Tutorials › Econometrics (Graduate) › Prediction for Censored and Truncated Outcomes

Censoring and Truncation · Tutorial 543 of 952

Prediction for Censored and Truncated Outcomes

Learn to match each prediction task to its target and calculate latent, censored, and conditionally selected outcome predictions.

Intermediate 10 min read

What You'll Learn

  • Distinguish a latent-outcome prediction from an expected recorded outcome.
  • Calculate censoring probabilities and expected outcomes under one- and two-limit censoring.
  • Predict the mean of outcomes observed only above a truncation threshold.
  • Derive the conditional mean for a normally distributed outcome truncated from below.
  • Identify why a prediction for selected observations does not equal a population prediction.
  • Check a prediction by stating its target, conditioning rule, and units.

One Model, Several Prediction Targets

A fitted limited-outcome regression can answer several different prediction questions. It might predict the mean of an underlying outcome, the mean of the value actually recorded after censoring, the chance that a value is censored, or the mean among cases that pass a truncation rule. These quantities are generally not equal. A useful first step is therefore to name the target before calculating a prediction.

For example, consider an outcome such as days of hospital stay that cannot be recorded below zero. The regression may describe a latent outcome that could mathematically take negative values, while the recorded outcome is set to zero whenever that latent value is nonpositive. The latent mean and the expected recorded value answer different questions, even when they are based on the same fitted parameters.

Model and assumptions: Conditional on covariates \(x\), let the latent outcome be \(Y^*=\mu(x)+\varepsilon\), where \(\mu(x)=x'\beta\) and \(\varepsilon\mid x\sim N(0,\sigma^2)\). For a two-limit censoring rule with known limits \(L<U\), the observed outcome is \(Y=L\) when \(Y^*\leq L\), \(Y=Y^*\) when \(L<Y^*<U\), and \(Y=U\) when \(Y^*\geq U\). The formulas below assume this conditional normal model and the stated observation rule. In an application, estimated parameters are substituted for \(\beta\) and \(\sigma\).

Write \(\phi(z)\) for the standard normal density and \(\Phi(z)\) for its cumulative distribution function. The latent conditional mean is \(\mu(x)\). It is not automatically the expected observed value: censoring changes the distribution of what is recorded by placing probability mass at one or both limits.

Left Censoring: Latent Mean, Observed Mean, and Censoring Probability

With left censoring at zero, the observation rule is \(Y=\max(0,Y^*)\). The latent conditional mean remains \(\mu\), but the probability of recording zero is \(P(Y^*\leq 0\mid x)=\Phi(-\mu/\sigma)\). The expected recorded value is

$$ E(Y\mid x)=\mu\Phi\left(\frac{\mu}{\sigma}\right) +\sigma\phi\left(\frac{\mu}{\sigma}\right). $$

This expectation accounts for both parts of the observed distribution: the mass at zero and the positive outcomes that are recorded without alteration. In particular, it is not obtained by simply replacing every negative latent prediction with zero; that operation is a point prediction, whereas the formula averages over the conditional distribution of \(Y^*\).

Worked Example: Expected Outcome With a Zero Censoring Limit

Suppose the fitted model for hypothetical hospital stays gives \(\mu=2\) days and \(\sigma=3\) days for a patient with specified covariates. Assume the latent outcome is normally distributed and recorded stays are left-censored at zero. Find the latent mean, the probability of recording zero, and the expected recorded stay.

The latent conditional mean is \(2\) days. The standardized censoring point is \((0-2)/3=-0.6667\), so

$$ P(Y=0\mid x)=\Phi(-0.6667)\approx0.2525. $$

Thus the model assigns about a 25.25% probability to a recorded stay of zero days. For the expected recorded stay, use \(z=\mu/\sigma=2/3\), with \(\Phi(0.6667)\approx0.7475\) and \(\phi(0.6667)\approx0.3194\):

$$ E(Y\mid x) =2(0.7475)+3(0.3194) =1.495+0.9582 \approx2.453\text{ days}. $$

The three answers target different quantities: \(2\) days is the latent mean, \(0.2525\) is the probability of a zero record, and \(2.453\) days is the expected recorded outcome. The calculations rely on the stated normal model and zero-censoring rule; changing either would generally change the prediction.

