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Binary and Nonlinear Outcome IV Methods · Tutorial 551 of 699

Why Nonlinear Outcomes Break Ordinary Two-Stage Substitution

Learn why first-stage fitted treatment values cannot generally be substituted into nonlinear outcome models, and how to identify the resulting mismatch.

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What You'll Learn

  • Distinguish a fitted treatment value from the treatment actually received.
  • Derive the mixture of binary-outcome probabilities conditional on an instrument.
  • Compare that mixture with the probability obtained by plugging in a fitted value.
  • Recognize when nonlinear substitution can agree and when it generally cannot.
  • Diagnose common failures and distinguish them from the linear 2SLS procedure.

Why Substitution Is Tempting—and Why It Fails

In linear instrumental-variables analysis, a familiar procedure uses the first stage to predict treatment and then uses the prediction in a second-stage calculation. It is tempting to carry the same idea over to a probit or logit outcome model: predict treatment, insert that prediction into the nonlinear outcome equation, and interpret the resulting coefficient as an IV effect. The difficulty is not merely that the outcome is binary. The key issue is that a nonlinear function of an average is generally not the average of the nonlinear function.

For a binary treatment, the first-stage fitted value is often a predicted treatment probability. A person with fitted value \(0.2\), for example, did not receive “0.2 units” of a binary treatment. Rather, \(0.2\) represents a conditional average or probability under the first-stage model. Substituting that value into a nonlinear outcome response function does not usually reproduce the average outcome probability across people with treatment values zero and one.

Definition: Ordinary predictor substitution in a nonlinear outcome model means estimating a first-stage conditional mean \(\hat D\) for treatment \(D\), then using \(\hat D\) in place of \(D\) as a predictor in a probit, logit, or other nonlinear outcome model. This operation is generally not a valid nonlinear analogue of linear two-stage least squares (2SLS).

To isolate the mathematical problem, let \(Z\) be a binary instrument and \(D\) a binary treatment. Suppose, for the algebra in the next example, that the conditional response restriction is \(P(Y=1\mid D,Z)=g(\alpha+\beta D)\), where \(g\) is a nonlinear response function such as the standard normal cumulative distribution function \(\Phi\) or the logistic function. This restriction explicitly says that, conditional on treatment, the response probability does not additionally depend on \(Z\). It is a simplifying condition for demonstrating the substitution problem; it is not a complete set of assumptions for causal identification.

The Mixture Versus the Plug-In Probability

Write \(p_z=P(D=1\mid Z=z)\). Because \(D\) is binary, conditioning on its two possible values gives the actual outcome probability at instrument value \(z\):

$$ \begin{aligned} P(Y=1\mid Z=z) &=P(Y=1\mid D=0,Z=z)P(D=0\mid Z=z)\\ &\quad+P(Y=1\mid D=1,Z=z)P(D=1\mid Z=z)\\ &=(1-p_z)g(\alpha)+p_zg(\alpha+\beta). \end{aligned} $$

Ordinary predictor substitution instead evaluates the response function at the fitted treatment probability: \(g(\alpha+\beta p_z)\). These are different expressions. The first averages two response probabilities, weighting them by the probabilities of the two treatment states. The second treats the average treatment value as if it were itself a treatment level to be passed through the nonlinear response function.

Key result: In general, \((1-p)g(\alpha)+pg(\alpha+\beta)\ne g(\alpha+\beta p)\). Equality is guaranteed when the response function is affine over the relevant range (as in a linear conditional-mean model), but can also occur in special cases such as \(p=0\), \(p=1\), or \(\beta=0\). Nonlinearity means equality should not be presumed.

This is an instance of the general expectation identity that fails for nonlinear functions: \(E[g(V)\mid Z]\ne g(E[V\mid Z])\) in general. Jensen’s inequality can sometimes describe the direction of the difference if the function is convex or concave over the entire relevant range. But probit and logit response functions are not globally convex or globally concave as functions of their index, so there is no universal direction for the substitution error.

Worked Binary-Outcome Examples

Worked Example: Probit With a Binary Treatment

Consider the hypothetical response restriction \(P(Y=1\mid D,Z)=\Phi(-1+1.5D)\). Suppose the first-stage treatment probabilities are \(p_0=0.2\) when \(Z=0\) and \(p_1=0.8\) when \(Z=1\). Calculate the outcome probabilities conditional on each instrument value, then compare them with ordinary probit substitution.

