Parameter-Dependent Truncation Normalizers

Why the Truncation Normalizer Matters

A truncated sample contains only records that satisfy a specified inclusion rule. For example, a hypothetical insurer might observe only losses exceeding a reporting threshold, or a study might retain only measurements below a fixed upper limit. If the chance of satisfying that rule depends on the model parameter, then the observed records have a conditional distribution whose density includes a parameter-dependent normalizer. That normalizer is part of the likelihood, not a constant to discard.

The key calculation is to differentiate the log of the conditional density. Its score contains a term from the ordinary density and a second term from the probability of inclusion. Omitting the second term generally changes the maximum likelihood estimate (MLE).

Definition: Let $X$ have density $f(x;\theta)$, where $\theta$ is an unknown parameter, and suppose a record is observed only when $X$ belongs to a specified set $A$. The inclusion probability is $\pi(\theta)=P_\theta(X\in A)$. For an observed value $x\in A$, the conditional density is $$ f_A(x;\theta)=\frac{f(x;\theta)}{\pi(\theta)}. $$ For independent observations $x_1,\ldots,x_n$ subject to inclusion sets $A_1,\ldots,A_n$, the conditional likelihood is the product of these conditional densities, with the appropriate $\pi_i(\theta)=P_\theta(X\in A_i)$ for each record.

This is the likelihood for a sample collected conditional on passing the inclusion rule. It differs from a likelihood that treats the records as an unselected sample. The inclusion sets may vary by record; even if they are fixed and known, their probabilities can still depend on $\theta$.

The Score Includes a Normalizer Derivative

For a common inclusion set $A$, the log likelihood from $n$ independent selected observations is

\ell(\theta)=\sum_{i=1}^n \log f(x_i;\theta)-n\log\pi(\theta).

Differentiating gives the score equation that must be used to find an interior MLE:

U(\theta)=\frac{\partial\ell}{\partial\theta} =\sum_{i=1}^n \frac{\partial}{\partial\theta}\log f(x_i;\theta) -n\frac{\partial}{\partial\theta}\log\pi(\theta)=0.

When the inclusion set differs by record, replace the final term by $-\sum_i \partial\log\pi_i(\theta)/\partial\theta$. The score equation is a necessary condition for an interior maximum, not by itself a guarantee of one. Check the likelihood, the second derivative or other evidence of a maximum, and any parameter-space boundaries.

Derivation: Write $\pi(\theta)=\int_A f(x;\theta)\,dx$. Under conditions that permit differentiating under the integral sign, $$ \frac{\partial}{\partial\theta}\log\pi(\theta) =\frac{\int_A \frac{\partial}{\partial\theta}f(x;\theta)\,dx}{\pi(\theta)} =\frac{\int_A f(x;\theta)\frac{\partial}{\partial\theta}\log f(x;\theta)\,dx}{\pi(\theta)}. $$ The final expression is the expected complete-data score conditional on $X\in A$. Thus the score from one selected observation is its ordinary density score minus the expected ordinary score among values that would be selected. This adjustment centers the score for the conditional sampling design.

The differentiation-under-the-integral step requires suitable regularity, such as a parameter-independent integration set and an integrable derivative. If the selection boundary itself depends on the parameter, differentiating the integral may also require a boundary term. In all cases, differentiate the actual inclusion probability in the specified model rather than assuming it is constant.

Worked Examples

Worked Example: Left-Truncated Exponential Losses

A hypothetical study observes exact losses only when they exceed $a=1$ dollar-unit. Assume the underlying loss $X$ is exponential with rate $\lambda>0$, so $f(x;\lambda)=\lambda e^{-\lambda x}$ for $x>0$. The three observed values are $1.4,\ 2.1,\ 3.5$. Find the conditional MLE of $\lambda$, and compare it with the estimate obtained by incorrectly treating the observations as an unselected sample.

Method and inclusion probability. The inclusion event is $X>1$. For this exponential model, $P_\lambda(X>1)=e^{-\lambda}$. Each selected observation therefore has density

f_A(x;\lambda)=\frac{\lambda e^{-\lambda x}}{e^{-\lambda}} =\lambda e^{-\lambda(x-1)},\qquad x>1.

