Likelihood Ratio Tests For Uniform Endpoints

Why a Uniform Endpoint Is a Special Testing Problem

In many likelihood ratio tests, the parameter changes the shape of a distribution while leaving its support fixed. A uniform distribution with an unknown upper endpoint behaves differently: changing the endpoint also changes which observations are possible. The largest observation therefore carries direct information about the endpoint, and the likelihood ratio depends on an extreme order statistic rather than on an approximately normal estimate.

This tutorial considers independent observations from a uniform distribution on the interval from zero to an unknown positive endpoint. It derives the likelihood ratio test of a specified endpoint against all positive endpoints, obtains its exact null distribution, and explains why Wilks’ usual chi-square approximation does not apply. The methods are aligned with the SOA VEE Mathematical Statistics syllabus.

Model and test: Let $X_1,\ldots,X_n$ be independent and identically distributed with $X_i\sim \operatorname{Uniform}(0,\theta)$, where $\theta>0$. We test $H_0:\theta=\theta_0$, for a specified $\theta_0>0$, against the unrestricted alternative $H_1:\theta>0,\ \theta\ne\theta_0$. The likelihood ratio is $\Lambda=L(\theta_0)/\sup_{\theta>0}L(\theta)$. Smaller values of $\Lambda$ provide evidence against $H_0$.

Write $X_{(n)}=\max(X_1,\ldots,X_n)$ for the sample maximum. The density of each observation is $1/\theta$ when $0\le X_i\le\theta$, and is zero otherwise. Thus, for a sample in the model’s support, the likelihood is

L(\theta)=\theta^{-n}\,\mathbf{1}\{X_{(n)}\le\theta\},\qquad \theta>0.

When $X_{(n)}>0$, the likelihood is zero for $\theta<X_{(n)}$ and decreases as $\theta^{-n}$ for $\theta\ge X_{(n)}$. Its maximum is therefore attained at $\widehat{\theta}=X_{(n)}$, with maximized likelihood $X_{(n)}^{-n}$. The qualification $X_{(n)}>0$ matters: if every observation is zero, then $L(\theta)=\theta^{-n}$ for every $\theta>0$, which is unbounded as $\theta$ decreases to zero, so no maximum likelihood estimate exists. Under the stated continuous model, this all-zero sample has probability zero.

Deriving the Likelihood Ratio

Under the null hypothesis, the likelihood is $\theta_0^{-n}$ if the sample maximum does not exceed $\theta_0$. If even one observation exceeds $\theta_0$, the null likelihood is zero, because that observation is impossible under $H_0$. For $X_{(n)}>0$, combining this constrained likelihood with the unrestricted maximum gives

\Lambda= \begin{cases} \left(\dfrac{X_{(n)}}{\theta_0}\right)^n, & 0<X_{(n)}\le\theta_0,\\[6pt] 0, & X_{(n)}>\theta_0. \end{cases}

A likelihood ratio test rejects for small $\Lambda$. For data that are possible under the null, this means that an unusually small sample maximum leads to rejection. Data with $X_{(n)}>\theta_0$ also lead to rejection: they have zero null likelihood. Although that second event has probability zero when $H_0$ is true, it is essential to describe the test correctly for observed data.

Worked Example: Calculating the Likelihood Ratio and Exact p-Value

Suppose $n=5$, $\theta_0=10$, and the observed sample is $2,4,6,8,9$. Test $H_0:\theta=10$ against all other positive endpoints, and calculate the likelihood ratio and exact p-value.

Step 1: Find the sample maximum. Here $X_{(n)}=9$, which is positive and no greater than the null endpoint. The unrestricted estimate is $\widehat{\theta}=9$.

Step 2: Calculate the two likelihoods. Under the null, $L(10)=10^{-5}=0.00001$. At the unrestricted maximum, $L(9)=9^{-5}=1/59049\approx0.000016935$.

Step 3: Form the ratio. Thus $\Lambda=9^{-5}/10^{-5}=(9/10)^5=0.59049$.

Step 4: Interpret the ratio and p-value. Under the null, the likelihood ratio is uniformly distributed on $(0,1)$, as derived below. Therefore the exact lower-tail p-value is $0.59049$. At significance level $0.05$, this is not sufficient evidence to reject $H_0$.

Check and mistake avoided. The ratio is at most one, as a null likelihood cannot exceed the unrestricted maximum. A common error is to reverse the ratio and report a value greater than one; the convention here is null likelihood divided by unrestricted maximum likelihood.

