Question

Let \( X, Y_1, Y_2 \) be independent random variables such that \( X \) has the probability density function \[ f(x) = \begin{cases} 2e^{-2x} & \text{if } x \geq 0, \\ 0 & \text{otherwise}, \end{cases} \] and \( Y_1 \) and \( Y_2 \) are identically distributed with probability density function \[ g(x) = \begin{cases} e^{-x} & \text{if } x \geq 0, \\ 0 & \text{otherwise}. \end{cases} \] For \( i = 1, 2 \), let \( R_i \) denote the rank of \( Y_i \) among \( X, Y_1, Y_2 \). Then \( E(R_1 + R_2) \) equals:

• A. \( \frac{13}{3} \)
• B. \( \frac{22}{5} \)
• C. \( \frac{21}{5} \)
• D. \( \frac{9}{2} \)

Hint:

For rank-related problems involving exponential random variables, the expected value is often computed by considering the probabilities of each possible ranking arrangement and integrating over the distributions.

Answer 1

The Correct Option is A

Step 1: Distribution of \( X \) and \( Y_i \)
\( X \sim \text{Exp}(2) \) (Exponential distribution with rate 2).
\( Y_1, Y_2 \sim \text{Exp}(1) \) (Exponential distribution with rate 1).
Step 2: Understanding the Rank Distribution
The ranks \( R_1 \) and \( R_2 \) are determined by how the random variables compare to each other. Since \( X \) has a larger mean (due to rate 2) compared to \( Y_1 \) and \( Y_2 \) (which have rate 1), the ranks will be based on the order of the realizations.
Step 3: Calculating \( E(R_1 + R_2) \)
Using the properties of ranks for exponential distributions, it can be shown that: \[ E(R_1 + R_2) = \frac{13}{3}. \] Thus, the correct answer is \( \boxed{\frac{13}{3}} \).