Finding one sided confidence intervals

[eruth]

1. Let X1,...,Xn be iid with cdf Fθ, where Fθ(x) = (x/θ)^β for x in [0, θ]. Here β>0 is a known constant and θ>0 is an unknown parameter. Let X(n)= max (X1,...,Xn). f(x|θ)=nβ(x^(nβ-1))/(θ^(nβ)) when x is in [0, θ].
Part one was to show that P= X(n)/θ is a pivot for θ. Which I did by showing its distribution doesn't depend on any unknowns.
Part two is to find the right-sided 95% confidence interval for θ.


2.
I know to make 1-alpha= Pr(Q(alphaL) ≤ P ≤ Q(alphaU))
= Pr(X(n)/Q(alphaU) ≤θ≤ X(n)/Q(alphaL))

But then I'm not sure where to go from there. I know it needs to be in the form of [L(X1,...,Xn), c] where c is some constant. I know I need to make alphaL and alphaU either .05 and 1 or 0 and .95...
Thanks so much any help would be great, as the next problem builds on this one.

 

[lippyka]

The solution is to calculate the upper and lower quantiles, Q(αL) and Q(αU), of the distribution of P. From this, you can construct the 95% confidence interval for θ as [X(n)/Q(αU), X(n)/Q(αL)].