Two-Limit Censoring

For lower and upper limits \(L\) and \(U\), define the standardized limits \(a=(L-\mu)/\sigma\) and \(b=(U-\mu)/\sigma\). The probabilities of recording the lower limit, an interior value, and the upper limit are respectively \(\Phi(a)\), \(\Phi(b)-\Phi(a)\), and \(1-\Phi(b)\). The expected observed outcome weights each boundary by its probability and averages the latent outcome over the uncensored interval:

$$ E(Y\mid x) =L\Phi(a) +\mu\bigl[\Phi(b)-\Phi(a)\bigr] +\sigma\bigl[\phi(a)-\phi(b)\bigr] +U\bigl[1-\Phi(b)\bigr]. $$

The middle two terms together are the contribution from cases with \(L<Y^*<U\). The formula is a prediction of the recorded outcome on its original scale, not a prediction of the latent mean. As a quick check, the three probabilities sum to one, and the resulting expectation must lie between \(L\) and \(U\).

Worked Example: Prediction Under Two-Limit Censoring

Suppose a hypothetical daily-outcome model has \(\mu=8\) units and \(\sigma=4\) units, with censoring at \(L=0\) and \(U=10\). Calculate the probability of each recording category and the expected recorded outcome. Assume the conditional normal model and the two-limit rule stated above.

The standardized limits are \(a=(0-8)/4=-2\) and \(b=(10-8)/4=0.5\). Using \(\Phi(-2)\approx0.02275\) and \(\Phi(0.5)\approx0.69146\), the probabilities are:

$$ P(Y=0\mid x)=0.02275,\quad P(0<Y<10\mid x)=0.69146-0.02275=0.66871,\quad P(Y=10\mid x)=1-0.69146=0.30854. $$

These add to approximately 1.00000. The expected recorded value uses \(\phi(-2)\approx0.05399\) and \(\phi(0.5)\approx0.35207\):

$$ \begin{aligned} E(Y\mid x) &=0(0.02275)+8(0.66871)+4(0.05399-0.35207)+10(0.30854)\\ &=5.34968-1.19232+3.08540\\ &\approx7.243\text{ units}. \end{aligned} $$

The latent mean is \(8\) units, while the expected recorded outcome is about \(7.243\) units. The difference follows from the censoring rule: a substantial fraction of latent outcomes above 10 are recorded as 10. The expected recorded value is within the allowed range from 0 to 10, as it must be.

Truncation: Predicting Among Observed Cases

Censoring records a boundary value for cases beyond a limit; truncation excludes those cases from the observed sample. These observation rules therefore imply different prediction targets. Under left truncation at a threshold \(c\), a case is observed only when \(Y^*>c\). The mean outcome among observed cases with covariates \(x\) is a conditional mean, \(E(Y^*\mid Y^*>c,x)\), rather than the unconditional latent mean \(\mu\).

Let \(a=(c-\mu)/\sigma\). Under the conditional normal model, the probability of being observed is \(1-\Phi(a)\), and the mean among observed cases is

$$ E(Y^*\mid Y^*>c,x) =\mu+\sigma\frac{\phi(a)}{1-\Phi(a)}. $$

Why the Truncated Mean Has an Adjustment

Standardize the latent outcome as \(Z=(Y^*-\mu)/\sigma\), so \(Z\sim N(0,1)\) and selection requires \(Z>a\). By the definition of a conditional expectation,

$$ E(Z\mid Z>a) =\frac{\int_a^\infty z\phi(z)\,dz}{P(Z>a)} =\frac{\int_a^\infty z\phi(z)\,dz}{1-\Phi(a)}. $$

