Why Three Treatment Arms Require Multinomial Probabilities
When treatment has three possible arms, a subject receives exactly one of them, but the probability model must describe all three possibilities. For example, a study might compare usual care, treatment A, and treatment B. The inverse-probability weight for a subject is based on the probability of the arm actually received—not on a binary-treatment odds and not on the sum of probabilities for the other arms.
This tutorial considers one treatment decision per subject. Let \(A\in\{0,1,2\}\) denote the assigned arm and let \(L\) denote the measured covariates used to model treatment assignment. Write \(e_a(L)=P(A=a\mid L)\) for the conditional probability of arm \(a\). A treatment model estimates the vector \((e_0(L),e_1(L),e_2(L))\), whose entries are nonnegative and sum to one. The un-stabilized inverse-probability treatment weight for a subject with observed arm \(A=a\) is the reciprocal of that arm’s estimated probability.
A standard model for nominal, unordered treatment categories is the multinomial logistic regression model. Taking arm 0 as the reference category, specify two linear predictors, \(\eta_1(L)=\beta_{10}+\beta_1^\mathsf{T}L\) and \(\eta_2(L)=\beta_{20}+\beta_2^\mathsf{T}L\). The resulting probabilities use the softmax transformation. Other treatment-probability estimation methods can also produce the three probabilities; whichever method is used, the weight calculation is the same once the fitted probabilities are available.
The model is estimated from observed pairs \((A_i,L_i)\). For independent observations, its likelihood is the product of the probability assigned to each observed arm; equivalently, its log-likelihood is the sum of the log probabilities for those arms. The model must be fitted using the observed arm labels, and predictions must retain one probability for each arm. Under the usual maximum-likelihood regularity conditions—identifiability, a correctly specified model for consistency, and a finite, well-behaved parameter value—the fitted parameters estimate the model coefficients. The weight itself is still a plug-in estimate, since it uses fitted rather than known probabilities.
From Fitted Probabilities to Individual Weights
The calculation has two distinct steps: first obtain all three fitted probabilities for a subject’s covariate values; then select the probability matching that subject’s observed arm and take its reciprocal. The unused probabilities are not added to the denominator, and the denominator is not the probability of “not receiving” the observed arm.
For each subject, evaluate \(\hat e_0(L_i),\hat e_1(L_i),\hat e_2(L_i)\), and check that the entries sum to one, up to rounding.
If \(A_i=2\), for instance, use \(\hat e_2(L_i)\), even though the model also predicts probabilities for arms 0 and 1.
Compute \(W_i=1/\hat e_{A_i}(L_i)\). A small selected probability produces a large weight and warrants a positivity and model-fit check.
Worked Example: Calculating Weights From a Probability Table
Suppose a fitted model gives the probabilities below for three subjects. The covariate pattern is summarized by \(L\); the table gives the observed arm and the fitted probabilities before the weight is calculated.
| Subject | \(L\) | Observed arm \(A\) | \(\hat e_0(L)\) | \(\hat e_1(L)\) | \(\hat e_2(L)\) |
|---|---|---|---|---|---|
| 1 | 0 | 0 | 0.50 | 0.30 | 0.20 |
| 2 | 1 | 1 | 0.20 | 0.50 | 0.30 |
| 3 | 1 | 2 | 0.20 | 0.50 | 0.30 |
Each row is a valid probability vector: for subject 1, \(0.50+0.30+0.20=1.00\); for subjects 2 and 3, \(0.20+0.50+0.30=1.00\).
Subject 1 received arm 0, so \(W_1=1/0.50=2\). Subject 2 received arm 1, so \(W_2=1/0.50=2\). Subject 3 received arm 2, so \(W_3=1/0.30=10/3\approx3.33\). In particular, subject 3’s weight is not \(1/0.20\): 0.20 is the probability of arm 0, which this subject did not receive.
Intercept-Only Multinomial Model
An intercept-only model is a useful hand calculation. It estimates the same treatment probability for every subject, regardless of covariates. If \(n_a\) subjects receive arm \(a\) among \(n\) subjects, its maximum-likelihood estimate of \(P(A=a)\) is the observed arm proportion \(n_a/n\). This simple case shows how the multinomial likelihood produces all three probabilities together.
Worked Example: Estimating Three Probabilities From Arm Counts
In a sample of \(n=10\), four subjects receive arm 0, three receive arm 1, and three receive arm 2. Under an intercept-only model, the likelihood for the three arm probabilities is proportional to \(e_0^4e_1^3e_2^3\), subject to \(e_0+e_1+e_2=1\).
The maximizing probabilities are \(\hat e_0=4/10=0.4\), \(\hat e_1=3/10=0.3\), and \(\hat e_2=3/10=0.3\). The weights are constant within each arm: a subject receiving arm 0 has weight \(1/0.4=2.5\); one receiving arm 1 has weight \(1/0.3=10/3\approx3.33\); and one receiving arm 2 also has weight \(10/3\approx3.33\). The weighted totals by arm are \(4(2.5)=10\), \(3(10/3)=10\), and \(3(10/3)=10\). This equality follows from the intercept-only estimates; it should not be assumed to hold exactly for every covariate-adjusted model.
