just needed some help on the following question. I've already attempted it twice but still can't get the answer out...
Let $X_i,$ $i=1,...,n$ denote a random sample of size n from a population with a uniform distribution on the interval $(0,\theta)$. Let $X_{(n)}=max{(X_1,...,X_n)}$ and define $U=(1/\theta)X_{(n)}$.
Part one required us to show $F_U(u)=u^n$ if $0\le u\le1$ which I managed to do.
Part two: Because the distribution of U does not depend on $\theta$, U is a pivotal quantity. Find the 95% lower confidence bound for $\theta$.
My attempt: I did a bivariate transformation with $U_1=\frac{X_{(n)}}{U}$ and $U_2=U$ and obtained a joint density function $$f_{U_1,U_2}(u_1,u_2)=n^2(\frac{u_1u_2}{\theta})^{n-1}\frac{1}{\theta}u_2^n$$
Then integrated over the range of $U_2$ to get the marginal density of $U_1$ $$f_{U_1}(u_1)=\frac{nu_1^{n-1}}{\theta^n}$$
I then integrated from 0 to a for the density function of $U_1=\frac{X_{(n)}}{U}=\theta$, equated this to 0.95 and came up with $$a=(1.90)^{1/n}\theta$$
However I'm pretty sure this is wrong. Any help would be much appreciated! Thanks!
You are making this question much more difficult than it actually is. If we wish to be $100(1-\alpha)\%$ confident that $\theta \in [L, \infty)$, then $L$ must satisfy $$\Pr[\theta \ge L] = 1-\alpha;$$ equivalently, $$1-\alpha = \Pr\biggl[\frac{X_{(n)}}{L} \ge \frac{X_{(n)}}{\theta}\biggr] = \Pr\biggl[U \le \frac{X_{(n)}}{L}\biggr] = \biggl(\frac{X_{(n)}}{L}\biggr)^{\!n}.$$ Note this immediately ensures that $L \ge X_{(n)}$. Solving for $L$ then yields $$L = \frac{X_{(n)}}{(1-\alpha)^{1/n}}.$$