The standard normal density satisfies \(\phi'(z)=-z\phi(z)\). Therefore the numerator is

$$ \int_a^\infty z\phi(z)\,dz =-\int_a^\infty \phi'(z)\,dz =-\bigl[\phi(z)\bigr]_a^\infty =\phi(a), $$

because \(\phi(z)\) tends to zero as \(z\) tends to infinity. Thus \(E(Z\mid Z>a)=\phi(a)/(1-\Phi(a))\). Since \(Y^*=\mu+\sigma Z\), multiplying the conditional expectation of \(Z\) by \(\sigma\) and adding \(\mu\) gives the stated formula. For a lower-truncated sample, the adjustment is positive: selected observations have a mean above the unconditional latent mean.

Worked Example: Mean Among Outcomes Above a Truncation Threshold

Suppose a hypothetical duration model has latent mean \(\mu=20\) days and standard deviation \(\sigma=10\) days. Durations are observed only when they exceed \(c=25\) days. Assume the latent outcome is conditionally normal and selection depends only on crossing this threshold. Find the selection probability and the mean duration among observed cases.

The standardized threshold is \(a=(25-20)/10=0.5\). Since \(1-\Phi(0.5)\approx0.30854\), about 30.85% of cases with these covariates are predicted to be observed. Also, \(\phi(0.5)\approx0.35207\), so the inverse Mills ratio for this lower truncation is

$$ \frac{\phi(0.5)}{1-\Phi(0.5)} =\frac{0.35207}{0.30854} \approx1.14108. $$

The predicted conditional mean is therefore

$$ E(Y^*\mid Y^*>25,x) =20+10(1.14108) \approx20+11.411 \approx31.41\text{ days}. $$

This is the model-implied mean for cases whose durations exceed 25 days, not the unconditional latent mean of 20 days and not the threshold itself. The selection probability and conditional mean are both predictions under the assumed normal distribution and truncation rule.

Choosing and Reporting the Target

A prediction should state what is being averaged, under what observation rule, and for which covariates. When parameters are estimated, each formula is evaluated using the fitted \(\hat{\beta}\) and \(\hat{\sigma}\); the resulting value is a plug-in prediction. That substitution does not make distinct targets interchangeable. For example, the latent conditional mean \(x'\hat{\beta}\) describes the underlying regression distribution, whereas the expected recorded outcome applies the censoring rule, and the selected-sample mean conditions on being observed.

The practical importance is easy to miss. A policy question about the average recorded payment may require an expected censored outcome. A question about the probability of hitting a cap requires an upper-censoring probability. A forecast for a sample that excludes all values below an eligibility threshold requires a conditional truncated mean. Reporting the latent mean for any of those questions can give a numerically plausible but substantively wrong answer.

Common pitfall: A wrong approach is to report \(x'\hat{\beta}\) as the predicted mean for every outcome under censoring or truncation. That is the latent conditional mean under the model. The correction is to identify the target first: apply the censoring rule and average for an expected recorded value, calculate a tail probability for a censoring event, or condition on passing the threshold for a truncated-sample mean.
1
Name the quantity.
Specify whether the request concerns the latent mean, an observed-outcome mean, a boundary probability, or a mean among selected cases.
2
Write the observation rule.
State the censoring limits or truncation threshold and whether excluded cases are missing from the sample.
3
Calculate and interpret.
Use the matching formula, check probabilities sum to one where appropriate, and report the outcome in its original units.

Check Your Understanding

For each question, identify the prediction target before choosing a formula.

  1. Under left censoring at zero, what does \(x'\beta\) predict, and why need it not equal \(E(Y\mid x)\)?
  2. For two-limit censoring, how would you calculate the probability that the recorded value equals the upper limit?
  3. A sample contains only outcomes above a known threshold. Why is its conditional mean generally above the unconditional latent mean under a normal model?
  4. In the two-limit worked example, what were the predicted probability of recording 10 units and the expected recorded outcome?
  5. Explain why a prediction for the mean among truncated observations should not be described as the population mean.
Clinical Biostats