Why the Support Must Be Part of the Likelihood
In many familiar models, the possible values of an observation do not depend on the unknown parameter. For example, a normal observation can lie anywhere on the real line for every permitted mean. In such a setting, maximizing the smooth part of the likelihood and solving a score equation can be useful. Uniform models often behave differently: changing the parameter changes which observations are possible. A parameter value that looks favorable after differentiating the density may give the observed data likelihood zero.
The central rule is simple: first find which parameter values make the observed sample possible, and then maximize the likelihood over those values. This is aligned with the SOA VEE Mathematical Statistics syllabus.
For a continuous model, the likelihood is formed from densities, not from the probability of observing the exact sample. A density can be positive at a sample point even though that exact point has probability zero. In a uniform model, the density is constant inside the support and zero outside it; the support condition must therefore remain in the likelihood.
Uniform Samples: Feasibility Before Differentiation
Suppose \(X_1,\ldots,X_n\) are independent and identically distributed as \(U(0,\theta)\), with \(\theta>0\). The density is \(1/\theta\) for \(0\le x\le\theta\), and zero otherwise. Write \(x_{(n)}=\max_i x_i\). If all observations are nonnegative, the likelihood is
The indicator is not decoration: it records whether the parameter allows the observed sample. When \(x_{(n)}>0\), the feasible set is \([x_{(n)},\infty)\). On that set, \(\theta^{-n}\) decreases as \(\theta\) increases, so its largest value is at the smallest feasible parameter. If every observation is zero, however, the feasible set is all \(\theta>0\), and \(\theta^{-n}\) grows without bound as \(\theta\) approaches zero. There is then no maximizer in the parameter space.
Why the Score Equation Can Fail
The score is the derivative of the log-likelihood with respect to the parameter, wherever that log-likelihood is defined and differentiable. In a model whose support depends on the parameter, the log-likelihood may be undefined outside the feasible set and may have a boundary at which the maximum occurs. A score equation describes stationary points in a differentiable region; it does not replace the task of checking the feasible set and its boundary.
For \(U(0,\theta)\), on the feasible region the log-likelihood is \(-n\log\theta\), with derivative \(-n/\theta\). This derivative is never zero. That does not mean there is no MLE: when \(x_{(n)}>0\), the maximum is at the boundary \(\theta=x_{(n)}\). The derivation is a direct consequence of feasibility and monotonicity:
The sample can have positive likelihood only if \(\theta\ge x_{(n)}\), with \(\theta>0\).
For \(x_{(n)}>0\), \(L(\theta)=\theta^{-n}\) there, and its derivative is \(-n\theta^{-n-1}<0\). Thus the likelihood strictly decreases over the feasible set.
The smallest permitted value is \(x_{(n)}\), so it maximizes the likelihood. If \(x_{(n)}=0\), there is no smallest positive feasible value, and the likelihood instead grows without bound as \(\theta\to0^+\).
This argument also shows why a routine “set the score equal to zero” instruction is not a universal MLE method. The maximum may occur at a support boundary, the likelihood may be flat across a feasible interval, or an apparent candidate may be outside the feasible set.
Worked Example: Estimating the Upper Endpoint of a Uniform Distribution
A hypothetical sensor records four independent readings from \(U(0,\theta)\), where the unknown upper endpoint satisfies \(\theta>0\). The readings are \(0.4,\ 1.1,\ 0.8,\ 1.5\). Find the MLE and explain why the score equation does not identify it.
Method and support check. The readings are nonnegative, and their maximum is \(x_{(n)}=1.5>0\). Positive likelihood therefore requires \(\theta\ge1.5\). On this feasible set,
Maximize. Since \(d(\theta^{-4})/d\theta=-4\theta^{-5}<0\) for \(\theta>0\), the likelihood decreases as \(\theta\) increases. Its maximum over \(\theta\ge1.5\) is attained at \(\theta=1.5\). The likelihood there is
Check. Every observed reading is at most \(1.5\), so the estimate is feasible. Any larger \(\theta\) gives a smaller likelihood. The score is \(-4/\theta\), which never equals zero; the maximum is a boundary maximum, not an interior stationary point. The common mistake avoided is concluding “no MLE” merely because the score equation has no solution.
When Both Endpoints Depend on the Parameter
The same procedure applies when the support has a parameter-dependent lower endpoint as well as an upper endpoint. For \(X_i\sim U(\theta,2\theta)\), with \(\theta>0\), the density is \(1/\theta\) on \([\theta,2\theta]\). Let \(x_{(1)}=\min_i x_i\) and \(x_{(n)}=\max_i x_i\). Every observation must satisfy both \(\theta\le x_i\) and \(x_i\le2\theta\). Combining these inequalities gives
The likelihood is \(\theta^{-n}\) on this interval and zero outside it. If the interval contains a positive permitted value, the likelihood decreases with \(\theta\), so the maximum is at its smallest feasible value. If the interval has no positive permitted value, no parameter assigns positive likelihood to the data. Checking that the feasible set is nonempty is essential; one must not report an endpoint from an empty interval.
