# Likelihood of Uniform Distribution Indicator Function

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!

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.
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 $$0.$$)