From Sequential Probabilities to Cumulative Weights
With treatment measured at two decision times, each person's weight is the product of two contributions. The first reweights treatment \(A_1\); the second reweights \(A_2\) conditional on the first treatment. This small setting makes it possible to audit every probability and multiplication, while also showing why stabilized and unstabilized weights need not assign the same relative importance to the observed records.
Let \(L_0\) denote baseline covariates and \(L_1\) covariates measured before the second treatment. Write \(A_1,A_2\in\{0,1\}\). The denominator models describe the probability of the treatment actually received, conditional on the observed covariate and treatment history. For this example, choose stabilized numerators that condition on prior treatment but not on \(L_0,L_1\).
These expressions require positive denominator probabilities for the observed treatment decisions. For the stabilized weight, the numerator probabilities must also be positive on the observed treatment paths. For a causal interpretation, the usual consistency, sequential exchangeability, and positivity assumptions are additionally needed; accurate probability calculations alone do not establish those assumptions. The focus here is the arithmetic of the weights and the summaries they produce.
A Small Dataset and Its Probability Inputs
Suppose the probabilities below are supplied by fitted treatment models. The columns \(d_1,d_2\) are denominator probabilities for the observed decisions; \(q_1,q_2\) are the corresponding numerator probabilities. They are not probabilities of a single common treatment value across rows: each is evaluated at the treatment actually received by that person.
| Person | \(A_1\) | \(A_2\) | \(d_1\) | \(q_1\) | \(d_2\) | \(q_2\) | Outcome \(Y\) |
|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 0.80 | 0.50 | 0.75 | 0.60 | 10 |
| 2 | 1 | 0 | 0.40 | 0.50 | 0.60 | 0.40 | 20 |
| 3 | 0 | 1 | 0.70 | 0.50 | 0.50 | 0.30 | 30 |
| 4 | 0 | 0 | 0.30 | 0.50 | 0.80 | 0.70 | 40 |
The first numerator is \(P(A_1)\), set to \(0.50\) for either observed value. At the second time, \(q_2=P(A_2\mid A_1)\): for \(A_1=1\), the probabilities of \(A_2=1\) and \(A_2=0\) are \(0.60\) and \(0.40\); for \(A_1=0\), they are \(0.30\) and \(0.70\). Each pair sums to one, as a valid conditional distribution must. The denominator probabilities also correspond to the observed values: person 2, for instance, has \(P(A_2=0\mid A_1,L_0,L_1)=0.60\).
Worked Example: Calculating Each Person's Two Weights
For person 1, the observed treatment probabilities are \(d_1=0.80\), \(d_2=0.75\), with numerators \(q_1=0.50\), \(q_2=0.60\). Therefore $$ W^U_1=\frac{1}{0.80}\frac{1}{0.75} =\frac{1}{0.60}=\frac{5}{3}\approx1.667, \qquad W^S_1=\frac{0.50}{0.80}\frac{0.60}{0.75} =0.625(0.80)=0.500. $$
For person 2, the observed second treatment is \(A_2=0\), so use the probability \(0.60\) of receiving zero, and its numerator \(0.40\): $$ W^U_2=\frac{1}{0.40}\frac{1}{0.60} =\frac{25}{6}\approx4.167, \qquad W^S_2=\frac{0.50}{0.40}\frac{0.40}{0.60} =1.25\left(\frac{2}{3}\right)=\frac{5}{6}\approx0.833. $$
For person 3, $$ W^U_3=\frac{1}{0.70}\frac{1}{0.50} =\frac{20}{7}\approx2.857, \qquad W^S_3=\frac{0.50}{0.70}\frac{0.30}{0.50} =\frac{5}{7}(0.60)=\frac{3}{7}\approx0.429. $$
For person 4, $$ W^U_4=\frac{1}{0.30}\frac{1}{0.80} =\frac{25}{6}\approx4.167, \qquad W^S_4=\frac{0.50}{0.30}\frac{0.70}{0.80} =\frac{5}{3}\frac{7}{8}=\frac{35}{24}\approx1.458. $$
The cumulative weights are therefore \(1.667,4.167,2.857,4.167\) without stabilization and \(0.500,0.833,0.429,1.458\) with stabilization. Neither weight is calculated from the outcome \(Y\); the outcome is used later in the weighted summary.
