Distributional Assumptions Behind Residual Inclusion

Why the First-Stage Residual Needs a Justification

Including a first-stage residual in an outcome equation is not automatically valid just because the residual was estimated correctly. The key question is whether that residual contains enough information about the structural error to make the outcome equation’s conditional mean correctly specified. The answer depends on assumptions about conditional means or, more strongly, conditional distributions.

Consider a structural equation with one endogenous regressor \(D\), an instrument \(Z\), and observed covariates \(X\):

Here \(V=D-m(Z,X)\) is the population first-stage disturbance, where \(m(Z,X)=E[D\mid Z,X]\) is the correctly specified conditional mean. In applications, \(m\) is estimated and a fitted residual is used. The structural error is \(\varepsilon\). A control-function approach seeks a function \(h\) such that

Since \(D=m(Z,X)+V\), conditioning on \(D,Z,X\) also determines \(V\). If the displayed restriction holds, then \(E[Y\mid D,Z,X]=\beta D+X'\gamma+h(V,X)\). Including the correct function of \(V\) can therefore account for the conditional-mean dependence between \(D\) and \(\varepsilon\). The restriction is substantive: it says the conditional mean of the structural error depends on \(D\) and \(Z\) only through the first-stage disturbance and \(X\).

Linear Conditional Expectations Do Not Require Normality

One route to residual inclusion directly assumes a linear conditional expectation, such as \(E[\varepsilon\mid V,X]=\lambda V\). This is a restriction on a conditional mean, not a claim that the variables have a normal distribution. Even this restriction must be stated carefully: by itself, it does not ensure \(E[\varepsilon\mid D,Z,X]=\lambda V\). The latter also requires that conditioning on \(Z\) supplies no further information about the conditional mean of \(\varepsilon\) once \(V\) and \(X\) are known.

A sufficient conditional-mean assumption is \(E[\varepsilon\mid V,Z,X]=\lambda V\), with a constant \(\lambda\). It implies \(E[\varepsilon\mid D,Z,X]=\lambda V\), because \(V\) is determined by \(D,Z,X\). Under this assumption, the outcome conditional mean has the form

No normality is needed for this conclusion. The error distribution may be discrete, skewed, or heavy-tailed, provided the conditional-mean restriction and the other model requirements hold. If the coefficient varies with covariates, the relevant conditional mean might instead be \(\lambda(X)V\); a single common residual coefficient would then be misspecified unless that variation is accommodated.

Joint Normality as a Sufficient Route

Joint normality can justify a linear conditional mean under specific conditions. Suppose that, conditional on \(X\), the pair \((\varepsilon,V)\) is jointly normal, has conditional means zero, and has conditional covariance and variance \(\operatorname{Cov}(\varepsilon,V\mid X)=\sigma_{\varepsilon V}(X)\) and \(\operatorname{Var}(V\mid X)=\sigma_V^2(X)>0\). Also require that the conditional distribution of this pair does not vary with \(Z\) once \(X\) is given. The normal conditional-expectation formula then gives

The ratio is the conditional linear projection coefficient. If it is constant in \(X\), a common linear residual term is justified. If it varies with \(X\), the conditional mean is still linear in \(V\) at each \(X\), but the model must allow the coefficient to vary. Conditional joint normality is sufficient for this linear form; it is not necessary. Nor does normality alone fix an incorrectly specified first-stage conditional mean or eliminate dependence on \(Z\) beyond \(V\).

Broader Conditional-Distribution Restrictions

A residual can be useful under assumptions broader than joint normality. A particularly clear condition is that the conditional distribution of \(\varepsilon\) given \(D,Z,X\) depends on \(D,Z\) only through \(V\) and \(X\). In notation, for every relevant set \(A\),

This distributional restriction implies the corresponding conditional-mean restriction whenever the conditional mean exists. The conditional distribution on the right need not be normal, and its mean need not be linear in \(V\). For example, it might yield a nonlinear function \(h(V,X)\). In that case, the residual may still contain the required information, but a model that includes only a linear residual term is not enough. A flexible function of the residual, or an explicitly justified nonlinear specification, would be needed for the conditional mean.

The important distinction is between what information the control contains and how that information enters the conditional mean. A scalar residual might be sufficient information but require a nonlinear function. Alternatively, the conditional distribution might depend on other features of \(D,Z\) even after \(V,X\) are known, in which case the residual itself is not sufficient.

When a Residual Does Not Summarize Dependence

Correctly specifying the first-stage conditional mean does not guarantee that its residual is sufficient. Even with \(V\) independent of \(Z\), the conditional mean of \(\varepsilon\) can vary with \(Z\) at a fixed value of \(V\). This violates the restriction that the conditional mean depends on \(D,Z\) only through \(V,X\).

How to Assess a Residual-Inclusion Argument

The assumptions behind residual inclusion should be separated rather than treated as a single package. First, the first-stage conditional mean must be appropriate for defining \(V\). Second, the conditional-mean or conditional-distribution restriction must say that \(V\) captures the relevant dependence of \(\varepsilon\) on \(D,Z\). Third, the outcome equation must use the correct function of \(V\), which may be nonlinear or vary with \(X\). Finally, the instrument and identification conditions needed for the structural parameter remain necessary; residual inclusion does not establish instrument validity by itself.

A common mistake is to argue, “The first stage is linear, so adding its residual controls for endogeneity.” Linearity of \(E[D\mid Z,X]\) defines a convenient residual; it does not imply that \(E[\varepsilon\mid D,Z,X]\) is linear in that residual, or even that it depends on the residual alone. The correction is to state and defend the relevant conditional-mean or conditional-distribution assumption, and then match the outcome equation to it. Joint normality can provide one sufficient justification under the conditions above, but it is not the only justification and should not be inferred from a linear first stage.