What Is the Minimal Sufficient Statistic for Uniform(θ, θ)?

The notation $\operatorname{Uniform}(\theta,\theta)$ gives the same value for the lower and upper endpoints. Taken literally, it describes a point mass at $\theta$, rather than a continuous uniform distribution over an interval of positive length. For a sample from that model, $X_1$ determines every observation, so it is a minimal sufficient statistic.

If one endpoint or the interval width was mistyped, there is no single answer without the intended model. The likelihood makes the distinction clear. Two common nondegenerate examples are worked out below.

Literal interpretation: both endpoints equal θ

Assume $X_1,\ldots,X_n$ are independent observations and each has the point-mass distribution at $\theta$. Then

Consequently, knowing $X_1$ gives the entire observed sample. The statistic $T=X_1$ is sufficient: conditional on $X_1$, the rest of the sample is already determined. It is minimal in the usual sense that, on the set of samples that can occur, the full data contain no information beyond $X_1$. Any measurable transformation of $X_1$ that is invertible with a measurable inverse on its image is an equivalent choice.

This is a degenerate model, and the usual continuous density and likelihood-ratio calculations for uniform distributions do not apply directly. In particular, there is no ordinary Lebesgue density on an interval of positive length. If the intended question concerned a continuous uniform distribution, check whether one endpoint or a width term is missing.

Example: Uniform(θ, θ + 1)

Suppose instead that $X_1,\ldots,X_n$ are independent and uniformly distributed on $[\theta,\theta+1]$. Write $X_{(1)}=\min_i X_i$ and $X_{(n)}=\max_i X_i$. The joint density is

As a function of $\theta$, this is nonzero exactly when

Thus the likelihood depends on the sample through the pair $(X_{(1)},X_{(n)})$. That pair is sufficient by the factorization theorem. It is also minimal sufficient: for possible samples, equal likelihood functions in $\theta$ require the same interval of allowed values, which determines both the minimum and maximum. Therefore, for this model, the minimal sufficient statistic is

Numerical illustration

For a sample $0.4,\ 0.8,\ 1.1$, the minimum is $0.4$ and the maximum is $1.1$. The likelihood is positive precisely when

The pair $(0.4,1.1)$ records this likelihood interval. The middle observation, $0.8$, does not change the interval, so it supplies no additional information about $\theta$ once the endpoints of the sample are known.

Another common model: Uniform(0, θ)

For independent observations from $\operatorname{Uniform}(0,\theta)$, with $\theta>0$, the joint density is

It depends on the data through the sample maximum, so $X_{(n)}$ is sufficient. The likelihood’s cutoff in $\theta$ is exactly the observed maximum, making $X_{(n)}$ minimal sufficient as well. This differs from the $\operatorname{Uniform}(\theta,\theta+1)$ model, where both the sample minimum and maximum are needed.

Common mistakes

• Assuming the parameterization: $\operatorname{Uniform}(\theta,\theta)$, $\operatorname{Uniform}(0,\theta)$, and $\operatorname{Uniform}(\theta,\theta+1)$ are different models and have different answers.
• Using only the maximum for every uniform model: For an interval that shifts with $\theta$, the minimum can also constrain the parameter.
• Confusing sufficient with minimal sufficient: A sufficient statistic may retain more sample detail than needed. Minimal sufficiency requires that no smaller data summary retain all the information about the parameter.
• Ignoring support: In uniform models, the likelihood depends on which parameter values make the observed sample possible, not just on the density’s numerical value inside the interval.

To identify the statistic for a different uniform parameterization, write its joint density, keep the indicator expressing the sample-support constraints, and determine which sample features set the boundaries of the allowed parameter values.