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 CI for \theta.