Question

Let \( X_1, X_2, X_3 \) be a random sample from a continuous distribution with the probability density function

\[ f(x) = \begin{cases} e^{-(x-\mu)}, & x \geq \mu \\ 0, & \text{otherwise} \end{cases} \]

Let \( X_{(1)} = \min\{X_1, X_2, X_3\} \) and \( c > 0 \) be a real number. Then \( (X_{(1)} - c, X_{(1)}) \) is a 97% confidence interval for \( \mu \), if \( c \) (round off to 2 decimal places) equals ................

Hint:

For confidence intervals involving order statistics, use the CDF of the order statistic and solve for the required probability. The exponential distribution’s memoryless property often simplifies the calculation.

Answer 1

Correct Answer: 1.12

Step 1: Identify the distribution of the minimum.
The minimum of \( X_1, X_2, X_3 \) is \( X_{(1)} = \min\{X_1, X_2, X_3\} \). Since the \( X_i \)'s are i.i.d. and follow an exponential distribution with parameter \( \mu \), the cumulative distribution function (CDF) of \( X_{(1)} \) is: \[ F_{X_{(1)}}(x) = 1 - (1 - F_X(x))^3 = 1 - e^{-3(x - \mu)} \quad \text{for} \quad x \geq \mu. \] Step 2: Use the 97% confidence interval.
For a 97% confidence interval, we want the probability that \( X_{(1)} \) lies within \( (X_{(1)} - c, X_{(1)}) \) to be 0.97. Therefore, we solve for \( c \) such that: \[ P(X_{(1)} \geq c) = 0.97. \] This leads to: \[ e^{-3c} = 0.03 \quad \Rightarrow \quad 3c = \ln\left(\frac{1}{0.03}\right) \quad \Rightarrow \quad c \approx 1.8. \] Final Answer: \[ \boxed{1.8}. \]