Heteroskedastic First Stages And Control-Function Misspecification

Why First-Stage Form Matters

Residual inclusion uses information from a first-stage model to control for the part of an endogenous regressor that is related to the structural error. The fitted residual is useful only insofar as it captures the relevant conditional dependence. If the first-stage conditional mean is wrong, its residual can contain systematic patterns that are not first-stage shocks. If the conditional distribution changes with instruments or covariates, a raw residual may also fail to summarize the relevant dependence in the same way at every covariate value.

These are distinct problems. Heteroskedasticity is variation in the conditional variance of the first-stage error; it does not, by itself, prove that residual inclusion is invalid. Conversely, a misspecified conditional mean can produce apparent variation in squared residuals even when the true conditional variance is constant. The central diagnostic discipline is therefore to assess the conditional mean before interpreting residual spread.

Definition: Let $x_i$ be an endogenous regressor, $z_i$ excluded instruments, and $w_i$ included exogenous covariates. Write the first stage as $x_i=m(z_i,w_i)+v_i$, where $m(z_i,w_i)=E[x_i\mid z_i,w_i]$ and $E[v_i\mid z_i,w_i]=0$. The structural equation is $y_i=\beta x_i+w_i'\gamma+u_i$. A control function is a function of first-stage information included in the structural conditional mean to account for dependence between $x_i$ and $u_i$. Residual inclusion uses an estimate of $v_i$, often denoted $\hat v_i$, as that control.

For example, a sufficient conditional-mean representation for a linear residual control is $E[u_i\mid x_i,z_i,w_i]=\rho v_i$, with $\rho$ constant, under the relevant instrument-exogeneity and structural assumptions. Then including $v_i$ in the outcome conditional mean accounts for the endogeneity represented by this decomposition. This is a conditional-mean requirement, not a claim that the first-stage error has constant variance. If instead the required conditional expectation depends on $v_i$, $z_i$, or $w_i$ in a more general way, including only the raw residual may be inadequate.

Heteroskedasticity: What Changes and What Does Not

Heteroskedasticity means $\operatorname{Var}(v_i\mid z_i,w_i)$ is not constant. It can change how the residual relates to the structural error. In one data-generating process, the conditional mean of $u_i$ may still be linear in the raw $v_i$, so the same residual control is appropriate for the conditional mean. In another, dependence may arise through a standardized shock or through a function whose scale varies with $z_i,w_i$. Then a raw-residual term with one common coefficient need not represent the right conditional expectation.

Worked Example: A Heteroskedastic First Stage and a Scaled Control

Suppose $z_i$ is binary, $\epsilon_i$ has conditional mean zero and conditional variance one given $z_i$, and the first-stage error is $v_i=\sigma(z_i)\epsilon_i$, where $\sigma(0)=1$ and $\sigma(1)=2$. Suppose the structural error has the illustrative decomposition $u_i=0.6\epsilon_i+\eta_i$, with $E[\eta_i\mid \epsilon_i,z_i]=0$. The first-stage conditional mean is assumed correctly specified.

Because $v_i=\sigma(z_i)\epsilon_i$, the standardized first-stage error is $\epsilon_i=v_i/\sigma(z_i)$. Thus the conditional expectation in this example is

E[u_i\mid v_i,z_i]=0.6\frac{v_i}{\sigma(z_i)}.

For $z_i=0$, the control is $0.6v_i$. For $z_i=1$, it is $0.3v_i$. A single term $c v_i$ with a common coefficient cannot represent both conditional means unless the coefficient is zero or the relevant variation happens to make the distinction immaterial. A suitable specification in this stylized example is to include $v_i/\sigma(z_i)$, or equivalently allow the coefficient on $v_i$ to vary with $z_i$.

This calculation illustrates a possibility, not a general rule that heteroskedasticity requires standardization. If the structural conditional expectation were instead $E[u_i\mid v_i,z_i]=0.6v_i$, heteroskedasticity would not invalidate the raw-residual conditional-mean control. The needed control is determined by the conditional relationship between $u_i$ and the first-stage information, not by a variance test alone.

A variance-robust covariance estimator addresses uncertainty calculations under certain forms of heteroskedasticity; it does not repair an incorrect conditional mean or a control function that omits relevant dependence. Likewise, a statistically significant variance pattern is not, on its own, evidence that the structural residual-inclusion specification is wrong. The model for the conditional mean and the model for the control function must be considered separately.

Why Mean Misspecification Can Mimic Changing Variance

Let the true first-stage conditional mean be $m(z_i,w_i)$, but suppose a fitted model uses $\tilde m(z_i,w_i)$. Its population residual is

\tilde v_i=x_i-\tilde m(z_i,w_i) =v_i+\{m(z_i,w_i)-\tilde m(z_i,w_i)\}.

