Random-Intercept Logit for Panel Outcomes

Why Repeated Binary Outcomes Need a Panel Model

Suppose an analyst records whether each person is employed in several years. A standard logit model can relate employment in one year to that year’s covariates, but treating every person-year as unrelated discards the fact that one person’s outcomes may resemble one another. Stable characteristics that are not measured—such as persistent preferences, health, or ability—can contribute to this resemblance.

A random-intercept logit represents that stable person-specific component with a latent intercept shared across the person’s observations. Conditional on the intercept, the model is an ordinary logit for each occasion. The intercept is then integrated out to obtain probabilities and likelihood contributions that account for the assumed distribution of unobserved heterogeneity.

Definition: Let $Y_{it}\in\{0,1\}$ denote the binary outcome for individual $i=1,\ldots,N$ on occasion $t=1,\ldots,T_i$, and let $x_{it}$ be a vector of observed covariates. A random-intercept logit model is $$ P(Y_{it}=1\mid x_{it},a_i)=\Lambda(x_{it}'\beta+a_i),\qquad \Lambda(u)=\frac{1}{1+\exp(-u)}, $$ where $a_i$ is an individual-specific random intercept shared across occasions and $\beta$ is a coefficient vector. The standard specification assumes $a_i\sim N(0,\sigma_a^2)$, independently across individuals and independently of the individual’s full covariate history.

The model also assumes that, conditional on the covariates and $a_i$, the outcomes $Y_{i1},\ldots,Y_{iT_i}$ are independent Bernoulli variables with the probabilities in the definition. This is conditional independence; it does not assert that repeated outcomes are independent without conditioning on $a_i$. The zero mean for $a_i$ is a location convention: an overall intercept can be included in $x_{it}'\beta$, while $\sigma_a^2$ describes the variance of the person-specific deviations.

Conditional Interpretation and the Independence Assumption

For two covariate values that differ by one unit in component $k$, holding the other covariates and the random intercept fixed, the conditional log odds change by $\beta_k$. The corresponding conditional odds ratio is $\exp(\beta_k)$. Thus $\beta_k$ describes a contrast among observations with the same value of the latent intercept. It is not, in general, the log-odds contrast after averaging over the random-intercept distribution.

Independence of $a_i$ and the entire covariate history is substantive. It says that, in the population represented by the model, people with different covariate histories do not systematically differ in the distribution of their latent intercepts. If, for example, people with more education tend to have higher unobserved propensity for employment, this assumption may fail. The conventional random-intercept model does not remove that dependence merely by including a random intercept. The later correlated-random-effects approach addresses a different specification; here, the independence assumption is stated as part of the model.

Worked Example: Interpreting a Conditional Coefficient

Consider a hypothetical panel model for whether a person is employed. Suppose the coefficient for an additional year of education is $\hat\beta_{\text{edu}}=0.70$, with education measured in years. Interpret the coefficient and calculate the conditional odds ratio.

Conditional log odds. Holding the other included covariates and the person’s random intercept fixed, one additional year of education changes the log odds of employment by $0.70$.

Conditional odds ratio. $$ \exp(\hat\beta_{\text{edu}})=\exp(0.70)=2.014. $$

Under this model, the conditional odds of employment are about $2.014$ times as large for a one-year increase in education, holding the stated quantities fixed. This is an odds comparison, not a statement that the probability of employment doubles. It also relies on the model’s conditional logit specification and does not by itself establish a causal effect.

Integrating the Individual Effect

For a fixed value $a_i$, conditional independence gives the probability of the person’s observed outcome history as a product of Bernoulli probabilities. But $a_i$ is unobserved. The likelihood contribution for that person is therefore the conditional likelihood averaged over the assumed distribution of $a_i$, not the conditional likelihood evaluated at $a_i=0$.

Individual likelihood contribution: Write $p_{it}(a)=\Lambda(x_{it}'\beta+a)$. Under the normal random-intercept specification and conditional independence, $$ L_i(\beta,\sigma_a)= \int_{-\infty}^{\infty} \left\{\prod_{t=1}^{T_i} p_{it}(a)^{Y_{it}}\,[1-p_{it}(a)]^{1-Y_{it}}\right\} \frac{1}{\sigma_a}\phi\!\left(\frac{a}{\sigma_a}\right)\, da, $$ where $\phi$ is the standard normal density. The sample likelihood is $\prod_{i=1}^{N}L_i(\beta,\sigma_a)$; its log is $\sum_{i=1}^{N}\log L_i(\beta,\sigma_a)$.

The product inside the integral is the conditional likelihood for one person. Integration averages that whole product over the person-effect distribution. In general, it is not equal to the product of separately integrated occasion probabilities: the same $a_i$ enters every occasion. That shared latent value is precisely what induces dependence in the observed outcome history.

Worked Example: Integrating a Two-Occasion Outcome Pattern

Consider a hypothetical person observed twice, with $Y_{i1}=1$ and $Y_{i2}=0$. The linear predictors excluding the random intercept are $1$ and $0.3$, respectively. Let $\sigma_a=0.8$. Approximate the normal integral using the three-node Gauss–Hermite rule for a standard normal: nodes $-\sqrt{3},0,\sqrt{3}$ with weights $1/6,2/3,1/6$. The corresponding random-intercept values are $-0.8\sqrt{3},0,0.8\sqrt{3}$, approximately $-1.386,0,1.386$.

