Why Factor a Joint Treatment Decision?
A treatment decision can contain several components assigned at the same visit: for example, a medication choice and a dose category, or two distinct interventions. Write the treatment vector as \(A=(A_1,\ldots,A_K)\), where each component takes values in a finite set. A joint treatment model assigns probabilities directly to the possible vectors. A sequential factorization instead models one component at a time, conditioning each component on the preceding components and the covariates available before treatment.
This factorization can make a complicated joint distribution easier to represent and estimate. It does not, by itself, assert that the components were assigned at different times or that one component causally affects another. The order is a way to factor a joint probability. If these factors are used to construct inverse probability weights, the resulting probabilities must still describe the intended treatment distribution and respect its support.
The same definition applies to a target or numerator distribution \(q(a\mid V)\), where \(V\) denotes the variables retained in that distribution. Its factors condition on \(V\) and on the preceding components in the chosen order. The values \(H\) and \(V\) need not be the same.
The Chain-Rule Factorization
For discrete treatment components, the probability chain rule gives an exact factorization. It requires no independence assumption. For a particular vector \(a=(a_1,\ldots,a_K)\), each conditional probability is defined along the path only when the conditioning history has positive probability under the distribution being factorized.
Proof. For \(j=1,\ldots,K\), let \(p_j=P(A_1=a_1,\ldots,A_j=a_j\mid H=h)\), with \(p_0=1\). Since \(f(a\mid h)=p_K>0\), every prefix probability \(p_j\) is positive: if any prefix had probability zero, its subset event \(A=a\) would also have probability zero. Thus the conditional probability for component \(j\) along this path is defined and equals \(p_j/p_{j-1}\). Multiplying these ratios gives $$ \prod_{j=1}^{K}\frac{p_j}{p_{j-1}}=\frac{p_K}{p_0}=p_K=f(a\mid h). $$ This proves the factorization. \(\square\)
The positive-probability qualification matters. If a prefix has probability zero, a conditional probability after that prefix is not determined by the joint distribution. In particular, it is not valid to use a product of conditional factors on every conceivable treatment vector without checking that the factors are defined.
Worked Example: Recovering a Two-Component Joint Distribution
Suppose \(A_1\) and \(A_2\) are binary, and at a fixed covariate value \(h\) the joint model gives the following probabilities:
| \((A_1,A_2)\) | Probability |
|---|---|
| (0,0) | 0.40 |
| (0,1) | 0.10 |
| (1,0) | 0.20 |
| (1,1) | 0.30 |
First, the marginal probability for the first component is \(P(A_1=1\mid h)=0.20+0.30=0.50\), and \(P(A_1=0\mid h)=0.50\). Given \(A_1=0\), the probability that \(A_2=1\) is \(0.10/0.50=0.20\), so \(P(A_2=0\mid A_1=0,h)=0.80\). Given \(A_1=1\), \(P(A_2=1\mid A_1=1,h)=0.30/0.50=0.60\), and \(P(A_2=0\mid A_1=1,h)=0.20/0.50=0.40\).
For the vector \((1,1)\), the sequential product is \(0.50\times0.60=0.30\), the given joint probability. For \((0,1)\), it is \(0.50\times0.20=0.10\). The remaining products are \(0.50\times0.80=0.40\) for \((0,0)\) and \(0.50\times0.40=0.20\) for \((1,0)\). All four probabilities are recovered.
If one instead multiplied the marginal probabilities of \(A_1\) and \(A_2\), one would be assuming conditional independence given \(h\). Here \(P(A_2=1\mid h)=0.10+0.30=0.40\), so that independence model would assign probability \(0.50\times0.40=0.20\) to \((1,1)\), not the correct 0.30.
Factoring a Treatment-Weight Ratio
Suppose the treatment weight compares a target distribution \(q(a\mid V)\) with the treatment distribution \(f(a\mid H)\). Applying the chain rule to each distribution gives a product of component-specific probability ratios—but only on vectors where the required conditional factors exist.
