Control Functions For Poisson And Count Outcomes

Why Include a First-Stage Residual?

A count outcome might record the number of hospital visits, insurance claims, or purchases. A Poisson or negative-binomial model can describe such outcomes through a conditional mean, often using an exponential function to keep fitted means positive. But if a regressor is endogenous, an ordinary count regression can still give a misleading estimate: the regressor may be related to unobserved determinants of the count.

A control-function approach addresses this problem by using the first-stage residual as an additional outcome-regression variable. The residual summarizes the part of the endogenous regressor not explained by the instruments and included exogenous covariates. Under a suitable conditional-dependence assumption, including it makes the remaining variation in the regressor conditionally unrelated to the outcome disturbance. The approach is not justified merely by the fact that a residual can be calculated; its validity depends on the first-stage and outcome-model assumptions stated below.

Setup: For observation $i$, let $Y_i$ be a count outcome, $x_i$ a continuous endogenous regressor, $w_i$ a vector of included exogenous covariates, and $z_i$ a vector of excluded instruments. A linear first stage is $$ x_i = z_i'\pi + w_i'\kappa + v_i, $$ where $v_i$ is the first-stage disturbance. Estimate this equation and define the generated residual $\hat v_i=x_i-z_i'\hat\pi-w_i'\hat\kappa$.

The instruments must predict $x_i$ conditional on $w_i$, and the excluded instruments must not have a direct effect on the outcome conditional on the structural regressors and the control function. In addition, the control function must adequately summarize the dependence between $x_i$ and the outcome disturbance. These are substantive identification and specification requirements, not properties guaranteed by a two-stage calculation.

From a Conditional Error Decomposition to a Count Mean

A useful way to see the logic is to suppose that the structural conditional mean before accounting for endogeneity has a multiplicative unobserved component:

E[Y_i\mid x_i,w_i,u_i]=\exp(x_i\beta+w_i'\gamma+u_i).

Here $u_i$ captures unobserved determinants of the outcome. If $x_i$ is endogenous, $u_i$ and the first-stage disturbance $v_i$ are conditionally dependent. A linear control-function specification assumes a decomposition $u_i=\lambda v_i+\varepsilon_i$, with the conditional expectation $E[\exp(\varepsilon_i)\mid v_i,w_i,z_i]$ constant. The constant can be absorbed into the intercept. Taking the conditional expectation over $\varepsilon_i$ then gives:

\begin{aligned} E[Y_i\mid x_i,w_i,v_i] &=\exp(x_i\beta+w_i'\gamma+\lambda v_i) E[\exp(\varepsilon_i)\mid v_i,w_i,z_i]\\ &=\exp(\beta_0+x_i\beta+w_i'\gamma+\lambda v_i), \end{aligned}

where $\beta_0$ includes the constant factor. This calculation motivates including $v_i$ in the exponential index. A suitable joint normal model for latent errors can imply a linear conditional correction of this form, but linearity is an assumption about the conditional dependence, not a universal consequence of endogeneity. More generally, the correction may be a nonlinear function $h(v_i)$; then a linear residual term may be misspecified.

Control-function count mean: Under the linear residual-inclusion specification, estimate $$ E[Y_i\mid x_i,w_i,v_i] =\mu_i =\exp(\beta_0+x_i\beta+w_i'\gamma+\lambda v_i). $$ In practice, $v_i$ is replaced by the estimated first-stage residual $\hat v_i$. The coefficient $\lambda$ captures the modeled association between the first-stage disturbance and the conditional outcome mean, after conditioning on $x_i$ and $w_i$.

Poisson Residual Inclusion

For a conditional Poisson model, the distributional assumption is $Y_i\mid x_i,w_i,v_i\sim\operatorname{Poisson}(\mu_i)$, with $\mu_i$ given by the control-function mean above. A Poisson random variable has conditional mean $\mu_i$ and conditional variance $\mu_i$. The likelihood contribution for an observed count $y_i$ is $e^{-\mu_i}\mu_i^{y_i}/y_i!$. The Poisson distribution is a full conditional-distribution assumption; if the target is only the conditional mean, Poisson quasi-maximum likelihood can instead be used with a correctly specified mean and an appropriate robust variance estimator.

