Stabilization with Multivalued Treatments

Why Multivalued Treatments Need Careful Stabilization

At a treatment decision, a patient may receive one of several categories: for example, no treatment, a standard dose, or an intensive dose. The inverse-probability weight for the observed category uses its probability under a denominator model, while a stabilized weight also includes a numerator probability. With more than two categories, each probability must be attached to the observed category; a probability for “not the observed treatment” is not a substitute.

Across repeated decision times, the observed-category ratios are multiplied. The construction is simple once the histories and conditional probabilities are defined, but a zero-probability category or a mismatched probability can make a calculation invalid. The examples below work through both the category-specific and longitudinal calculations.

Definition and Conditions

Definition: At decision time $t$, let $A_t$ take values in the finite treatment set $\mathcal A=\{0,1,\ldots,K-1\}$. Let $V$ denote baseline covariates, let $\bar A_{t-1}=(A_1,\ldots,A_{t-1})$, and let $\bar L_t$ denote the measured time-varying covariate history available before the decision at time $t$. Write $H_t=(V,\bar L_t,\bar A_{t-1})$ for the denominator history and $R_t=(V,\bar A_{t-1})$ for one common numerator history. Define $g_t(a\mid H_t)=P(A_t=a\mid H_t)$ and $q_t(a\mid R_t)=P(A_t=a\mid R_t)$, where each is a probability distribution over $\mathcal A$. For an individual whose observed treatment at time $t$ is $A_t$, the stabilized treatment weight through time $T$ is $SW_T=\prod_{t=1}^{T}q_t(A_t\mid R_t)/g_t(A_t\mid H_t)$.

This numerator choice preserves dependence on baseline covariates and prior treatment while omitting time-varying covariates from the numerator. Other numerator specifications are possible, but the probabilities must correspond to the numerator model actually chosen. The denominator conditions on the covariate and treatment history used to model treatment assignment. For a causal interpretation, additional assumptions such as consistency and sequential exchangeability are needed; the calculations here concern construction of the weights, not proof of those assumptions.

Both $g_t(\cdot\mid H_t)$ and $q_t(\cdot\mid R_t)$ must sum to one over the treatment categories. The support condition for the ratio is $q_t(a\mid R_t)>0$ implies $g_t(a\mid H_t)>0$, for histories under consideration. In words, the numerator must not assign positive probability to a category that has zero probability under the denominator. If $g_t(a\mid H_t)=0$, the observed category cannot equal $a$ at that history with positive conditional probability. When both probabilities are zero, the ratio can be assigned the value zero by convention; it is not a possible observed-category contribution at that history.

Calculating Category-Specific Ratios

Worked Example: One Decision With Three Treatment Categories

Suppose the categories are $0=$ none, $1=$ standard, and $2=$ intensive. For one history, let the denominator probabilities be $g=(0.50,0.30,0.20)$ and the numerator probabilities be $q=(0.60,0.25,0.15)$. An individual receives category $2$. The relevant probabilities are $g(2)=0.20$ and $q(2)=0.15$, so the stabilized component is $0.15/0.20=0.75$.

The other category-specific ratios at this history would be $0.60/0.50=1.20$ for category $0$ and $0.25/0.30=0.8333\ldots$ for category $1$. These are alternative possible observed-category contributions, not three factors to multiply together. As a check, their denominator-weighted average is $0.50(1.20)+0.30(0.8333\ldots)+0.20(0.75)=0.60+0.25+0.15=1$.

That check follows from a general identity. It is important to sum only over categories with positive denominator probability: writing a term $g_t(a\mid H_t)\,q_t(a\mid R_t)/g_t(a\mid H_t)$ for a category with $g_t=0$ would create an undefined $0/0$ expression if $q_t=0$ as well.

