A Three-Visit Factorization Worked Example

From a Dropout Pattern to an Observed-Data Contribution

In a three-visit study with monotone dropout, a participant who misses a visit does not return at a later visit. The possible records are therefore simple to enumerate, but the corresponding likelihood contributions require care: each contribution includes the probability of that response pattern and the distribution of the outcomes that were observed under it. Outcomes that were not observed must be integrated out, not treated as if their values were known.

Let $Y_1,Y_2,Y_3$ be the outcomes that would be measured at visits 1, 2, and 3. Let $R$ denote the last visit observed, with $R=0$ indicating that no visit outcome was observed. Thus $R=1$ means only $Y_1$ was observed, and $R=3$ means all three outcomes were observed. This coding assumes the monotone pattern: if $Y_k$ is missing, then all later visit outcomes are also missing.

Definition: Write $\pi_r=P(R=r)$ for the probability of pattern $r$, and $f_r(y_1,y_2,y_3)$ for the joint density of the three outcomes conditional on $R=r$. For discrete outcomes, read a density as a probability mass function and replace integration with summation. The pattern-mixture factorization of the complete-data law is $$ P(R=r,Y_1\in dy_1,Y_2\in dy_2,Y_3\in dy_3) =\pi_r f_r(y_1,y_2,y_3)\,dy_1\,dy_2\,dy_3. $$ The probabilities satisfy $\pi_r\geq 0$ and $\sum_{r=0}^3\pi_r=1$; each $f_r$ is a normalized joint distribution.

The densities $f_r$ describe the full outcome vector within each pattern, including outcomes that are missing for members of that pattern. A sequential representation is available for each $r$: $f_r(y_1,y_2,y_3)=f_r(y_1)f_r(y_2\mid y_1)f_r(y_3\mid y_1,y_2)$. Here and below, conditioning on $R=r$ is understood in each factor. This is a factorization of the same pattern-specific joint density, not an assertion that the outcomes are independent.

Enumerating All Four Patterns

The observed-data contribution for a particular participant is the joint density or mass of the response pattern and the outcomes actually recorded. For $R=r$, the observed outcomes are $Y_1,\ldots,Y_r$; integrate the remaining coordinates out of $f_r$. In particular, the pattern probability $\pi_r$ stays outside that integral.

\begin{aligned} R=0:\quad &\pi_0\int\!\!\int\!\!\int f_0(y_1,y_2,y_3)\,dy_3\,dy_2\,dy_1=\pi_0,\\ R=1:\quad &\pi_1\int\!\!\int f_1(y_1,y_2,y_3)\,dy_3\,dy_2 =\pi_1 f_1(y_1),\\ R=2:\quad &\pi_2\int f_2(y_1,y_2,y_3)\,dy_3 =\pi_2 f_2(y_1,y_2),\\ R=3:\quad &\pi_3 f_3(y_1,y_2,y_3). \end{aligned}

The equalities follow from normalization and marginalization. For instance, integrating $f_2$ over $y_3$ gives the marginal density of $(Y_1,Y_2)$ conditional on $R=2$. If outcomes are discrete, the corresponding contribution for $R=2$ is $\pi_2\sum_{y_3}f_2(y_1,y_2,y_3)$. For $R=0$, there are no recorded outcomes, so the contribution is just the probability of that pattern.

These expressions require a well-defined joint law for $R$ and the three outcomes, normalized pattern-specific distributions, and the stated monotone coding of what was observed. For a sample of independent participants, the observed-data likelihood is the product of their individual contributions. No missing-at-random condition is needed to write this factorization. Any additional restrictions linking pattern probabilities or pattern-specific outcome distributions would be extra model assumptions, not consequences of the factorization itself.

Worked Example: Binary Outcomes in All Four Patterns

Consider a hypothetical study in which each $Y_k$ is binary. Suppose the pattern probabilities are $(\pi_0,\pi_1,\pi_2,\pi_3)=(0.20,0.30,0.25,0.25)$. They are nonnegative and sum to 1. Conditional on each pattern, suppose the outcomes are independent Bernoulli variables with the following success probabilities:

Pattern $r$	$P(Y_1=1\mid R=r)$	$P(Y_2=1\mid R=r)$	$P(Y_3=1\mid R=r)$
0	0.40	0.50	0.50
1	0.60	0.40	0.30
2	0.70	0.55	0.20
3	0.50	0.60	0.45

