Why Constraints Change Maximum Likelihood
A score equation identifies possible maxima only among parameter values where the likelihood is differentiable and the parameter is interior to its allowed range. In actuarial models, a parameter may be restricted by its definition, by stated bounds, or by the support of the distribution. A valid maximum likelihood estimate must satisfy all those restrictions. It may occur at an endpoint even when the score equation has no solution there.
The practical task is therefore to describe the feasible parameter values, find any interior stationary points, and compare their likelihoods with the likelihood at the allowable boundaries. For models whose support depends on the parameter, the support condition is part of the likelihood—not a detail to check after maximizing.
For a parameter restricted to a closed interval \([a,b]\), the candidate set typically includes any interior points where the derivative is zero, plus \(a\) and \(b\). A boundary can be a maximum even though the derivative there is not zero. If the likelihood is not differentiable, or its positive region changes with \(\theta\), inspect the likelihood directly on its feasible set. If every allowed parameter assigns likelihood zero to the data, there is no meaningful MLE for that model and data set.
Uniform Upper-Endpoint Estimation
Suppose \(X_1,\ldots,X_n\) are independent and identically distributed complete observations from a uniform distribution on \([0,\theta]\), where \(\theta>0\) is unknown. Write \(m=\max(x_1,\ldots,x_n)\). The density is \(1/\theta\) between zero and \(\theta\), and zero outside that interval. Thus, the likelihood is positive only when \(\theta\ge m\).
When the data are nonnegative, the log-likelihood on the feasible region is \(\ell(\theta)=-n\log\theta\). Its derivative is \(-n/\theta\), which is negative for every feasible positive \(\theta\). The likelihood therefore decreases as \(\theta\) increases. The smallest feasible endpoint, \(\theta=m\), maximizes it.
The endpoint result follows from two requirements working together. Every observed value must be inside the distribution's support, so \(\theta\) cannot be less than the largest observation. Once that condition is met, increasing \(\theta\) lowers the density of every observation, multiplying the likelihood by the factor \(\theta^{-n}\). The largest observation is therefore a support boundary and the likelihood maximum.
Worked Example: Estimating an Unrestricted Uniform Endpoint
A hypothetical coverage model treats four complete losses, measured in thousands of dollars, as independent observations from a uniform distribution on \([0,\theta]\). The recorded values are \(3, 5, 7,\) and \(9\). Estimate \(\theta\) by maximum likelihood. No interest rate or mortality assumption is involved; the specified distribution is uniform with unknown upper endpoint \(\theta>0\).
Step 1: Find the support restriction. The largest observation is \(m=9\). A value \(\theta<9\) gives zero likelihood because the loss of 9 would fall outside \([0,\theta]\). Therefore, the feasible values satisfy \(\theta\ge 9\).
Step 2: Write the likelihood. There are \(n=4\) observations, so for feasible \(\theta\), \(L(\theta)=\theta^{-4}\). The log-likelihood is \(\ell(\theta)=-4\log\theta\), with derivative \(\ell'(\theta)=-4/\theta<0\).
Step 3: Maximize over feasible values. Since the likelihood decreases throughout \([9,\infty)\), its maximum is at the smallest feasible value, \(\hat{\theta}=9\) thousand dollars. The maximized likelihood is \(L(9)=9^{-4}=1/6561\).
Check and method. At \(\theta=10\), for comparison, \(L(10)=1/10000<1/6561\), consistent with the decreasing likelihood. The method is to enforce support, then maximize on that feasible region. The estimate is the sample maximum.
Restrictions Can Put the Estimate at a Different Boundary
An additional parameter restriction may change which feasible value maximizes the likelihood. For the same uniform model, suppose \(\theta\) is required to lie in \([a,b]\), where \(0<a<b\). The support condition still requires \(\theta\ge m\). If \(m\le a\), the likelihood decreases over the entire allowed interval, so the maximum is at \(a\). If \(a<m\le b\), it decreases over the feasible interval starting at \(m\), so the maximum is at \(m\). If \(m>b\), no allowed value gives the observations positive likelihood.
The first case is a maximum at a parameter restriction boundary: the data would allow a smaller endpoint, but the model does not. In the second case, the estimate is at the data-support boundary. These are distinct reasons for an endpoint estimate, although they can coincide.
Worked Example: A Lower Parameter Bound Binds
A hypothetical severity model assumes three independent complete losses follow a uniform distribution on \([0,\theta]\). The parameter is restricted to \(5\le\theta\le12\), in thousands of dollars, and the observations are \(1.2, 2.6,\) and \(4.1\). Find the MLE. No interest rate or mortality assumption is involved; \(\theta\) is the only unknown distribution parameter.
Step 1: Check the observed maximum against the parameter bounds. Here \(m=4.1\), which is below the allowed lower bound \(5\). Every allowed \(\theta\) is therefore large enough to include all three losses in its support.
Step 2: Form the likelihood over the allowed interval. With \(n=3\), \(L(\theta)=\theta^{-3}\) for \(5\le\theta\le12\). Its log-likelihood derivative is \(-3/\theta<0\), so the likelihood decreases throughout this interval.
Step 3: Select the maximizing boundary. The smallest allowed value is \(\theta=5\), so \(\hat{\theta}=5\) thousand dollars. At the two endpoints, \(L(5)=1/125=0.008\) and \(L(12)=1/1728\), which confirms \(L(5)>L(12)\).