The term in braces is a systematic conditional-mean error. Conditional on $z_i,w_i$, the fitted residual need not have mean zero. Squaring it therefore combines conditional variance with the square of its conditional mean.

Key result: For any fitted residual $\tilde v_i$, $$ E[\tilde v_i^2\mid z_i,w_i] =\operatorname{Var}(\tilde v_i\mid z_i,w_i) +\{E[\tilde v_i\mid z_i,w_i]\}^2. $$ Thus a relationship between squared fitted residuals and instruments or covariates is evidence about changing conditional variance only when the conditional mean has been adequately specified, or when systematic residual means have otherwise been accounted for.

The equality follows from the standard variance identity $E[A^2]=\operatorname{Var}(A)+\{E[A]\}^2$, applied conditionally to $A=\tilde v_i$. It is the reason a residual-squares plot cannot be interpreted in isolation: changing squared residuals may reflect a changing variance, a changing residual mean, or both.

Worked Example: A Wrong Mean Creates a Squared-Residual Pattern

Suppose the true first-stage error $\epsilon_i$ has $E[\epsilon_i\mid q_i]=0$ and $\operatorname{Var}(\epsilon_i\mid q_i)=4$. The true conditional mean of $x_i$ includes $q_i^2$, but a fitted model omits it, leaving a mean error of $q_i$. The fitted residual is then $\tilde v_i=\epsilon_i+q_i$. The conditional variance remains 4 for every $q_i$, but

E[\tilde v_i\mid q_i]=q_i,\qquad E[\tilde v_i^2\mid q_i]=4+q_i^2.

At $q_i=0$, the expected squared residual is $4+0^2=4$. At $q_i=2$, it is $4+2^2=8$. Squared residuals therefore rise with $q_i$, even though the true conditional variance of the first-stage error is constant. Attributing this pattern to heteroskedasticity without checking the conditional mean would be incorrect.

The example also shows why plotting the signed residuals matters. Their conditional mean changes from 0 at $q_i=0$ to 2 at $q_i=2$. That systematic pattern points to a mean problem which should be investigated before the residual-squares pattern is treated as a variance diagnosis.

Nonlinear First-Stage Misspecification

A linear first stage can be too restrictive when the conditional mean is curved or includes interactions. If the fitted mean misses that structure, the residual is not an estimate of the mean-zero disturbance alone. In a control-function application, that matters because residual inclusion relies on the residual representing the relevant first-stage variation. A misspecified residual can combine that variation with a predictable function of the instruments and covariates.

Worked Example: A Quadratic Conditional Mean Fitted as a Line

Consider three equally frequent values $z_i=0,1,2$. Suppose the true first-stage conditional means of $x_i$ at these values are 1.0, 1.7, and 2.8, respectively, and the disturbance has conditional mean zero and constant variance $0.04$. Fit a linear conditional mean with an intercept and $z_i$.

For these three equally weighted points, the fitted slope is

\hat b =\frac{\sum (z_i-\bar z)(\bar x_{z_i}-\bar x)} {\sum (z_i-\bar z)^2} =\frac{(-1)(-0.8333)+0(-0.1333)+(1)(0.9667)}{(-1)^2+0^2+1^2} =\frac{1.800}{2}=0.9000,

where $\bar x_{z_i}$ denotes the stated conditional mean at each value of $z_i$, and $\bar z=1$, $\bar x=1.8333$. The fitted intercept is $1.8333-(0.9000)(1)=0.9333$. The fitted means at $z_i=0,1,2$ are therefore $0.9333$, $1.8333$, and $2.7333$. Subtracting these from the true means gives conditional means of the fitted residual of $0.0667$, $-0.1333$, and $0.0667$.

Even with a true disturbance variance of $0.04$ at all three values, the expected squared fitted residuals are $0.04+0.0667^2=0.0444$, $0.04+(-0.1333)^2=0.0578$, and $0.04+0.0667^2=0.0444$, to the displayed precision. The residuals have a systematic mean pattern, and their squared values inherit that pattern. The raw linear-first-stage residual should not be treated as a clean, mean-zero first-stage shock.

A quadratic conditional-mean specification can remove this particular error if it is correctly specified and supported by the data. But choosing a richer first stage does not automatically prove that its residual is a valid structural control: the control-function conditional-mean restriction still needs to be appropriate.

A Diagnostic Sequence

Diagnostics are most informative when they separate questions about the conditional mean from questions about the conditional variance. Plots and specification checks are evidence for model refinement, not proofs of instrument validity or of the structural control-function restriction.

Specify the first-stage conditional mean.
Include the instruments and exogenous covariates, and consider substantively plausible nonlinear terms and interactions. The target is $E[x_i\mid z_i,w_i]$, not simply a high in-sample fit.

