Score Equations for Truncated Models

Why Truncation Changes the Score

When a severity is recorded only if it falls within a specified region, the observed values do not have the original density \(f_\theta\). They have a conditional density, which divides the original density by the probability of being observed. If that probability depends on the parameter, its derivative contributes to the score. Omitting this term generally changes the maximum-likelihood estimate.

Here, \(X\) is a continuous loss with density \(f_\theta(x)\), distribution function \(F_\theta(x)\), and survival function \(S_\theta(x)=1-F_\theta(x)\). The parameter \(\theta\) may be a scalar or a vector. Assume independent observations, fixed known truncation thresholds, and a model for which the density and normalizing probabilities are differentiable in the parameter. The observed sample consists only of values that satisfy the stated truncation condition.

Definition: Under left truncation at \(a\), a value is observed only when \(X>a\), so its conditional density is \(f_\theta(x)/S_\theta(a)\) for \(x>a\). Under right truncation at \(b\), a value is observed only when \(X<b\), so its conditional density is \(f_\theta(x)/F_\theta(b)\) for \(x<b\). The denominators are the corresponding truncation normalizers.

These formulas concern truncation: values outside the observation region do not enter the sample as recorded values. They are not formulas for a censored record, where an observation is retained but its exact value is not known. A likelihood for a truncated sample must condition on inclusion.

Deriving the Normalizer Correction

For a general fixed observation region \(A\), let \(G_\theta=P_\theta(X\in A)\). The density of an observed value \(x\in A\), conditional on inclusion, is \(f_\theta(x)/G_\theta\). Its log-density is therefore the original log-density minus the log of the inclusion probability.

\ell_{\mathrm{obs}}(\theta;x) =\log f_\theta(x)-\log G_\theta, \qquad U_{\mathrm{obs}}(\theta;x) =\frac{\partial}{\partial\theta}\log f_\theta(x) -\frac{\partial}{\partial\theta}\log G_\theta.

The first term in the score is the complete-data score, the derivative that would arise from the density \(f_\theta(x)\) alone. The second term corrects for the parameter-dependent chance of entering the observed sample. For a scalar parameter, the correction is \(G_\theta'/G_\theta\); for a vector parameter, it is the gradient \(\nabla_\theta\log G_\theta\).

Justification: Since \(G_\theta\) is the integral of the density over the observation region, differentiating the conditional log-density gives \[ \frac{\partial}{\partial\theta}\log\!\left(\frac{f_\theta(x)}{G_\theta}\right) =\frac{\partial}{\partial\theta}\log f_\theta(x) -\frac{G_\theta'}{G_\theta}. \] When differentiation under the integral is valid, \(G_\theta'=\int_A \partial f_\theta(t)/\partial\theta\,dt\). Thus the normalizer term reflects how the model changes the probability of being observed; it is not an optional adjustment.

For \(n\) independent left-truncated observations \(x_i>a_i\), with known thresholds \(a_i\), the conditional log-likelihood and score are

\ell_L(\theta) =\sum_{i=1}^n\left[\log f_\theta(x_i)-\log S_\theta(a_i)\right], \qquad U_L(\theta) =\sum_{i=1}^n\left[ \frac{\partial}{\partial\theta}\log f_\theta(x_i) -\frac{\partial}{\partial\theta}\log S_\theta(a_i) \right].

For right truncation at known thresholds \(b_i\), replace \(S_\theta(a_i)\) by \(F_\theta(b_i)\):

\ell_R(\theta) =\sum_{i=1}^n\left[\log f_\theta(x_i)-\log F_\theta(b_i)\right], \qquad U_R(\theta) =\sum_{i=1}^n\left[ \frac{\partial}{\partial\theta}\log f_\theta(x_i) -\frac{\partial}{\partial\theta}\log F_\theta(b_i) \right].

A maximum-likelihood estimate in the interior of the parameter space must satisfy the score equation \(U(\theta)=0\). A root is a candidate, not by itself proof of a global maximum: the likelihood may have other roots or a larger boundary value. The examples show how to check the conclusion.

