Likelihood Contributions Under Censoring

Why the Observation Rule Determines the Likelihood

A censored outcome does not always reveal the value of the underlying response. It may tell us only that the response fell below a threshold, exceeded a threshold, or lay between two limits. The likelihood must reflect what was actually observed: a density for an exact continuous value, but a probability for the set of values compatible with a censored record.

This distinction matters in econometric models because using a threshold as though it were an exact response assigns the observation information it does not contain. The resulting likelihood is not merely an approximation to the correct one; it describes a different observation process.

Model and Observation Assumptions

Let \(Y_i\) be the underlying outcome for unit \(i\), and let \(x_i\) denote its observed covariates. Suppose the conditional distribution of \(Y_i\), given \(x_i\), has density \(f(y\mid x_i;\theta)\) and cumulative distribution function \(F(y\mid x_i;\theta)\), where \(\theta\) is the vector of unknown model parameters. For the formulas below, take \(Y_i\) to be continuous. The sample units are independent conditional on their covariates, so their likelihood contributions multiply.

The censoring status and any applicable limits are recorded. We assume the observation rule is known and that, conditional on the covariates and recorded censoring information, censoring is noninformative for the outcome parameters: the censoring mechanism does not introduce additional unknown parameters or depend on the unobserved outcome in a way that must be modeled. Under this assumption, the contribution for a record is the model density or probability of the values consistent with that record. If censoring is informative, its mechanism may need to be included in the likelihood.

Definition: A likelihood contribution is the density or probability, as a function of \(\theta\), of the observation actually recorded. For a continuous exact value, the contribution is a density evaluated at that value. For a censored record, it is the probability that the outcome belongs to the set allowed by the record.

Four Basic Contributions

For a continuous outcome, an exact value \(y_i\) contributes \(f(y_i\mid x_i;\theta)\). A left-censored record at \(c_i\), meaning that only \(Y_i\leq c_i\) is known, contributes the probability \(F(c_i\mid x_i;\theta)\). A right-censored record at \(c_i\), meaning that only \(Y_i>c_i\) is known, contributes the upper-tail probability \(1-F(c_i\mid x_i;\theta)\). An interval-censored record, known to satisfy \(l_i<Y_i\leq u_i\), contributes the probability between the limits.

\begin{aligned} \text{Exact at } y_i: \quad & f(y_i\mid x_i;\theta),\\ \text{Left-censored at } c_i: \quad & F(c_i\mid x_i;\theta),\\ \text{Right-censored at } c_i: \quad & 1-F(c_i\mid x_i;\theta),\\ \text{Interval-censored in } (l_i,u_i]: \quad & F(u_i\mid x_i;\theta)-F(l_i\mid x_i;\theta). \end{aligned}

The intervals and inequalities must match the observation rule. For a continuous distribution, including or excluding a single endpoint does not change the probability because a single point has probability zero. For a distribution with point masses, endpoint conventions can matter, and the likelihood must use the probability of the exact recorded event.

Worked Example: Contribution From an Exact Value

Suppose a hypothetical outcome has the model \(Y\sim N(\mu,\sigma^2)\), with \(\mu=8\) outcome units and \(\sigma=2\) outcome units. One unit’s outcome is observed exactly as \(y=10\). Find its likelihood contribution.

Check the observation type. The recorded value is exact, so the contribution is a density, not the probability of observing precisely 10. Under the normal model,

f(y;\mu,\sigma)=\frac{1}{\sigma\sqrt{2\pi}} \exp\left[-\frac{1}{2}\left(\frac{y-\mu}{\sigma}\right)^2\right].

Substituting the values gives

f(10;8,2) =\frac{1}{2\sqrt{2\pi}}\exp\left[-\frac{1}{2} \left(\frac{10-8}{2}\right)^2\right] =\frac{1}{2\sqrt{2\pi}}e^{-1/2} \approx 0.120985.

Thus the contribution is approximately \(0.120985\) per outcome unit. Its units are the inverse of the outcome’s units, as is appropriate for a density. It is not the probability that \(Y=10\), which is zero for a continuous distribution.

Worked Example: Contribution From a Left-Censored Value

In a separate hypothetical case, suppose \(Y\sim N(10,2^2)\), and the record says only that the outcome is at most 9 units. Find the contribution.

Check the observation type. The exact value is unknown; the record identifies the region \(Y\leq 9\). The contribution is therefore a cumulative probability. Standardizing the threshold gives

P(Y\leq 9) =F(9) =\Phi\left(\frac{9-10}{2}\right) =\Phi(-0.5) \approx 0.308538.

The likelihood contribution is approximately \(0.308538\). The value 9 is not substituted into the normal density: the observation does not say that the outcome equals 9, only that it is no greater than 9.

Worked Example: Contribution From a Right-Censored Value

Suppose a hypothetical outcome again follows \(Y\sim N(10,2^2)\), but this time the record says that it exceeds 11 units. Determine the contribution.

Check the observation type. The compatible outcomes form the upper tail \(Y>11\), so the contribution is one minus the CDF at 11:

P(Y>11) =1-F(11) =1-\Phi\left(\frac{11-10}{2}\right) =1-\Phi(0.5) \approx 1-0.691462 =0.308538.

The contribution is approximately \(0.308538\). It is a probability for an upper-tail region, not a density at the censoring threshold. For a continuous distribution, writing \(Y\geq 11\) instead would yield the same probability.

Worked Example: Contribution From an Interval-Censored Value

Now suppose a hypothetical observation from \(Y\sim N(10,2^2)\) is known to lie in the interval \(9<Y\leq 11\). Find its likelihood contribution.

