Nonseparable Models And Limits Of Scalar Control Functions

Why a Residual May Not Be Enough

Residual inclusion uses information from the first stage to control for dependence between an endogenous regressor and unobserved determinants of the outcome. Its justification is not simply that a residual has been calculated and added to an equation. The key question is whether that control captures the features of the unobservables that matter for the outcome, conditional on the endogenous regressor and the instruments.

This issue becomes especially clear in a nonseparable structural model. If the outcome is a nonlinear function of an unobserved variable, its conditional mean can depend on more than the conditional mean of that variable. It may depend on its variance, higher moments, or its entire conditional distribution. A scalar residual is sufficient only under restrictions that make it an adequate index of the relevant conditional distribution or features.

Definition: Let \(Y=h(X,U)\) be a structural outcome equation, where \(X\) is endogenous and \(U\) is an unobserved determinant. A control \(C\) is sufficient for the conditional mean if the conditional distribution of \(U\) given \(X,Z\) is determined by \(X,C\). In particular, if \(F_{U\mid X,Z}=F_{U\mid X,C}\), then the conditional mean of \(Y\) given \(X,Z\) can be written as a function of \(X,C\), provided the relevant expectation exists. A first-stage residual is not automatically such a control.

The notation \(F_{U\mid X,Z}\) denotes the conditional distribution of \(U\) given \(X,Z\). Under sufficiency, every conditional feature of \(U\) needed to calculate the outcome mean can be recovered from \(X,C\). In some familiar models, a scalar residual has this property because the first stage and distributional assumptions make the latent outcome disturbance’s conditional law depend on the residual alone. In other models, the residual records only one aspect of the first-stage variation and misses other relevant features.

A Nonseparable Outcome Can Still Have a Scalar Control

Nonseparability alone does not invalidate scalar control functions. If the conditional distribution of the outcome unobservable is fully indexed by a scalar residual, that residual can control for endogeneity even when the structural outcome function is nonlinear. The following example makes the required distributional restriction explicit.

Worked Example: A Scalar Residual Controls a Nonlinear Outcome

Suppose the first stage is \(X=Z+V\), where \(Z\) is an instrument and \(V\sim N(0,1)\). Let \(U=0.6V+0.8\eta\), where \(\eta\sim N(0,1)\) is independent of \(V,Z\). The structural outcome is \(Y=\exp(XU)\). Calculate \(E[Y\mid X=x,V=v]\) at \(x=1.2\) and \(v=0.5\), and explain why \(V\) is sufficient here.

Conditional on \(V=v\), the distribution of \(U\) is normal with mean \(0.6v\) and variance \(0.8^2=0.64\). Since \(\eta\) is independent of \(V,Z\), conditioning additionally on \(Z\) supplies no further information about \(U\). Thus \(V\) indexes the conditional distribution needed for the outcome mean.

For a normal random variable \(U\) with mean \(m\) and variance \(s^2\), its moment-generating function gives \(E[\exp(tU)]=\exp(tm+t^2s^2/2)\). Therefore,

E[Y\mid X=x,V=v] = \exp\left(0.6xv+\frac{1}{2}x^2(0.64)\right).

Substituting the specified values,

E[Y\mid X=1.2,V=0.5] = \exp\left(0.6(1.2)(0.5)+\frac{1}{2}(1.2)^2(0.64)\right) = \exp(0.8208) \approx 2.272.

The outcome is nonseparable in \(X\) and \(U\), but the scalar residual works because it determines the conditional distribution of \(U\). The conclusion depends on the stated joint normal construction and independence, not on a general property of residuals.

The calculation also shows why the control term need not be linear or additive. The conditional mean involves both \(xv\) and \(x^2\); a specification that adds only a linear term in \(v\) would not reproduce this conditional mean. Sufficiency of a control and correct functional form for the outcome regression are separate requirements.

