What Is the MLE for a Uniform(θ, 0) Distribution?

If $\operatorname{Uniform}(\theta,0)$ means a uniform distribution on the interval from $\theta$ to $0$, then standard endpoint notation requires $\theta<0$. For an independent sample, the maximum-likelihood estimate is the smallest observed value. The reason is that the support restricts how large $\theta$ can be, and the likelihood increases as $\theta$ moves up to that restriction.

The notation can be easy to reverse: if you meant $\operatorname{Uniform}(0,\theta)$, the estimate is the sample maximum instead. The derivation below uses $\operatorname{Uniform}(\theta,0)$.

Likelihood for the sample

Let $X_1,\ldots,X_n$ be independent observations from $\operatorname{Uniform}(\theta,0)$, where $\theta<0$. The density of one observation is

Write the smallest and largest observed values as $X_{(1)}=\min_i X_i$ and $X_{(n)}=\max_i X_i$. Multiplying the individual densities gives the likelihood

where the indicator is 1 when all observations lie in the proposed interval and 0 otherwise. For a sample compatible with the model, the restriction on the parameter is therefore $\theta\le x_{(1)}$.

Why the sample minimum maximizes it

Among feasible values of $\theta$, the likelihood is $(-\theta)^{-n}$. As $\theta$ increases while remaining negative, $-\theta$ decreases, so $(-\theta)^{-n}$ increases. Equivalently,

Thus the likelihood is largest at the greatest allowed value of $\theta$, which is $x_{(1)}$. For an ordinary sample with $x_{(1)}<0$ and $x_{(n)}\le 0$,

This is a constrained maximum at the edge of the allowable parameter range. You should not expect the derivative of the log-likelihood to equal zero there; the maximum occurs at a boundary, not at an interior stationary point.

Worked example

Suppose the observations are $-2.4$, $-1.1$, $-3.0$, and $-1.8$. Their minimum is $-3.0$, so the parameter must satisfy $\theta\le -3.0$ for all four observations to be in the support. The likelihood for a feasible parameter is $(-\theta)^{-4}$.

At $\theta=-3$, the likelihood is $3^{-4}=1/81$. At $\theta=-4$, it is $4^{-4}=1/256$, which is smaller. Values such as $\theta=-2.5$ are not feasible because the observation $-3.0$ lies below the proposed interval. Therefore the MLE is $-3.0$.

If the endpoints are written the other way around

For $X_i\sim\operatorname{Uniform}(0,\theta)$ with $\theta>0$, the likelihood is $\theta^{-n}$ when $\theta\ge X_{(n)}$ and all observations are nonnegative. This likelihood decreases as $\theta$ increases, so the smallest feasible value maximizes it:

This is why checking which endpoint contains the unknown parameter matters: the MLE is the sample minimum when the parameter is the lower endpoint, and the sample maximum when it is the upper endpoint.

Common mistakes

• Using the sample maximum automatically. That is correct for $\operatorname{Uniform}(0,\theta)$, not for $\operatorname{Uniform}(\theta,0)$.
• Ignoring the support constraint. A candidate parameter is impossible if even one observation falls outside its interval; its likelihood is zero.
• Looking only for a zero derivative. The likelihood is monotone over the feasible range, so its maximum is at the boundary.
• Assuming the MLE is unbiased. Here, $\mathbb{E}[X_{(1)}]=n\theta/(n+1)$, so the MLE is biased for finite $n$. If an unbiased estimator is needed, $(n+1)X_{(1)}/n$ is unbiased for $\theta$.