Deriving the Inverse Mills Ratio Correction

Why Selection Changes the Conditional Mean

In a sample-selection model, an outcome is observed only for units who meet a selection rule. If the unobserved determinants of selection are related to the unobserved determinants of the outcome, the outcome errors among selected units need not average to zero—even when their population mean is zero. The observed outcomes can therefore have a conditional mean that differs systematically from the outcome equation’s index.

The Gaussian sample-selection model gives a precise expression for this difference. Under joint normality, it is proportional to the inverse Mills ratio: the standard normal density divided by the probability of being selected. This tutorial derives that expression and shows how to interpret and calculate it. It does not by itself establish that a correction is empirically identified or explain how to estimate the model.

Model and Assumptions

Let $S_i$ indicate whether unit $i$'s outcome is observed. Suppose selection occurs when a latent selection index is positive, and the outcome is observed only when $S_i=1$:

S_i^* = z_i'\gamma + u_i,\qquad S_i = \mathbf{1}\{S_i^*>0\},\qquad Y_i = x_i'\beta+\varepsilon_i\quad\text{when }S_i=1.

Here $x_i$ and $z_i$ are observed covariate vectors, $\beta$ and $\gamma$ are parameter vectors, and $\varepsilon_i$ and $u_i$ are unobserved errors. The outcome equation is meaningful for the population of interest, even though $Y_i$ is recorded only for selected units. Treat the covariates as fixed when taking the conditional expectations below.

Assumptions for the derivation: Conditional on the covariates, $(\varepsilon_i,u_i)$ is jointly normal with mean zero and a covariance matrix that does not depend on $x_i,z_i$: $$ \begin{pmatrix}\varepsilon_i\\u_i\end{pmatrix} \sim N\left[ \begin{pmatrix}0\\0\end{pmatrix}, \begin{pmatrix}\sigma_\varepsilon^2&\rho\sigma_\varepsilon\sigma_u\\ \rho\sigma_\varepsilon\sigma_u&\sigma_u^2\end{pmatrix} \right], \qquad \sigma_\varepsilon>0,\quad \sigma_u>0,\quad -1\leq\rho\leq1. $$ Selection is exactly the threshold rule $S_i=1$ when $z_i'\gamma+u_i>0$. In particular, the outcome error has mean zero before conditioning on selection, and its dependence on the selection error is represented by $\rho$.

The scale of the selection error is not separately identified in the binary selection rule: multiplying the latent index and its error by the same positive constant leaves the rule unchanged. In the probit normalization one sets $\sigma_u=1$. Keeping $\sigma_u$ in the derivation makes clear where standardization enters.

Deriving the Inverse Mills Ratio

Write $\eta_i=z_i'\gamma/\sigma_u$, and standardize the selection error as $V_i=u_i/\sigma_u$. Then $V_i\sim N(0,1)$, and selection is equivalent to $V_i>-\eta_i$. The probability of selection conditional on $z_i$ is

P(S_i=1\mid z_i)=P(V_i>-\eta_i)=\Phi(\eta_i),

where $\phi$ and $\Phi$ denote the standard normal density and cumulative distribution function. To find the mean of the standardized error among selected units, integrate its density over the selection region:

E[V_i\mid S_i=1,z_i] =\frac{\int_{-\eta_i}^{\infty}v\phi(v)\,dv}{\Phi(\eta_i)} =\frac{\phi(\eta_i)}{\Phi(\eta_i)}.

For the last equality, use $\phi'(v)=-v\phi(v)$. Thus $v\phi(v)=-\phi'(v)$, and $\int_{-\eta_i}^{\infty}v\phi(v)\,dv =-[\phi(v)]_{-\eta_i}^{\infty} =\phi(-\eta_i)=\phi(\eta_i)$. The denominator is the probability of the same selection event, so the integral has been divided by the appropriate probability to form a conditional mean.

Definition: For a real number $t$, the inverse Mills ratio for upper-tail selection is $$ \lambda(t)=\frac{\phi(t)}{\Phi(t)}. $$ With the selection rule $S_i=1$ when $z_i'\gamma+u_i>0$, the relevant argument is $\eta_i=z_i'\gamma/\sigma_u$, and $E[V_i\mid S_i=1,z_i]=\lambda(\eta_i)$.