What the Product Shows
Stabilization changes each person's weight by multiplying the unstabilized weight by the observed numerator probabilities. In this example, the two-time-point relationship is exact, not an approximation:
To verify it, substitute \(W^U=1/\{P(A_1\mid L_0)P(A_2\mid A_1,L_0,L_1)\}\). Multiplication by \(P(A_1)P(A_2\mid A_1)\) gives $$ \frac{P(A_1)P(A_2\mid A_1)} {P(A_1\mid L_0)P(A_2\mid A_1,L_0,L_1)}, $$ which is the definition of \(W^S\). The numerator factors can differ from person to person, so stabilization does not generally multiply every person's weight by the same constant. It can therefore change a weighted mean, not just the scale of all weights.
Worked Example: Comparing the Weighted Outcome Means
A weighted mean of \(Y\) is \(\sum_i W_iY_i/\sum_i W_i\). Using the unstabilized weights, the total weight is $$ \sum_i W^U_i=\frac{5}{3}+\frac{25}{6}+\frac{20}{7}+\frac{25}{6} =\frac{90}{7}. $$
The weighted outcome total is $$ \sum_i W^U_iY_i =\frac{5}{3}(10)+\frac{25}{6}(20)+\frac{20}{7}(30)+\frac{25}{6}(40) =\frac{50}{3}+\frac{250}{3}+\frac{600}{7}+\frac{500}{3} =\frac{7400}{21}. $$ Thus the unstabilized weighted mean is $$ \frac{7400/21}{90/7}=\frac{740}{27}\approx27.407. $$
With stabilized weights, the total is $$ \sum_i W^S_i=\frac{1}{2}+\frac{5}{6}+\frac{3}{7}+\frac{35}{24} =\frac{541}{168}. $$ The weighted outcome total is $$ \sum_i W^S_iY_i =\frac{1}{2}(10)+\frac{5}{6}(20)+\frac{3}{7}(30)+\frac{35}{24}(40) =5+\frac{50}{3}+\frac{90}{7}+\frac{175}{3} =\frac{650}{7}. $$ Consequently, the stabilized weighted mean is $$ \frac{650/7}{541/168}=\frac{15600}{541}\approx28.835. $$
The two finite-sample summaries differ. This is not an arithmetic error: each person receives a different numerator multiplier, and the four observed outcomes are not all equal. Nor does the difference alone establish which estimate is preferable. Its interpretation depends on the target estimand and the assumptions behind the treatment models.
Checks, Assumptions, and Interpretation
A useful theoretical check is that, for correctly specified probability distributions, the expected stabilized weight is one under the observed-data law. Let \(H_t\) denote the observed history available before treatment \(A_t\), including the relevant covariates, and let \(p_t(a\mid H_t)\) be the denominator treatment probability. Let \(q_t(a\mid A_{t-1})\) be a normalized numerator distribution, so that \(\sum_{a\in\{0,1\}}q_t(a\mid A_{t-1})=1\). Conditional on \(H_t\), the expected factor at time \(t\) is $$ E\left[\left.\frac{q_t(A_t\mid A_{t-1})}{p_t(A_t\mid H_t)}\right|H_t\right] =\sum_{a\in\{0,1\}}p_t(a\mid H_t) \frac{q_t(a\mid A_{t-1})}{p_t(a\mid H_t)} =1, $$ provided \(p_t(a\mid H_t)>0\) whenever the numerator assigns positive probability. Iterating this conditional expectation over the two decision times gives \(E[W^S]=1\). The same argument applies when the numerator also conditions on baseline variables, as long as it is normalized given the conditioning history.