To verify the estimate, the log-likelihood, omitting a constant not involving the probabilities, is \(\ell=\sum_{a=0}^2 n_a\log(e_a)\). Maximize it subject to \(\sum_a e_a=1\) using a Lagrange multiplier \(\lambda\). The first-order equation for each arm is \(n_a/e_a=\lambda\), so \(e_a=n_a/\lambda\). Summing these equations over arms gives \(1=\sum_a e_a=\sum_a n_a/\lambda=n/\lambda\), hence \(\lambda=n\) and \(e_a=n_a/n\). Because the log-likelihood is strictly concave in positive \(e_a\) when all \(n_a>0\), this stationary point is the unique maximum. The arm-specific total weight is then \(n_a(n/n_a)=n\). \(\square\)
Why the Weights Rebalance Treatment Assignment
The interpretation of inverse-probability weighting can be stated as an expectation identity. It depends on positivity: for each covariate value under consideration, every arm whose weighted mean is targeted must have positive probability. It also uses the treatment probability conditional on the covariates in the assignment mechanism. With estimated probabilities, the corresponding calculation is a plug-in approximation; an estimated model does not guarantee exact balance in a finite sample.
Condition on \(L\). Since \(h(L)/e_a(L)\) is fixed given \(L\), and \(P(A=a\mid L)=e_a(L)\), its conditional expectation is \(E[I(A=a)h(L)/e_a(L)\mid L]=h(L)E[I(A=a)\mid L]/e_a(L)=h(L)e_a(L)/e_a(L)=h(L)\). Positivity makes the division valid. Taking expectations of both sides and using iterated expectation gives the stated identity. Integrability ensures the expectations are defined. This is a population identity for the true probabilities; replacing \(e_a\) by estimated probabilities does not make it an exact identity for a finite data set. \(\square\)
The identity explains why an arm-specific weighted sample can represent the covariate distribution in the full population: weighting an observed member of arm \(a\) by the reciprocal of that arm’s assignment probability corrects for the probability of entering that arm, conditional on \(L\). It does not, by itself, guarantee causal interpretation of an outcome contrast. That interpretation additionally requires the causal assumptions appropriate to the study, including adequate control of confounding and consistency.
Worked Example: A Covariate Pattern With a Large Weight
Suppose a fitted model predicts \((\hat e_0,\hat e_1,\hat e_2)=(0.70,0.25,0.05)\) for a subject with covariates \(L\). The subject received arm 2. The selected probability is 0.05, so \(W=1/0.05=20\). If the same subject had received arm 0, the weight would instead be \(1/0.70\approx1.43\); if the subject had received arm 1, it would be \(1/0.25=4\).
The weight of 20 is mathematically correct given that fitted probability, but it is a warning that this arm is uncommon for this covariate pattern. Check whether the pattern is represented in the data, whether the fitted probabilities are plausible, and whether a few observations dominate weighted summaries. If the true probability is zero for a relevant covariate pattern, positivity fails and inverse weighting cannot create observations in that arm for that pattern. A very large estimated weight can reflect near-violation of positivity, model misspecification, or ordinary sampling variability; the number alone cannot distinguish these causes.
Diagnostics, Alternatives, and Common Mistakes
At minimum, verify for every subject that all predicted probabilities lie between zero and one and that their sum is one within numerical tolerance. Inspect the distribution of the selected probabilities and weights, and compare them across arms. A probability close to zero leads directly to a large reciprocal. Also check whether predicted arm probabilities are compatible with observed arm frequencies and covariate patterns; good overall calibration does not rule out poor predictions in sparse regions.
The multinomial model is appropriate when the arms are distinct categories without an imposed ordering. A collection of separate binary models is an alternative only if its predictions are reconciled into a valid probability vector; independently fitted probabilities need not sum to one. A different probability-estimation method may be preferable if the multinomial functional form fits poorly, but it must still estimate probabilities for all arms and respect their joint probability constraint. Continuous treatment requires a density-based approach rather than this discrete reciprocal-probability calculation, and is a separate problem.
- Wrong: “The subject received arm 1, so use the probability of arm 0 because it is the reference category.” Correction: The reference category is a parameterization choice, not a weight rule; use \(\hat e_1(L)\) for a subject who received arm 1.
- Wrong: “For a subject in arm 2, use \(1-\hat e_2(L)\), the probability of not receiving arm 2.” Correction: The weight is \(1/\hat e_2(L)\), not the reciprocal of the probability of the complementary event.
- Wrong: “All three probabilities are used in the subject’s denominator.” Correction: They form a probability vector, but the individual weight uses only the entry for the observed arm.
- Wrong: “A finite fitted probability proves positivity.” Correction: A model can assign a small positive value even where the data provide little support. Positivity is a condition on the treatment mechanism in the target population, not merely on numerical model output.
Inverse-probability weighting is therefore not a way to repair absent treatment support. It can also be sensitive to model errors and sampling variation in small probabilities. These limitations motivate careful diagnostics and, where needed, alternative modeling or design decisions; any change to the analysis must be justified rather than treated as an automatic correction.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A subject received arm 1, and the fitted probabilities are \((0.2,0.5,0.3)\). What is the weight?
- In an intercept-only sample of 12 subjects, arm counts are 6, 4, and 2. What are the fitted probabilities and the arm-specific weights?
- For a covariate pattern, the true probability of arm 2 is zero. Can inverse-probability weighting recover the full-population covariate distribution within arm 2?
- State the expectation identity for arm \(a\), and identify the condition that permits its denominator.
Solutions. (1) Use the arm 1 probability: \(W=1/0.5=2\). (2) The probabilities are \(6/12=0.5\), \(4/12=1/3\), and \(2/12=1/6\). Their reciprocal weights are 2, 3, and 6, respectively; the weighted total in each arm is 12. (3) No. The reciprocal of zero is undefined, and that pattern has no arm 2 observations to reweight; positivity fails. (4) For integrable \(h(L)\), \(E[I(A=a)h(L)/e_a(L)]=E[h(L)]\), provided \(e_a(L)>0\) almost surely on the target population where the identity is used.