From Selection Error to Outcome Error

Joint normality connects the conditional mean of the outcome error to that of the selection error. The conditional expectation of $\varepsilon_i$ given $u_i$ is its linear normal-regression expression:

E[\varepsilon_i\mid u_i] =\frac{\operatorname{Cov}(\varepsilon_i,u_i)}{\operatorname{Var}(u_i)}u_i =\frac{\rho\sigma_\varepsilon}{\sigma_u}u_i.

A useful way to see why conditioning on selection preserves this relation is to write $\varepsilon_i=(\rho\sigma_\varepsilon/\sigma_u)u_i+\xi_i$, where $\xi_i$ is jointly normal and uncorrelated with $u_i$. Joint normality makes $\xi_i$ independent of $u_i$, so its conditional expectation remains zero after conditioning on the event defined by $u_i$. Taking expectations over selected units therefore gives

E[\varepsilon_i\mid S_i=1,z_i] =\frac{\rho\sigma_\varepsilon}{\sigma_u}E[u_i\mid S_i=1,z_i] =\rho\sigma_\varepsilon E[V_i\mid S_i=1,z_i] =\rho\sigma_\varepsilon\lambda(\eta_i).

Consequently, the conditional mean of the observed outcome is

E[Y_i\mid S_i=1,x_i,z_i] =x_i'\beta+\rho\sigma_\varepsilon\lambda(\eta_i).

The second term is the selection correction. Its magnitude reflects both the association between the two errors, through $\rho\sigma_\varepsilon$, and the location of the selection threshold, through $\lambda(\eta_i)$. If $\rho=0$, joint normality implies that selection does not shift the conditional mean of the outcome error, so the correction is zero. A positive $\rho$ gives a positive correction under this selection rule; a negative $\rho$ gives a negative correction.

Worked Example: A Positive Correction With High Selection Probability

Suppose the outcome index for a hypothetical individual is $x_i'\beta=30$ minutes, $\rho=0.7$, and $\sigma_\varepsilon=4$ minutes. Let $\eta_i=1.5$. The selection probability is $\Phi(1.5)\approx0.933193$, so the model implies a high probability of selection. Assume the joint-normality and threshold assumptions above hold.

Using $\phi(1.5)\approx0.129518$ and $\Phi(1.5)\approx0.933193$, the inverse Mills ratio is

\lambda(1.5)=\frac{\phi(1.5)}{\Phi(1.5)} \approx\frac{0.129518}{0.933193} \approx0.13878975.

The correction is calculated using the unrounded ratio:

\rho\sigma_\varepsilon\lambda(\eta_i) =0.7(4)\frac{\phi(1.5)}{\Phi(1.5)} \approx0.3886113\text{ minutes}.

Thus the selected-sample conditional mean is $30+0.3886113=30.3886113$ minutes, or approximately $30.39$ minutes. The positive adjustment is consistent with the positive error correlation: units selected at these covariate values have, on average, positive outcome errors under the model.

Worked Example: A Negative Correction

Consider a different hypothetical setting with outcome index $x_i'\beta=12$ hours, $\rho=-0.5$, $\sigma_\varepsilon=6$ hours, and $\eta_i=0$. The probability of selection is $\Phi(0)=0.5$. Suppose the same model assumptions hold. Since $\phi(0)=1/\sqrt{2\pi}\approx0.39894228$, the inverse Mills ratio is

\lambda(0)=\frac{\phi(0)}{\Phi(0)} =\frac{0.39894228}{0.5} \approx0.79788456.

The correction and conditional mean are

(-0.5)(6)(0.79788456)\approx-2.3936537\text{ hours}, \qquad E[Y_i\mid S_i=1,x_i,z_i]\approx12-2.3936537 =9.6063463\text{ hours}.

The selected-sample mean is therefore about $9.61$ hours. This direction is appropriate for the assumed negative error correlation: selection tends to be associated with lower outcome errors. It is not a claim that selection causes an individual’s outcome to fall; it is a model-based conditional-mean difference.

