Question

Suppose that X is a discrete random variable with the following probability mass function: \[ P(X = 0) = \frac{1}{2}(1 + e^{-1}), P(X = k) = \frac{e^{-1}}{2 k!}  for  k = 1, 2, 3, .... \] Which of the following statements is/are true?

• A. \( E(X) = 1 \)
• B. \( E(X)<1 \)
• C. \( E(X | X>0)<\frac{1}{2} \)
• D. \( E(X | X>0)>\frac{1}{2} \)

Hint:

- When calculating expectations, use the PMF and sum over all possible values of \( X \). For conditional expectations, adjust the probabilities accordingly. - The expected value \( E(X) \) can be computed as a weighted average of all possible outcomes, weighted by their probabilities.

Answer 1

The Correct Option is B

1) Understanding the problem:
We are given a discrete random variable \( X \) with a probability mass function (PMF), and we need to compute expected values and conditional expectations. 2) Analyzing each option:
- (A) \( E(X) = 1 \): This is false. We need to calculate the expected value of \( X \). Since the PMF involves an exponential decay, \( E(X) \) will be less than 1.
- (B) \( E(X)<1 \): This is true. We calculate the expected value \( E(X) \) as: \[ E(X) = \sum_{k=0}^{\infty} k P(X = k) = \frac{1}{2} (1 + e^{-1}) + \sum_{k=1}^{\infty} \frac{e^{-1}}{2 k!} = \frac{1}{2}(1 + e^{-1}) + \frac{1}{2} (e^{-1} + 1) \] This value is less than 1.
- (C) \( E(X | X>0)<\frac{1}{2} \): This is false. The conditional expectation given \( X>0 \) is greater than \( \frac{1}{2} \), as the value tends to be skewed towards higher values of \( X \).
- (D) \( E(X | X>0)>\frac{1}{2} \): This is true. The conditional expectation \( E(X | X>0) \) is greater than \( \frac{1}{2} \) because most of the mass is at higher values of \( X \), such as 1, 2, 3, etc.
Thus, the correct answer is (B) and (D).