Question

Let \( X_1, X_2, \dots, X_n \), where \( n \geq 5 \), be a random sample from a distribution with the probability density function \[ f(x; \theta) = \begin{cases} e^{-(x - \theta)}, & x \geq \theta, \\ 0, & \text{otherwise}, \end{cases} \] where \( \theta \in \mathbb{R} \) is the unknown parameter. Then which of the following statement(s) is (are) true? 
 

• A. A 95% confidence interval of \( \theta \) has to be of finite length
• B. \( \left( \min{\{X_1, X_2, \dots, X_n\}} + \frac{1}{n} \ln(0.05), \min{\{X_1, X_2, \dots, X_n\}} \right) \) is a 95% confidence interval of \( \theta \)
• C. A 95% confidence interval of \( \theta \) can be of length 1
• D. A 95% confidence interval of \( \theta \) can be of length 2

Hint:

For exponential distributions, the minimum of the sample is a good estimator for the parameter \( \theta \) and leads to a finite confidence interval.

Answer 1

The Correct Option is B, C, D

Step 1: Find the distribution of the minimum

Let $X_{(1)} = \min{X_1, X_2, ..., X_n}$. For $x \geq \theta$: $$P(X_{(1)} > x) = P(X_1 > x, ..., X_n > x) = [P(X_1 > x)]^n = [e^{-(x-\theta)}]^n = e^{-n(x-\theta)}$$

Therefore, $X_{(1)} - \theta \sim \text{Exponential}(n)$ with parameter $n$.

The CDF of $X_{(1)}$ is: $$P(X_{(1)} \leq x) = 1 - e^{-n(x-\theta)}, \quad x \geq \theta$$

Step 2: Find a pivotal quantity

Let $Y = n(X_{(1)} - \theta)$. Then $Y \sim \text{Exponential}(1)$.

For a 95% confidence interval, we need: $$P(a \leq Y \leq b) = 0.95$$

For the exponential distribution with rate 1: $$P(Y \leq y) = 1 - e^{-y}$$

We can choose: $P(Y \leq b) = 0.95$, which gives $1 - e^{-b} = 0.95$, so $e^{-b} = 0.05$, thus $b = -\ln(0.05) = \ln(20)$.

And $a = 0$ (since $Y \geq 0$).

Therefore: $$P(0 \leq n(X_{(1)} - \theta) \leq \ln(20)) = 0.95$$

Rearranging: $$P(X_{(1)} - \frac{\ln(20)}{n} \leq \theta \leq X_{(1)}) = 0.95$$

So a 95% confidence interval is: $\left(X_{(1)} - \frac{\ln(20)}{n}, X_{(1)}\right)$

The length of this interval is $\frac{\ln(20)}{n} \approx \frac{2.996}{n}$.

Option (A): A 95% confidence interval of $\theta$ has to be of finite length

Yes, any valid confidence interval must have finite length. The interval we constructed has length $\frac{\ln(20)}{n}$, which is finite.

Option (A) is TRUE 

Option (B): $\left(\min{X_1,X_2,...,X_n} + \frac{1}{n}\ln(0.05), \min{X_1,X_2,...,X_n}\right)$ is a 95% CI

Note that $\ln(0.05) = \ln(1/20) = -\ln(20)$.

So this interval is: $\left(X_{(1)} - \frac{\ln(20)}{n}, X_{(1)}\right)$

This matches our derived 95% confidence interval!

Option (B) is TRUE 

Option (C): A 95% confidence interval of $\theta$ can be of length 1

The length of our interval is $\frac{\ln(20)}{n} \approx \frac{2.996}{n}$.

For length = 1: $\frac{\ln(20)}{n} = 1$, which gives $n = \ln(20) \approx 2.996 < 5$.

Since we're given $n \geq 5$, we have $\frac{\ln(20)}{n} \leq \frac{2.996}{5} \approx 0.599 < 1$.

However, we could construct different confidence intervals. For instance, using: $$P(c \leq n(X_{(1)} - \theta) \leq d) = 0.95$$

with $d - c = n$, we could get a length of 1. For example, if we choose $c$ and $d$ such that: $$e^{-c} - e^{-d} = 0.95$$

and $d = c + n$, then the interval length would be $\frac{d-c}{n} = 1$.

For $n = 5$: We need $e^{-c} - e^{-(c+5)} = 0.95$, which gives $e^{-c}(1 - e^{-5}) \approx 0.9933 e^{-c} = 0.95$, so $e^{-c} \approx 0.9565$, giving $c \approx 0.0445$.

This is valid, so a 95% CI can have length 1.

Option (C) is TRUE 

Option (D): A 95% confidence interval of $\theta$ can be of length 2

Similarly, we can construct a confidence interval of length 2 by choosing appropriate $c$ and $d$ such that $d - c = 2n$ and the probability condition is satisfied.

Option (D) is TRUE 

Answer: (B), (C), and (D)