Finding aconfidence interval for $\theta$ of the uniform distribution on $(0, \frac{1}{\theta})$

Question

Suppose $X_1,\ldots, X_n$ are i.i.d. random variables $Uniform (0, 1/ \theta)$. Find a 95% confidence interval for $\theta$.

What I tried:

$f_{X} (x) = \frac{1}{1/\theta}=\theta, F_{X} (x) = \frac{x-0}{1/\theta - 0}=\theta x$

$f_{X_{(n)}} (x) = n \theta^n x^{n-1}$

then $F_{X_{(n)}} (x) = \theta^n x^{n}$

Let Y= $\theta X_{(n)}$, $F_Y(y)=P(Y\le y)= P(\theta X_{(n)}\le y)= P(X_{(n)}\le y/\theta )=y^n$

Thus, $f_Y(y)=ny^{n-1}=\frac{\gamma(n+1)}{\gamma(n)\gamma(1)} y^{n-1}(1-y)^{1-1}$

So, $Y= \theta X_{(n)} \sim Beta(n,1)$ , and it can be used as a pivot.

I am not sure if it is right and how to compute the interval. Any ideas?

Answer 1

First of all, $$ F_Y(y) = F_{X_{(n)}}(y/\theta^n) = \theta^n \left(y/\theta^n\right)^n \neq y^n. $$ You should take $Y=\theta X_{(n)}$ instead. Then $F_Y(y)=y^n$ for $0<y<1$.

To construct confidence interval chose $a_1$ and $a_2$ such that $\mathbb P(a_1 \leq Y \leq a_2)=0.95$. Say, you can take $a_2=1$ and $F_Y(a_1)=a_1^n=0.05$. In this case $$ 0.95 = \mathbb P(a_1 \leq \theta X_{(n)} \leq 1). $$ Solve the inequalities with respect to $\theta$.