Why Check the Observed-Data Likelihood?
A selection factorization describes the outcome distribution first and then models the missingness pattern conditional on the outcomes. A pattern-mixture factorization describes the pattern probabilities and then the outcome distribution conditional on the pattern. These are two ways to factor a joint law, not automatically two identical statistical models. To compare their likelihoods, the factors must describe the same joint distribution, and the missing outcomes must be marginalized consistently.
The key comparison is between the observed-data contribution for one individual under the two factorizations. For a discrete outcome, the missing values are summed out; for a continuous outcome, they are integrated out. The equality is exact when the corresponding joint-density sections agree over the missing outcomes. If equality of joint densities is known only almost everywhere in the full data space, the resulting equality is guaranteed for almost every observed value, not necessarily for every fixed observed value.
The Algebraic Equivalence
Suppose the outcome is discrete. The selection-factor contribution for pattern \(r\) is \( \sum_{y_{\mathrm{mis}}} p(y_{\mathrm{obs}},y_{\mathrm{mis}})P(r\mid y_{\mathrm{obs}},y_{\mathrm{mis}}) \). If the two factorizations describe the same joint law, then for each complete outcome and pattern, \( p(y)P(r\mid y)=\pi_r p(y\mid r) \). Substitution gives the pattern-mixture contribution \( \pi_r\sum_{y_{\mathrm{mis}}}p(y_{\mathrm{obs}},y_{\mathrm{mis}}\mid r) \). For continuous missing outcomes, replace the sum by an integral with respect to the appropriate dominating measure. For mixed outcomes, sum over discrete components and integrate over continuous components.
In the continuous case, the same algebra applies to integrals if the joint-density sections agree almost everywhere over the missing-outcome values for the observed value under consideration. A common sufficient condition is that the two joint densities are equal almost everywhere in the full data space. By Fubini's theorem, their sections then agree for almost every observed value, and the marginalized contributions agree for almost every observed value. That argument does not establish equality for every fixed observed value: a single observed-value slice can have measure zero in the full space.
This result concerns equality of probability laws and their observed-data distributions. A likelihood written as a density at a particular continuous data value depends on the chosen density version on null sets. Such versions yield the same probability law, even if their numerical values differ at exceptional points.
Worked Comparisons
Worked Example: A Discrete Outcome and Two Patterns
Let \(Y\) be binary and let \(R=1\) mean \(Y\) is observed; \(R=0\) means it is missing. Suppose the selection factors are \(P(Y=1)=0.6\), \(P(Y=0)=0.4\), \(P(R=1\mid Y=1)=0.5\), and \(P(R=1\mid Y=0)=0.25\). Find the observed-data contributions and express them using a compatible pattern-mixture factorization.
Under selection, the joint probabilities for \(R=1\) are \(P(Y=1,R=1)=0.6(0.5)=0.30\) and \(P(Y=0,R=1)=0.4(0.25)=0.10\). Thus \(\pi_1=P(R=1)=0.30+0.10=0.40\), and \(P(Y=1\mid R=1)=0.30/0.40=0.75\).
For \(R=0\), the joint probabilities are \(P(Y=1,R=0)=0.6(0.5)=0.30\) and \(P(Y=0,R=0)=0.4(0.75)=0.30\). Thus \(\pi_0=0.60\), and \(P(Y=1\mid R=0)=0.30/0.60=0.50\). These pattern probabilities and conditional outcome distributions form the compatible pattern-mixture factorization.
For an observed \(Y=1,R=1\), the selection contribution is \(0.6(0.5)=0.30\); the pattern-mixture contribution is \(0.40(0.75)=0.30\). For \(Y=0,R=1\), the two expressions are \(0.4(0.25)=0.10\) and \(0.40(0.25)=0.10\), respectively. When \(R=0\), \(Y\) is not observed, so its values are summed: selection gives \(0.30+0.30=0.60\), and pattern-mixture gives \(\pi_0=0.60\sum_yP(Y=y\mid R=0)=0.60(1)=0.60\).
