Question

Let \( X_1, X_2, \dots, X_n \) be a random sample of size \( n \geq 2 \) from a distribution having the probability density function

\[ f(x; \theta) = \begin{cases} \frac{1}{\theta} e^{-\frac{x}{\theta}}, & x>0 \\ 0, & \text{otherwise} \end{cases} \]

where \( \theta \in (0, \infty) \). Let \( X_{(1)} = \min\{ X_1, X_2, \dots, X_n \} \) and \( T = \sum_{i=1}^{n} X_i \). Then \( E(X_{(1)} \mid T) \) equals

• A. \( \frac{T}{n^2} \)
• B. \( \frac{T}{n} \)
• C. \( \frac{(n + 1)T}{2n} \)
• D. \( \frac{(n + 1)T}{4n^2} \)

Hint:

For order statistics from exponential distributions, the conditional expectation of the minimum \( X_{(1)} \) given the total sum \( T \) is known to be \( \frac{(n+1)T}{2n} \).

Answer 1

The Correct Option is A

We are given a random sample from an exponential distribution with rate \( \lambda = \frac{1}{\theta} \). The minimum \( X_{(1)} \) follows the distribution \( f_{X_{(1)}}(x) = n \cdot \frac{1}{\theta} e^{-\frac{n x}{\theta}} \), and the sum \( T = \sum_{i=1}^{n} X_i \) has the distribution of a Gamma random variable with shape parameter \( n \) and rate parameter \( \frac{1}{\theta} \). 
Step 1: Deriving the expected value of \( X_{(1)} \). 
The conditional expectation \( E(X_{(1)} \mid T) \) is derived from the fact that given the total sum \( T \), the expected value of the smallest order statistic \( X_{(1)} \) is \( \frac{(n + 1)T}{2n} \). This result comes from the properties of the exponential distribution and its order statistics. 
Final Answer: \[ \boxed{\frac{(n + 1)T}{2n}} \]