Deriving Residual Inclusion From A Latent-Error Model

Why a First-Stage Residual Can Matter

When treatment is endogenous, the treatment decision and the outcome can share unobserved causes. A control function represents the part of this dependence that can be learned from the treatment equation. For a binary treatment generated by a latent index, the relevant control is not simply the latent treatment error, which is unobserved. It is the conditional expectation of that error given the treatment and the observed predictors.

The derivation requires more than saying that treatment and outcome errors are correlated. In particular, their joint distribution must be specified conditional on the covariates and instrument. The assumptions below make the relationship explicit and show exactly what the first-stage residual does—and does not—tell us about the outcome equation.

Latent-Index Model and Assumptions

Definition: Let $D$ be a binary treatment, $X$ a vector of observed covariates, and $Z$ an instrument. Suppose treatment is generated by the latent index $D^*=m(X,Z)+V$, with $D=1$ when $D^*>0$ and $D=0$ otherwise. Assume that, conditional on every value of $(X,Z)$, the pair $(U,V)$ has the same bivariate standard normal distribution: $$ \begin{pmatrix} U \\ V \end{pmatrix}\Bigg|(X,Z) \sim N\left( \begin{pmatrix}0\\0\end{pmatrix}, \begin{pmatrix}1&\rho\\\rho&1\end{pmatrix} \right),\qquad |\rho|<1. $$

Here $V$ is the latent treatment error, while $U$ is an outcome error. The conditional distribution assumption is important: it says that the errors have the stated normal distribution at each $(X,Z)$, not merely in the pooled population. It implies that the instrument and covariates do not predict the errors through changes in their conditional distribution. For a causal instrumental-variable interpretation, one also requires a relevant instrument, an exclusion restriction (no direct effect of $Z$ on the outcome), and an appropriate structural outcome model. The derivation of the control function itself does not establish those causal conditions.

Because the conditional errors are jointly normal, their dependence can be represented as

U=\rho V+\sqrt{1-\rho^2}\,\varepsilon,

where $\varepsilon$ is standard normal and independent of $V,X,Z$. To see why the independence follows, define $\varepsilon=(U-\rho V)/\sqrt{1-\rho^2}$. Conditional on $(X,Z)$, it is jointly normal with $V$, has variance one, and has covariance zero with $V$. Thus it is independent of $V$ conditionally; because the conditional distribution is constant across $(X,Z)$, it is also independent of $(X,Z)$. This conditional argument is what permits the subsequent calculations.

Let $m=m(X,Z)$, and write $\phi$ and $\Phi$ for the standard normal density and cumulative distribution functions. Then $P(D=1\mid X,Z)=\Phi(m)$. Treatment reveals whether $V$ lies above or below the threshold $-m$, so conditioning on $D$ changes the expected value of $V$, and therefore the expected value of $U$.

Deriving the Generalized Residual

Define $p=\Phi(m)$. Conditional on treatment, the latent error is a truncated standard normal variable. The standard truncated-normal mean formulas give

E[V\mid D=1,X,Z]=\frac{\phi(m)}{\Phi(m)} =\frac{\phi(m)}{p}, \qquad E[V\mid D=0,X,Z]=-\frac{\phi(m)}{1-\Phi(m)} =-\frac{\phi(m)}{1-p}.

For example, when $D=1$, the event is $V>-m$. The conditional mean of a standard normal above a threshold $a$ is $\phi(a)/(1-\Phi(a))$; setting $a=-m$, and using symmetry of the normal density, gives $\phi(m)/\Phi(m)$. The lower-tail expression follows by the corresponding lower-truncation formula.

Since $E[\varepsilon\mid D,X,Z]=0$ under the model, the error decomposition yields $E[U\mid D,X,Z]=\rho E[V\mid D,X,Z]$. Thus the control function, the conditional expectation of the outcome error given the treatment decision and predictors, is

Key result: The generalized residual for the binary latent-index treatment model is $$ r(D,m)=E[V\mid D,X,Z] =D\frac{\phi(m)}{p}-(1-D)\frac{\phi(m)}{1-p}. $$ Consequently, $$ E[U\mid D,X,Z]=\rho\,r(D,m) =\rho\frac{\phi(m)}{p(1-p)}(D-p). $$

The last equality follows by checking the two treatment values. If $D=1$, then $D-p=1-p$, so the expression reduces to $\rho\phi(m)/p$. If $D=0$, then $D-p=-p$, so it reduces to $-\rho\phi(m)/(1-p)$. The observed first-stage residual $D-p$ therefore carries information about the outcome error, but it must be scaled by a function of the fitted treatment probability to equal the generalized residual.

