How to Find a Confidence Interval for a Uniform Distribution

A confidence interval for a uniform distribution depends on which parameter is unknown. The common textbook model is independent observations from $\operatorname{Uniform}(0,\theta)$, where the lower endpoint is known to be 0 and the upper endpoint $\theta$ is unknown. In that case, the sample maximum gives an exact interval.

Below, $n$ is the sample size, $M$ is the largest observed value, and the confidence level is $1-\alpha$. If you instead mean a uniform distribution with both endpoints unknown, or an interval for its mean, see the alternatives near the end.

The sample maximum as a pivot

Let $X_1,\ldots,X_n$ be independent and uniformly distributed between 0 and $\theta$, and define

$$M=\max(X_1,\ldots,X_n).$$

For $0\le m\le\theta$, all $n$ observations are at most $m$ only when each falls in an interval of probability $m/\theta$. Independence therefore gives

$$\Pr(M\le m)=\left(\frac{m}{\theta}\right)^n.$$

Equivalently, the ratio $M/\theta$ has a distribution that does not depend on the unknown parameter:

$$\Pr\left(\frac{M}{\theta}\le t\right)=t^n,\qquad 0\le t\le1.$$

This ratio is called a pivot. Its known distribution lets you choose probability bounds and solve them for $\theta$.

Exact two-sided confidence interval

For an equal-tailed interval with confidence level $1-\alpha$, put probability $\alpha/2$ below the pivot and probability $\alpha/2$ above it. The pivot quantiles are $(\alpha/2)^{1/n}$ and $(1-\alpha/2)^{1/n}$, so

$$\Pr\left((\alpha/2)^{1/n}\le\frac{M}{\theta}\le(1-\alpha/2)^{1/n}\right)=1-\alpha.$$

Rearranging gives the exact interval

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

For a 95% interval, set $\alpha=0.05$:

$$\boxed{\left[\frac{M}{0.975^{1/n}},\;\frac{M}{0.025^{1/n}}\right]}.$$

“Exact” means that, under the stated model, the interval contains the true $\theta$ with probability exactly 0.95 over repeated samples. It does not mean there is a 95% probability that a fixed parameter lies in this particular interval.

A 95% example

Suppose $n=10$ observations are taken and their maximum is $M=8$. The equal-tailed interval is

$$\left[\frac{8}{0.975^{1/10}},\;\frac{8}{0.025^{1/10}}\right] \approx [8.02,\;11.57].$$

Thus, the exact equal-tailed 95% confidence interval for the upper endpoint is approximately $[8.02,11.57]$. The observed maximum cannot exceed $\theta$, and the interval’s lower bound is slightly above 8 because this construction allocates some error probability to each tail.

One-sided 95% upper bound

Sometimes the practical goal is only to put an upper limit on $\theta$. Since $M\le\theta$ always, use

$$\Pr\left(M\le\theta\le\frac{M}{\alpha^{1/n}}\right)=1-\alpha.$$

For the example, this gives

$$\left[8,\frac{8}{0.05^{1/10}}\right]\approx[8,10.79].$$

This is a one-sided 95% confidence interval, not the equal-tailed two-sided interval. The choice is about the question you need the interval to answer; neither formula should be presented without identifying which type of interval it is.

If the uniform distribution has other unknowns

If the model is $\operatorname{Uniform}(a,b)$ with both endpoints unknown, the formula above does not apply directly. There is no single confidence interval for “the distribution” until you specify the target, such as $a$, $b$, the width $b-a$, or the mean $(a+b)/2$. For example, when $n\ge2$, the sample range $R=M-m$, where $m$ is the sample minimum, can be used to form an interval for the width. Its normalized distribution has cumulative distribution function

$$\Pr\left(\frac{R}{b-a}\le r\right)=nr^{n-1}-(n-1)r^n,\qquad 0\le r\le1.$$

For an interval for the mean of $\operatorname{Uniform}(0,\theta)$, use the interval for $\theta$ and divide both endpoints by 2, since the mean is $\theta/2$. If the support endpoints of a uniform distribution are known, its mean is already known as their midpoint; confidence intervals for means in other models are different problems.

Common mistakes

• Using the sample mean as if it were the maximum. The exact pivot above comes from the sample maximum; replacing it with the sample mean does not give these limits.
• Forgetting the sample size. The exponent $1/n$ is essential. A larger sample maximum generally gives more information about $\theta$.
• Confusing one-sided and two-sided intervals. The interval $[M,M/\alpha^{1/n}]$ has confidence level $1-\alpha$ but is not equal-tailed.
• Assuming both endpoints are unknown. The displayed formulas require a known lower endpoint of 0. If both endpoints are unknown, first specify which parameter you want to estimate.