Question

Let \( X \) be a random variable having the Poisson distribution with mean \( 1 \). Let \( g: \mathbb{N} \cup \{0\} \to \mathbb{R} \) be defined by
\[g(x) = \begin{cases} 1 & \text{if } x \in \{0, 2\} \\ 0 & \text{if } x \notin \{0, 2\} \end{cases}\]
Then \( E(g(X)) \) is equal to:

• A. \( e^{-1} \)
• B. \( 2e^{-1} \)
• C. \( \frac{5}{2} e^{-1} \)
• D. \( \frac{3}{2} e^{-1} \)

Answer 1

The Correct Option is D

1. Expectation of \( g(X) \): - The expectation is: \[ \mathbb{E}(g(X)) = \sum_{x=0}^\infty g(x) \cdot P(X = x). \] - Since \( g(x) = 1 \) for \( x \in \{0, 2\} \), and \( g(x) = 0 \) otherwise: \[ \mathbb{E}(g(X)) = P(X = 0) + P(X = 2). \] 2. Poisson Probabilities: - For a Poisson random variable with mean 1, the probability mass function is: \[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad \text{with } \lambda = 1. \] - Substituting \( \lambda = 1 \): \[ P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} = e^{-1}, \quad P(X = 2) = \frac{e^{-1} \cdot 1^2}{2!} = \frac{e^{-1}}{2}. \] 3. Combine Results: - Add the probabilities: \[ \mathbb{E}(g(X)) = e^{-1} + \frac{e^{-1}}{2} = \frac{3}{2}e^{-1}. \]