Suppose I have $n$ uniform random variables (iid) over $[0,\theta]$ and I want to build a confidence interval for $\theta$. Let $\bar{X_n} = \frac{1}{n} \sum_{k=1}^{n} X_k$, I am asked to show that $\frac{\bar{X_n}}{\theta}$ does not depend on $\theta$, but I don't visualize this intuitively. I tried to write $$\frac{\bar{X_n}}{\theta} = \frac{1}{n} \sum_{k=1}^{n} \frac{X_k}{\theta}$$ But I don't see how $\frac{X_k}{\theta}$ is independent of $\theta$.
Thank you
Intuitively - if $X \sim U[0, \theta]$, then dividing by the length - $\theta$ simply "standardize" the variable, namely restrict it to $[0,1]$. Formally, let $Y = X/\theta$, then $$ F_Y(y)=\mathbb{P}(X/\theta\le y) = \mathbb{P}(X\le\theta y)=F_X(\theta y)=\frac{\theta y}{\theta} = y, \quad \forall y\in(0,1]. $$ Namely, $Y \sim U[0,1]$, that is independent of $\theta$.