Form the log likelihood. The excesses above the threshold are $1.4-1=0.4$, $2.1-1=1.1$, and $3.5-1=2.5$; their sum is $4.0$. Thus

\ell(\lambda)=3\log\lambda-\lambda(0.4+1.1+2.5) =3\log\lambda-4\lambda.

Differentiate and solve. The score is $U(\lambda)=3/\lambda-4$. Setting it to zero gives $\widehat{\lambda}=3/4=0.75$. The second derivative is $-3/\lambda^2<0$ for every $\lambda>0$, so this stationary point is a maximum.

Comparison and check. Ignoring truncation would give $3/(1.4+2.1+3.5)=3/7\approx0.4286$, not $0.75$. The correct likelihood depends on the excesses over the entry threshold because the conditional density is exponential in $x-1$. The mistake avoided here is using the ordinary, unselected-sample likelihood after the sample has been selected.

Worked Example: Left-Truncated Normal Measurements

A hypothetical measurement study retains only values above zero. Assume $X\sim N(\mu,1)$, with known variance $1$ and unknown mean $\mu$. Two exact selected values are both $r=\sqrt{2/\pi}\approx0.7978846$. Find the MLE of $\mu$, using the conditional likelihood.

Method and inclusion probability. The model specifies a normal distribution with variance $1$. Since $P_\mu(X>0)=\Phi(\mu)$, where $\Phi$ is the standard normal cumulative distribution function, and $\phi$ is its density, the conditional log likelihood, apart from a constant, is

\ell(\mu)=-\frac12\sum_{i=1}^2(x_i-\mu)^2-2\log\Phi(\mu).

Differentiate the normalizer as well as the density. Because $\partial\log\Phi(\mu)/\partial\mu=\phi(\mu)/\Phi(\mu)$, the score is

U(\mu)=\sum_{i=1}^2(x_i-\mu)-2\frac{\phi(\mu)}{\Phi(\mu)}.

At $\mu=0$, each $x_i=r$, so the sum in the first term is $2r$. Also, $\phi(0)/\Phi(0)=(1/\sqrt{2\pi})/(1/2)=\sqrt{2/\pi}=r$. Therefore $U(0)=2r-2r=0$.

Check that this is a maximum. Set $q(\mu)=\phi(\mu)/\Phi(\mu)$. Differentiation gives $q'(\mu)=-q(\mu)(\mu+q(\mu))$, and hence

\ell''(\mu)=-2+2q(\mu)(\mu+q(\mu)).

At zero, $q(0)=r$, so $\ell''(0)=-2+2r^2=-2+4/\pi\approx-0.7268<0$. In fact, $1-q(\mu)(\mu+q(\mu))$ is the variance of a standard normal variable conditional on exceeding $-\mu$, and is positive. Thus the log likelihood is strictly concave here; the score root is the unique maximum. The conditional MLE is $\widehat{\mu}=0$.

Common mistake avoided. Using only the normal density score would set $\sum_i(x_i-\mu)=0$, producing $\mu=r$. That is not the MLE under the stated selection rule: the derivative of $-2\log\Phi(\mu)$ must also be included.

Worked Example: Right-Truncated Exponential Data and the Shift in the Estimate

A hypothetical data process records exact losses only when $0<X<1$. Assume $X$ has an exponential distribution with rate $\lambda>0$. The observed losses are $0.3000$ and $0.5360$, with sum $0.8360$. Find the conditional MLE and compare it with the estimate from the untruncated likelihood.

Method and normalizer. The inclusion probability is $P_\lambda(0<X<1)=1-e^{-\lambda}$. Thus

f_A(x;\lambda)=\frac{\lambda e^{-\lambda x}}{1-e^{-\lambda}},\qquad 0<x<1.

The conditional log likelihood and score are

\ell(\lambda)=2\log\lambda-\lambda(0.8360)-2\log(1-e^{-\lambda}), \qquad U(\lambda)=\frac{2}{\lambda}-0.8360-\frac{2}{e^\lambda-1}.