Result: Suppose $\mathcal A$ is finite, $g_t(\cdot\mid H_t)$ and $q_t(\cdot\mid R_t)$ are probability distributions, and $q_t(a\mid R_t)>0$ implies $g_t(a\mid H_t)>0$. Then, conditional on $H_t$, $$ E\left[\left.\frac{q_t(A_t\mid R_t)}{g_t(A_t\mid H_t)}\right|H_t\right]=1, $$ where the ratio is defined as zero for categories with $q_t=g_t=0$.

Proof. Conditional on $H_t$, the probability that $A_t=a$ is $g_t(a\mid H_t)$. Expand the conditional expectation over only categories with positive denominator probability: $$ \sum_{a:g_t(a\mid H_t)>0}g_t(a\mid H_t)\frac{q_t(a\mid R_t)}{g_t(a\mid H_t)} =\sum_{a:g_t(a\mid H_t)>0}q_t(a\mid R_t). $$ For every omitted category, $g_t(a\mid H_t)=0$. The support condition then implies $q_t(a\mid R_t)=0$, so the displayed sum includes all the numerator probability mass. It therefore equals $\sum_{a\in\mathcal A}q_t(a\mid R_t)=1$. This proves the identity without evaluating a ratio on a zero-probability category. $\square$

Products Across Decision Times

At each time, use the probabilities for the individual’s observed category, evaluated at that time’s histories. The cumulative weight is the product of these component ratios; a probability from one time or category cannot be reused at another time unless the model gives it that value.

Worked Example: Two Times and Three Categories

Suppose the observed sequence is $A_1=2$, followed by $A_2=1$. At time 1, the denominator probabilities for categories $0,1,2$ are $(0.50,0.30,0.20)$, and the numerator probabilities are $(0.60,0.25,0.15)$. Thus the observed-category component is $q_1(2)/g_1(2)=0.15/0.20=0.75$.

At time 2, conditional on this individual’s relevant histories, suppose the denominator probabilities are $(0.20,0.50,0.30)$, and the numerator probabilities are $(0.40,0.35,0.25)$. Since the observed category is $1$, the second component is $q_2(1)/g_2(1)=0.35/0.50=0.70$. The cumulative stabilized weight is therefore $$ SW_2=0.75\times0.70=0.525. $$

Both probability vectors sum to one: $0.50+0.30+0.20=1$, $0.60+0.25+0.15=1$, $0.20+0.50+0.30=1$, and $0.40+0.35+0.25=1$. Each observed-category denominator probability is positive, and each numerator probability is no larger than one. The product $0.525$ is the weight for this treatment path, not a probability that the path occurred.

The conditional mean-one property also extends to the product when the histories are properly nested. Specifically, suppose the past weights are measurable given $H_t$, and the stated support and probability-distribution conditions hold at every time. Then $E[SW_T]=1$. This is a population identity, not a claim that the average of estimated weights in a finite sample must equal exactly one.

Derivation. Write $W_t=\prod_{s=1}^{t}q_s(A_s\mid R_s)/g_s(A_s\mid H_s)$, with $W_0=1$. Since $W_{t-1}$ is determined by the history available at time $t$, it is measurable given $H_t$. Applying the conditional identity just proved gives $$ E[W_t]=E\left[E\left[\left.W_{t-1}\frac{q_t(A_t\mid R_t)}{g_t(A_t\mid H_t)}\right|H_t\right]\right] =E[W_{t-1}\cdot1]=E[W_{t-1}]. $$ Repeating this equality for $t=T,T-1,\ldots,1$ yields $E[W_T]=E[W_0]=1$. The argument assumes the relevant expectations exist; for nonnegative weights, the conditional calculation establishes the equality when the stated distributions and support conditions hold. $\square$

Changing the Numerator Changes the Target Distribution

Worked Example: Two Valid Numerators Give Different Weights

At a particular history, suppose denominator probabilities for categories $0,1,2$ are $(0.40,0.40,0.20)$. Consider two possible numerator distributions: $q^{(1)}=(0.50,0.30,0.20)$ and $q^{(2)}=(0.25,0.50,0.25)$. Both sum to one and both satisfy the support condition because all denominator probabilities are positive.