The independence assumption is part of this hypothetical model, so $f_r$ is the product of its three Bernoulli mass functions. For a participant with $R=1$ and $Y_1=1$, the missing outcomes must be summed over:

\begin{aligned} L_1 &=0.30\sum_{y_2=0}^1\sum_{y_3=0}^1 P(Y_1=1,Y_2=y_2,Y_3=y_3\mid R=1)\\ &=0.30(0.60)\{(0.40+0.60)(0.30+0.70)\}\\ &=0.30(0.60)(1)(1)=0.1800. \end{aligned}

The factors in braces sum the two missing binary outcomes; each sums to 1 because its Bernoulli probabilities are normalized. For a participant with $R=2$, $Y_1=1$, and $Y_2=0$, only $Y_3$ is missing:

\begin{aligned} L_2 &=0.25\sum_{y_3=0}^1 P(Y_1=1,Y_2=0,Y_3=y_3\mid R=2)\\ &=0.25(0.70)(1-0.55)\{0.20+(1-0.20)\}\\ &=0.25(0.70)(0.45)(1)=0.07875. \end{aligned}

For $R=3$, all outcomes are observed. If the record is $(Y_1,Y_2,Y_3)=(1,0,1)$, its contribution is $0.25(0.50)(1-0.60)(0.45)=0.02250$. For $R=0$, the contribution is $0.20$, because there are no observed outcomes and summing the normalized $f_0$ over all three binary outcomes gives 1. The arithmetic illustrates an important distinction: the outcome probabilities used for a pattern are those conditional on that pattern.

A Continuous-Outcome Calculation

With continuous outcomes, the same marginalization uses integrals, and an observed-data contribution is generally a density value rather than the probability of observing an exact numerical outcome. The integrals are often straightforward when the pattern-specific distribution is specified in a convenient form.

Worked Example: Integrating Normal Outcomes

Suppose, hypothetically, that conditional on $R=1$, the three outcomes are independent with $Y_1\sim N(0,1)$, $Y_2\sim N(2,1)$, and $Y_3\sim N(-1,1)$. The notation $N(\mu,\sigma^2)$ uses variance as its second argument. Let $\pi_1=0.30$. For an observed $Y_1=1$, the contribution is

\begin{aligned} L_1 &=0.30\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \phi(1;0,1)\phi(y_2;2,1)\phi(y_3;-1,1)\,dy_3\,dy_2\\ &=0.30\phi(1;0,1)(1)(1)\\ &=0.30\left\{\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}\right)\right\}\\ &\approx 0.30(0.24197)=0.07259. \end{aligned}

Here $\phi(y;\mu,\sigma^2)$ denotes the $N(\mu,\sigma^2)$ density evaluated at $y$. Each integral over a missing outcome is 1 because its conditional density is normalized.

Now suppose that conditional on $R=2$, the outcomes are independent with $Y_1\sim N(0,1)$, $Y_2\sim N(1,1)$, and $Y_3\sim N(3,1)$, and let $\pi_2=0.25$. For observed values $Y_1=0.5$ and $Y_2=1$, the missing third outcome integrates out:

\begin{aligned} L_2 &=0.25\int_{-\infty}^{\infty} \phi(0.5;0,1)\phi(1;1,1)\phi(y_3;3,1)\,dy_3\\ &=0.25(0.35207)(0.39894)(1)\\ &\approx 0.03511. \end{aligned}

This calculation checks the assumptions used: the stated normal densities are normalized, the conditional independence assumption permits their product form, and $Y_3$ is unobserved for $R=2$. Without conditional independence, the joint density would not generally be that product, although the rule of integrating the missing coordinate would remain the same.

Building the Sample Likelihood

For independent participants indexed by $i=1,\ldots,n$, let $r_i$ be the observed pattern and let $y_{i,\mathrm{obs}}$ denote that participant’s recorded outcomes. The likelihood is the product of the appropriate pattern-specific contributions. If $n_r$ participants have pattern $r$, the pattern-probability part of that likelihood is $\prod_{r=0}^3\pi_r^{n_r}$; the outcome-density terms must also be included for participants with observed outcomes.