Find and check the root. At $\lambda=1$, $U(1)=2-0.8360-2/(e-1)\approx0.0000466$, a small positive value. The derivative of the score is $U'(\lambda)=-2/\lambda^2+2e^\lambda/(e^\lambda-1)^2$, which is negative for $\lambda>0$. Numerically solving $U(\lambda)=0$ gives $\widehat{\lambda}\approx1.000294$, just above $1$, which rounds to $1.000$ to three decimal places. The decreasing score changes sign at its root, confirming a maximum.

Compare with the incorrect estimate. If the observations were treated as untruncated, the exponential MLE would be $2/0.8360\approx2.3923$. That estimate is not the conditional MLE: it omits the derivative of the inclusion normalizer. The small difference between the correct root and $1$ is consistent with the small, but nonzero, score at $1$; rounding the data sum does not make that score exactly zero.

Practical Checks and Common Traps

The normalizer depends on both the model and the sampling rule. For left truncation at a fixed threshold $a$, the inclusion probability is $P_\theta(X>a)$. For right truncation at a fixed upper threshold $b$, it is $P_\theta(X<b)$. For interval truncation, it is $P_\theta(a<X<b)$. These probabilities are not interchangeable: use the event that actually determines whether a record appears.

Before solving a score equation, check that every observation satisfies its inclusion rule and that the denominator matches that rule. Then verify the sign and scale of the normalizer derivative, solve within the parameter space, and assess whether the root is a maximum. If truncation probabilities are constant in the parameter, their derivatives are zero; otherwise, dropping them is generally unjustified.

Wrong: Write a likelihood using $f(x;\theta)$ alone for a sample selected by $X\in A$. Correction: Use $f(x;\theta)/P_\theta(X\in A)$ for each conditionally observed record.
Wrong: Keep the normalizer in the likelihood but treat it as constant when differentiating. Correction: Differentiate its log whenever its inclusion probability depends on the unknown parameter.
Wrong: Assume every score root is an MLE. Correction: Check curvature or compare likelihood values, and consider boundaries of the parameter space.

Key takeaway: A truncated-data likelihood is conditional on inclusion. Its score is the ordinary density score minus the derivative of the log inclusion probability. Retaining that parameter-dependent term is essential for a valid MLE.

Check Your Understanding

For each problem, write the conditional likelihood before solving or interpreting the score.

A hypothetical exponential loss model has rate $\lambda$ and records only losses above $2$. The observed losses are $2.5,\ 3.5,\ 4.0$. Find the conditional MLE of $\lambda$. (Answer: $0.7500$.)
An exponential model with rate $\lambda$ is observed only on $0<X<1$. Two observations are $0.2500000$ and $0.5860466$. Use the conditional score to estimate $\lambda$, and state why the untruncated estimate differs. (Answer: $\widehat{\lambda}\approx1.0000$; the untruncated estimate is approximately $2.3922$, because it omits the interval-inclusion normalizer.)
A hypothetical study observes only $X>0$, where $X\sim N(\mu,4)$ and the variance is known. Two selected observations both equal $\sqrt{2/\pi}\,2$. Find the conditional MLE of $\mu$. (Answer: $\widehat{\mu}=0$.)
For independent observations with record-specific inclusion sets $A_i$, write the log likelihood and its score in terms of $f(x_i;\theta)$ and $\pi_i(\theta)=P_\theta(X\in A_i)$. (Answer: $\ell(\theta)=\sum_i[\log f(x_i;\theta)-\log\pi_i(\theta)]$, and $U(\theta)=\sum_i[\partial_\theta\log f(x_i;\theta)-\partial_\theta\log\pi_i(\theta)]$.)
Suppose an inclusion probability is constant in $\theta$. What happens to its contribution to the score, and what condition must still be checked before calling a score root an MLE? (Answer: Its log-derivative is zero; still check that the root is a maximum and consider parameter-space boundaries.)

What You'll Learn