For an individual receiving category $1$, the first numerator gives the ratio $0.30/0.40=0.75$; the second gives $0.50/0.40=1.25$. The difference is not an arithmetic error: the numerator models specify different treatment distributions to which the weighting construction relates the data. Select the numerator deliberately and use it consistently across the corresponding calculation.

The denominator-weighted ratios still average to one under either choice. For $q^{(1)}$, the check is $0.40(0.50/0.40)+0.40(0.30/0.40)+0.20(0.20/0.20)=0.50+0.30+0.20=1$. For $q^{(2)}$, it is $0.40(0.25/0.40)+0.40(0.50/0.40)+0.20(0.25/0.20)=0.25+0.50+0.25=1$.

Diagnostics, Comparisons, and Limitations

Before using a multivalued stabilized weight, check the probability calculations themselves. For each time and model, verify that probabilities across all categories sum to one, that the numerator and denominator refer to the same category, and that the denominator probability for each observed category is positive. A small observed-category denominator probability produces a large component ratio when the numerator probability is appreciable. Across time, several moderately large ratios can multiply into a very large cumulative weight.

Check category alignment. Confirm that the ordering of categories in the probability output matches the treatment coding. A swapped category can yield plausible-looking but incorrect ratios.
Check histories and timing. The denominator probability must use covariates observed before the treatment decision. Using a post-decision variable changes the model’s conditioning set.
Check support and sparsity. Tabulate observed categories by relevant histories. A category that is absent or nearly absent in some histories may signal a positivity problem or an unstable fitted probability.
Check components and products. Inspect component ratios as well as cumulative weights by time. A reasonable first component does not ensure a well-behaved product.

The unstabilized treatment weight uses $1/g_t(A_t\mid H_t)$ at each time; the stabilized construction multiplies by the chosen numerator probability as well. Multivalued treatment does not call for a separate “one versus all others” denominator when the observed treatment category is modeled with a categorical probability distribution. Nor does stabilization guarantee that weights are bounded, that fitted probabilities are accurate, or that a finite sample has mean weight exactly one. Extreme weights can remain, especially with sparse categories, near-violations of positivity, or long treatment histories. Weight truncation and its consequences require separate decisions and do not repair an incorrectly specified probability model.

Common Mistake

Pitfall: Treating the treatment as a collection of binary indicators and using, for example, the probability of “category 2 versus not category 2” when the observed treatment is category 1. That is the wrong probability for the observed category. Instead, model the treatment distribution over all categories and use $q_t(A_t\mid R_t)/g_t(A_t\mid H_t)$ for the category actually received. Also, never simplify a zero-probability term as $0\times(0/0)=0$; the ratio is undefined. Sum only over categories with $g_t>0$, or explicitly define the ratio to be zero when both probabilities are zero.

Check Your Understanding

Answer each question, then compare with the solutions.

At one decision, $g=(0.60,0.25,0.15)$ and $q=(0.50,0.30,0.20)$. What is the stabilized component for an individual receiving category 1?
If $g(a)=0$ and $q(a)>0$ for some category, which condition fails, and why is the ratio unusable?
For two observed treatment categories, the component ratios are $0.8$ and $1.5$. What is the cumulative weight?
If $g=(0.5,0.5,0)$ and $q=(0.4,0.4,0.2)$, is the support condition satisfied? Explain.
Why need not the sample average of estimated stabilized weights equal one exactly?

Solutions. (1) The observed category is 1, so the component is $0.30/0.25=1.2$. (2) The support condition fails because the numerator assigns positive mass to a category the denominator says cannot occur; $q(a)/g(a)$ divides by zero. (3) Multiply the components: $0.8\times1.5=1.2$. (4) No. The third category has $q(2)=0.2>0$ but $g(2)=0$. (5) The mean-one statement is a population expectation under the specified probabilities; estimated probabilities, finite sampling variation, and model estimation mean a sample average need not equal that expectation exactly.