The Shared-Parameter Factorization

Why Introduce a Shared Random Effect?

An outcome and its response pattern can be associated even when the model does not make response depend directly on the outcome. For example, people with a latent tendency toward poorer health might both have different outcome distributions and be more likely to miss a measurement. A shared-parameter factorization represents this by introducing a latent random effect that affects both the outcome and the response pattern. The random effect is shared in the model; it is not an additional observed outcome or a response indicator.

Let $Y$ denote the full outcome vector for one unit, $R$ its response-pattern vector, and $B$ a latent random effect. The aim is to specify the joint law of $Y$ and $R$ through their distributions conditional on $B$, together with the distribution of $B$. This tutorial uses no covariates; the same structure can be extended when covariates are included.

Definition: A shared-parameter factorization specifies the joint law as $$ f_{Y,R,B}(y,r,b)=f_B(b)f_{Y\mid B}(y\mid b)p_{R\mid B}(r\mid b). $$ It assumes that $Y$ and $R$ are conditionally independent given $B$. The marginal joint law is obtained by summing over $b$ when $B$ is discrete, or integrating over $b$ when $B$ is continuous: $$ f_{Y,R}(y,r)=\int f_B(b)f_{Y\mid B}(y\mid b)p_{R\mid B}(r\mid b)\,db. $$

Here $f_B$ is the distribution of the latent effect, $f_{Y\mid B}$ is the conditional outcome density or probability mass function, and $p_{R\mid B}$ is the conditional response-pattern probability. Each conditional distribution must be a valid probability law for every possible $b$: it must be nonnegative and sum or integrate to one over its possible values. The mixing distribution $f_B$ must also be a valid probability law.

The conditional independence assumption is important: at each fixed value of $B$, the model gives the joint conditional law as $f_{Y\mid B}(y\mid b)p_{R\mid B}(r\mid b)$. It does not say that $Y$ and $R$ are marginally independent. Mixing over $B$ can induce a marginal association, as the examples show. If conditional dependence remains after accounting for $B$, this product form is not the appropriate model without further modification.

From the Conditional Model to the Marginal Law

The joint law includes $B$, but $B$ is unobserved. The law of $Y$ and $R$ alone is therefore found by eliminating $B$, summing or integrating its joint probability contribution across all possible values. For discrete $B$, with possible values $b_1,\ldots,b_K$ and probabilities $\pi_k=P(B=b_k)$, the formula is

P(Y=y,R=r)=\sum_{k=1}^{K}\pi_k f_{Y\mid B}(y\mid b_k)p_{R\mid B}(r\mid b_k).

For continuous $B$, the sum becomes an integral with respect to its density. This is a mixture: each value of $B$ supplies a conditional joint contribution, weighted by how probable that value is. One must average the product of the two conditional contributions, not multiply their separately averaged values.

Justification: The marginal joint law is obtained by summing or integrating the full joint law over $B$. For continuous $B$, $$ f_{Y,R}(y,r)=\int f_{Y,R,B}(y,r,b)\,db =\int f_B(b)f_{Y\mid B}(y\mid b)p_{R\mid B}(r\mid b)\,db. $$ This is a valid joint distribution: integrating over $y$ and summing over $r$ gives $$ \int\sum_r f_{Y,R}(y,r)\,dy =\int f_B(b)\left[\int f_{Y\mid B}(y\mid b)\,dy\right] \left[\sum_r p_{R\mid B}(r\mid b)\right]db =\int f_B(b)\,db=1. $$ The bracketed terms equal one because they are conditional probability laws.

Worked Examples

Worked Example: A Two-Class Latent Effect

Hypothetical model. Let $B$ have two classes, with $P(B=0)=0.6$ and $P(B=1)=0.4$. Let $Y$ be a binary outcome and $R=1$ indicate that it is observed. Given $B=0$, suppose $P(Y=1\mid B=0)=0.2$ and $P(R=1\mid B=0)=0.8$. Given $B=1$, suppose the corresponding probabilities are 0.7 and 0.3. Assume $Y$ and $R$ are conditionally independent given $B$.

The joint probability of a positive outcome and an observed response is the mixture of the two class-specific products: $$ P(Y=1,R=1) =(0.6)(0.2)(0.8)+(0.4)(0.7)(0.3) =0.096+0.084=0.180. $$

The marginal probabilities are $P(Y=1)=(0.6)(0.2)+(0.4)(0.7)=0.400$ and $P(R=1)=(0.6)(0.8)+(0.4)(0.3)=0.600$. If $Y$ and $R$ were marginally independent, their joint probability would be $(0.400)(0.600)=0.240$, not 0.180. Thus conditional independence within each latent class does not imply marginal independence after the classes are mixed.

Worked Example: Integrating a Continuous Random Effect

Hypothetical model. Suppose $B\sim\text{Uniform}(0,1)$. Conditional on $B=b$, let $Y$ and $R$ be binary and conditionally independent, with $P(Y=1\mid B=b)=b$ and $P(R=1\mid B=b)=(1+b)/2$. These are valid probabilities throughout $0\leq b\leq1$.