The two treatment-specific outcome probabilities are \(\Phi(-1)\approx0.1586553\) for \(D=0\), and \(\Phi(0.5)\approx0.6914625\) for \(D=1\). Under \(Z=0\), the mixture probability is

$$ \begin{aligned} P(Y=1\mid Z=0) &=(1-0.2)\Phi(-1)+0.2\Phi(0.5)\\ &=0.8(0.1586553)+0.2(0.6914625)\\ &\approx0.2652167. \end{aligned} $$

Substitution instead gives \(\Phi(-1+1.5(0.2))=\Phi(-0.7)\approx0.2419637\). The plug-in probability is about \(0.02325\) lower than the mixture probability.

For \(Z=1\), the mixture is

$$ \begin{aligned} P(Y=1\mid Z=1) &=(1-0.8)\Phi(-1)+0.8\Phi(0.5)\\ &=0.2(0.1586553)+0.8(0.6914625)\\ &\approx0.5849010. \end{aligned} $$

The substituted probability is \(\Phi(-1+1.5(0.8))=\Phi(0.2)\approx0.5792597\), about \(0.005641\) lower. Both calculations use the stated conditional response restriction, which is what allows the mixture probabilities to be computed from the two treatment-specific probabilities.

Worked Example: A Logit Mixture and Its Plug-In Value

Suppose, hypothetically, that \(P(Y=1\mid D,Z)=\operatorname{logit}^{-1}(-2+3D)\), where \(\operatorname{logit}^{-1}(a)=1/(1+e^{-a})\). At an instrument value where \(P(D=1\mid Z)=0.25\), compare the actual mixture with the substituted logit probability.

The response probabilities at \(D=0\) and \(D=1\) are \(\operatorname{logit}^{-1}(-2)\approx0.1192029\) and \(\operatorname{logit}^{-1}(1)\approx0.7310586\). Therefore,

$$ \begin{aligned} P(Y=1\mid Z) &=0.75(0.1192029)+0.25(0.7310586)\\ &\approx0.2721668. \end{aligned} $$

The plug-in calculation is \(\operatorname{logit}^{-1}(-2+3(0.25)) =\operatorname{logit}^{-1}(-1.25)\approx0.2227001\). It is about \(0.04947\) below the mixture probability. The discrepancy is substantial even though both calculations use the same intercept, treatment coefficient, and fitted treatment probability.

A Short Derivation for Continuous Treatment

The same issue is not confined to binary treatment. A simple probit calculation makes the averaging operation explicit. Suppose, as a hypothetical distributional setup, that \(D\mid Z\sim N(\mu,\sigma^2)\) and \(Y=1\{\alpha+\beta D+\varepsilon>0\}\), where \(\varepsilon\sim N(0,1)\) is independent of \(D\) conditional on \(Z\). Then \(P(Y=1\mid D,Z)=\Phi(\alpha+\beta D)\). To obtain the probability conditional only on \(Z\), write \(D=\mu+\sigma U\), where \(U\sim N(0,1)\) is independent of \(\varepsilon\). The latent index becomes \(\alpha+\beta\mu+\beta\sigma U+\varepsilon\), whose random part is normal with variance \(1+\beta^2\sigma^2\). Thus:

$$ \begin{aligned} P(Y=1\mid Z) &=P(\alpha+\beta D+\varepsilon>0\mid Z)\\ &=\Phi\left(\frac{\alpha+\beta\mu} {\sqrt{1+\beta^2\sigma^2}}\right). \end{aligned} $$

Plugging the conditional mean \(\mu\) into the response function gives \(\Phi(\alpha+\beta\mu)\), which omits the variance term in the denominator. Unless that omission has no effect in a special case, the two probabilities differ. The derivation depends on the stated normality and independence conditions; it is an illustration of averaging, not a general correction formula for IV estimation.

Worked Example: Residual Variation Changes a Probit Probability

Let \(\alpha=-0.5\), \(\beta=1\), \(\mu=1\), and \(\sigma=0.8\) in the continuous-treatment setup just described. The plug-in calculation is \(\Phi(-0.5+1)\ =\Phi(0.5)\approx0.6914625\).

The actual probability after averaging over the conditional treatment distribution is

$$ \begin{aligned} P(Y=1\mid Z) &=\Phi\left(\frac{-0.5+1}{\sqrt{1+1^2(0.8^2)}}\right)\\ &=\Phi\left(\frac{0.5}{\sqrt{1.64}}\right)\\ &=\Phi(0.3904)\approx0.6519. \end{aligned} $$

Here the plug-in value is approximately \(0.0396\) higher. The gap arises because treatment varies around its conditional mean and the nonlinear response must be averaged across that variation.

Why Linear 2SLS Is Different

In a linear conditional-mean model, the response function is affine: \(g(v)=v\), or more generally a linear expression in the predictors. For any random variable \(V\), \(E[V\mid Z]\) is exactly the conditional mean. Consequently, taking a conditional expectation of a linear expression replaces the random treatment by its conditional mean without the nonlinear averaging discrepancy shown above.

