Confidence intervals from pivots
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 \( \theta\).
This was originally published here: https://calvinmccarter.wordpress.com/2013/11/06/confidence-intervals-from-pivots/