Exact Calibration Under the Null

Under $H_0$, each observation is uniform on $(0,\theta_0)$. For $0\le m\le\theta_0$, independence gives the distribution of the sample maximum:

P_{\theta_0}\!\left(X_{(n)}\le m\right) =\prod_{i=1}^n P_{\theta_0}(X_i\le m) =\left(\frac{m}{\theta_0}\right)^n.

Since $X_{(n)}\le\theta_0$ with probability one under the null, the likelihood ratio is $\Lambda=(X_{(n)}/\theta_0)^n$. For any $0\le c\le1$, its null distribution is

P_{\theta_0}(\Lambda\le c) =P_{\theta_0}\!\left(X_{(n)}\le\theta_0 c^{1/n}\right) =\left(\frac{\theta_0c^{1/n}}{\theta_0}\right)^n =c.

Thus $\Lambda$ is exactly uniform on $(0,1)$ under $H_0$, for every sample size $n$. A size-$\alpha$ likelihood ratio test rejects when $\Lambda\le\alpha$. Equivalently, it rejects when the sample maximum is below a critical endpoint, or when the data are impossible under the null:

Exact likelihood ratio test: For $0<\alpha<1$, reject $H_0:\theta=\theta_0$ if $$ X_{(n)}\le\theta_0\alpha^{1/n} \quad\text{or}\quad X_{(n)}>\theta_0. $$ Under $H_0$, the second event has probability zero, and the first has probability exactly $\alpha$.

For a sample maximum within the null support, the exact p-value is the observed value of $\Lambda$, namely $(X_{(n)}/\theta_0)^n$. If the observed maximum exceeds $\theta_0$, the null likelihood is zero and the likelihood ratio is zero, so the test rejects. The likelihood ratio test does not spread its rejection probability across two ordinary tails: within the null support, only a very small maximum produces a small ratio.

Worked Example: Finding the Exact Critical Value

A hypothetical quality-control study models $n=3$ independent measurements as uniform from zero to an unknown endpoint. Test $\theta=8$ at significance level $\alpha=0.10$. The observed measurements are $1.2,2.8,3.5$. Find the critical value, calculate the likelihood ratio, and state the decision.

Step 1: Calculate the critical maximum. The rejection threshold is $8(0.10)^{1/3}$. Since $(0.10)^{1/3}\approx0.464159$, the threshold is approximately $3.7133$.

Step 2: Compare the observed maximum with the threshold. The sample maximum is $3.5$, and $3.5\le3.7133$, so the test rejects at level $0.10$.

Step 3: Verify using the likelihood ratio. The likelihood ratio is $(3.5/8)^3=(7/16)^3=343/4096\approx0.08374$. This is less than $0.10$, confirming the decision. The exact p-value is approximately $0.08374$.

Check and mistake avoided. The critical value is a cutoff for the sample maximum, not for each observation separately. Applying the threshold to all three measurements as if each had to be below it would not give the size-$0.10$ likelihood ratio test.

Worked Example: An Observation Impossible Under the Null

Suppose four independent measurements follow the uniform endpoint model, and the null endpoint is $\theta_0=6$. The observed values are $1.1,3.0,4.2,6.4$. Determine the likelihood ratio test result.

Step 1: Identify the maximum and unrestricted fit. The maximum is $X_{(n)}=6.4$, so the unrestricted maximum likelihood estimate is $6.4$. The maximized likelihood is $6.4^{-4}=1/1677.7216\approx0.0005960$.

Step 2: Evaluate the null likelihood. Since $6.4>6$, the observed sample is outside the support under $H_0$. Therefore $L(6)=0$.

Step 3: Form the ratio and decide. The likelihood ratio is $0/6.4^{-4}=0$. The likelihood ratio test rejects $H_0$, regardless of the positive significance level.

Check and mistake avoided. It is incorrect to substitute $6.4$ into the null likelihood as if the endpoint could be exceeded. The support restriction makes the null likelihood exactly zero.

Why Wilks’ Theorem Does Not Apply

Wilks’ theorem gives a familiar chi-square approximation for $-2\log\Lambda$ in regular likelihood problems. Its usual conditions include a parameter-independent support and a maximum likelihood estimate that is asymptotically normal at the square-root-of-sample-size rate. The uniform endpoint model violates these requirements. As $\theta$ changes, the support $[0,\theta]$ changes as well, and the maximum observation sits directly at the boundary of the likelihood’s feasible parameter values.

The rate of the maximum can be seen directly. Under $H_0$, for fixed $t\ge0$ and $n>t$,

P_{\theta_0}\!\left(\frac{n(\theta_0-X_{(n)})}{\theta_0}>t\right) =P_{\theta_0}\!\left(X_{(n)}<\theta_0\left(1-\frac{t}{n}\right)\right) =\left(1-\frac{t}{n}\right)^n \longrightarrow e^{-t}.