For the event $Y=1,R=1$, the marginal probability is $$ P(Y=1,R=1)=\int_0^1 b\frac{1+b}{2}\,db =\frac{1}{2}\left[\frac{b^2}{2}+\frac{b^3}{3}\right]_0^1 =\frac{5}{12}\approx0.4167. $$

The other cells can be checked in the same way. For example, $$ P(Y=1,R=0)=\int_0^1 b\frac{1-b}{2}\,db =\frac{1}{12}, \qquad P(Y=0,R=1)=\int_0^1 (1-b)\frac{1+b}{2}\,db =\frac{1}{3}. $$ The remaining probability is $P(Y=0,R=0)=1/6$; the four cells sum to one. Also, $P(Y=1)=\int_0^1 b\,db=1/2$, while $P(R=1)=\int_0^1(1+b)/2\,db=3/4$. Their product, $3/8$, differs from the joint probability $5/12$. The example makes explicit that integration averages the conditional product across the distribution of $B$.

Worked Example: A Two-Visit Response Pattern

Hypothetical model. Let $Y$ be a two-visit binary outcome vector and let $R$ record the response pattern. Focus on the outcome event $A=\{Y=(1,0)\}$ and the response pattern $C=\{R=(1,0)\}$. Suppose $P(B=0)=P(B=1)=0.5$. Set $P(A\mid B=0)=0.3$, $P(A\mid B=1)=0.6$, $P(C\mid B=0)=0.8$, and $P(C\mid B=1)=0.2$. Assume the full vectors $Y$ and $R$ are conditionally independent given $B$. The remaining outcome and response probabilities can be distributed over their other possible values so each conditional law sums to one.

The marginal joint probability of this outcome event and response pattern is $$ P(A,C)=(0.5)(0.3)(0.8)+(0.5)(0.6)(0.2) =0.120+0.060=0.180. $$ Separately, $P(A)=(0.5)(0.3)+(0.5)(0.6)=0.450$, and $P(C)=(0.5)(0.8)+(0.5)(0.2)=0.500$. Their product is 0.225, so the marginal event probabilities are not independent. The calculation applies to a specific vector outcome and response pattern; the same mixture formula applies to every pair of possible vectors.

Assumptions, Diagnostics, and Alternatives

Before using this factorization, check that the model is internally coherent. The distribution assigned to $B$ must be normalized. For every $b$, the conditional outcome probabilities or density must normalize over all possible $Y$, and the response probabilities must sum to one over all possible patterns $R$. The same latent effect must enter both conditional models for the representation to be shared-parameter. Finally, the product form requires the stated conditional independence assumption.

Useful warning signs include conditional probabilities outside $[0,1]$, response-pattern probabilities that do not sum to one, or an outcome model that is only defined for observed outcomes when the stated target is the full outcome vector. A poor fit of the implied marginal distribution can also indicate that the chosen latent-effect distribution or conditional models are inadequate. Observed association between outcomes and response does not, by itself, verify the shared-effect explanation: other models can produce the same association.

The shared-parameter representation differs in emphasis from the selection factorization and the pattern-mixture factorization. Those express the joint law through different conditional or marginal components; the shared-parameter approach introduces an unobserved variable and mixes over it. These are ways to represent a joint law, not automatic guarantees that the law is empirically determined. A latent distribution and conditional models must be justified for the application, and the observed data do not necessarily determine every feature of the latent construction.

A key limitation is that conditional independence given $B$ can be a strong assumption. If $Y$ and $R$ remain associated after conditioning on the proposed effect, the product $f_{Y\mid B}p_{R\mid B}$ omits that dependence. Another limitation is interpretive: a latent effect is not directly measured, so its meaning depends on how the model defines it. As in other missing-data analyses, a specified factorization is a model, not proof that its assumptions hold.

Common pitfall: It is incorrect to replace the mixture of conditional products with the product of marginal probabilities: $$ \int f_B(b)f_{Y\mid B}(y\mid b)p_{R\mid B}(r\mid b)\,db \ne \left[\int f_B(b)f_{Y\mid B}(y\mid b)\,db\right] \left[\int f_B(b)p_{R\mid B}(r\mid b)\,db\right] $$ in general. The right-hand side would assert marginal independence for that pair of events. The shared-parameter model assumes conditional independence given $B$, not marginal independence.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

In the shared-parameter factorization, what role does $B$ play, and what conditional independence assumption gives the joint law its product form?
Suppose $B$ has two equally likely values. Given $B=0$, $P(Y=1)=0.1$ and $P(R=1)=0.9$; given $B=1$, the probabilities are 0.5 and 0.5. Under conditional independence, calculate $P(Y=1,R=1)$.
Why is it generally incorrect to multiply the marginal probabilities of $Y=y$ and $R=r$ to obtain their joint probability?
Name two checks that can reveal an invalid specification of the conditional distributions.

Solutions. (1) $B$ is a latent random effect that enters both the conditional outcome and response models; the product form assumes $Y$ and $R$ are conditionally independent given $B$. (2) The mixture is $(0.5)(0.1)(0.9)+(0.5)(0.5)(0.5)=0.045+0.125=0.170$. (3) Mixing over $B$ can induce marginal association, so the joint law is the average of conditional products, not generally the product of separate averages. (4) For example, check that all conditional probabilities are nonnegative and that they sum to one over all possible outcomes or response patterns for each value of $B$.

The Shared-Parameter Factorization

What You'll Learn