Worked Examples

Worked Example: Exponential Losses With Individual Left-Truncation Thresholds

Suppose \(X\sim\operatorname{Exponential}(\lambda)\), with density \(f_\lambda(x)=\lambda e^{-\lambda x}\) for \(x>0\), where \(\lambda>0\) is measured per loss unit. Three independent records are observed above their respective fixed thresholds: \((a_1,x_1)=(1,2)\), \((a_2,x_2)=(2,5)\), and \((a_3,x_3)=(4,7)\). Find the maximum-likelihood estimate of \(\lambda\).

Method. Under left truncation, the normalizer is \(S_\lambda(a_i)=e^{-\lambda a_i}\). Therefore the conditional density for record \(i\) is \[ \frac{\lambda e^{-\lambda x_i}}{e^{-\lambda a_i}} =\lambda e^{-\lambda(x_i-a_i)},\qquad x_i>a_i. \] The excesses above threshold are \(2-1=1\), \(5-2=3\), and \(7-4=3\), whose sum is \(7\).

The log-likelihood is \[ \ell(\lambda)=3\log\lambda-\lambda(1+3+3) =3\log\lambda-7\lambda. \] Its score is \(U(\lambda)=3/\lambda-7\). Setting the score to zero gives \(3/\lambda=7\), hence \(\widehat{\lambda}=3/7\), or approximately \(0.4286\) per loss unit.

The second derivative is \(\ell''(\lambda)=-3/\lambda^2<0\), so the log-likelihood is strictly concave. It tends to \(-\infty\) as \(\lambda\) approaches zero or infinity; thus this stationary point is the unique global maximum. Check: \(U(3/7)=3/(3/7)-7=7-7=0\). The common mistake is to use the unconditioned observations and estimate \(3/(2+5+7)=3/14\); that ignores the three truncation thresholds.

Worked Example: A Global Maximum Under Right Truncation

A hypothetical sample has three independent exponential losses, all observed only when \(X<c\), where \(X\sim\operatorname{Exponential}(\alpha)\), \(\alpha>0\), and \(c=\log 2\). The three observed values are each \(x=1-\log 2\). Find the maximum-likelihood estimate and establish that it is global.

The right-truncation normalizer is \(F_\alpha(c)=1-e^{-\alpha c}\). For \(n=3\), the log-likelihood and score are \[ \ell(\alpha)=3\left[\log\alpha-\alpha x-\log(1-e^{-\alpha c})\right], \qquad U(\alpha)=3\left[\frac1\alpha-x-\frac{c}{e^{\alpha c}-1}\right]. \] The sign of the score is the sign of \(m(\alpha)-x\), where \[ m(\alpha)=\frac1\alpha-\frac{c}{e^{\alpha c}-1}. \]

To establish the maximum rather than merely identify a stationary point, note that \(m(\alpha)\) is the mean of an exponential random variable with rate \(\alpha\), conditional on lying in \((0,c)\). Differentiating this conditional mean gives \(m'(\alpha)=-\operatorname{Var}_\alpha(X\mid X<c)<0\): equivalently, differentiating the ratio of the integrals for the conditional mean produces minus the conditional variance. Thus \(m\) is strictly decreasing. As \(\alpha\) decreases to zero, the conditional density approaches the uniform density on \((0,c)\), so \(m(\alpha)\) approaches \(c/2\); as \(\alpha\) increases without bound, \(m(\alpha)\) approaches zero.

Here \(x=1-c\), and \(0<x<c/2\), since \(c=\log2\) gives \(x\approx0.3069\) and \(c/2\approx0.3466\). At \(\alpha=1\), \[ m(1)=1-\frac{c}{2-1}=1-c=x. \] Strict decrease means this is the unique solution of \(m(\alpha)=x\). The score is positive before \(\alpha=1\) and negative after it, so the log-likelihood increases up to \(1\) and decreases thereafter. In addition, as \(\alpha\to\infty\), each conditional density at the positive observed value \(x\) tends to zero, so \(\ell(\alpha)\to-\infty\). The lower-bound limit is finite, and the likelihood increases from that limit to \(\alpha=1\). Therefore \(\widehat{\alpha}=1\) is the global maximum. Check: the score at \(1\) is \(3[1-(1-c)-c]=0\). The trap is to stop after checking a zero score and negative local curvature; the monotonicity argument rules out another, better maximum.

