An Exact Interval From a Uniform Sample Maximum

Why the Sample Maximum Is Informative

Suppose a hypothetical risk study records independent observations from a population that is uniform between zero and an unknown upper endpoint. The largest observation is informative about that endpoint: it cannot exceed the endpoint, and with more observations it is increasingly likely to lie close to it. This makes the sample maximum a natural starting point for an exact confidence interval.

The method depends on the sampling model, not on a large-sample approximation. In particular, it gives a confidence procedure with its stated coverage for every sample size, provided the model assumptions hold.

Model and notation: Let \(X_1,\ldots,X_n\) be independent and identically distributed, with \(X_i\sim\operatorname{Uniform}(0,\theta)\), where the unknown endpoint satisfies \(\theta>0\). Write \(M=X_{(n)}=\max(X_1,\ldots,X_n)\) for the sample maximum. The interval derived here assumes the lower endpoint is known to be zero and all observations are sampled independently from this same distribution.

Under this model, one observation has distribution function \(F_X(x)=x/\theta\) for \(0\leq x\leq\theta\), with probability zero below zero and probability one above \(\theta\). The sample maximum must be no larger than \(\theta\). If the lower endpoint is unknown, the observations are dependent, or the population is not uniform, the derivation below does not apply as stated.

Distribution of the Sample Maximum

To find the distribution of \(M\), consider a value \(m\) between zero and \(\theta\). The event that the maximum is at most \(m\) occurs exactly when every observation is at most \(m\). Independence lets us multiply those probabilities.

\begin{aligned} F_M(m) &=P(M\leq m)\\ &=P(X_1\leq m,\ldots,X_n\leq m)\\ &=\prod_{i=1}^n P(X_i\leq m)\\ &=\left(\frac{m}{\theta}\right)^n,\qquad 0\leq m\leq\theta. \end{aligned}

Thus \(M\) has density \(f_M(m)=n m^{n-1}/\theta^n\) for \(0<m<\theta\), obtained by differentiating its distribution function. More importantly for inference, dividing the maximum by the unknown endpoint removes the parameter from its distribution.

Key result: The pivot \(M/\theta\) has distribution function \(P(M/\theta\leq r)=r^n\) for \(0\leq r\leq1\). Its distribution depends on the sample size \(n\), but not on \(\theta\). Therefore, its quantiles can be used to construct an exact confidence interval.

Inverting the Pivot to Obtain an Interval

Let the desired confidence level be \(1-\alpha\), where \(0<\alpha<1\). Allocate probability \(\gamma_L\) to the lower tail of the pivot and \(\gamma_U\) to its upper tail, with \(\gamma_L+\gamma_U=\alpha\). Since the pivot distribution function is \(r^n\), the two pivot quantiles are \(\gamma_L^{1/n}\) and \((1-\gamma_U)^{1/n}\).

P\left(\gamma_L^{1/n}\leq\frac{M}{\theta}\leq(1-\gamma_U)^{1/n}\right) =(1-\gamma_U)-\gamma_L =1-\alpha.

Both bounds on the pivot are positive, so we can solve the inequalities for \(\theta\). The direction of each inequality matters because \(\theta\) is in the denominator: solving the lower pivot inequality gives an upper bound for \(\theta\), while solving the upper pivot inequality gives a lower bound.

\gamma_L^{1/n}\leq\frac{M}{\theta}\leq(1-\gamma_U)^{1/n} \quad\Longleftrightarrow\quad \frac{M}{(1-\gamma_U)^{1/n}}\leq\theta\leq\frac{M}{\gamma_L^{1/n}}.

The resulting interval is exact: under the stated model, the probability that it contains the fixed parameter \(\theta\), calculated over repeated samples, is exactly \(1-\alpha\). The endpoints are random before sampling because they depend on \(M\). Once the data have been observed, the interval is fixed; its confidence level describes the long-run coverage of the procedure.

Find the maximum.
Compute \(M\) from the observed sample and record the sample size \(n\).

Choose the tail allocation.
Set \(\gamma_L+\gamma_U=\alpha\). Equal tails use \(\gamma_L=\gamma_U=\alpha/2\).

Calculate the pivot quantiles.
Use \(\gamma_L^{1/n}\) and \((1-\gamma_U)^{1/n}\).

Invert and check.
Form the interval with lower endpoint \(M/(1-\gamma_U)^{1/n}\) and upper endpoint \(M/\gamma_L^{1/n}\), then verify the coverage from the pivot probabilities.

For equal-tailed confidence level \(1-\alpha\), the formula simplifies to

\left[ \frac{M}{(1-\alpha/2)^{1/n}}, \frac{M}{(\alpha/2)^{1/n}} \right].

Because both pivot quantiles are less than one, the lower confidence limit is greater than \(M\). This is not a contradiction: the observed maximum is a lower bound on the true endpoint, and a confidence interval for that endpoint should lie above the observed maximum.

Worked Example: A 90% Equal-Tailed Interval

In a hypothetical study of a bounded risk measure, two independent observations are \(5\) and \(8\) units. Assume each observation follows \(\operatorname{Uniform}(0,\theta)\), with an unknown \(\theta\). Find a 90% equal-tailed confidence interval for \(\theta\), and check its coverage.

Method. Here \(n=2\), \(M=\max(5,8)=8\), and \(\alpha=0.10\). Equal tails assign \(0.05\) to each tail of the pivot.

The lower and upper pivot quantiles are

q_L=0.05^{1/2}=0.2236068,\qquad q_U=0.95^{1/2}=0.9746794.

Inverting the pivot bounds gives

\left[\frac{8}{0.9746794},\frac{8}{0.2236068}\right] =[8.2078,35.7771].

