Likelihood function when $X\sim U(0,\theta)$

Question

Let $X_1, ..., X_n$ be $i.i.d$ random variables, uniformly distributed over $(0,\theta)$. Derive the likelihood function given the sample $x_1, ..., x_n$.

Answer

The likelihood function is:

\begin{align} \mathcal{L}(\theta|x_1, ..., x_ n ) &= \prod_{i=1}^{n}f(x_i|\theta)\\ &= \frac{1}{\theta^n}\mathbb{1}(X_1, ..., X_n \in [0,\theta])\\ &= \frac{1}{\theta^n}\mathbb{1}(\max(X_1, ..., X_n) \leq \theta) \end{align}

The second equality is clear to me, that the likelihood will be equal to $0$ if at least one observation $X_i$ will fall outside the interval $[0, \theta]$.

My question is:

Why it is the same as the maximum among observations being less than $\theta$, implied by the the third equality? (how can we justify the third equality, what is the intuition?)

Why the third equality is not: $\frac{1}{\theta^n}\mathbb{1}(\min(X_1, ..., X_n)\geq 0, \max(X_1, ..., X_n) \leq \theta)$

Edited: changed min and max to \min and \max respectively.

Answer 1

The fact that all the observable values $X_1,...,X_n$ are all non-negative is generally considered to be implicit throughout the working (since this is the support of the original distribution) and so it is not stated explicitly. You can certainly add it in explicitly if you wish, which is what you are doing in the condition $\min (X_1,...,X_n) \geqslant 0$.

Answer 2

You are correct that the third equality could use the event $\{\min(X_1, \ldots, X_n)\geq 0, \max(X_1, \ldots, X_n)\leq\theta\}$.

The more compact notation is used because for any value of $\theta$, $\min(X_1, \ldots, X_n)\geq 0$ almost surely, so that event is not informative. The only useful event is $\max(X_1, \ldots, X_n)\leq\theta$.

Answer 3

Just to stress a point, namely that the notion of likelihood is one of a function of $\theta$ indexed by the sample $(x_1,\ldots,x_n)$, instead of the density as a function of $(x_1,\ldots,x_n)$ indexed by the parameter $\theta$. This means that, as a function of $\theta$, and for a given sample $(x_1,\ldots,x_n)$ from that family of Uniform distributions there is no more relevance in stating that $$\mathbb{I}_{\min(x_1,\ldots,x_n)>0}=1$$ than in stating that, say, $3$ is larger than $2$. Both statements are true but irrelevant.