Sequential Factorization of Response Indicators

Why Factor the Response Indicators?

In a longitudinal study, the response pattern records which outcomes were observed at each visit. A pattern such as $ (1,1,0) $ means the first two outcomes were observed and the third was not. The probability of a complete pattern can be expressed as a sequence of conditional probabilities, one for each visit. This sequential expression makes clear what information each response probability is conditioned on and allows response behavior at later visits to depend on earlier response indicators.

The factorization is an application of the probability chain rule. It is an identity for a joint distribution, not by itself a model for why people miss visits. In particular, it does not require response indicators to be independent, and it does not assert that response at a visit is independent of unobserved outcomes.

Definition: For an individual with $T$ scheduled visits, let $Y=(Y_1,\ldots,Y_T)$ denote the outcome vector, $R=(R_1,\ldots,R_T)$ the response-indicator vector, and $X$ the baseline covariates. $R_t=1$ indicates that $Y_t$ is observed, and $R_t=0$ indicates that it is missing. For each visit $t$, define the visit-specific conditional response probability $$ q_t(r_{1:t-1};y,x) = P(R_t=1\mid R_{1:t-1}=r_{1:t-1},Y=y,X=x). $$ For $t=1$, there are no earlier response indicators, so $q_1(y,x)=P(R_1=1\mid Y=y,X=x)$. The probability of a response pattern is obtained by multiplying the corresponding conditional probabilities for its observed and missing indicators.

The conditioning on $Y=y$ in this definition is on the full outcome vector, including outcomes that may be missing for the response pattern under consideration. That makes the expression a mathematical description of the joint law. It does not mean the full outcome vector is available to an analyst for every individual. Unless additional assumptions or a model are imposed, these conditional probabilities need not be directly estimable from observed data alone.

The Sequential Chain-Rule Factorization

Write $r=(r_1,\ldots,r_T)$ for a particular response pattern. At each visit, the factor for that pattern is the conditional probability of the indicator value actually recorded: use $q_t$ if $r_t=1$, and $1-q_t$ if $r_t=0$. Thus the full pattern probability is

P(R=r\mid Y=y,X=x) = \prod_{t=1}^{T} q_t(r_{1:t-1};y,x)^{r_t} \left\{1-q_t(r_{1:t-1};y,x)\right\}^{1-r_t}.

In each factor, the earlier-indicator history is the history specified by the same pattern $r$. For example, the factor at visit 3 for pattern $ (1,0,1) $ conditions on $R_1=1,R_2=0$, not on a generic or averaged response history.

Key result: The sequential factorization holds for any joint distribution of the response indicators conditional on $Y$ and $X$, provided the conditional probabilities are defined for histories with positive probability. It does not require independence across visits, monotone dropout, or any particular missingness mechanism.

The reason is the chain rule. For a particular pattern, the probability of the first indicator is multiplied by the probability of the second indicator given the first, then by the probability of the third given the first two, and so on. In full notation,

\begin{aligned} P(R_1=r_1,\ldots,R_T=r_T\mid Y=y,X=x) &=P(R_1=r_1\mid y,x)\\ &\quad{}\times P(R_2=r_2\mid R_1=r_1,y,x)\\ &\quad{}\times\cdots\\ &\quad{}\times P(R_T=r_T\mid R_{1:T-1}=r_{1:T-1},y,x). \end{aligned}

This follows by repeatedly applying $P(A,B\mid C)=P(A\mid C)P(B\mid A,C)$. Applying that identity first to $R_1$ and the remaining indicators, then to $R_2$ and those after it, yields the displayed product. No factor is dropped: every indicator in the pattern appears exactly once, conditioned on all earlier indicators and on the same $Y,X$.

Worked Example: A Three-Visit Pattern

Suppose that, at fixed outcome vector $y$ and baseline covariates $x$, the conditional probabilities relevant to pattern $ (1,1,0) $ are $P(R_1=1\mid y,x)=0.80$, $P(R_2=1\mid R_1=1,y,x)=0.75$, and $P(R_3=1\mid R_1=1,R_2=1,y,x)=0.30$. Find the probability of this response pattern conditional on $y,x$.