Check and method. The estimate is allowed and satisfies \(5\ge4.1\), so it supports every observation. The method is direct constrained maximization: the lower parameter bound, rather than the sample maximum, determines the estimate in this case.
Interior Candidates and Boundary Comparisons
The uniform endpoint example has no interior stationary point, but the general comparison rule also applies when an interior score solution exists. A useful illustration is a normal model with known variance and a restricted mean. For independent observations from \(N(\mu,\sigma^2)\), where \(\sigma^2\) is known and \(\mu\in[a,b]\), the log-likelihood, apart from a constant, is \(-\sum_i(x_i-\mu)^2/(2\sigma^2)\). Its derivative is \(\sum_i(x_i-\mu)/\sigma^2\), which vanishes at the sample mean \(\bar{x}\).
The second derivative is \(-n/\sigma^2<0\), so the log-likelihood is strictly concave. If \(\bar{x}\) is inside the interval, it is the maximum. If \(\bar{x}<a\), the log-likelihood is decreasing over the allowed interval and the maximum is at \(a\). If \(\bar{x}>b\), it is increasing over the allowed interval and the maximum is at \(b\). Equivalently, the constrained estimate is the sample mean restricted to the allowed interval.
Worked Example: A Normal Mean Above Its Allowed Range
Four hypothetical claim amounts, in thousands of dollars, are modeled as independent observations from \(N(\mu,4)\), with known variance \(\sigma^2=4\) and restricted mean \(0\le\mu\le4\). The observations are \(2,4,6,\) and \(8\). Find the constrained MLE. No interest rate or mortality assumption is involved; the distribution is normal with known variance 4 and unknown mean \(\mu\).
Step 1: Calculate the unrestricted score solution. The sample mean is \(\bar{x}=(2+4+6+8)/4=20/4=5\). This is not allowed because \(5>4\), so it cannot be the constrained estimate.
Step 2: Check the score throughout the allowed range. The log-likelihood derivative is \(\ell'(\mu)=\sum_i(x_i-\mu)/4=(20-4\mu)/4=5-\mu\). For every \(\mu\in[0,4]\), \(5-\mu\ge1>0\); hence the log-likelihood is increasing across the full allowed interval.
Step 3: Choose the boundary and verify it. The maximum is at the upper boundary, \(\hat{\mu}=4\). The residual sum of squares there is \((2-4)^2+(4-4)^2+(6-4)^2+(8-4)^2=4+0+4+16=24\). At the other boundary, \(\mu=0\), it is \(4+16+36+64=120\). Since the log-likelihood is a constant minus the residual sum of squares divided by \(2\sigma^2=8\), the log-likelihood at 4 exceeds the one at 0 by \((120-24)/8=12\).
Check and method. The estimate \(\hat{\mu}=4\) is feasible, and the likelihood comparison confirms that the upper boundary is preferable to the lower one. The method is to solve the unrestricted score equation, reject its infeasible solution, and maximize over the allowed interval. It avoids reporting the sample mean when the model prohibits it.
Exam Traps and a Reliable Procedure
A frequent error is to treat a score equation as the entire MLE method. For the uniform model, solving \(-n/\theta=0\) produces no solution, but concluding that no MLE exists is wrong: the derivative is negative, and the maximum occurs at the smallest feasible endpoint. Another error is to maximize over the stated parameter interval without first checking whether that value can support the observations. A proposed \(\theta\) below the observed maximum has likelihood zero, not \(\theta^{-n}\).
A reliable check is to separate the restrictions and then examine the likelihood. First identify the parameter space specified by the model. Next identify any additional support condition imposed by the observations. Then find interior stationary points, if they exist, and compare them with all feasible boundaries. Finally check that the selected estimate is feasible and that its likelihood is at least as large as the alternatives.
Record all restrictions on the unknown parameter before differentiating.
For parameter-dependent supports, determine which parameter values assign positive likelihood to the observed data.
Use score solutions only as interior candidates; evaluate feasible boundary values as well.
Confirm feasibility and compare likelihoods or use a justified monotonicity or concavity argument.
Check Your Understanding
For each problem, identify the feasible region before selecting the maximum likelihood estimate.
- Three independent complete observations come from a uniform distribution on \([0,\theta]\), with \(\theta>0\). The observations are \(2,5,6\). Find the MLE of \(\theta\). (Answer: \(\hat{\theta}=6\).)
- For a uniform distribution on \([0,\theta]\), the parameter is restricted to \(4\le\theta\le10\). Complete observations are \(1,2,3\). State the maximizing endpoint and the MLE. (Answer: the lower endpoint; \(\hat{\theta}=4\).)
- A uniform model on \([0,\theta]\) has \(4\le\theta\le9\), and the complete observations are \(3,5,8\). Find the MLE and explain which condition determines it. (Answer: \(\hat{\theta}=8\), determined by the observed-support boundary.)
- Independent observations from \(N(\mu,9)\) have known variance 9, with \(0\le\mu\le8\). The observations are \(7,9,11\). Find the constrained MLE. (Answer: \(\hat{\mu}=8\).)
- Independent observations from \(N(\mu,4)\) have known variance 4, with \(2\le\mu\le5\). The observations are \(3,5,7,9\). Find the unrestricted score solution and the constrained MLE. (Answer: the unrestricted solution is \(\bar{x}=6\); the constrained MLE is \(\hat{\mu}=5\).)