Worked Example: Pattern Counts and the Likelihood

Suppose a hypothetical sample of 20 independent participants has 2 records with $R=0$, 5 with $R=1$, 6 with $R=2$, and 7 with $R=3$. These counts add to 20. Using the pattern probabilities from the binary example, the pattern-only likelihood factor is

\begin{aligned} L_{\mathrm{pattern}} &=(0.20)^2(0.30)^5(0.25)^6(0.25)^7\\ &=(0.20)^2(0.30)^5(0.25)^{13},\\ \log L_{\mathrm{pattern}} &=2\log(0.20)+5\log(0.30)+13\log(0.25)\\ &\approx -3.2189-6.0199-18.0218=-27.2606. \end{aligned}

This is only the pattern-probability factor, not the full observed-data likelihood. The full likelihood multiplies it by the appropriate observed outcome mass for each participant with $R=1,2,$ or $3$, as in the preceding binary example. A participant with $R=0$ contributes only $\pi_0$. This separation makes it easier to check whether every record contributes both the correct pattern probability and, when available, the correct observed-outcome term.

Derivation, Checks, and Alternatives

The observed-data formulas follow directly from the complete-data law. For example, for $R=2$, integrate the joint law over the unrecorded $Y_3$:

\begin{aligned} p(R=2,Y_1\in dy_1,Y_2\in dy_2) &=\int p(R=2,Y_1\in dy_1,Y_2\in dy_2,Y_3\in dy_3)\\ &=\pi_2\left\{\int f_2(y_1,y_2,y_3)\,dy_3\right\}dy_1\,dy_2\\ &=\pi_2 f_2(y_1,y_2)\,dy_1\,dy_2. \end{aligned}

The other patterns follow by integrating over the coordinates not observed: three coordinates for $R=0$, two for $R=1$, and none for $R=3$. The integration domain is the support of the missing outcomes. For discrete outcomes, replacing the integrals by sums gives the same result.

A useful diagnostic is to check the bookkeeping before calculating: the four patterns must be exactly $(---),(+--),(++-),(+++)$, where a plus denotes an observed visit and a minus a missing visit. Then check that the pattern probabilities sum to 1, that each $f_r$ integrates or sums to 1, and that the contribution for each row retains only that pattern’s observed outcomes. If a purported $R=1$ record contains $Y_3$, the coding or data layout is inconsistent with monotone dropout. If a missing coordinate remains in the final contribution, marginalization is incomplete; if a normalized missing-outcome density contributes a factor other than 1 when integrated over its full support, check the density or integration limits.

The pattern-mixture form writes the joint law as a pattern probability times an outcome distribution conditional on pattern. A selection factorization instead starts with an outcome distribution and models the response pattern conditional on outcomes. These are alternative ways to factor a joint law, as in the earlier material on likelihood equivalence across factorizations. The present calculation focuses on what each pattern contributes after unobserved outcomes are marginalized. The factorization itself does not determine a unique model: restrictions such as common outcome distributions across patterns would be additional assumptions. Nor does writing the observed-data likelihood alone establish that every component or scientific target can be estimated from the records.

Common Pitfall

Common mistake: “A participant with $R=2$ contributes $\pi_2 f_2(y_1,y_2,y_3)$, with $y_3$ set to zero because it is missing.” This is incorrect: a missing outcome is not an observed zero, and its value is not supplied by the record. The contribution is $\pi_2\int f_2(y_1,y_2,y_3)\,dy_3$, or a sum over possible $y_3$ for a discrete outcome. Only under an explicitly specified model could a particular value be substituted, and that would not generally represent marginalization.

The same caution applies to $R=0$: there is no observed outcome vector to plug into a density. Integrating the full conditional density gives 1, leaving the pattern probability. Correct pattern coding and integration are therefore central to reading a joint factorization, not merely algebraic details.

Check Your Understanding

Answer each question using the last-observed-visit coding $R$. Short solutions follow the questions.

For a record with only $Y_1$ observed, write its observed-data contribution.
If $f_2$ is a joint density for all three outcomes conditional on $R=2$, which coordinate is integrated out to obtain the contribution for observed $Y_1=y_1,Y_2=y_2$?
Suppose $(\pi_0,\pi_1,\pi_2,\pi_3)=(0.1,0.2,0.3,0.4)$. What is the contribution for a record with $R=0$?
A participant’s record has $Y_1$ missing, $Y_2$ observed, and $Y_3$ missing. Is it one of the four monotone patterns coded here?

Solutions. (1) The contribution is $\pi_1 f_1(y_1)$, where $f_1(y_1)$ is the marginal density or mass after integrating or summing over $Y_2,Y_3$. (2) Integrate $Y_3$: the contribution is $\pi_2\int f_2(y_1,y_2,y_3)\,dy_3$. (3) It is $\pi_0=0.1$, because the conditional outcome density sums or integrates to 1. (4) No. A later visit is observed after an earlier visit is missing, so this record is nonmonotone and cannot be coded as any of the four patterns in this tutorial.