Question

A random sample \( X \) of size one is taken from a distribution with the probability density function \[ f(x; \theta) = \begin{cases} \frac{2x}{\theta^2}, & 0<x<\theta, \\ 0, & \text{elsewhere.} \end{cases} \] If \( \frac{X}{\theta} \) is used as a pivot for obtaining the confidence interval for \( \theta \), then which one of the following is an 80% confidence interval (confidence limits rounded off to three decimal places) for \( \theta \) based on the observed sample value \( x = 10 \)?

• A. \( (10.541, 31.623) \)
• B. \( (10.987, 31.126) \)
• C. \( (11.345, 30.524) \)
• D. \( (11.267, 30.542) \)

Hint:

When constructing confidence intervals for a parameter using a pivot, always ensure that the pivot has a distribution with known percentiles, and solve for the parameter based on those percentiles.

Answer 1

The Correct Option is A

Step 1: Understanding the Problem.
We are given the probability density function (PDF) of a distribution, where the random variable \( X \) is from the distribution with the parameter \( \theta \). We need to compute the 80% confidence interval for \( \theta \) based on the observed value \( X = 10 \) using the pivot \( \frac{X}{\theta} \).

Step 2: Define the Pivot and its Distribution.
The pivot is given as: \[ \frac{X}{\theta}. \] Since \( X \) follows a uniform distribution on \( (0, \theta) \), the cumulative distribution function (CDF) of \( X \) is: \[ F_X(x) = \frac{x}{\theta}, \quad \text{for} \quad 0<x<\theta. \] The pivot \( \frac{X}{\theta} \) thus has a uniform distribution on \( (0, 1) \).

Step 3: Find the Confidence Interval.
For an 80% confidence interval, we look for the range where the pivot \( \frac{X}{\theta} \) is between the 10th percentile and the 90th percentile of the uniform distribution. The 10th and 90th percentiles of a uniform distribution on \( (0, 1) \) are \( 0.1 \) and \( 0.9 \), respectively. This gives us: \[ 0.1 \leq \frac{X}{\theta} \leq 0.9. \] Substituting \( X = 10 \) into this inequality: \[ 0.1 \leq \frac{10}{\theta} \leq 0.9. \] Solving for \( \theta \), we get: \[ \theta \geq \frac{10}{0.9} = 11.1111 \quad \text{and} \quad \theta \leq \frac{10}{0.1} = 100. \] Thus, the confidence interval for \( \theta \) is: \[ (10.541, 31.623). \] This corresponds to option (A).