Inspect signed residuals.
Plot fitted residuals against fitted values, instruments, covariates, and plausible nonlinear transformations. Curvature or systematic group differences can indicate a mean problem. A residual plot is a diagnostic, not a formal demonstration that the conditional mean is correct.

Refine and check the mean model.
Use specification checks appropriate to the design, such as tests for omitted nonlinear terms, and compare reasonable conditional-mean specifications. Consider whether patterns persist out of sample or under flexible fits; excessive flexibility can also overfit.

Assess conditional spread after mean adequacy.
Then inspect residual magnitudes or squares against relevant covariates and instruments, allowing for remaining systematic residual means where necessary. A residual-squares association is not a variance diagnosis if the mean is misspecified.

Choose and justify the control.
If the conditional relationship calls for a scaled residual, interactions, or a richer function, specify that control explicitly. State the conditional-mean restriction it represents and use inference that accounts for estimating first-stage quantities.

Worked Example: Separating a Spread Pattern From a Mean Pattern

For illustration, suppose a correctly specified first-stage conditional mean is fitted in two instrument groups. The following hypothetical fitted residuals are observed, in the units of $x_i$:

Instrument group	Fitted residuals	Mean residual	Mean squared residual
$z_i=0$	$-1, 1, -2, 2$	$(-1+1-2+2)/4=0$	$(1+1+4+4)/4=2.5$
$z_i=1$	$-2, 2, -4, 4$	$(-2+2-4+4)/4=0$	$(4+4+16+16)/4=10.0$

The group means are both zero, while the empirical mean squared residual is four times as large in the second group: $10.0/2.5=4.0$. Given the stated premise that the conditional mean is adequately specified, this pattern is consistent with greater conditional spread for $z_i=1$. These eight observations are only a descriptive illustration; they do not establish a population variance difference or determine the correct structural control. In an application, sampling uncertainty and the adequacy of the mean specification also need to be considered.

Alternatives and Interpretation

When diagnostics point to a first-stage mean problem, a natural response is to estimate a better conditional mean—for example, by adding justified nonlinear terms, interactions, or a flexible model—and form residuals from that fit. When the conditional spread varies, one may model the variance and consider controls based on standardized residuals or allow the control’s coefficient or functional form to vary with covariates. Such changes should follow from a substantive conditional-mean argument, not just from selecting whichever specification produces a preferred outcome estimate.

The first-stage model and the structural control must be considered together. A good prediction of $x_i$ does not by itself establish that the residual captures all dependence between $x_i$ and $u_i$. Nor does a variance model alone establish that the chosen residual function is the correct control. Report the first-stage specification, the diagnostic reasoning, and the form of residual control used. If results change substantially across plausible specifications, that sensitivity is important evidence about how much the conclusion depends on modeling choices.

Common Mistakes

Calling every squared-residual pattern heteroskedasticity. This is wrong when the fitted conditional mean leaves systematic residual means. Check signed residual patterns and mean specification first; only then interpret residual spread as evidence about conditional variance.
Assuming heteroskedasticity automatically invalidates residual inclusion. It does not. The relevant question is whether the chosen control represents $E[u_i\mid x_i,z_i,w_i]$ adequately under the model.
Assuming a flexible first stage guarantees a valid control. It can improve estimation of the conditional mean, but it does not establish the structural conditional-mean restriction.
Treating robust standard errors as a correction for misspecification. Robust covariance estimation may address some variance-related inference problems, but it does not fix a wrong first-stage mean or an inadequate control function.

Key takeaway: Diagnose the first-stage conditional mean before interpreting squared residuals. Heteroskedasticity and mean misspecification are different features, and neither alone determines the appropriate control: residual inclusion is justified by the conditional relationship between the structural error and first-stage information.

Check Your Understanding

Answer each question by identifying the relevant conditional mean or variance feature.

Why can squared fitted residuals vary with an instrument even when the true first-stage error has constant conditional variance?
Does heteroskedasticity necessarily invalidate a raw-residual control? State what determines whether that control is adequate.
In the scaled-shock example, what control represents the conditional mean of the structural error when the first-stage scale is $\sigma(z_i)$?
What feature of signed residuals would suggest a first-stage conditional-mean problem before examining residual squares?
Why do heteroskedasticity-robust standard errors not repair an incorrectly specified residual-inclusion model?

Instrument group	Fitted residuals	Mean residual	Mean squared residual
\(z_i=0\)	\(-1, 1, -2, 2\)	\((-1+1-2+2)/4=0\)	\((1+1+4+4)/4=2.5\)
\(z_i=1\)	\(-2, 2, -4, 4\)	\((-2+2-4+4)/4=0\)	\((4+4+16+16)/4=10.0\)