Worked Example: Pareto Losses With Left Truncation

Let \(X\) have a Pareto Type I distribution with fixed minimum \(x_m=1\) and unknown shape \(\alpha>0\): \[ f_\alpha(x)=\alpha x^{-(\alpha+1)},\qquad x\ge1. \] Three independent records are observed above known thresholds \(a_i\ge1\): \((a_1,x_1)=(1,2)\), \((a_2,x_2)=(2,4)\), and \((a_3,x_3)=(4,8)\). Estimate \(\alpha\).

For \(a_i\ge1\), the normalizer is \(S_\alpha(a_i)=a_i^{-\alpha}\). Dividing the density by that probability gives the conditional density \(\alpha a_i^\alpha x_i^{-(\alpha+1)}\). The log-likelihood is therefore \[ \ell(\alpha)=3\log\alpha+\alpha\sum_{i=1}^3\log a_i -(\alpha+1)\sum_{i=1}^3\log x_i =3\log\alpha-\alpha\sum_{i=1}^3\log(x_i/a_i)-\sum_{i=1}^3\log x_i. \] Each ratio \(x_i/a_i\) is \(2\), so \(\sum_i\log(x_i/a_i)=3\log2\). The score equation is \[ U(\alpha)=\frac{3}{\alpha}-3\log2=0, \] which gives \(\widehat{\alpha}=1/\log2\approx1.4427\).

The second derivative is \(-3/\alpha^2<0\), and the log-likelihood tends to \(-\infty\) at both parameter boundaries: the \(3\log\alpha\) term dominates near zero, while \(-3\alpha\log2\) dominates at infinity. Hence the root is the unique global maximum. Check: \(3/\widehat{\alpha}=3\log2\), so the score is zero. Using the unconditioned Pareto density would omit the \(\alpha\log a_i\) terms and would not give the correct score equation.

Reading and Checking Truncated Score Equations

A useful check is to keep the normalizer visible until after taking the derivative. For left truncation, the correction is \(-\partial\log S_\theta(a)/\partial\theta\); for right truncation, it is \(-\partial\log F_\theta(b)/\partial\theta\). The correction is subtracted because the log conditional density is \(\log f_\theta(x)-\log G_\theta\). A sign error here can shift the estimate or even produce an implausible score.

If thresholds differ by record, use each record's own normalizer in the sum. If the parameter is a vector, set each component of the score vector to zero and then assess the likelihood and parameter-space boundaries. The score equation is useful for finding candidates, but it does not replace checks of existence, uniqueness, and global optimality. These calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Key takeaway: A truncated-data score equals the complete-data score minus the derivative of the log inclusion probability. The inclusion probability must match the truncation rule for each record.

Check Your Understanding

For each problem, form the conditional likelihood before solving or evaluating its score.

An exponential severity has rate \(\lambda\). Two independent losses are observed above thresholds \(3\) and \(4\), with observed values \(4\) and \(6\), respectively. Find the conditional maximum-likelihood estimate of \(\lambda\). (Answer: \(\widehat{\lambda}=2/3\) per loss unit.)
An exponential severity has rate \(\lambda\) and is right-truncated at \(b=\log2\). Two observed values are \(1/4\) and \(1/2\). Calculate the score at \(\lambda=1\). (Answer: \(U(1)=2-3/4-2\log2\approx-0.1363\).)
A Pareto Type I severity has fixed minimum \(1\) and unknown shape \(\alpha\). Records \((a_i,x_i)\) are \((2,4)\), \((3,6)\), and \((5,10)\), each observed conditional on exceeding \(a_i\). Find the maximum-likelihood estimate of \(\alpha\). (Answer: \(\widehat{\alpha}=1/\log2\approx1.4427\).)
For a continuous model with density \(f_\theta\), observations are retained only when \(a<X<b\), for fixed \(a,b\). Write the one-observation conditional score in terms of \(F_\theta\), and identify its normalizer correction. (Answer: \(\partial_\theta\log f_\theta(x)-\partial_\theta\log[F_\theta(b)-F_\theta(a)]\); the correction is the negative derivative of the log interval probability.)
An exponential severity with rate \(\alpha\) is right-truncated at \(c=\log3\). Two observed values both equal \(x=1-(\log3)/2\). Show that the conditional score has a unique zero and identify the global maximum. (Answer: \(\widehat{\alpha}=1\); the truncated exponential mean \(m(\alpha)\) decreases strictly from \(c/2\) to \(0\), and \(m(1)=x\), so the score changes from positive to negative there.)

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