Confidence intervals from pivots

November 26, 2022

This is an example of using a pivot to find a confidence interval.

\[X_{1},...,X_{n} \sim \text{Uniform}(0,\theta).\]

1. Find a pivot.

Let \( Q=X_{(n)}/\theta\).

2. Find its distribution:

\[ P(Q \le t)= P(X_i \le t\theta)^n = t^n .\]

3. Find an expression involving an upper and lower bound on the pivot:

\[ P(a \le Q \le b) = b^n-a^n.\]

This implies

\[ P(a \le Q \le 1) = 1-a^n \]

4. Substitute the expression for the pivot from Step 1, and set the RHS to \(1-\alpha \).

\[ P(a \le X_{(n)}/\theta \le 1)=1-a^n \]

\[ P(1/a \ge \theta/X_{(n)} \ge 1) = 1-a^n \]

\[ P( X_{(n)} \le \theta \le \frac{X_{(n)}}{a} ) = 1-a^n \]

Let \( 1-\alpha = 1-a^n\). Then \(a=\alpha^{1/n}\).

\[P(X_{(n)} \le \theta \le \frac{X_{(n)}}{\alpha^{1/n}})=1-\alpha \]

This gives us \( [X_{(n)},\frac{X_{(n)}}{\alpha^{1/n}}]\) as a \( 1-\alpha\) confidence interval for \( \theta\).