one constant preserves it, but multiplying treated and control weights by different constants generally changes it.

Diagnostics, Alternatives, and Limitations

A reliable workflow records the original stabilized weights and the normalization constants. After marginal normalization, verify that the overall weight mean is 1 (or that the total is the chosen target). After within-arm normalization, verify that each arm’s weight mean is 1 separately. Then recompute the treated and control outcome means and confirm that they match the values calculated before rescaling, up to rounding. A changed arm mean usually signals that the wrong units were scaled, that the numerator and denominator were not treated consistently, or that a non-ratio statistic was compared.

Normalization is also not a remedy for extreme weights or a violation of positivity. A common positive multiplier leaves relative weights unchanged, so it does not remove the influence of observations with unusually large weights. The effective sample size, often summarized as \((\sum_i w_i)^2/\sum_i w_i^2\), is invariant to common rescaling: replacing each weight by \(cw_i\) multiplies both numerator and denominator by \(c^2\). The same is true within an arm if the effective sample size is calculated separately there. Scaling can make values more convenient to display, but it does not create information or improve overlap.

Marginal normalization is useful when a common scale is desired for display or for a calculation that uses all units together. Within-arm normalization is useful for presenting each treatment group’s weighted mean weight as 1. Neither choice is universally preferable for every downstream calculation: any estimator that uses the weights outside arm-specific ratios must be defined with its scale convention in mind. In particular, a pooled mean, a weighted total, or a model that combines arms need not share the invariance of the two separate Hájek means.

Common Mistakes

  • Assuming every weighted summary is unchanged. Arm-specific weighted means and their difference are unchanged by positive rescaling within each arm; a pooled weighted mean generally is not.
  • Adding arm-specific weighted outcome totals incorrectly. Calculate each product and sum it before dividing. For example, \(0.5(5)+1.5(7)=13\), not 12.5.
  • Treating normalization as a solution to extreme weights. Rescaling changes the units of the weights, not their relative size or the overlap in the data.
  • Calling a pooled observed-outcome mean the ATE. The ATE estimate here is the difference between the treated and control weighted potential-outcome means, not the pooled mean across observed outcomes.
Key takeaway: Multiplication by a common positive constant leaves a weighted ratio unchanged. Multiplication by different positive constants in the two arms also leaves each arm’s ratio—and therefore their difference—unchanged, but it can change pooled summaries and weighted totals.

Check Your Understanding

Answer each question, then check the solutions below.

  1. Why does dividing every weight by the overall mean weight leave each treatment-arm weighted mean unchanged?
  2. A treatment arm has weighted outcome total 30 and weight total 5. What is its weighted mean after all weights in that arm are multiplied by 4?
  3. Two arm-specific weighted means are 8 and 5. If each arm is independently normalized to have mean weight 1, what happens to their difference?
  4. Explain why an overall pooled weighted outcome mean may change after within-arm normalization even though the treatment contrast does not.

Solutions. (1) Dividing all weights by the same positive number divides both the arm’s weighted outcome total and its weight total by that number, so their ratio is unchanged. (2) The new totals are \(4(30)=120\) and \(4(5)=20\), and the mean is \(120/20=6\), the original \(30/5=6\). (3) The contrast remains \(8-5=3\), because each arm’s own weighted mean is invariant to its arm-specific multiplier. (4) Within-arm normalization can change the relative total weight assigned to treated and control units. The pooled mean combines the arms using those totals, whereas the treatment contrast subtracts two separately normalized arm means.

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