Likelihood of Uniform Distribution Indicator Function

Question

It is well known that the likelihood function for the uniform distribution on $[0,\theta]$ is given by

$$\frac{1}{\theta^n} \mathbf{1}_{\max(x_1,\ldots,x_n)\leq \theta}$$

Where the reason for this indicator is that the likelihood will be equal to $0$ if one of our observations $x_i$ exceeds $\theta$. But why do we not impose a similar condition on an observation being less than $0$? That is, also including $\mathbf{1}_{\min{(x_1,\ldots,x_n)\geq0}}$?

Sorry if I've misunderstood anything, please feel free to correct me!

Answer 1

No matter what $\theta$ is, your actual observations $x_i$ will be nonnegative (under the assumption that it comes from a uniform distribution on $[0, \theta]$). So technically, yes, you could include an indicator for $x_i \ge 0$, but that inequality will always hold.

Answer 2

The point is just that the family of distributions you are considering has different values of $\theta$ for different distributions within the family, but the lower bound of the interval remains the same for all of them.

(The likelihood function is a function of $\theta$ with $x_1,\ldots,x_n$ fixed, so it is $$ L(\theta) = \begin{cases} 1/\theta^n & \text{for } \theta\ge\text{something}, \\ 0 & \text{otherwise.} \end{cases} $$ Thus the m.l.e. of $\theta$ is the lowest value that $\theta$ can attain without $L(\theta)$ being $0.$)