At a given value $a$, the conditional probability of this outcome pattern is $\Lambda(1+a)[1-\Lambda(0.3+a)]$. At $a=1.386$, the two probabilities of success are $\Lambda(2.386)\approx0.9158$ and $\Lambda(1.686)\approx0.8437$, so the pattern probability is $0.9158(1-0.8437)\approx0.1431$.

At $a=0$, the pattern probability is $\Lambda(1)[1-\Lambda(0.3)]\approx0.3111$. At $a=-1.386$, it is $\Lambda(-0.386)[1-\Lambda(-1.086)]\approx0.4047(0.7476)=0.3026$. The integrated approximation is therefore $$ L_i\approx \frac{1}{6}(0.14317)+\frac{2}{3}(0.31111)+\frac{1}{6}(0.3026) =0.2817. $$

This is an approximation to the probability of the person’s observed two-outcome pattern under the specified model. In particular, it is not obtained by multiplying the two marginal probabilities of the outcomes; both occasions share the same random intercept in the integral.

Marginal Probabilities and Induced Dependence

A population-averaged probability for a covariate vector $x$ is obtained by integrating the conditional probability itself: $$ P(Y_{it}=1\mid x_{it}=x)= \int_{-\infty}^{\infty}\Lambda(x'\beta+a) \frac{1}{\sigma_a}\phi\!\left(\frac{a}{\sigma_a}\right)\, da. $$

Because the logistic function is nonlinear, this integral is generally not $\Lambda(x'\beta)$, which is the conditional probability at $a=0$. For the same reason, the population-averaged log odds need not differ across covariate values by exactly $\beta_k$. A conditional odds ratio and a marginal odds ratio answer different questions. Marginal probabilities can be computed numerically, and marginal effects require accounting for the integration as well as the change in covariates.

Worked Example: Marginal Probabilities and Dependence

Use the hypothetical two-occasion model from the previous example, with $\sigma_a=0.8$, linear predictors $1$ and $0.3$, and the same three-node approximation. First calculate the marginal success probability at each occasion.

For occasion 1, the conditional probabilities at the positive, zero, and negative nodes are approximately $0.9158$, $0.7311$, and $0.4048$. Thus $$ P(Y_{i1}=1)\approx\frac{1}{6}(0.9158)+\frac{2}{3}(0.7311)+\frac{1}{6}(0.4048)=0.7075. $$ For occasion 2, the corresponding probabilities are $0.8437$, $0.5744$, and $0.2524$, giving $$ P(Y_{i2}=1)\approx\frac{1}{6}(0.8437)+\frac{2}{3}(0.5744)+\frac{1}{6}(0.2525)=0.5656. $$

The integrated probability of $Y_{i1}=1,Y_{i2}=1$ is approximately $\frac{1}{6}(0.9158)(0.8437)+\frac{2}{3}(0.7311)(0.5744)+\frac{1}{6}(0.4048)(0.2524)=0.4258$. The product of the marginal probabilities is approximately $0.7075(0.5656)=0.4002$, which differs from $0.4258$. The covariance is approximately $0.4258-0.4002=0.0256$. In this example the shared intercept induces positive dependence: high values of $a_i$ raise both conditional probabilities, and low values lower both.

The direction and magnitude of dependence are not a universal numerical constant; they depend on the covariates and model parameters. The calculation demonstrates why conditional independence should not be confused with unconditional independence.

Why the Model Matters—and a Common Pitfall

Integrating over the random intercept gives a coherent likelihood for repeated binary responses while representing persistent, unobserved differences between people. It also makes explicit which variation the model attributes to stable person-level heterogeneity. The normal distribution is a modeling assumption, not a fact guaranteed by having repeated measurements. Inference and predictions depend on the assumed distribution and on the independence of that effect from the covariate history.

A common pitfall is to plug $a_i=0$ into the logit and call the resulting probability the population probability. That quantity is the conditional probability for an individual whose random intercept is at its mean, not the average probability across the random-intercept distribution. The correction is to integrate $\Lambda(x'\beta+a)$ over the distribution of $a$, as in the marginal probability formula. A related error is to claim repeated outcomes are independent because the model assumes conditional independence; after integration, the shared unobserved effect generally makes them dependent.

Key takeaway: The random-intercept logit specifies a logit model conditional on a shared individual effect. Its individual likelihood contribution integrates the conditional outcome-history likelihood over that effect’s distribution. Coefficients are conditional log-odds effects, while population-averaged probabilities require their own integration.

Check Your Understanding

Answer each question using the model and assumptions stated above.

What does $\exp(\beta_k)$ compare when the random intercept is held fixed?
Why must the conditional likelihood for one person be integrated as a product over occasions, rather than integrating each occasion’s probability separately?
What does the assumption that $a_i$ is independent of the full covariate history rule out?
Why is $\Lambda(x'\beta)$ generally not the population-averaged probability?
Can repeated outcomes be conditionally independent and still be dependent after the random intercept is integrated out? Explain.