Worked Example: Constructing and Using a Poisson Control Function

Suppose a hypothetical study models the number of annual service visits $Y_i$. The first-stage estimate for visits’ endogenous predictor $x_i$ is $\hat x_i=0.8+0.5z_i+0.3w_i$. For one individual, $z_i=2$, $w_i=1$, and the observed $x_i=2.6$. The fitted first-stage value and residual are:

\hat x_i=0.8+0.5(2)+0.3(1)=2.1, \qquad \hat v_i=2.6-2.1=0.5.

Suppose the estimated Poisson conditional mean is $\hat\mu_i=\exp(-0.4+0.25x_i+0.2w_i+0.6\hat v_i)$. Substituting the observed values gives:

\hat\mu_i =\exp[-0.4+0.25(2.6)+0.2(1)+0.6(0.5)] =\exp(0.75) \approx 2.117.

Under the fitted model, this individual’s conditional expected number of annual visits is about 2.117. The model’s conditional Poisson variance for this individual is also 2.117. Holding $x_i$ and $w_i$ fixed, a one-unit increase in the residual changes the fitted mean by the factor $\exp(0.6)\approx1.822$. That is a model-based conditional association, not a causal effect of changing a residual.

The calculation requires more than a good numerical fit. It relies on the instrument exclusion and relevance conditions, a valid first-stage residual, and the assumption that the linear residual term adequately controls the relevant dependence. If the conditional Poisson distribution is assumed, that distributional specification is also required; for a mean-only Poisson fit, the conditional mean must still be correctly specified.

Negative-Binomial Residual Inclusion

A negative-binomial model keeps the same control-function conditional mean but allows more conditional variance than the Poisson model. Use the NB2 parameterization, in which $Y_i\mid x_i,w_i,v_i$ has a negative-binomial distribution with mean $\mu_i$ and dispersion parameter $\alpha>0$:

E[Y_i\mid x_i,w_i,v_i]=\mu_i, \qquad \operatorname{Var}(Y_i\mid x_i,w_i,v_i)=\mu_i+\alpha\mu_i^2.

The conditional mean can be specified as $\exp(\beta_0+x_i\beta+w_i'\gamma+\lambda v_i)$, with $\hat v_i$ used in estimation. The additional dispersion parameter allows the conditional variance to exceed the conditional mean. It does not, by itself, correct endogeneity: the control-function assumptions are still needed. A standard distributional interpretation of NB2 is a Poisson count conditional on a gamma-distributed rate, with the gamma variation independent of the covariates and control function given the specified conditional mean.

Worked Example: Negative-Binomial Mean and Variance

Consider a hypothetical model for the number of monthly claims, with estimated conditional mean $\hat\mu_i=\exp(-0.2+0.3x_i-0.1w_i+0.5\hat v_i)$ and estimated NB2 dispersion $\hat\alpha=0.4$. For an individual with $x_i=1.5$, $w_i=2$, and $\hat v_i=-0.4$, the index is:

$$ -0.2+0.3(1.5)-0.1(2)+0.5(-0.4) =-0.2+0.45-0.2-0.2=-0.15. $$

Thus the fitted mean is $\hat\mu_i=\exp(-0.15)\approx0.8607$ claims per month. The NB2 conditional variance is:

\widehat{\operatorname{Var}}(Y_i\mid x_i,w_i,\hat v_i) =\hat\mu_i+\hat\alpha\hat\mu_i^2 =0.8607+0.4(0.8607^2) \approx0.8607+0.2963 =1.157.

The fitted variance, about 1.157 squared claims, exceeds the fitted mean of 0.8607 claims because the estimated dispersion is positive. Holding $w_i$ and $\hat v_i$ fixed, a one-unit increase in $x_i$ multiplies the fitted mean by $\exp(0.3)\approx1.350$, or increases it by about 35.0%. This is a conditional mean comparison under the model, not evidence that the coefficient is causal without the identifying assumptions.

