Mixed Outcome Types in Pattern-Specific Distributions

Why Mixed-Type Joint Distributions Matter

A repeated-outcome trajectory may contain both measurements, such as a laboratory value, and categorical responses, such as whether an adverse event occurred. Modeling each outcome separately describes its marginal distribution but does not, by itself, describe how the outcomes occur together. Within a pattern-mixture model, that joint description is made separately for each dropout pattern. It can represent, for example, whether a categorical response is more likely at higher measurement values and whether the later measurement distribution differs between response categories.

The central task is to specify a valid joint distribution for the outcomes of interest within each pattern. A useful construction is a chain of conditional distributions. This allows mixed types to coexist and makes the dependence assumptions explicit.

Definition: Let $G$ denote dropout pattern, with pattern probabilities $\pi_g=P(G=g)$. For a continuous outcome $X$, a categorical outcome $B$, and a continuous outcome $Y$, the pattern-specific joint density-mass function is $$ f_g(x,b,y)=f_{X\mid G}(x\mid g)\,P(B=b\mid X=x,G=g)\,f_{Y\mid X,B,G}(y\mid x,b,g). $$ The first and third factors are densities; the middle factor is a probability mass. Their product is a density in the continuous coordinates and a mass in the categorical coordinate, not generally the probability of observing the exact continuous values $x$ and $y$.

Factorization Encodes Dependence

The factorization above follows the multiplication rule for conditional distributions. It does not require $X$, $B$, and $Y$ to be independent. In fact, the conditional probability for $B$ can vary with $X$, while the conditional distribution of $Y$ can vary with both $X$ and $B$. Reversing the factorization order is also possible, provided the factors are specified as a coherent set of conditional distributions.

For example, a common specification uses a normal model for $X$, a logistic regression for the binary outcome $B$, and a conditional normal model for $Y$. The logistic regression captures how the probability of the category changes with $X$; the conditional mean of $Y$ can then depend on both $X$ and $B$. Independence is a special, more restrictive case: it would require the relevant conditional distributions not to depend on the other outcomes.

Worked Example: Reading Dependence From Conditional Models

Suppose that, within one hypothetical dropout pattern $g$, $X$ is a continuous measurement, $B$ indicates whether a symptom occurred ($B=1$ if it did), and $Y$ is a later continuous measurement. Specify $$ \operatorname{logit}P(B=1\mid X=x,G=g)=-0.5+0.06x, \qquad Y\mid X=x,B=b,G=g\sim N\bigl(2.2+0.1(x-12)+b,16\bigr). $$ At $x=10$, the log-odds for $B=1$ are $-0.5+0.06(10)=0.1$, so $$ P(B=1\mid X=10,G=g)=\frac{1}{1+e^{-0.1}}\approx0.5249792. $$

For $B=1$, the conditional mean of $Y$ at $X=10$ is $2.2+0.1(10-12)+1=3.0$, and its standard deviation is $4$, since the second argument of $N$ is the variance. Thus the conditional density at $Y=5$ is $$ f_{Y\mid X,B,G}(5\mid10,1,g) =\frac{1}{4\sqrt{2\pi}}\exp\left(-\frac{(5-3)^2}{2(16)}\right) \approx0.0880163. $$

This model also implies conditional dependence between $B$ and $Y$ at a fixed $X$. Write $Y=m(X)+B+\varepsilon$, where $m(x)=2.2+0.1(x-12)$ and $\varepsilon$ has conditional mean zero and is independent of $B$ given $X,G$. Then $$ \operatorname{Cov}(B,Y\mid X=10,G=g) =\operatorname{Cov}(B,B+\varepsilon\mid X=10,G=g) =\operatorname{Var}(B\mid X=10,G=g) =p(1-p)\approx0.249376, $$ where $p=0.5249792$. This covariance is in the units of $Y$, because $B$ is an indicator. The value is positive: at the same $X$, the model assigns a one-unit higher conditional mean of $Y$ when the symptom is present.

Pattern Weights and the Overall Distribution

A pattern-mixture model combines the within-pattern distributions with the probabilities of the patterns. For the mixed outcomes above, the overall density-mass function is $$ f(x,b,y)=\sum_g \pi_g f_g(x,b,y), \qquad \pi_g\geq0,\quad\sum_g\pi_g=1. $$ Each component must be a valid density-mass function on the relevant support. The weights are probabilities of patterns, not weights chosen to make the outcome distributions look similar. At a specified continuous point, the resulting quantity remains a density-mass value, not the probability of that exact point.

