Question

Let \( \{X_n\}_{n \geq 1} \) be a sequence of independent random variables with \( X_n \) having the probability density function as \[ f_n(x) = \begin{cases} \dfrac{1}{2^{n/2} \Gamma\left( \frac{5}{2} \right)} e^{\tfrac{-x^2}{2}} x^{(\tfrac{n}{2}-1)} , & x > 0, \\ 0, & \text{otherwise}. \end{cases} \] Then \[ \lim_{n \to \infty} \left[ P(X_n > \frac{3n}{4}) + P(X_n > n + 2 \sqrt{2n}) \right] \] equals 
 

• A. 1 + \( \Phi(2) \)
• B. 1 - \( \Phi(2) \)
• C. \( \Phi(2) \)
• D. 2 - \( \Phi(2) \)

Hint:

For large \( n \), sequences of random variables with Gaussian-like distributions can be approximated by standard normal probabilities using \( \Phi(x) \).

Answer 1

The Correct Option is D

Step 1: Analyze the limiting behavior. 
As \( n \to \infty \), the random variable \( X_n \) behaves like a standard normal variable. The given probability density function approaches the standard normal distribution.

Step 2: Use the standard normal distribution. 
Using the cumulative distribution function \( \Phi(x) \) for the standard normal distribution, the limiting probability can be written as: \[ P(X_n > \frac{3n}{4}) + P(X_n > n + 2 \sqrt{2n}) = 2 - \Phi(2). \]

Step 3: Conclusion. 
The correct answer is (D) \(2 -  \Phi(2) \).