Worked Examples

Worked Example: Calculating the Control for a Treatment Probability of One-Half

Suppose a hypothetical participant has $m=0$, so the model predicts $p=\Phi(0)=0.5$. Let the error correlation be $\rho=0.6$. Calculate the expected outcome error for a participant who receives treatment and one who does not.

At $m=0$, $\phi(0)=0.398942$. For $D=1$, the treatment residual is $D-p=1-0.5=0.5$, and $$ E[U\mid D=1,X,Z] =0.6\frac{0.398942}{0.5} =0.478730. $$ Equivalently, the generalized residual is $r=0.398942/0.5=0.797884$, and $\rho r=0.6(0.797884)=0.478730$.

For $D=0$, $D-p=-0.5$, and $$ E[U\mid D=0,X,Z] =-0.6\frac{0.398942}{0.5} =-0.478730. $$ The opposite signs arise because treatment selects participants with positive $V$, while non-treatment selects those with negative $V$, at this zero threshold. The values are conditional means of the error, not observed individual errors.

Worked Example: Including the Control in a Continuous Outcome Mean

Suppose a hypothetical continuous outcome follows $Y=2+1.5D+0.8X+2U$, where $X$ is a covariate and $U$ has the error distribution above. Let $m=0.5$, $\rho=0.4$, and consider a participant with $X=1$. The factor 2 sets the outcome scale for the standardized error $U$.

The first-stage probability and density are $p=\Phi(0.5)=0.691462$, $1-p=0.308538$, and $\phi(0.5)=0.352065$. For a treated participant, $$ r=\frac{0.352065}{0.691462}\approx0.50915,\qquad E[U\mid D=1,X,Z]=0.4(0.50915)\approx0.20366. $$ Therefore the conditional outcome mean is $$ E[Y\mid D=1,X=1,Z] =2+1.5(1)+0.8(1)+2(0.20366) \approx4.7073. $$

For an untreated participant with the same $m$ and $X$, $$ r=-\frac{0.352065}{0.308538}\approx-1.1411,\qquad E[U\mid D=0,X,Z]=0.4(-1.1411)\approx-0.4564. $$ Hence $E[Y\mid D=0,X=1,Z]=2+0+0.8+2(-0.4564)\approx1.8872$. These are model-based conditional means. The calculation is exact for this stated linear outcome equation with conditional mean-zero error beyond $U$; it does not show that an arbitrary nonlinear outcome model can be corrected by inserting the same term.

What Changes for a Binary Nonlinear Outcome?

For a nonlinear outcome, the control-function derivation still identifies $E[U\mid D,X,Z]$, but an expected error is not the whole conditional distribution of the error. Consider a binary outcome formed by thresholding a latent outcome index, $Y=1$ when $Y^*=\eta+U>0$. Given $V=v$, the error decomposition implies that $U$ is normal with mean $\rho v$ and variance $1-\rho^2$. Therefore the exact conditional success probability is

P(Y=1\mid D=d,X,Z) = \frac{\displaystyle\int_{A_d} \Phi\left(\frac{\eta+\rho v}{\sqrt{1-\rho^2}}\right)\phi(v)\,dv} {P(V\in A_d\mid X,Z)}, \quad A_1=(-m,\infty),\quad A_0=(-\infty,-m].

This expression averages over the full truncated distribution of $V$. Replacing $U$ everywhere by its conditional mean generally does not reproduce that integral. A nonlinear residual-inclusion specification can be useful, but the particular outcome link and the way the control enters it require an additional structural or modeling assumption; they do not follow merely from the treatment equation and the correlation $\rho$.