Worked Example: A Two-Pattern Mixed Joint Distribution

Consider a hypothetical setting with two dropout patterns, $G=1$ and $G=2$, having probabilities $\pi_1=0.7$ and $\pi_2=0.3$. In both patterns, $X$ and $Y$ are continuous and $B$ is binary. We calculate the overall density-mass at $x=10,b=1,y=5$.

In pattern 1, let $X\mid G=1\sim N(10,\sigma_1^2)$, with $\sigma_1=1/\{0.12\sqrt{2\pi}\}$, so its density at $10$ is $0.12$. Let $B\mid G=1$ be Bernoulli with probability $0.5$, independent of $X$, and let $Y\mid X,B,G=1\sim N(5,16)$. The density of $Y$ at 5 is $1/(4\sqrt{2\pi})\approx0.0997356$. Therefore, $$ f_1(10,1,5) =0.12(0.5)\left\{\frac{1}{4\sqrt{2\pi}}\right\} \approx0.0059841. $$ The conditional models here imply independence of these outcomes within pattern 1.

In pattern 2, let $X\mid G=2\sim N(10,\sigma_2^2)$, where $\sigma_2=1/\{0.1064827\sqrt{2\pi}\}$, giving $f_{X\mid G}(10\mid2)=0.1064827$. Use $$ \operatorname{logit}P(B=1\mid X=x,G=2)=-0.5+0.06x, \qquad Y\mid X=x,B=b,G=2\sim N\bigl(2.2+0.1(x-12)+b,16\bigr). $$ At $x=10$, the probability of $B=1$ is $1/(1+e^{-0.1})\approx0.5249792$. At $x=10,b=1$, the conditional mean of $Y$ is $3.0$, so its density at 5 is $0.0880163$, as calculated above. Hence $$ f_2(10,1,5) \approx(0.1064827)(0.5249792)(0.0880163) \approx0.0049202. $$

Weighting the two pattern-specific values gives $$ \begin{aligned} f(10,1,5) &=0.7f_1(10,1,5)+0.3f_2(10,1,5)\\ &\approx0.7(0.0059841)+0.3(0.0049202)\\ &\approx0.0056649. \end{aligned} $$ This is the overall density-mass at the stated coordinates. It is not a probability that $X$ equals 10 and $Y$ equals 5; probabilities for continuous outcomes require intervals or regions and integration.

A Moment Derivation Shows What Dependence Changes

A joint model matters because dependence affects summaries of combinations of outcomes. The law of total covariance gives a concise justification. For a binary $B$ and continuous $Y$, within a fixed pattern, $$ \operatorname{Cov}(B,Y\mid G=g) =E\{\operatorname{Cov}(B,Y\mid B,G=g)\mid G=g\} +\operatorname{Cov}\{E(B\mid B,G=g),E(Y\mid B,G=g)\mid G=g\}. $$ Since $B$ is fixed when conditioning on itself, the first term is zero, and $E(B\mid B,G=g)=B$. Thus the covariance is determined by how the conditional mean of $Y$ varies across the categories of $B$. This form is particularly useful when the conditional distribution of $Y$ is specified separately by category.

Worked Example: Binary-Continuous Dependence and Marginal Moments

In a hypothetical pattern, suppose $P(B=1\mid G=g)=0.3$, $Y\mid B=0,G=g\sim N(0,1)$, and $Y\mid B=1,G=g\sim N(2,1)$. The two conditional distributions have the same variance but different means. The pattern-specific mean of $Y$ is $$ E(Y\mid G=g)=0.7(0)+0.3(2)=0.6. $$ By the law of total variance, $$ \operatorname{Var}(Y\mid G=g) =E\{\operatorname{Var}(Y\mid B,G=g)\mid G=g\} +\operatorname{Var}\{E(Y\mid B,G=g)\mid G=g\} =1+0.3(0.7)(2-0)^2=1.84. $$

For dependence, $E(BY\mid G=g)=0.3E(Y\mid B=1,G=g)=0.3(2)=0.6$. Therefore, $$ \operatorname{Cov}(B,Y\mid G=g) =E(BY\mid G=g)-E(B\mid G=g)E(Y\mid G=g) =0.6-(0.3)(0.6)=0.42. $$ The positive covariance reflects the higher mean of $Y$ in the $B=1$ category. If one replaced the joint model by a product of its marginals, this dependence would be set to zero, and the conditional structure—and generally the variance of combined outcomes—would be misrepresented.

