Question

Let \( X \) be a random variable with the moment generating function \[ M_X(t) = \frac{6}{\pi^2} \sum_{n \geq 1} \frac{e^{t^2/2n}}{n^2}, t \in \mathbb{R}. \] Then \( P(X \in \mathbb{Q}) \), where \( \mathbb{Q} \) is the set of rational numbers, equals 
 

• A. 0
• B. \( \dfrac{1}{4} \)
• C. \( \dfrac{1}{2} \)
• D. \( \dfrac{3}{4} \)

Hint:

For continuous random variables, the probability that the variable takes any specific value is always zero.

Answer 1

The Correct Option is A

Step 1: Understand the moment generating function.
The moment generating function \( M_X(t) \) is given, but the probability that \( X \) takes a rational value depends on the nature of the distribution.

Step 2: Probability of rational values.
For continuous distributions, the probability of the random variable \( X \) taking any specific value, including a rational value, is always 0. This is because the probability of any single point in a continuous distribution is zero.

Step 3: Conclusion.
Thus, \( P(X \in \mathbb{Q}) = 0 \), so the correct answer is (A) 0.