For hypothetical independent individuals with three records \((Y=1,R=1)\), two records \((Y=0,R=1)\), and four records with \(R=0\), the likelihood contribution is \(0.30^3(0.10)^2(0.60)^4\) under either factorization. The equality follows record by record, so multiplying the contributions preserves it.
Worked Example: Integrating a Continuous Missing Outcome
Let \(Y_1,Y_2\) be independent Uniform\((0,1)\) outcomes. Pattern \(A\) observes \(Y_1\) and misses \(Y_2\). Suppose \(P(R=A\mid Y_1=y_1,Y_2=y_2)=(1+y_2)/3\). Verify that the observed density for \(Y_1=x,R=A\) agrees under the two factorizations.
The complete-outcome density is \(p(y_1,y_2)=1\) on the unit square. The pattern probability is \( \pi_A=\int_0^1\int_0^1 (1+y_2)/3\,dy_1\,dy_2 =\int_0^1(1+y_2)/3\,dy_2=1/2 \). The conditional complete-outcome density in pattern \(A\) is therefore \(p(y_1,y_2\mid A)=p(y_1,y_2)P(A\mid y_1,y_2)/\pi_A=2(1+y_2)/3\).
Under selection, the observed density at \(Y_1=x,R=A\), for \(0<x<1\), is \( \int_0^1 1\cdot(1+y_2)/3\,dy_2 =[y_2/3+y_2^2/6]_0^1=1/2 \). Under the pattern-mixture factorization it is \( \pi_A\int_0^1 p(x,y_2\mid A)\,dy_2 =(1/2)\int_0^1 2(1+y_2)/3\,dy_2 =(1/2)(1)=1/2 \). Both are densities per unit of \(Y_1\). Here the section-wise equality holds throughout the unit interval, so the integrals agree for every \(x\) in that interval.
Worked Example: Why Almost-Everywhere Equality Needs Care
Let \(X\) and \(Z\) each range over \([0,1]\), and suppose the joint density is \(f(x,z)=1\) on the unit square. Define another version \(g\) to equal 1 everywhere except when \(x=0\), where \(g(0,z)=2\). Compare the integrated densities at \(x=0\).
The functions \(f\) and \(g\) differ only on the line \(x=0\), which has two-dimensional Lebesgue measure zero. Therefore they are versions of the same joint density and give the same joint probability law. But their sections at \(x=0\) give \( \int_0^1 f(0,z)\,dz=\int_0^1 1\,dz=1 \) and \( \int_0^1 g(0,z)\,dz=\int_0^1 2\,dz=2 \). Thus full-space almost-everywhere equality alone does not imply equality of these section integrals at this particular observed value. The discrepancy occurs at a null observed value and does not change the observed-data probability law. For a fixed-value comparison, choose compatible sections or verify their equality directly.
How to Check Equivalence Reliably
The algebra is simple once the relevant equality is established. The practical work is checking that the factors are normalized, correspond to the same joint law, and use the same pattern definition and observed components. A disciplined comparison is:
Identify the selection expression \(p(y)P(r\mid y)\) and the pattern-mixture expression \(\pi_rp(y\mid r)\), including their parameter values or restrictions.
Verify that the expressions represent the same joint distribution, including normalization and support. Do not infer compatibility just because each expression separately looks like a valid factorization.
For each pattern, sum or integrate over exactly \(Y_{\mathrm{mis}}(r)\). Keep the observed value and pattern fixed in the contribution.
For discrete outcomes, compare the sums. For continuous outcomes, establish section-wise equality for a fixed observed value, or report equality almost everywhere when that is all the density argument guarantees.
