This is an example from Bertsekas' Introduction to Probability 2nd edition example 8.2
Consider now a variation involving the first $n$ dates. Assume that Juliet is late by random amounts $$X_1, \ldots, X_n$$ which given $\Theta = \theta$ are uniformly distributed in the interval $[0, \theta]$ and conditionally independent. Let $$X = (X_1, \ldots, X_n)$$ and $$x = (x_1, \ldots, x_n)$$
The parameter $\theta$ is unknown and is modeled as the value of a random variable $\theta$ uniformly distributed between zero and one hour.
Why is
$$f_{X \mid \theta}(x\mid \theta) = \begin{cases} \frac{1}{\theta^n} & \bar{x} \leq \theta \leq 1 \\ 0 & \text{otherwise} \end{cases} $$
$$\bar{x} = \max(x_1, \ldots, x_n)$$
shouldn't $f_{X \mid \theta}(x\mid \theta)$ base on the distribution of $X$ instead of taking the max?
If $n$ random variables $X_1,\ldots,X_n$ are independent, then their joint density $f_{X_1,\ldots,X_n}(x_1,\ldots,x_n)$ is equal to $f_{X_1}(x_1)\cdots f_{X_n}(x_n)$. In this case, for every index $i\in\{1,\ldots,n\}$, and every value of $x>0$, we have $$ f_{X_i}(x) = \begin{cases} 1/\theta & \text{if }x<\theta, \\ 0 & \text{if }x>\theta. \end{cases} $$ Multiplying those gives $(1/\theta)^n$ if $x_1,\ldots,x_n$ are all $<\theta$. But if even one $x_i$ is $>\theta$, then you're multiplying a $0$ by the other factors and getting $0$. To say that for every $i$ we have $x_i<\theta$ is to say that $\max\{x_1,\ldots,x_n\}<\theta$.