Factorizing a Latent Class Joint Model

Why Factorize a Joint Model?

A latent class joint model represents unobserved subpopulations whose longitudinal outcome patterns and event-time distributions may differ. Its central structural claim is that class membership accounts for the association between a subject’s repeated outcomes and event time: once the subject’s class is specified, the longitudinal record and event time are independent. That claim makes the model more structured than an unrestricted joint distribution, but it is an assumption to assess, not a consequence of using latent classes.

The earlier material on latent classes as unobserved trajectory subpopulations motivates treating class membership as unobserved and subject-specific. Here the focus is how that membership enters the joint distribution. The factorization is useful only if its assumptions are scientifically plausible and the component distributions can, in principle, be distinguished from the available information.

Definition and Assumptions

Definition: For subject $i$, let $Z_i$ denote a latent class taking one of $K$ values, $Y_i=(Y_{i1},\ldots,Y_{in_i})$ the vector of repeated longitudinal outcomes, and $T_i$ the event time. Write $\pi_k=P(Z_i=k)$, $f_k(y)$ for the conditional joint distribution of the full outcome vector given $Z_i=k$, and $h_k(t)$ for the conditional event-time distribution given $Z_i=k$. A latent class joint model with conditional independence assumes $Y_i\perp T_i\mid Z_i$, and therefore $$ P(Z_i=k, Y_i\in dy, T_i\in dt)=\pi_k f_k(y)h_k(t)\,dy\,dt. $$ For discrete variables, the corresponding probability mass functions replace the densities.

The probabilities satisfy $\pi_k\geq 0$ and $\sum_{k=1}^{K}\pi_k=1$. The classes are mutually exclusive and exhaustive: each subject belongs to one class, even though that class is not observed. The factorization applies to the whole repeated-outcome vector. It does not require the components $Y_{i1},\ldots,Y_{in_i}$ to be independent of one another within a class. Their dependence over time can be represented inside $f_k(y)$.

The model also requires subjects to be independent sampling units, or requires the sampling dependence to be handled separately. The event time $T_i$ is the underlying time to the event, not necessarily the time recorded in a censored data set. If censoring is present, assumptions about the censoring process are needed to connect observed follow-up to the event-time distribution. Conditional independence of $Y_i$ and $T_i$ alone does not specify those assumptions.

The key factorization is conditional, not marginal. Summing over the unobserved class gives the population joint distribution:

p(y,t)=\sum_{k=1}^{K}\pi_k f_k(y)h_k(t).

This is generally not the product of the marginal distributions of $Y_i$ and $T_i$. Class membership can induce marginal association because the same latent class affects both distributions. Thus finding an association between longitudinal outcomes and event time does not by itself contradict conditional independence within classes.

What the Factorization Implies

A short calculation makes the distinction between conditional and marginal independence precise. Let $A$ be an indicator for a specified longitudinal outcome pattern and $B$ an indicator for an event by a specified time. In a two-class model, suppose $A$ and $B$ are independent within each class. Write $a_k=P(A=1\mid Z=k)$, $b_k=P(B=1\mid Z=k)$, and $\pi=P(Z=1)$. Then

\begin{aligned} \operatorname{Cov}(A,B) &=P(A=1,B=1)-P(A=1)P(B=1)\\ &=\pi a_1b_1+(1-\pi)a_2b_2\\ &\quad-\{\pi a_1+(1-\pi)a_2\}\{\pi b_1+(1-\pi)b_2\}\\ &=\pi(1-\pi)(a_1-a_2)(b_1-b_2). \end{aligned}

The final expression can be nonzero even though the within-class covariance is zero. Its sign depends on whether the two class-specific probabilities move in the same or opposite directions. If class differences in the two domains point in opposite directions, the marginal association is negative; if they point in the same direction, it is positive.