Useful diagnostics include checking that \(\sum_rP(r\mid y)=1\) for each complete outcome, that \(\sum_r\pi_r=1\), and that each conditional distribution \(p(y\mid r)\) integrates or sums to one. Then compare the joint expressions pattern by pattern. A numerical check at selected values can reveal arithmetic or coding errors, but it does not prove equality over the full support. For continuous models, quadrature should also be checked for adequate accuracy and consistent integration limits.
| Approach | What it supplies | What to verify |
|---|---|---|
| Selection factorization | Outcome law and pattern probabilities conditional on the complete outcome | Pattern probabilities sum to one at each complete outcome |
| Pattern-mixture factorization | Pattern probabilities and outcome laws conditional on each pattern | Pattern probabilities sum to one and each conditional outcome law is normalized |
| Observed-data comparison | Likelihood contribution after missing outcomes are marginalized | Both factorizations use the same joint law and the same marginalization; continuous fixed-value claims require section-wise care |
Common Pitfalls and Limits
- Assuming any two valid factorizations are equivalent. Each can define a valid joint model without defining the same one. The equality requires compatibility of their joint laws.
- Comparing complete-data factors but not observed contributions. The likelihood for a pattern with missing outcomes uses a sum or integral, not the joint density evaluated at an invented value for a missing outcome.
- Claiming pointwise equality from full-space almost-everywhere equality. Fubini's theorem yields section equality for almost every observed value. It does not settle every fixed slice; check the section or state the almost-everywhere conclusion.
- Ignoring support or normalization. If a pattern has probability zero, its conditional outcome distribution may not be defined by the usual ratio. Equivalence claims for that pattern require separate treatment and cannot be obtained by dividing by \(\pi_r=0\).
The comparison is also limited in scope: equal observed-data likelihoods follow from the same joint law, but the algebra does not establish that a chosen factorization is scientifically appropriate. Nor does numerical agreement at a few data points establish equality of models. The main alternative is to work directly from the observed-data distribution; that can avoid displaying complete-data factors, but the same marginalization and version issues remain.
Check Your Understanding
Answer each question, then compare with the solutions below.
- In a discrete model, state the selection and pattern-mixture expressions for a pattern-specific observed-data contribution.
- Why does equality of joint densities almost everywhere imply equality of marginalized densities only for almost every observed value without an additional condition?
- Suppose \(\pi_r=0\). Why can the usual conditional density \(p(y\mid r)\) not be obtained by dividing the joint density by \(\pi_r\)?
- For a hypothetical binary outcome, \(P(Y=1)=0.7\), \(P(R=1\mid Y=1)=0.2\), and \(P(R=1\mid Y=0)=0.5\). Compute the contribution for \(Y=1,R=1\), and the compatible pattern-mixture expression for it.
- What must be checked before concluding that two separately specified factorizations have the same observed-data likelihood?
Solutions. (1) The selection contribution is \(\sum_{y_{\mathrm{mis}}}p(y_{\mathrm{obs}},y_{\mathrm{mis}})P(r\mid y_{\mathrm{obs}},y_{\mathrm{mis}})\); the pattern-mixture contribution is \(\pi_r\sum_{y_{\mathrm{mis}}}p(y_{\mathrm{obs}},y_{\mathrm{mis}}\mid r)\). (2) A fixed observed-value slice can have measure zero in the full space, so full-space almost-everywhere equality does not control the density values on that slice. Fubini gives section equality for almost every observed value. (3) Division by zero is undefined, and a conditional distribution on a zero-probability pattern is not determined by the ordinary joint-to-marginal ratio. (4) Selection gives \(0.7(0.2)=0.14\). Also \(\pi_1=0.7(0.2)+0.3(0.5)=0.29\), and \(P(Y=1\mid R=1)=0.14/0.29\). Thus the pattern-mixture expression is \(0.29(0.14/0.29)=0.14\). (5) Check that the factors define the same joint law, have consistent pattern definitions and support, are normalized, and marginalize the same missing components; for a fixed continuous observed value, also check section-wise equality or limit the claim to almost every value.