The first two indicators equal 1, while the third equals 0. The third-visit factor is therefore the probability of nonresponse, $1-0.30=0.70$, conditional on the two earlier responses. The chain rule gives

\begin{aligned} P(R=(1,1,0)\mid y,x) &=0.80\times0.75\times(1-0.30)\\ &=0.80\times0.75\times0.70\\ &=0.420. \end{aligned}

Thus the conditional probability of this particular response pattern is $0.420$, or 42.0%, at the stated $y,x$. The calculation does not say that the three indicators are independent. In fact, the second and third conditional probabilities explicitly refer to earlier indicators.

Reading and Using the Factors

Each factor describes a probability at a specific point in the sequence, given a particular earlier response history. In the unrestricted factorization, it may also depend on any component of the full outcome vector and on baseline covariates. Therefore, the sequential form does not require that the probability at visit $t$ depend only on outcomes observed before visit $t$. Such restrictions would be additional assumptions about the response mechanism, not consequences of the chain rule.

For example, sequential MAR places restrictions on which outcomes may affect response probabilities after conditioning on the relevant observed history. The chain rule itself does not impose those restrictions. A factor $q_t$ may depend on an outcome that is unobserved for a given person; including that dependence is part of the unrestricted conditional representation, not evidence that the dependence can be learned from the observed records without further assumptions.

Worked Example: Two Visits and All Four Patterns

At fixed $y,x$, suppose $P(R_1=1\mid y,x)=0.70$, $P(R_2=1\mid R_1=1,y,x)=0.80$, and $P(R_2=1\mid R_1=0,y,x)=0.25$. Calculate all four response-pattern probabilities and verify that they sum to one.

For pattern $ (1,1) $, both responses occur:

P(1,1\mid y,x)=0.70(0.80)=0.560.

For pattern $ (1,0) $, the second visit is missing after a response at the first:

P(1,0\mid y,x)=0.70(1-0.80)=0.70(0.20)=0.140.

For pattern $ (0,1) $, the second-visit response probability is conditioned on $R_1=0$:

P(0,1\mid y,x)=(1-0.70)(0.25)=0.30(0.25)=0.075.

For pattern $ (0,0) $, the second visit is missing given that the first was missing:

P(0,0\mid y,x)=(1-0.70)(1-0.25)=0.30(0.75)=0.225.

The four probabilities add to $0.560+0.140+0.075+0.225=1.000$, as required for the exhaustive two-visit patterns. Notice that $P(R_2=1\mid R_1=1,y,x)=0.80$ differs from $P(R_2=1\mid R_1=0,y,x)=0.25$. The response indicators are not independent in this example, but the chain-rule factorization still holds.

Worked Example: Outcome and Covariate Dependence in a Factor

Consider a two-visit setting with baseline treatment indicator $X$ and a binary outcome $Y_1$. For illustration, suppose a model assigns $P(R_1=1\mid Y_1=1,X=1)=0.60$ and $P(R_1=1\mid Y_1=0,X=1)=0.90$. Suppose also that for the specified history $R_1=1$, the second-visit response probability is $P(R_2=1\mid R_1=1,Y,X)=0.70$. Calculate the probabilities of complete response, $R=(1,1)$, for the two outcome values when $X=1$.

When $Y_1=1$, the chain-rule factors give

P(R=(1,1)\mid Y_1=1,X=1) =0.60(0.70)=0.420.

When $Y_1=0$, the first factor changes, while the stated second factor remains $0.70$:

P(R=(1,1)\mid Y_1=0,X=1) =0.90(0.70)=0.630.

These are hypothetical conditional probabilities, not estimates from a real study. They illustrate that outcome and baseline-covariate dependence can be represented in the factors. If $Y_1$ is missing for an individual, its value cannot simply be inserted into an observed-data calculation; how missing outcomes are handled requires additional modeling or assumptions.

Assumptions, Diagnostics, and Common Pitfalls

The chain-rule identity needs no independence assumption. Its technical requirement is that each conditional probability be defined for the history being considered. If an earlier response history has probability zero at particular $y,x$, then a conditional probability given that history is undefined there; the joint probability of any pattern containing that impossible history is zero. Values assigned to conditional factors on zero-probability histories do not change the joint law.

