Integrating Shared Random Effects From the Joint Law

Why Integrate Out a Shared Random Effect?

A shared-parameter model represents the association between outcomes and response by giving them a common latent random effect. The random effect may describe an unobserved subject characteristic that influences both the outcome process and the chance or pattern of response. The model is often specified conditionally on that effect, but the marginal law for the outcomes and response pattern is obtained only after averaging over its distribution.

The calculation is straightforward to state but important to interpret: multiply the conditional outcome and response factors specified by the model, weight that product by the random-effect distribution, and integrate over the random effect. The result is a joint distribution for outcomes and response patterns that no longer contains the latent variable.

The product \(f(y\mid b)q(r\mid b)\) is the conditional factorization assumed by this model. The integration step itself does not establish that factorization; it takes the specified conditional law as its starting point. For each fixed \(b\), \(f(y\mid b)\) must be a valid outcome density or mass function, and \(q(r\mid b)\) must be a valid response-pattern probability distribution. The mixing distribution \(h\) must also be a valid density or mass function.

For longitudinal data, \(Y\) and \(R\) are vectors. For example, with two visits, \(Y=(Y_1,Y_2)\) and \(R=(R_1,R_2)\), where \(R_j=1\) indicates that \(Y_j\) is observed. The integral produces the joint law of the full outcome vector and response pattern. It does not by itself integrate out unobserved outcomes or produce a likelihood contribution based only on observed data.

The Marginalization Calculation

If \(B\) is discrete, with possible values \(b_1,\ldots,b_K\) and probabilities \(w_k=P(B=b_k)\), the calculation is a finite mixture:

If \(B\) is continuous, the corresponding weighted average is an integral:

In either case, the value of \(b\) is not observed in the marginal law. Each possible value contributes according to its probability under the random-effect distribution and the conditional likelihood of the specified \(y\) and \(r\).

A Continuous Example and a Short Derivation

For a continuous random effect, marginalization may require evaluating an integral. A useful example also shows how the integral can sometimes be simplified analytically. Suppose \(B\sim N(0,1)\), the outcome satisfies \(Y\mid B=b\sim N(b,1)\), and the response indicator has probit probability \(P(R=1\mid B=b)=\Phi(b)\), where \(\Phi\) is the standard normal cumulative distribution function. Use the specified conditional product for the joint law of \(Y\) and \(R\).

At an outcome value \(y\), the joint density for \(Y=y\) and \(R=1\) is

where \(\phi\) is the standard normal density. To evaluate this at \(y=0\), first combine the prior density of \(B\) and the outcome likelihood. Completing the square gives

Thus the integral can be written as \(f_Y(y)E\{\Phi(B)\mid Y=y\}\), where \(f_Y\) is the marginal density of \(Y\). Here \(Y=B+\varepsilon\) with independent \(B,\varepsilon\sim N(0,1)\), so \(Y\sim N(0,2)\). At \(y=0\), \(f_Y(0)=1/\sqrt{4\pi}\approx0.2821\), and \(B\mid Y=0\sim N(0,1/2)\).

For \(Z\sim N(0,1)\) independent of \(B\), \(E\{\Phi(B)\}=P(Z\leq B)\). When \(B\sim N(m,v)\), \(B-Z\sim N(m,v+1)\), so \(P(Z\leq B)=\Phi\left(m/\sqrt{v+1}\right)\). Applying this with \(m=0\) and \(v=1/2\) gives \(E\{\Phi(B)\mid Y=0\}=0.5\). Therefore,

This is a joint density at \(Y=0\), not a probability that \(Y\) equals exactly zero; its units are inverse units of \(Y\). By symmetry, \(f(0,R=0)\) is also approximately \(0.1410\), and the two response-specific densities add to \(f_Y(0)\approx0.2821\).

Checks, Alternatives, and Limitations

Several checks help catch errors in derivation or computation:

• Check the conditional distributions. For each fixed \(b\), the outcome probabilities or density must be valid, and the response probabilities over all patterns must sum to 1.
• Check marginal normalization. Summing over all discrete outcomes and response patterns, or integrating over continuous components as appropriate, must give 1.
• Check what was integrated. The calculation above removes \(B\), not missing components of \(Y\). Marginalizing unobserved outcomes is a distinct operation.
• Check units and type. A result involving continuous outcomes is generally a density, not a point probability. A result for wholly discrete variables is a probability mass.
• Check numerical integration. If the integral has no closed form, compare quadrature results at increasing accuracy or with an independent numerical method. Instability, a value outside the permitted probability range for a discrete joint mass, or a failed normalization check signals a problem.

The shared-parameter calculation is one way to represent the joint law. The selection factorization orders the joint law around the outcome and response mechanism; the pattern-mixture factorization orders it around response patterns and their outcome distributions. These are alternative ways to express or model a joint distribution, not instructions to omit the integration over \(B\) when the specified shared-parameter model includes it. The earlier material on observed-data likelihood marginalization concerns which unobserved quantities must be removed to form an observed-data likelihood; that is a separate step from deriving the marginal law of \(Y\) and \(R\) over the random effect.

A common mistake is to treat the conditional product as if it were already the marginal joint law: \(f(y,r)=f(y\mid b)q(r\mid b)\). That expression still depends on \(b\), so it is conditional on the latent effect. The correction is to weight by \(h(b)\) and integrate over all possible \(b\). A related mistake is to integrate only the outcome factor or only the response factor and then multiply the two resulting marginal distributions. In general, that loses the shared mixing structure; the correct operation integrates the product.

The marginalization is only as credible as the model used inside the integral. A convenient random-effect distribution or simple conditional response model does not guarantee that the chosen specification is appropriate. The integral may also be difficult to evaluate, and different latent distributions or conditional models can lead to different marginal laws. Numerical accuracy does not resolve uncertainty about those modeling choices.