I am given the following.
$$X_1, X_2,...,X_n \sim U(0,\theta)$$ and I want to get an exact confidence interval of $\theta$ without using normal approximation.
What I know so far is that $Y_n=max(x_1,x_2,...,x_n)$ is the MLE for $\theta $ and $\frac{n+1}{n}Y_n$ is the unbiased estimator.
I have been instructed to find the confidence interval based on this estimator and this is what I have tried.
Let $W=\frac{n+1}{n}Y_n/\theta$ then
$$Pr[l<W<u]=1-\alpha$$ $$Pr[W<l]=\alpha/2$$ $$Pr[W<u]=1-\alpha/2$$ where $l$ and $u$ represent the lower and the upper bound of the confidence interval.
Now here is what confuses me.
When I solve for the confidence interval I get $$\frac{Y_n}{^n\sqrt{1-\alpha/2 }}<\theta<\frac{Y_n}{^n\sqrt{\alpha/2 }}$$ and algebraically the $\frac{n+1}{n}$ became irrelevant.
even though I got $l=\frac{n+1}{n}^n\sqrt{\alpha/2}$ and $u=\frac{n+1}{n}^n\sqrt{1-\alpha/2}$
Trying to clarify my question as much as possible, what I probably want to say is that if the unbiased estimator $\frac{n+1}{n}Y_n$ is superior to just the MLE, $Y_n$, how is it that the confident interval is not centered around that unbiased estimator?
I know that this is a very odd question and I could not find something similar to my argument online, so I would really appreciate your help.