Consequently, $n(\theta_0-X_{(n)})/\theta_0$ converges in distribution to an exponential random variable with mean one. The endpoint estimate approaches the true endpoint at rate $n$, not at the usual square-root-of-$n$ rate. This extreme-order-statistic behavior is another sign that the regular approximation behind Wilks’ theorem is unavailable.

In fact, the likelihood ratio has an exact, non-Wilks distribution under the null. Because $\Lambda$ is uniform on $(0,1)$, for $T=-2\log\Lambda$ and $t\ge0$,

P_{\theta_0}(T\ge t) =P_{\theta_0}(\Lambda\le e^{-t/2}) =e^{-t/2}.

This is the upper-tail probability of a chi-square random variable with two degrees of freedom. Thus $T$ has a chi-square distribution with two degrees of freedom exactly under the null, for every $n$. A routine application of Wilks’ theorem based only on the one parameter being tested would instead suggest one degree of freedom, which is not correct here. The two-degree-of-freedom result is a feature of this model’s exact likelihood ratio distribution; it is not a general replacement rule for nonregular models.

Check the support.
The uniform endpoint $\theta$ determines both the density and the largest possible observation, so the support depends on the parameter.

Identify the estimator’s behavior.
The unrestricted estimate is the sample maximum when that maximum is positive. Its error is governed by an extreme order statistic and has rate $n$, not $\sqrt n$.

Calibrate from the exact null law.
The likelihood ratio is uniform under the null, so the exact test and p-value follow directly without invoking a Wilks approximation.

Common Mistakes and Practical Checks

The unusual behavior of this model makes several otherwise familiar shortcuts unsafe. Before reporting a test result, check the support, the direction of the ratio, and whether the likelihood maximum actually exists.

Using the wrong likelihood ratio convention. Here $\Lambda$ is the null likelihood divided by the unrestricted maximum, so $0\le\Lambda\le1$ whenever the denominator is finite. Reject for small values.
Ignoring the support. If $X_{(n)}>\theta_0$, the null likelihood is zero, not $\theta_0^{-n}$.
Using a routine one-degree-of-freedom Wilks cutoff. The support depends on $\theta$, and the estimator has a nonstandard rate. Use the exact distribution derived here.
Claiming an MLE exists for every sample. For an all-zero sample, the likelihood is unbounded as $\theta\downarrow0$; there is no MLE. Under the continuous model this event has probability zero, but the qualification should still be stated.
Confusing the individual observations with the maximum. The likelihood ratio test’s lower-tail cutoff applies to $X_{(n)}$, the sample maximum.

Key takeaway: For independent observations from $\operatorname{Uniform}(0,\theta)$, the likelihood ratio for a specified endpoint is determined by the sample maximum. Under the null it is exactly uniform, giving an exact test; the parameter-dependent support and the maximum’s nonstandard rate rule out the usual Wilks approximation.

Check Your Understanding

For each problem, use the exact likelihood ratio results for the uniform endpoint model.

For $n=2$, test $\theta=5$ using observations $3.0$ and $4.5$. Calculate the likelihood ratio and exact p-value, and decide at level $0.05$. (Answer: $\Lambda=(4.5/5)^2=0.81$, so the exact p-value is $0.81$ and the test does not reject.)
For $n=4$, the null endpoint is $12$, the observed maximum is $7.2$, and $\alpha=0.05$. Find the critical maximum and decide using both the threshold and the p-value. (Answer: The critical maximum is $12(0.05)^{1/4}\approx5.6745$; $\Lambda=(7.2/12)^4=0.1296$, so do not reject.)
For $n=6$, test $\theta=9$ at level $0.01$, given a sample maximum of $4.0$. Calculate the exact likelihood ratio and state the decision. (Answer: $\Lambda=(4/9)^6=4096/531441\approx0.007707$; since this is below $0.01$, reject.)
For a sample of five observations, the null endpoint is $7$, but the observed maximum is $7.3$. What are the null likelihood and likelihood ratio, and what is the decision? (Answer: The null likelihood is zero, the likelihood ratio is zero, and the test rejects.)
Explain why a one-degree-of-freedom Wilks approximation is not justified for testing one uniform endpoint, and state the exact null distribution of $T=-2\log\Lambda$. (Answer: The support depends on the endpoint and the maximum likelihood estimate converges at rate $n$, not $\sqrt n$; exactly under the null, $T\sim\chi^2_2$, not the usual Wilks $\chi^2_1$.)

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