Question

Let \( X \) be a random variable having Poisson distribution such that \( E(X^2) = 110 \). Then which one of the following statements is NOT true?

• A. \( E(X^n) = 10 E\left( (X+1)^n - 1 \right) \), for all \( n = 1, 2, 3, \dots \)
• B. \( P(X \text{ is even}) = \frac{1}{4} (1 + e^{-20}) \)
• C. \( P(X = k)<P(X = k+1), \quad \text{for} \, k = 0, 1, \dots, 8 \)
• D. \( P(X = k)>P(X = k+1), \quad \text{for} \, k = 10, 11, \dots \)

Hint:

For Poisson distributions, the probability of the event being even or odd can be computed using the probability generating function.

Answer 1

The Correct Option is B

We are given that \( X \) follows a Poisson distribution with \( E(X^2) = 110 \). For a Poisson distribution, we know that \( E(X) = \lambda \) and \( \text{Var}(X) = \lambda \). Thus, \( E(X^2) = \lambda + \lambda^2 \), which gives \( \lambda = 10 \). Hence, the Poisson distribution is \( X \sim \text{Poisson}(10) \). Step 1: Evaluate each option.
- Option (A) is correct because \( E(X^n) \) for Poisson distribution satisfies this relation.
- Option (B) is incorrect because the formula for \( P(X \text{ is even}) \) for a Poisson distribution is different, and \( \frac{1}{4}(1 + e^{-20}) \) is not the correct expression.
- Option (C) is correct because for a Poisson distribution with \( \lambda = 10 \), the probability increases for small \( k \) and decreases for large \( k \).
- Option (D) is correct because the probability decreases for large \( k \).
Final Answer: \[ \boxed{P(X \text{ is even}) = \frac{1}{4} (1 + e^{-20})}. \]