Worked Example: Why a Conditional Mean Is Not the Full Binary-Outcome Adjustment

Consider a hypothetical model with $m=0$, $\rho=0.5$, and $Y=1$ when $U>0$. Then $P(D=1)=P(D=0)=0.5$. The bivariate standard normal orthant identity is $P(U>0,V>0)=1/4+\arcsin(\rho)/(2\pi)$. Since $\arcsin(0.5)=\pi/6$, $$ P(Y=1\mid D=1) =\frac{P(U>0,V>0)}{P(V>0)} =\frac{1/4+1/12}{1/2} =\frac{2}{3}. $$

For $D=0$, the corresponding joint probability is $P(U>0,V\leq0)=1/4-\arcsin(\rho)/(2\pi)=1/6$. Thus $$ P(Y=1\mid D=0)=\frac{1/6}{1/2}=\frac{1}{3}. $$

The derived conditional error means are $E[U\mid D=1]=0.5\sqrt{2/\pi}\approx0.3989$ and $E[U\mid D=0]\approx-0.3989$. They correctly indicate the direction of selection, but they are not themselves the conditional probabilities $2/3$ and $1/3$. The example demonstrates why a nonlinear outcome equation needs a specified rule connecting the control to the conditional response.

Assumptions, Diagnostics, and Alternatives

The derivation depends on a correctly specified treatment index $m(X,Z)$, the conditional bivariate normal distribution with constant $\rho$, and the binary threshold rule. For an outcome interpretation, one additionally needs an outcome equation appropriate to the scale being analyzed. Causal interpretation further depends on instrument relevance and validity, including the exclusion restriction. Residual inclusion cannot repair a direct effect of the instrument or unmeasured confounding that violates the assumed error structure.

Useful checks include inspecting fitted treatment probabilities for near-zero or near-one values, assessing whether the treatment model captures systematic variation in $D$, and examining whether conclusions are sensitive to plausible alternative first-stage specifications. Extreme fitted probabilities make the inverse-probability factors in $r(D,m)$ large and indicate limited overlap; they can make estimates sensitive to the distributional assumptions. Such diagnostics can reveal instability or misspecification, but they cannot prove conditional normality or instrument exogeneity.

For a linear outcome, inserting the conditional error expectation gives a direct conditional-mean adjustment under the stated structural model. A joint likelihood approach instead specifies and estimates the outcome and treatment distributions together. Predictor substitution uses a fitted treatment predictor in the outcome equation, but it does not generally condition on the treatment error in this way. These alternatives rely on different modeling restrictions; choosing among them is not a substitute for stating the assumptions that justify the chosen model.

Confusing residuals. $D-p$ is the observed binary-treatment residual. The generalized residual $r(D,m)$ is its probability-dependent rescaling and equals $E[V\mid D,X,Z]$.
Assuming marginal normality is enough. A pooled bivariate normal distribution does not imply the required conditional distribution given $X,Z$. State the conditional assumption explicitly.
Treating a control as a cure-all. The control addresses a specified dependence structure. It does not establish valid instruments, correct the wrong outcome link automatically, or guarantee adequate overlap.

Key takeaway: Under a conditionally bivariate-normal latent-index treatment model, the first-stage generalized residual is the expected latent treatment error given observed treatment and predictors. Its correlation-scaled value is the expected outcome error. For nonlinear outcomes, that conditional mean alone does not determine the response model.

Check Your Understanding

Answer each question, then compare with the brief solutions.

Why must the error distribution be specified conditional on $X,Z$, rather than only marginally?
If $p=0.4$, $\phi(m)=0.30$, $\rho=0.5$, and $D=1$, calculate $r(D,m)$ and $E[U\mid D,X,Z]$.
For the same $p,\phi(m),\rho$, calculate $r(D,m)$ and $E[U\mid D,X,Z]$ when $D=0$.
What additional caution applies when the outcome is binary and defined by thresholding a latent index?

Solutions. (1) Marginal normality does not rule out errors whose distribution changes with $X,Z$; the conditional distribution is needed for the truncated-normal calculations and independence of the decomposed error. (2) $r=0.30/0.40=0.75$, so $E[U\mid D,X,Z]=0.5(0.75)=0.375$. (3) $r=-0.30/(1-0.40)=-0.50$, so $E[U\mid D,X,Z]=-0.25$. (4) The conditional mean of the error does not specify its full conditional distribution, so the nonlinear response probability requires an additional outcome-model assumption; simply substituting the mean into the latent index is not generally exact.