Why the Residual’s Scale Can Matter

A common residual is the difference between \(X\) and its conditional mean given the instrument. That residual can fail to summarize the relevant latent variation if its conditional distribution changes with \(Z\). In that case, the same residual value can correspond to different values or distributions of the structural unobservable across instrument values. A nonlinear outcome can then have different conditional means at the same \(X\) and residual.

Worked Example: Heteroskedastic First-Stage Errors

Let \(Z\) take values \(A\) and \(B\), independently of \(\varepsilon\), and suppose \(X=\sigma(Z)\varepsilon\), where \(\sigma(A)=1\), \(\sigma(B)=2\), and \(E[\varepsilon\mid Z]=0\). The first-stage residual from the conditional-mean first stage is \(V=X\). The structural outcome is \(Y=\exp(X\varepsilon)\). Compare the conditional outcome means at \(X=0.5\) for the two instrument values.

Given \(X=x\) and \(Z=z\), the first-stage equation implies \(\varepsilon=x/\sigma(z)\). Thus,

E[Y\mid X=x,Z=z] = \exp\left(\frac{x^2}{\sigma(z)}\right).

For \(Z=A\), \(X=0.5\), and \(V=0.5\),

E[Y\mid X=0.5,Z=A] = \exp\left(\frac{(0.5)^2}{1}\right) = \exp(0.25) \approx 1.284.

For \(Z=B\), the same \(X=0.5\) also gives \(V=0.5\), but

E[Y\mid X=0.5,Z=B] = \exp\left(\frac{(0.5)^2}{2}\right) = \exp(0.125) \approx 1.133.

The raw residual is the same in both cases, but the conditional outcome means differ. Here \(V\) alone does not capture the relevant latent value: \(\varepsilon=V/\sigma(Z)\). Under the specified first stage, a standardized residual \(C=V/\sigma(Z)\) recovers \(\varepsilon\) and can serve as a control. This conclusion relies on the scale specification being correct; in applications, the conditional scale may need to be modeled or otherwise restricted.

This example is a specific warning, not a claim that every heteroskedastic first stage defeats residual inclusion. If the control is constructed to capture the changing scale, or if restrictions imply that the outcome-relevant conditional distribution is still indexed by the chosen control, a control-function representation may remain possible. The important check is whether the proposed control contains the information needed about \(U\), not merely whether a residual has been included.

Nonlinear Means Can Require More Than a Conditional Mean

Even if a scalar control summarizes the conditional mean of an unobservable, that need not be enough for a nonseparable outcome. The reason follows directly from conditional expectation. For a general structural function \(h\),

E[Y\mid X=x,Z=z] = \int h(x,u)\,dF_{U\mid X,Z}(u\mid x,z).

The integral depends on the conditional distribution wherever \(h(x,u)\) is nonlinear in \(u\). If \(h(x,u)=a(x)+b(x)u\), it depends on the conditional mean of \(U\). But for a function involving \(U^2\), it may require a conditional second moment; for a function involving \(U^4\), it may require a conditional fourth moment. Knowing one moment does not, in general, determine the others.

Worked Example: Identical Conditional Means, Different Outcome Means

Consider two conditioning states, \(A\) and \(B\), with the same \(X=2\) and the same scalar location control \(C=E[U\mid X,Z]=0\). In state \(A\), let \(U\) equal \(-1\) or \(1\), each with probability \(1/2\). In state \(B\), let \(U\) equal \(-2\) or \(2\), each with probability \(1/2\). The structural outcome is \(Y=XU^2\). Find the conditional outcome mean in each state.

Both conditional distributions are symmetric around zero, so their conditional means are \(0\). But their second moments differ:

E[U^2\mid X=2,Z=A] = \frac{1}{2}(1^2)+\frac{1}{2}((-1)^2)=1, \qquad E[U^2\mid X=2,Z=B] = \frac{1}{2}(2^2)+\frac{1}{2}((-2)^2)=4.