Proof. Because \(q(a\mid v)>0\), every prefix of \(a\) has positive probability under \(q(\cdot\mid v)\). The chain-rule theorem therefore gives \(q(a\mid v)=\prod_{j=1}^{K}q_j(a_j\mid a_1,\ldots,a_{j-1},v)\), with all these factors defined. The same argument using \(f(a\mid h)>0\) gives the corresponding factorization of \(f\). Since the denominator joint probability is positive, division is valid. Dividing the two products term by term yields the displayed identity. \(\square\)
If \(q(a\mid v)=0\), the joint ratio is zero when \(f(a\mid h)>0\), but a numerator conditional factor after a zero-probability prefix may be undefined. The product identity is therefore not asserted for that vector. This distinction is important even when the target distribution assigns zero probability to vectors outside its intended support.
Worked Example: Computing a Weight From Sequential Factors
Let \(A=(A_1,A_2,A_3)\) be binary and consider the vector \(a=(1,0,1)\). Under the denominator treatment model, suppose \(P(A_1=1\mid H)=0.40\), \(P(A_2=0\mid A_1=1,H)=0.25\), and \(P(A_3=1\mid A_1=1,A_2=0,H)=0.60\). The joint denominator probability is \(f(1,0,1\mid H)=0.40\times0.25\times0.60=0.06\).
Suppose the target distribution has sequential factors \(q_1(1\mid V)=0.50\), \(q_2(0\mid A_1=1,V)=0.25\), and \(q_3(1\mid A_1=1,A_2=0,V)=0.50\). Then \(q(1,0,1\mid V)=0.50\times0.25\times0.50=0.0625\). The weight ratio is \(0.0625/0.06=25/24\), approximately \(1.0417\).
The factor ratios give the same result: \((0.50/0.40)\times(0.25/0.25)\times(0.50/0.60) =1.25\times1\times(5/6)=25/24\). Each conditional factor is evaluated at the preceding components of the observed vector. A factor for a different history would describe a different path and cannot be substituted.
Three Components and the Meaning of the Order
For \(K=3\), the chain rule reads \(f(a_1,a_2,a_3\mid H)=f_1(a_1\mid H)f_2(a_2\mid a_1,H)f_3(a_3\mid a_1,a_2,H)\). Changing the component order changes which conditional distributions appear, but not the underlying joint distribution when the factors are exact. For example, one could factor in the order \(A_2,A_1,A_3\) instead. The fitted models may nevertheless differ if their functional forms or covariate terms are restricted.
Worked Example: Checking a Three-Component Factorization
Let \(A_1,A_2,A_3\) be binary. Suppose \(P(A_1=1\mid H)=0.40\), \(P(A_2=1\mid A_1=1,H)=0.75\), and \(P(A_2=1\mid A_1=0,H)=0.20\). For \(A_3\), let the probabilities of 1 given \((A_1,A_2)=(0,0),(0,1),(1,0),(1,1)\) be \(0.10,0.30,0.60,0.80\), respectively.
The probability of prefix \((0,0)\) is \(0.60\times0.80=0.48\). Thus \(P(0,0,0\mid H)=0.48\times0.90=0.432\), and \(P(0,0,1\mid H)=0.48\times0.10=0.048\). For prefix \((0,1)\), the probability is \(0.60\times0.20=0.12\); the two full-vector probabilities are \(0.12\times0.70=0.084\) for \((0,1,0)\) and \(0.12\times0.30=0.036\) for \((0,1,1)\).
For prefix \((1,0)\), the probability is \(0.40\times0.25=0.10\), giving \(P(1,0,0\mid H)=0.10\times0.40=0.040\) and \(P(1,0,1\mid H)=0.10\times0.60=0.060\). For prefix \((1,1)\), the probability is \(0.40\times0.75=0.30\), giving \(P(1,1,0\mid H)=0.30\times0.20=0.060\) and \(P(1,1,1\mid H)=0.30\times0.80=0.240\).
The eight probabilities sum to \(0.432+0.048+0.084+0.036+0.040+0.060+0.060+0.240=1.000\). This checks that the factorization defines a normalized joint distribution and illustrates how conditional models encode dependence among components.