Worked Example: A Uniform Distribution With Two Moving Endpoints
A hypothetical calibration study yields three independent observations from \(U(\theta,2\theta)\), where \(\theta>0\): \(1.2,\ 1.6,\ 2.0\). Find the MLE and compare the likelihood at the two ends of the feasible interval.
Method and support check. The sample minimum is \(x_{(1)}=1.2\), and the maximum is \(x_{(n)}=2.0\). The support conditions require
For three independent observations, the likelihood on that interval is \(L(\theta)=\theta^{-3}\). It decreases as \(\theta\) increases. At the lower endpoint, \(L(1.0)=1\); at the upper endpoint,
Therefore \(\widehat{\theta}_{\mathrm{MLE}}=1.0\). At this value the support is \([1.0,2.0]\), which includes all three observations. Check. The feasible interval is nonempty, and the likelihood at its smaller endpoint exceeds the likelihood at its larger endpoint. The common mistake avoided is using only the upper-support restriction \(\theta\ge x_{(n)}/2\) and forgetting the lower-support restriction \(\theta\le x_{(1)}\).
Flat Likelihoods and Invalid Score Candidates
Not every parameter-dependent-support model has a density factor that changes with the parameter. For \(U(\theta,\theta+1)\), with \(\theta\in\mathbb{R}\), the density is one inside an interval of width one. The sample is feasible precisely when every observation lies in that interval, or equivalently when
If this interval is nonempty, the likelihood is one throughout it and zero outside it. Every feasible parameter is an MLE. The score inside the feasible interval is zero, but it gives no information about which values are feasible; outside, the likelihood is zero and the log-likelihood is not defined. Thus a score calculation that omits the support can wrongly declare every real number an MLE.
Worked Example: Rejecting an Infeasible Score Candidate
Three independent observations from \(U(\theta,\theta+1)\) are \(2.2,\ 2.7,\ 3.0\), with \(\theta\in\mathbb{R}\). Find the full set of MLEs, then assess the candidate \(\theta=2.5\).
Method and support check. Here \(x_{(n)}=3.0\) and \(x_{(1)}=2.2\), so feasibility requires
For every feasible value, the density of each observation is \(1/1=1\), so the joint likelihood is \(1^3=1\). Outside the interval the likelihood is zero. Hence every \(\theta\in[2.0,2.2]\) is an MLE. In particular, the score within the feasible interval is the derivative of the constant log-likelihood \(\log 1=0\), which is zero.
Check the proposed candidate. At \(\theta=2.5\), the support is \([2.5,3.5]\). The observation \(2.2\) is outside this support, so the likelihood is zero. Although differentiating the density factor alone gives zero score for any \(\theta\), \(2.5\) is not an MLE. The common mistake avoided is treating a zero score as sufficient without verifying that the candidate gives positive likelihood to the data.
Common Exam Traps
- Differentiating before imposing feasibility. First find the parameter values compatible with every observation; only then maximize the likelihood on that set.
- Assuming every MLE is an interior score solution. A maximum can occur at a support boundary, as in \(U(0,\theta)\), or the likelihood can be constant on a feasible interval.
- Reporting an endpoint from an empty feasible set. In \(U(\theta,2\theta)\), the interval \([x_{(n)}/2,x_{(1)}]\) must contain a permitted parameter value before an MLE can be selected.
- Forgetting the parameter space. For \(U(0,\theta)\), \(\theta=0\) is not permitted. If all observations are zero, the likelihood has no maximizer: it becomes arbitrarily large as \(\theta\to0^+\).
Check Your Understanding
For each problem, identify the feasible parameter set before deciding what the likelihood implies.
- An independent sample from \(U(0,\theta)\), \(\theta>0\), is \(0.6,\ 1.4,\ 0.9\). Find the MLE and the likelihood at that estimate. (Answer: \(\widehat{\theta}_{\mathrm{MLE}}=1.4\); \(L(1.4)=1/(1.4)^3\approx0.3644\).)
- An independent sample from \(U(\theta,3\theta)\), \(\theta>0\), is \(1.0,\ 1.4,\ 2.1\). Find the feasible set and the MLE. (Answer: The feasible set is \([0.7,1.0]\), and \(\widehat{\theta}_{\mathrm{MLE}}=0.7\).)
- An independent sample from \(U(\theta,\theta+2)\), \(\theta\in\mathbb{R}\), is \(4.1,\ 5.0,\ 5.8\). Find all MLEs and their likelihood. (Answer: Every \(\theta\in[3.8,4.1]\) is an MLE, with likelihood \(1/8\).)
- An independent sample from \(U(0,\theta)\), \(\theta>0\), consists of three zeros. Does an MLE exist? (Answer: No; \(L(\theta)=\theta^{-3}\) is unbounded as \(\theta\to0^+\).)
- An independent sample from \(U(\theta,2\theta)\), \(\theta>0\), is \(1.1,\ 2.4,\ 2.6\). Determine whether any parameter gives positive likelihood. (Answer: No; feasibility would require \(1.3\le\theta\le1.1\), an empty set.)