This identity is a statement about the probability model, not an exact constraint on a realized sample. Here the sample average stabilized weight is $$ \frac{1}{4}\sum_i W^S_i=\frac{541}{672}\approx0.805, $$ not one. Estimated probabilities and sampling variability both matter in practice. A sample average away from one warrants checking the calculation and models, but it does not by itself disprove the identity.
Worked Example: Normalizing the Finite-Sample Weights
If the four stabilized weights are divided by their sample mean, each is multiplied by the same factor: $$ c=\frac{4}{541/168}=\frac{672}{541}\approx1.242. $$ The normalized weights sum to \(c(541/168)=4\), so their sample mean is exactly one by construction. Their weighted mean outcome is unchanged because $$ \frac{\sum_i(cW^S_i)Y_i}{\sum_i cW^S_i} =\frac{c\sum_iW^S_iY_i}{c\sum_iW^S_i} =\frac{\sum_iW^S_iY_i}{\sum_iW^S_i}. $$ This rescaling is weight normalization, not the stabilization that produced \(W^S_i\). Stabilization altered relative weights through treatment-specific numerator probabilities; normalization applies a common multiplier after the weights have been formed.
Several practical checks catch common failures before fitting a marginal structural model:
- Match every probability to the observed decision. For a person with \(A_2=0\), use the model probability of zero, not the probability of one.
- Check numerator distributions. The possible numerator probabilities must sum to one for each value of their conditioning history. In this example, the second-time numerators sum to one separately for \(A_1=1\) and \(A_1=0\).
- Check positivity and scale. A very small denominator probability produces a large reciprocal factor. Such values can dominate summaries and may signal sparse treatment histories or limited overlap.
- Check the product as well as its factors. The cumulative weight must include both decision times. Omitting the second factor does not calculate a two-time-point weight.
- Keep the target in view. Stabilized and unstabilized weighted estimates can differ in a finite sample. A weighted mean is not automatically a causal effect just because weights have been calculated.
A frequent mistake is to expect stabilization to leave every weighted summary unchanged, reasoning that it merely makes the weights numerically smaller. The correction is to inspect the multiplier \(P(A_1)P(A_2\mid A_1)\): it varies across observed treatment histories. Only a common multiplier cancels from a ratio such as a weighted mean. Another mistake is to assume the theoretical mean-weight identity forces a realized sample mean of exactly one; the conditional-expectation proof concerns the population distribution, whereas an observed sample is finite.
Compared with unstabilized weights, stabilized weights retain numerator factors that can moderate the weight scale while preserving treatment-history structure appropriate to the chosen numerator model. Compared with simple weight normalization, they need not preserve relative weights. These calculations do not resolve model misspecification, violations of exchangeability, or poor overlap, and they do not guarantee that the observed weighted sample reproduces a target distribution exactly.
Check Your Understanding
Answer each question, then compare with the short solutions.
- For a person with \(P(A_1=0\mid L_0)=0.25\), \(P(A_1=0)=0.50\), \(P(A_2=1\mid A_1=0,L_0,L_1)=0.40\), and \(P(A_2=1\mid A_1=0)=0.30\), calculate both weights for observed history \((A_1,A_2)=(0,1)\).
- Why can stabilized and unstabilized weighted means differ, even when all their component probabilities are calculated correctly?
- For the four-person data, what is the stabilized sample mean weight? Does its value contradict the theoretical mean-weight identity?
- What probability should be used in the second denominator for a person whose observed second treatment is zero?
- If every stabilized weight is multiplied by the same positive constant, what happens to the weighted mean outcome?
Solutions. 1. \(W^U=(1/0.25)(1/0.40)=10\); \(W^S=(0.50/0.25)(0.30/0.40)=1.5\). 2. The numerator product varies across treatment histories, changing relative weights and potentially the weighted outcome total relative to the total weight. 3. It is \(541/672\approx0.805\). No: the identity is an expectation under the probability model, not a requirement that each finite sample average equal one. 4. Use \(P(A_2=0\mid A_1,L_0,L_1)\), evaluated at that person's history. 5. Nothing; the common constant cancels between the weighted outcome total and the total weight.