Worked Example: Marginal Association From Two Independent Class-Specific Outcomes

Suppose a hypothetical population has two classes. Class 1 has probability $0.60$, and class 2 has probability $0.40$. Let $A=1$ indicate a high longitudinal measurement, and let $B=1$ indicate an event by one year. Within each class, assume $A$ and $B$ are independent. Suppose $P(A=1\mid Z=1)=0.80$, $P(A=1\mid Z=2)=0.30$, $P(B=1\mid Z=1)=0.20$, and $P(B=1\mid Z=2)=0.70$.

The joint probability of a high measurement and an event by one year is the sum of the class-specific products, weighted by class probabilities:

$$ P(A=1,B=1) =(0.60)(0.80)(0.20)+(0.40)(0.30)(0.70) =0.096+0.084=0.180. $$

The marginal probabilities are $P(A=1)=(0.60)(0.80)+(0.40)(0.30)=0.60$ and $P(B=1)=(0.60)(0.20)+(0.40)(0.70)=0.40$. If $A$ and $B$ were marginally independent, their joint probability would be $(0.60)(0.40)=0.240$, not $0.180$. The marginal covariance is $0.180-0.240=-0.0600$. This negative association is consistent with the assumed within-class independence: the class with more high measurements has fewer events by one year.

Reading the Joint Distribution

The factorization also clarifies what is and is not being modeled. Each term $\pi_k f_k(y)h_k(t)$ describes one class’s contribution to the joint distribution. The mixture adds those contributions because the class is unknown when considering a randomly selected subject. The class-specific outcome distribution concerns the full repeated record, while the class-specific event-time distribution concerns the event process. Their product is justified by the conditional independence assumption.

Worked Example: Evaluating a Joint Density

Consider a hypothetical two-class model with $\pi_1=0.70$ and $\pi_2=0.30$. For one specified pair of longitudinal measurements $y$, suppose the class-specific joint outcome densities are $f_1(y)=0.120$ per squared outcome unit and $f_2(y)=0.0400$ per squared outcome unit. At event time $t=2$ years, suppose the event-time densities are $h_1(2)=0.100$ per year and $h_2(2)=0.300$ per year. Assume the longitudinal vector and event time are independent conditional on class.

The joint density at this outcome vector and event time is

\begin{aligned} p(y,2) &=(0.70)(0.120)(0.100)+(0.30)(0.0400)(0.300)\\ &=0.00840+0.00360\\ &=0.0120 \end{aligned}

The result has units of inverse squared outcome units per year, as required for a joint density over two continuous outcome measurements and time. It is a density value, not the probability of observing exactly those continuous values. This calculation uses the joint density of the pair of outcomes in each class; it does not assume the two repeated outcomes are independent.

Worked Example: Diagnosing a Failed Conditional Independence Assumption

Suppose a hypothetical study has already defined a particular class. Among 100 subjects in that class, a high longitudinal outcome occurs for 50, and an event by a specified follow-up time occurs for 50. If the two indicators are conditionally independent, their joint probability must be $0.50\times0.50=0.25$, corresponding to 25 of 100 subjects in expectation.

Instead, suppose the class-specific table has 40 subjects with both indicators, 10 with only the high outcome, 10 with only the event, and 40 with neither. The margins remain 50 for each indicator, but the joint proportion is $40/100=0.40$. The within-class covariance is $0.40-(0.50)(0.50)=0.15$, rather than zero. Thus this table is incompatible with conditional independence for that class. A more flexible within-class dependence structure, or a revised class specification, would be needed to represent this pattern; simply retaining the product $f_k(y)h_k(t)$ cannot reproduce it.

Identifiability in Principle

A model is identifiable in principle when different parameter values cannot produce exactly the same distribution of observable data, apart from arbitrary relabeling of latent classes. The factorization supplies a useful structure, but it does not guarantee identifiability by itself. For the mixture components to be distinguishable, class-specific distributions must differ in informative ways, and the measurement and event-time information must be rich enough to separate their contributions. If two classes have identical distributions for both domains, combining or splitting them leaves the same observable distribution. If the available outcomes are too limited, different mixtures may also be observationally indistinguishable.