Assumptions, Diagnostics, and Alternatives

A pattern-specific factorization is a model, not a consequence of merely having mixed outcome types. Its assumptions should be stated before fitting or interpreting it. The conditional probability model must assign probabilities between zero and one and sum to one over categories. Each continuous conditional model must be a normalized density on its support. The factorization order must cover every outcome included in the joint distribution, with conditional distributions defined for relevant combinations of predictors and categories. Pattern weights must be nonnegative and sum to one.

Diagnostics should examine each component as well as the resulting joint behavior. Check whether observed categorical proportions vary with continuous predictors in a way the chosen link can represent. Compare continuous outcomes with their fitted conditional distributions across categories and patterns. Verify that estimated pattern weights and conditional probabilities are sensible, and that the implied joint density-mass integrates or sums to one. Warning signs include very sparse pattern-category combinations, near-zero or near-one fitted probabilities, unstable conditional parameters, or substantial mismatches between observed and model-implied conditional means or variability. A plausible marginal fit alone does not establish a plausible joint fit.

The main alternative is to fit separate marginal models and combine them as if outcomes were independent. That product construction can be appropriate when conditional independence is scientifically defensible, but it otherwise removes dependence by assumption. A multivariate Gaussian model can represent dependence among continuous outcomes, but it does not directly provide a coherent model for binary or other categorical components. Pattern-specific conditional factorizations are more flexible for mixed outcomes, at the cost of additional modeling choices and parameters. A selection model uses a different factorization, modeling pattern membership conditional on outcomes; it is a distinct framework rather than a substitute factor in the pattern-specific joint distribution.

Pattern-specific distributions also have limits. If an outcome is not observed for members of a pattern, the distribution of that unobserved outcome within the pattern cannot be learned directly from those observations alone. Any specification for it relies on modeling assumptions and available information. Thus a carefully factored joint distribution makes assumptions transparent; it does not make unobserved components automatically identified.

Common pitfall: Multiplying the marginal density of $X$, the marginal probability of $B$, and the marginal density of $Y$ is not a general formula for their joint distribution. It imposes independence. The correction is to use a valid conditional factorization, such as $f_X(x)P(B=b\mid X=x)f_Y(y\mid X=x,B=b)$, and justify any conditional independences by setting the relevant factors not to depend on other outcomes.

Key takeaway: A mixed-type pattern-specific distribution combines densities and category probabilities through conditional distributions. Dependence is represented by those conditional relationships; pattern probabilities then combine the pattern-specific distributions into an overall mixture.

Check Your Understanding

Answer each question, then compare with the solutions.

For continuous $X,Y$ and binary $B$, what does $f_g(x,1,y)$ represent: a probability at an exact point or a density-mass value?
In $f_g(x,b,y)=f_{X\mid G}(x\mid g)P(B=b\mid X=x,G=g)f_{Y\mid X,B,G}(y\mid x,b,g)$, which factors can encode dependence?
Two patterns have weights 0.6 and 0.4. At a specified mixed-outcome point their density-mass values are 0.02 and 0.05. What is the overall value?
If $Y\mid B=0$ and $Y\mid B=1$ have the same mean, does that model imply nonzero covariance between $B$ and $Y$? Explain under the binary-category mixture setup.

Solutions. (1) It is a density in $x,y$ and a mass at $B=1$, not the probability of exact continuous values. (2) The probability for $B$ can depend on $X$, and the conditional density for $Y$ can depend on $X$ and $B$. (3) $0.6(0.02)+0.4(0.05)=0.032$. (4) No, if the two conditional means are equal, then $E(Y\mid B)$ does not vary with $B$, so $\operatorname{Cov}(B,Y)=0$ under this mixture calculation, though other forms of dependence may still exist beyond the conditional means.