Sequential Models Compared With a Joint Model
A single joint model treats each possible treatment vector as a category. It can represent dependence directly and, if sufficiently flexible, estimate a probability for every vector within covariate strata. A sequential model instead estimates the first-component distribution and then conditional distributions for later components. Both approaches can represent the same joint distribution when their component probabilities are correctly specified.
| Feature | Sequential conditional models | Single joint model |
|---|---|---|
| Probability representation | Product of conditional probabilities in a chosen order | Probability assigned directly to each treatment vector |
| Dependence | Represented by conditioning later components on earlier ones | Represented directly among the joint categories |
| Potential advantage | Can use simpler component-specific models and share structure across histories | Directly targets the full joint distribution without selecting a factorization order |
| Potential difficulty | Later conditional models can become sparse as the number of preceding histories grows | The number of treatment categories can grow rapidly with the number of components |
The equality of the two representations is a probability identity, not a guarantee that two fitted procedures will agree in a finite sample. If the sequential factors are restricted—for example, if a later component model omits an important interaction with earlier components—the product can be misspecified. Conversely, a joint model with too few parameters may impose inappropriate restrictions on the vector probabilities. Compare models by examining the implied joint probabilities, not just the fit of each component model in isolation.
Diagnostics, Assumptions, and Common Pitfalls
The component distributions must be defined for the histories used in estimation, and their predicted probabilities must be valid: each conditional distribution is nonnegative and sums to one over the possible values of its component. For weighting, the observed treatment vectors must also have adequate probability under the denominator model wherever the target distribution assigns positive probability. If that support condition fails, the target-to-treatment ratio is not finite for those vectors.
- Check whether observed combinations are rare or absent, especially among histories defined by several preceding components. Sparse histories can produce unstable conditional estimates.
- Inspect calibration and predicted probabilities for each component conditional on its preceding treatment history and covariates. A good marginal fit for each component alone does not establish a good joint fit.
- Multiply fitted factors for relevant vectors and compare the implied joint probabilities with observed joint frequencies or with a suitable joint-model fit. Check that the implied probabilities sum to one.
- Examine the resulting weights for extreme values. Very small denominator probabilities can create large ratios, even when each individual component probability appears plausible.
Common mistake: multiplying marginal component probabilities. One might reason that a vector probability is \(P(A_1=a_1\mid H)P(A_2=a_2\mid H)\), then continue similarly for more components. That product is correct only under conditional independence of all components given \(H\). The chain rule instead conditions each later component on the preceding ones: \(P(A_2=a_2\mid A_1=a_1,H)\), and so on. The two-component worked example showed that omitting this conditioning changed the probability of \((1,1)\) from 0.30 to 0.20.
Another pitfall is to interpret the factor order as the actual chronology of treatment assignment. The chain rule permits any ordering of the components; a different order is not a different causal story. The order can be chosen for modeling convenience, but model restrictions can make the resulting estimates order-sensitive. Finally, factorization alone does not establish the causal assumptions needed for a marginal structural model. It only describes how to represent the treatment distribution used in the weight.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- For a three-component treatment, write the chain-rule factorization in the order \(A_1,A_2,A_3\). Which factors condition on preceding components?
- Suppose \(P(A_1=1\mid H)=0.30\) and \(P(A_2=1\mid A_1=1,H)=0.40\). What is \(P(A_1=1,A_2=1\mid H)\)?
- Why is a product of marginal component probabilities generally not the chain-rule factorization?
- State the support condition under which the sequential factorization of \(q(a\mid V)/f(a\mid H)\) is defined along the vector \(a\).
- Does changing the factorization order change the true joint probability? Why might fitted estimates still change?
Solutions. (1) The factorization is \(f_1(a_1\mid H)f_2(a_2\mid a_1,H)f_3(a_3\mid a_1,a_2,H)\); the second and third factors condition on the preceding components. (2) The joint probability is \(0.30\times0.40=0.12\). (3) Multiplying marginal probabilities omits dependence between components and is valid only under the required conditional independence. (4) Require \(q(a\mid V)>0\) and \(f(a\mid H)>0\), so every prefix along the vector has positive probability under both distributions and all factors are defined. (5) The exact chain rule gives the same joint probability in any order; restricted or misspecified conditional models can yield different fitted joint probabilities under different orders.