This algebraic fact does not mean that every informal “run a regression twice” procedure has correct standard errors or causal interpretation. Linear 2SLS has a specific estimator and inference theory, and it uses instruments to construct variation in treatment. In a nonlinear second stage, simply inserting \(\hat D\) into a probit or logit likelihood does not inherit those linear properties. It may yield a numerical coefficient, but that coefficient generally does not estimate the structural treatment coefficient in the model for \(Y\mid D\).

The distinction is also important for interpretation. A coefficient in a probit or logit model is a change on the latent index or log-odds scale, respectively; it is not itself a change in outcome probability. The nonlinear probability scale depends on the covariates and the treatment value. A plug-in procedure introduces the additional problem that its fitted probabilities need not equal the relevant averages over treatment values.

Assumptions, Diagnostics, and Failure Modes

The mixture identity requires a response model that specifies \(P(Y=1\mid D,Z)\), not merely \(P(Y=1\mid D)\). If the instrument has a direct association with the outcome conditional on treatment, the two treatment-specific probabilities may depend on \(Z\), and the simple expression using \(g(\alpha)\) and \(g(\alpha+\beta)\) does not follow. In a causal IV analysis, relevance, independence, and exclusion are distinct design conditions; the algebraic substitution problem neither establishes nor replaces them. A correctly specified outcome response function and an identified causal estimand are additional concerns.

Several practical checks can reveal that a nonlinear substitution analysis is on shaky ground:

  • Inspect the treatment scale. With binary \(D\), fitted values between zero and one are probabilities or conditional means, not observed treatment states.
  • Compare mixture and plug-in predictions. If the response model and treatment distribution are specified, calculate both quantities at representative covariate and instrument values. A difference directly displays the nonlinear averaging gap.
  • Check conditional response assumptions. A formula that averages treatment-specific probabilities must use probabilities conditional on the relevant instrument and covariates, or justify why those probabilities are invariant to them.
  • Separate model fit from identification. Good prediction of \(Y\) does not by itself show that the treatment coefficient has a causal IV interpretation.
  • Do not treat standard errors as the only issue. Correcting a variance formula cannot make an estimator consistent for a target it does not generally estimate.

A common mistake is to reason, “The first stage estimates the expected treatment, so putting that expectation into the outcome model gives the expected outcome.” That step silently assumes \(E[g(V)\mid Z]=g(E[V\mid Z])\). The correction is to average the response probabilities over the conditional distribution of treatment, or to use an estimation strategy justified for the nonlinear structural model and the causal estimand of interest.

How to Choose an Approach

ApproachWhat it doesMain caution
Linear 2SLS for a linear outcome modelUses instrument-generated treatment variation for a linear structural relationship.Its justification does not automatically extend to nonlinear second-stage models.
Ordinary nonlinear predictor substitutionUses a first-stage fitted treatment value as a predictor in probit, logit, or another nonlinear model.Generally substitutes a conditional mean where an average of nonlinear response probabilities is required.
Nonlinear IV methods matched to the estimandSpecify an appropriate nonlinear structural or causal model and use an estimator designed for it.Require their own identification, model, and inference assumptions; no single method is universally appropriate.

The choice should begin with the estimand and assumptions, not with whichever model is easiest to fit. Nonlinear IV methods differ in what effect they identify and in what functional-form or distributional restrictions they require. The central lesson here is narrower but essential: an ordinary nonlinear regression on first-stage fitted treatment is not justified simply by analogy with linear 2SLS.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. For binary \(D\), write the conditional outcome probability in terms of \(p=P(D=1\mid Z)\), \(g(\alpha)\), and \(g(\alpha+\beta)\).
  2. In the probit worked example, what are the mixture and plug-in probabilities when \(p=0.2\), and which is larger?
  3. Why does a fitted value of \(0.35\) for a binary treatment not mean that the observed treatment equals \(0.35\)?
  4. Under what broad condition does replacing a random predictor by its conditional mean commute with applying the response function?

Solutions. (1) Conditioning on \(D=0\) and \(D=1\) gives \((1-p)g(\alpha)+pg(\alpha+\beta)\). (2) The mixture is \(0.8\Phi(-1)+0.2\Phi(0.5)\approx0.2652167\); the plug-in value is \(\Phi(-0.7)\approx0.2419637\). The mixture is larger by about \(0.02325\). (3) For binary treatment, each observed value is zero or one; \(0.35\) is a fitted conditional mean or probability, not a treatment state. (4) The equality holds generally when the response function is affine over the relevant range. Particular distributions or parameter values can also produce equality, but it should not be assumed for nonlinear probit or logit functions.

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