In practical work, the factorization is often paired with models for the visit-specific probabilities. Those models introduce assumptions beyond the identity, such as a chosen functional form or a restriction on which covariates and outcomes enter a factor. Diagnostics then concern the adequacy and support of those modeled conditional probabilities, not whether the chain rule itself is true.

Check time alignment. The factor for visit $t$ must condition on indicators from visits $1$ through $t-1$, not on later indicators.
Check the conditioning history. A probability for $R_t=1$ after a particular history must use that history. Pooling histories changes the conditional probability unless the relevant probabilities happen to be equal.
Check support and sparse histories. Some earlier-indicator histories may be rare or absent. Conditional response estimates for sparse histories can be unstable, and estimates for absent histories cannot be learned directly from those data.
Check probability calibration. When conditional factors are estimated, compare predicted response proportions with observed proportions across visits and histories, where those comparisons are supported by the data. Large discrepancies can indicate a poor model or omitted predictors.
Separate algebra from identification. The chain rule expresses a joint distribution; it does not establish that every factor involving unobserved outcomes is identifiable from observed records.

A common mistake is to read the product as evidence that the response indicators are independent. That conclusion would require each conditional factor to be unchanged by the earlier response history (and other relevant conditioning variables). The correct reasoning is that the chain rule always conditions each later indicator on earlier ones; dependence is allowed and is represented in those conditional probabilities.

How This Factorization Differs from Alternatives

The Selection Factorization expresses the joint law by placing the outcome distribution first and then describing response conditional on outcomes and covariates. The sequential response factorization expands that conditional response distribution into visit-specific factors. It therefore gives a more detailed account of how response probabilities vary with earlier response history, while remaining a representation of the same conditional response law.

A joint law for response patterns alone, such as $P(R=r\mid X=x)$, instead averages over outcomes. It does not display how response depends on the full outcome vector. These representations answer different descriptive questions: the sequential factorization here makes conditioning on $Y$, $X$, and earlier indicators explicit. The pattern-mixture approach uses a different ordering of outcome and response-pattern components; the chain-rule identity does not select one approach as universally preferable.

The main limitation is practical as well as conceptual: an unrestricted factor may condition on outcomes that are missing, so the displayed conditional probabilities are not automatically estimable from observed data. Parsimonious models, restrictions such as MAR, or other assumptions may be used in applications, but each changes what is being assumed. The factors should not be interpreted as observed conditional rates unless the conditioning variables are in fact observed for the records used to calculate those rates.

Key takeaway: For any fixed outcome vector and baseline covariates, the probability of a response pattern is the product of visit-specific conditional probabilities for the indicator values in that pattern, given the full earlier indicator history. This is a chain-rule identity; assumptions about dependence, MAR, model form, and estimability are separate.

Check Your Understanding

Try each question, then compare your reasoning with the solutions below.

For three visits, write the sequential factorization of $P(R=(0,1,0)\mid Y=y,X=x)$.
Suppose $P(R_1=1\mid y,x)=0.60$, $P(R_2=1\mid R_1=0,y,x)=0.40$, and $P(R_3=1\mid R_1=0,R_2=1,y,x)=0.50$. Calculate $P(R=(0,1,0)\mid y,x)$.
Does the sequential factorization imply that $R_1$ and $R_2$ are independent? Explain.
Why does a factor conditioned on the full outcome vector not automatically provide an estimable probability for every observed record?

Solutions. (1) The pattern has $R_1=0,R_2=1,R_3=0$, so its probability is $P(R_1=0\mid y,x)P(R_2=1\mid R_1=0,y,x)P(R_3=0\mid R_1=0,R_2=1,y,x)$. (2) Substitute the values, using the complement probabilities for the first and third indicators: $ (1-0.60)(0.40)(1-0.50)=0.40(0.40)(0.50)=0.0800$. (3) No. The chain rule holds whether or not the indicators are independent; dependence may appear through different conditional probabilities for different earlier histories. (4) Some components of $Y$ may be missing, so their values are unavailable for conditioning in that record. The factorization is a joint-law identity, not a guarantee that its full-outcome conditional probabilities can be calculated or identified from observed data alone.