statistics – Exact confidence interval for the uniform parameter

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 the normal approximation.

What I know up to now 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 was asked to find the confidence interval based on this estimator and that's what I tried.

Let $$W = frac {n + 1} {n} Y_n / theta$$ then

$$Pr[l Pr[W Pr[W or l and u represent the lower and upper bound of the confidence interval.$$

Now, here is what troubles 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 $$frac {n + 1} {n}$$ has become irrelevant.

even though I have $$l = frac {n + 1} {n} ^ n sqrt { alpha / 2}$$ and $$u = frac {n + 1} {n} ^ n sqrt {1- alpha / 2}$$

In 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 greater than the MLE, $$Y_n$$how is it that the confidence interval is not centered on this unbiased estimator?

I know that this question is very strange and that I could not find something similar to my argument online, I would really appreciate your help.

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