Consequently, “identifiable in principle” is a statement about whether the model and data structure can distinguish parameters under suitable conditions, not a claim that every fitted latent class model has a unique, stable solution. A good specification should make the number of classes and component distributions scientifically interpretable, while avoiding redundant classes. Numerical convergence alone does not establish identifiability.

Define the latent structure.
Specify the possible class values and their probabilities, with nonnegative probabilities summing to one. Each subject has one class throughout the modeled follow-up.

Specify the two class-conditional distributions.
Use a joint distribution for the repeated-outcome vector and a distribution for event time in each class. Do not impose independence among repeated measurements unless separately justified.

State the conditional independence claim.
Check whether the scientific model supports $Y_i\perp T_i\mid Z_i$. If not, the product of the two class-conditional distributions is not an adequate joint model.

Assess distinguishability and observation assumptions.
Consider whether the component distributions are meaningfully distinct and whether censoring or missing outcomes require additional assumptions beyond the class factorization.

Alternatives, Diagnostics, and Limitations

The main alternatives express different dependence assumptions. A single-class model has no latent mixture and therefore cannot attribute cross-domain differences to unobserved subpopulations. A latent class model with within-class dependence retains classes but relaxes the product of the class-specific longitudinal and event-time distributions. Shared random-effects models use continuous latent subject-specific quantities rather than a finite set of classes; they represent a different dependence structure and answer a different modeling question. These approaches are not interchangeable just because each can induce association between repeated outcomes and event time.

Useful diagnostics include checking whether observed outcome patterns and event experience are reasonably represented across the fitted subpopulations, examining whether conclusions change under plausible class specifications, and investigating signs of residual association between the two domains after conditioning on class. In practice, class membership is unobserved, so such checks rely on model-based assessment rather than directly observed class-stratified tables. Sparse classes, weakly distinct components, and heavy censoring can make the assessment especially uncertain.

The central limitation is that conditional independence is a substantive restriction. If an unmodeled subject-level feature still links longitudinal outcomes to event risk within a class, the factorization can misrepresent their joint distribution. Conversely, adding classes to absorb every association can create components that are difficult to distinguish or interpret. Class membership should therefore be viewed as a modeling device for heterogeneous subpopulations, not proof that natural, sharply separated groups exist.

Common pitfall: “The model has latent classes, so longitudinal outcomes and event time must be independent.” This is incorrect: the assumption is independence within each class, and it must be stated and defended. The marginal distributions can remain associated because class membership affects both. Also, a valid factorization does not by itself prove that the class mixture is identifiable.

Check Your Understanding

Answer each question, then compare with the solutions below.

For three classes, what must the class probabilities satisfy?
Does the factorization require repeated measurements to be independent of one another within each class?
In a two-class model, $P(A=1\mid Z=1)=0.70$, $P(A=1\mid Z=2)=0.20$, $P(B=1\mid Z=1)=0.60$, $P(B=1\mid Z=2)=0.10$, and $P(Z=1)=0.40$. Assuming conditional independence, calculate $P(A=1,B=1)$.
Why can a model satisfy $Y\perp T\mid Z$ but still have marginal dependence between $Y$ and $T$?
Does conditional independence alone ensure that a latent class model is identifiable? Give one reason for your answer.

Solutions. (1) Each probability must be nonnegative and the three must sum to one. (2) No. The factorization uses the joint distribution of the entire repeated-outcome vector within each class; it does not factor that vector into separate measurements. (3) The result is $(0.40)(0.70)(0.60)+(0.60)(0.20)(0.10)=0.168+0.012=0.180$. (4) Class membership affects both domains, so mixing class-specific independent distributions can create marginal association. (5) No. For example, classes with identical component distributions cannot be distinguished from the observable joint distribution, even when the conditional independence restriction holds.