Since \(Y=XU^2\), the conditional mean is \(X E[U^2\mid X,Z]\). Substituting \(X=2\) gives

E[Y\mid X=2,Z=A]=2(1)=2, \qquad E[Y\mid X=2,Z=B]=2(4)=8.

The scalar location control has the same value in both states, yet it cannot distinguish conditional outcome means of \(2\) and \(8\). A control for the conditional second moment is sufficient for this particular mean calculation because the outcome is affine in \(U^2\). It would not generally suffice for an outcome such as \(Y=XU^4\), which requires \(E[U^4\mid X,Z]\). The required control features depend on the structural outcome function.

Deriving the Sufficiency Condition

The control-function logic can be justified directly. Suppose the structural outcome is \(Y=h(X,U)\), and the conditional distribution of \(U\) given \(X,Z\) is the same as its conditional distribution given \(X,C\). Then conditional expectation gives

\begin{aligned} E[Y\mid X=x,Z=z] &= \int h(x,u)\,dF_{U\mid X,Z}(u\mid x,z)\\ &= \int h(x,u)\,dF_{U\mid X,C}(u\mid x,c)\\ &= E[h(X,U)\mid X=x,C=c]. \end{aligned}

The second equality follows from the sufficiency condition on the conditional distributions. Therefore, the conditional mean can be represented using \(X\) and \(C\), without an additional dependence on \(Z\), as far as this conditional-mean calculation is concerned. If the condition fails, this derivation fails at that equality: the distribution inside the integral can still vary with \(Z\) after conditioning on \(X,C\).

This is a condition for a control representation, not a complete identification result. Recovering structural effects additionally requires a correctly specified outcome relation, appropriate exclusion and instrument assumptions, and enough variation to distinguish the structural parameters. A control function does not, by itself, establish those requirements.

Choosing a Control and Avoiding Overclaims

When a scalar residual appears inadequate, the remedy is not automatically to add arbitrary functions until the outcome regression fits. The proposed richer control must be motivated by a model of how the unobservables enter the first stage and how they affect the outcome. Depending on that model, useful controls may include a standardized residual, a scale or variance index, multiple residual components, or conditional moments required by the outcome function. Each choice brings assumptions that should be stated and assessed.

A practical reasoning sequence is:

State the structural outcome.
Identify which unobservables enter \(h(X,U)\) and whether the function is affine or nonlinear in them.

Describe the first-stage information.
Specify what the residual measures and whether its conditional scale or other features vary with the instruments.

Check conditional sufficiency.
Ask whether \(X,C\) determine the distributional features of \(U\) needed for the conditional outcome mean.

State the added restriction.
If the original residual is inadequate, explain why the richer control is informative and what assumptions make it valid.

Common pitfall: “The first-stage residual is included, so endogeneity has been controlled.” This is not justified merely by including a residual. If two observations with the same \(X\) and residual can have different conditional distributions of outcome-relevant unobservables, the residual may not be sufficient. A related mistake is to assume that a control for \(E[U\mid X,Z]\) automatically controls every nonlinear function of \(U\). That works for outcomes affine in \(U\), but not generally for outcomes involving powers, exponentials, or other nonlinear transformations.

Key takeaway: A scalar residual is sufficient only when the model’s restrictions make it an adequate index of the conditional distribution—or of all conditional features needed by the outcome mean. Nonseparability does not rule out scalar control, but it makes the sufficiency condition substantive and outcome-specific.

Check Your Understanding

Use the conditional-distribution argument, not the fact that a residual appears in an equation, to assess each case.

In the first worked example, which distributional assumptions make \(V\) sufficient for the conditional mean of the nonlinear outcome?
In the heteroskedastic example, why do equal values of \(X\) and the raw residual not imply equal conditional outcome means?
For an outcome affine in \(U\), which conditional feature of \(U\) is needed to calculate the conditional mean?
For \(Y=XU^4\), explain why a control for \(E[U^2\mid X,Z]\) is not generally sufficient.
Does a sufficient control function alone establish identification of a structural effect? State one reason for your answer.