Estimation, Interpretation, and a Residual Check

A basic two-step implementation estimates the first-stage regression of $x_i$ on $z_i$ and $w_i$, saves $\hat v_i$, and then estimates the Poisson or negative-binomial outcome model using $x_i$, $w_i$, and $\hat v_i$. The structural coefficient $\beta$ describes the conditional log-mean slope for $x_i$, holding the included covariates and control function fixed. In particular, a one-unit increase in $x_i$ multiplies the model’s conditional mean by $\exp(\beta)$, under the specified model. The residual coefficient $\lambda$ is not the effect of an intervention on the residual; it is a parameter in the correction for conditional dependence.

Testing $H_0:\lambda=0$ can provide a diagnostic for endogeneity within this particular specification. A nonzero estimate is evidence of residual dependence conditional on the included variables, if the model and inference are valid. A failure to reject does not establish exogeneity: the test may have limited power, and misspecified instruments, first stages, or control functions can undermine its interpretation. Standard errors must account for the fact that $\hat v_i$ was estimated in the first stage. Treating it as a directly observed covariate and using an ordinary second-stage variance formula can understate or otherwise misstate uncertainty; use inference appropriate for estimated control functions.

Worked Example: Testing the Residual Term

Suppose a hypothetical Poisson residual-inclusion fit gives $\hat\lambda=0.45$ with a standard error that accounts for first-stage estimation of $0.18$. To test $H_0:\lambda=0$ against $H_a:\lambda\ne0$, the large-sample Wald statistic is:

z=\frac{\hat\lambda-0}{SE(\hat\lambda)} =\frac{0.45}{0.18}=2.50.

Using a standard normal reference distribution, the two-sided p-value is $2[1-\Phi(2.50)]\approx0.0124$. At significance level $\alpha=0.05$, reject $H_0$ for this fitted specification. An approximate 95% Wald interval is:

0.45\pm1.96(0.18) =0.45\pm0.3528 =[0.0972,\;0.8028].

The interval is in units of the model’s log conditional mean per unit of first-stage residual. It excludes zero, consistent with the test. The result indicates evidence that the residual term contributes to the specified conditional mean; it does not prove that the instruments are valid, that the linear control function is correct, or that endogeneity has been completely removed.

Why Fitted-Value Substitution Is Not the Same Procedure

A common error is to replace $x_i$ by its first-stage fitted value $\hat x_i$ in the nonlinear outcome model and treat the resulting regression as a solution to endogeneity. In nonlinear models, this generally changes the conditional mean being modeled and does not reproduce the control-function correction. Residual inclusion instead retains the observed $x_i$ and adds $\hat v_i$, relying on the specified conditional-dependence relationship. This distinction matters because the exponential mean is nonlinear: substituting a fitted regressor is not generally equivalent to conditioning on the residual.

The residual-inclusion approach is also not automatic protection against misspecification. The first stage must capture the conditional mean of $x_i$ sufficiently well; instruments must be relevant and excluded from the structural outcome equation; and the assumed residual correction must be adequate. If the relationship between the residual and the conditional mean is nonlinear, a linear term $\lambda\hat v_i$ may not suffice. A negative-binomial dispersion term can accommodate overdispersion, but it cannot repair a failed exclusion restriction or an inadequate control function.

Key takeaway: For a count outcome with one continuous endogenous regressor, residual inclusion retains the observed regressor and adds its estimated first-stage residual to the conditional-mean index. The Poisson or negative-binomial distribution governs conditional count variation; the control-function assumptions, not the choice of count distribution, do the work of addressing endogeneity.

Check Your Understanding

Work through these questions using the definitions and assumptions in this tutorial.

Write a linear first-stage equation for an endogenous continuous regressor $x_i$, and define its estimated residual.
In a Poisson residual-inclusion model, what does $\exp(\beta)$ represent for a one-unit increase in $x_i$, holding the other index terms fixed?
Under the NB2 parameterization, calculate the conditional variance when $\mu_i=2$ and $\alpha=0.25$.
Why does a significant estimate of $\lambda$ not establish that the excluded instruments are valid?
Explain why a negative-binomial dispersion parameter does not, by itself, address endogeneity.