Worked Example: A Low Selection Probability

Suppose a hypothetical individual's outcome index is $8$ thousand dollars, $\rho=0.4$, $\sigma_\varepsilon=5$ thousand dollars, and $\eta_i=-1$. The selection probability is $\Phi(-1)\approx0.158655$. Under the stated assumptions, the inverse Mills ratio is

\lambda(-1)=\frac{\phi(-1)}{\Phi(-1)} \approx\frac{0.241971}{0.158655} \approx1.5251353.

Therefore the correction is

(0.4)(5)(1.5251353)\approx3.0502706\text{ thousand dollars}, \qquad E[Y_i\mid S_i=1,x_i,z_i]\approx8+3.0502706 =11.0502706\text{ thousand dollars}.

The selected-sample conditional mean is about $11.05$ thousand dollars. Here the selection probability is low, and the selected units must have relatively high values of the selection error to cross the threshold. Since the errors are positively correlated, their outcome errors also tend to be positive. The units not selected do not have observed outcomes under this observation rule.

What the Correction Does—and Does Not—Say

The formula shows why an outcome regression restricted to selected observations can have an error term with a nonzero conditional mean. In the Gaussian model, the conditional mean is not an arbitrary function: it has the specific form $\rho\sigma_\varepsilon\lambda(\eta_i)$. It is important, however, to keep the interpretation tied to the assumptions. If the joint distribution is not normal, or if the selection rule is misspecified, this exact inverse Mills ratio expression need not describe the conditional mean.

The inverse Mills ratio is positive for every finite argument, because both $\phi(t)$ and $\Phi(t)$ are positive. The sign of the correction in the outcome equation therefore comes from $\rho$, not from the ratio. The ratio’s size depends on the selection index. For example, at $\eta=-1$ it is relatively large because selection is relatively unlikely; at $\eta=1.5$ it is smaller because selection is likely. This is why a correction is not generally a single common shift for all selected observations: it varies with covariates through $\eta_i$.

The derivation concerns a conditional expectation, not an estimated coefficient or a causal effect. It does not say that changing a covariate that predicts selection necessarily changes the outcome through the correction term. The term records how conditioning on selection changes the distribution of the unobserved outcome error under the specified model.

Common Mistake

A frequent error is to use $\phi(\eta_i)/\Phi(-\eta_i)$ for the selection rule in this tutorial. That denominator is the probability of the opposite tail: $P(V_i\leq-\eta_i)=\Phi(-\eta_i)$, not the probability of selection $P(V_i>-\eta_i)=\Phi(\eta_i)$. For selection when the latent index is positive, the correct ratio is $\phi(\eta_i)/\Phi(\eta_i)$. Always translate the selection rule into an inequality for the standardized error before choosing the tail probability. Reversing the selection rule requires deriving the corresponding conditional mean again; it is not safe to change the denominator by memory alone.

Key takeaway: Under the Gaussian threshold-selection model, the selected-sample outcome mean is $x_i'\beta+\rho\sigma_\varepsilon\lambda(z_i'\gamma/\sigma_u)$, where $\lambda(t)=\phi(t)/\Phi(t)$. The ratio arises from the mean of a standard normal variable truncated below at $-z_i'\gamma/\sigma_u$.

Check Your Understanding

Answer each question using the selection rule and assumptions stated in this tutorial.

For $S_i=1$ when $z_i'\gamma+u_i>0$, express the selection event in terms of $V_i=u_i/\sigma_u$ and $\eta_i=z_i'\gamma/\sigma_u$.
Why is $E[V_i\mid S_i=1,z_i]$ equal to $\phi(\eta_i)/\Phi(\eta_i)$, rather than just the integral of $v\phi(v)$ over the selected region?
If $\rho=0$, what does the model imply about $E[\varepsilon_i\mid S_i=1,z_i]$, and which assumption is essential to this implication?
Suppose $\rho<0$ and $\sigma_\varepsilon>0$. What is the sign of the selected-sample correction for finite $\eta_i$? Explain.
Explain why $\phi(\eta_i)/\Phi(-\eta_i)$ is not the inverse Mills ratio for the upper-tail selection rule used here.