Rounded only at the final step, the interval is \([8.2078,35.7771]\) units. Its coverage is \(P(q_L\leq M/\theta\leq q_U)=0.95-0.05=0.90\), as required. The endpoints are both greater than the observed maximum of \(8\).

Unequal Tails and One-Sided Bounds

Tail allocation does not have to be equal. Choosing different values for \(\gamma_L\) and \(\gamma_U\), while keeping their sum equal to \(\alpha\), still gives exact coverage. The choice changes the two endpoints, but not the coverage probability. This can be useful when a particular side of the interval matters more for the analysis.

Worked Example: Allocating More Error to One Tail

A hypothetical reliability measure is modeled by independent observations from \(\operatorname{Uniform}(0,\theta)\). A sample of \(n=3\) has maximum \(M=10\). Construct an 85% interval by allocating \(0.10\) to the lower pivot tail and \(0.05\) to the upper pivot tail.

Method. The total tail probability is \(\alpha=0.15\), since \(0.10+0.05=0.15\). Thus \(\gamma_L=0.10\) and \(\gamma_U=0.05\). The pivot quantiles are

q_L=0.10^{1/3}=0.4641589,\qquad q_U=0.95^{1/3}=0.9830476.

Inverting these bounds gives

\left[\frac{10}{0.9830476},\frac{10}{0.4641589}\right] =[10.1724,21.5443].

Rounded to four decimal places, the interval is \([10.1724,21.5443]\). The pivot probability between the selected quantiles is \(0.95-0.10=0.85\), confirming the exact 85% coverage. The construction is valid even though the two tail probabilities differ.

A one-sided upper confidence bound follows by using just one pivot cutoff. For example, choose \(c=\alpha^{1/n}\). Then \(P(M/\theta\geq c)=1-c^n=1-\alpha\). Rearranging the event gives \(\theta\leq M/c\), so an exact one-sided \(100(1-\alpha)\%\) confidence set is \([M,M/\alpha^{1/n}]\). The lower endpoint \(M\) is included because the model itself implies \(\theta\geq M\).

Worked Example: A 99% One-Sided Upper Bound

A hypothetical bounded claim-cost indicator is modeled as an independent sample from \(\operatorname{Uniform}(0,\theta)\). For \(n=4\) observations, the maximum is \(M=30\). Find a 99% one-sided upper confidence bound for \(\theta\).

Method. Here \(\alpha=0.01\). The pivot cutoff is \(c=\alpha^{1/n}\), so

c=0.01^{1/4}=0.3162278,\qquad U=\frac{M}{c}=\frac{30}{0.3162278}=94.8683.

The confidence set is \([30,94.8683]\), rounded to four decimal places at the upper endpoint. To check coverage, \(\theta\leq U\) is equivalent to \(M/\theta\geq0.3162278\). Its probability is \(1-(0.3162278)^4=1-0.01=0.99\). This is a one-sided bound, not a 99% equal-tailed two-sided interval.

Assumptions, Interpretation, and an Exam Trap

The exactness here is a consequence of the specified distribution and independence, not a general property of maximum-based intervals. If observations have a different distribution, the maximum may still be informative, but its pivot distribution and interval formula will generally change. If the lower endpoint is an unknown parameter too, the ratio \(M/\theta\) alone does not remove all unknown parameters. Check the model before applying the formula.

A common mistake is to multiply the observed maximum by the pivot quantiles and report \([M q_L,M q_U]\). This treats the pivot bounds as if they were bounds on \(\theta\), even though they bound \(M/\theta\). For the first worked example, that incorrect calculation would give \([8(0.2236068),8(0.9746794)]=[1.7889,7.7974]\). The interval is below the observed maximum, which cannot contain a valid endpoint \(\theta\) under this model. The correction is to solve the pivot inequalities for \(\theta\), producing \([M/q_U,M/q_L]\).

Key takeaway: For an independent sample from \(\operatorname{Uniform}(0,\theta)\), the maximum divided by \(\theta\) has distribution function \(r^n\) on \([0,1]\). Find the desired pivot quantiles first, then invert the inequalities carefully. The resulting interval has exact finite-sample coverage under the model assumptions.

Check Your Understanding

For each problem, use the uniform model stated and check the pivot tail probabilities as well as the interval or bound.

A sample of \(n=2\) independent observations from \(\operatorname{Uniform}(0,\theta)\) has maximum \(M=6\). Find the 80% equal-tailed interval. (Answer: \([6/0.9^{1/2},\,6/0.1^{1/2}]\), approximately \([6.3246,18.9737]\).)
A sample of \(n=4\) from \(\operatorname{Uniform}(0,\theta)\) has maximum \(M=15\). Find the 90% one-sided upper confidence bound. (Answer: \(15/0.1^{1/4}\approx26.6742\), giving the confidence set \([15,26.6742]\).)
A sample of \(n=3\) from \(\operatorname{Uniform}(0,\theta)\) has maximum \(M=9\). Form a 90% interval with lower pivot-tail probability \(0.02\) and upper pivot-tail probability \(0.08\). (Answer: \([9/0.92^{1/3},\,9/0.02^{1/3}]\), approximately \([9.2537,33.1563]\).)
For a sample of \(n=5\) from \(\operatorname{Uniform}(0,\theta)\), someone proposes the confidence set \([M,2M]\). Derive its coverage probability and give the numerical coverage. (Answer: \(P(\theta\leq2M)=P(M/\theta\geq1/2)=1-(1/2)^5=31/32=0.96875\).)
Three independent observations from \(\operatorname{Uniform}(0,\theta)\) are \(4,7,11\). Find the 95% equal-tailed interval. (Answer: \([11/0.975^{1/3},\,11/0.025^{1/3}]\), approximately \([11.0932,37.6195]\).)

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