Check the observation type. The record identifies a range rather than an exact value. The probability of the range is the CDF at the upper endpoint minus the CDF at the lower endpoint:

\begin{aligned} P(9<Y\leq 11) &=F(11)-F(9)\\ &=\Phi(0.5)-\Phi(-0.5)\\ &\approx 0.382925. \end{aligned}

Thus the contribution is approximately \(0.382925\). It is the probability that the outcome falls anywhere within the recorded interval, not the density at its midpoint or at either endpoint.

From Individual Contributions to the Sample Likelihood

Let \(\mathcal{E}\), \(\mathcal{L}\), \(\mathcal{R}\), and \(\mathcal{I}\) denote the sets of indices with exact, left-censored, right-censored, and interval-censored records. Under conditional independence, the sample likelihood is the product of the appropriate contribution for every record:

L(\theta) = \prod_{i\in\mathcal{E}} f(y_i\mid x_i;\theta) \prod_{i\in\mathcal{L}} F(c_i\mid x_i;\theta) \prod_{i\in\mathcal{R}} \left[1-F(c_i\mid x_i;\theta)\right] \prod_{i\in\mathcal{I}} \left[F(u_i\mid x_i;\theta)-F(l_i\mid x_i;\theta)\right].

The log-likelihood is the logarithm of this product, so it is the sum of the log contributions:

\begin{aligned} \ell(\theta) ={}&\sum_{i\in\mathcal{E}}\log f(y_i\mid x_i;\theta) +\sum_{i\in\mathcal{L}}\log F(c_i\mid x_i;\theta)\\ &+\sum_{i\in\mathcal{R}}\log\left[1-F(c_i\mid x_i;\theta)\right]\\ &+\sum_{i\in\mathcal{I}} \log\left[F(u_i\mid x_i;\theta)-F(l_i\mid x_i;\theta)\right]. \end{aligned}

This expression gives a general construction, not a particular parametric regression model. To obtain a model-specific likelihood, substitute that model’s conditional density and CDF. Each term depends on \(\theta\), and parameter estimation seeks values that make the complete observed record comparatively likely under the assumed model.

Worked Example: Combining Four Observation Types

Consider four independent hypothetical observations from the same \(N(10,2^2)\) model: one exact value of 10, one left-censored at 9, one right-censored at 11, and one known to lie in \((9,11]\). Write the likelihood at these specified parameter values.

The exact record contributes \(0.199471\). The left-censored and right-censored records each contribute approximately \(0.308538\), and the interval-censored record contributes approximately \(0.382925\). Independence means that the sample likelihood is the product:

\begin{aligned} L(10,2) &=f(10;10,2)\,F(9;10,2)\, \left[1-F(11;10,2)\right]\, \left[F(11;10,2)-F(9;10,2)\right]\\ &\approx 0.199471(0.308538)(0.308538)(0.382925)\\ &\approx 0.00727. \end{aligned}

This is the likelihood value at \(\mu=10\) and \(\sigma=2\) for this four-record observation pattern. The calculation uses a density for the exact record and probabilities for the three records that specify regions.

Why the Formula Follows From What Was Observed

For an exact continuous value, a density arises from the probability of a small neighborhood around that value. For a small width \(h>0\), the probability of \(y\leq Y<y+h\) is

P(y\leq Y<y+h)=F(y+h)-F(y) =\int_y^{y+h} f(t)\,dt.

If the density is continuous at \(y\), the average density over this shrinking interval approaches \(f(y)\):

\lim_{h\to 0^+}\frac{P(y\leq Y<y+h)}{h} =\lim_{h\to 0^+}\frac{1}{h}\int_y^{y+h}f(t)\,dt =f(y).

Thus an exact continuous observation contributes a density, understood through probabilities in arbitrarily small neighborhoods. By contrast, a censored record already specifies a region with nonzero width, or a tail. Its contribution is the probability of that whole region: the CDF for a lower region, the survival probability for an upper region, or a difference of CDF values for an interval.

Common Mistakes and Practical Checks

Using the density at a censoring threshold. A record \(Y\leq c\) does not report \(Y=c\). Use \(F(c)\), not \(f(c)\).
Using the lower tail for right censoring. If the record says \(Y>c\), the contribution is \(1-F(c)\), not \(F(c)\).
Using only one CDF value for an interval. For \(l<Y\leq u\), subtract the probability below the lower endpoint: \(F(u)-F(l)\).
Calling a density a probability. An exact continuous value has probability zero. Its density can exceed 1 and has inverse-outcome units; it is not itself a point probability.
Ignoring the observation mechanism. The formulas presume censoring is noninformative under the stated assumptions. If who is censored depends on unobserved outcomes in a way not accounted for, these contributions alone may not describe the observed data.

Key takeaway: Translate the record into the set of outcome values it permits. An exact continuous value contributes a density; a censored record contributes the probability of its permitted region. Multiply those record-specific terms to form the likelihood.

Check Your Understanding

For each question, identify the event represented by the record before choosing its likelihood contribution.

A continuous outcome is observed exactly as \(y\). Should its contribution be \(f(y)\) or \(F(y)\), and why?
A record reports only \(Y\leq c\). Write its contribution using the CDF.
A record reports only \(Y>c\). Write its contribution using the CDF.
An outcome is known to lie in \((a,b]\). Explain why its contribution is a difference of two CDF values.
Why does an exact observation from a continuous distribution not contribute the probability \(P(Y=y)\)?
Under what independence and censoring assumptions do the individual contributions multiply to give the sample likelihood?