Question

Let \( X \) be a Poisson random variable with mean \( \frac{1}{2} \). Then \( E((X + 1)!) \) equals

• A. \( 2e^{-\frac{1}{2}} \)
• B. \( 4e^{-\frac{1}{2}} \)
• C. \( 4e^{-1} \)
• D. \( 2e^{-1} \)

Hint:

For Poisson random variables, the expected value of functions of \( X \) can often be computed by using standard results for sums involving \( e^{-\lambda} \).

Answer 1

The Correct Option is B

Step 1: Recall the Poisson distribution.
For a Poisson distribution with mean \( \lambda \), the probability mass function is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, k = 0, 1, 2, \dots \] Here, \( \lambda = \frac{1}{2} \).

Step 2: Use the definition of expected value.
The expected value of \( (X + 1)! \) is: \[ E((X + 1)!) = \sum_{k=0}^{\infty} (k + 1)! \frac{\left(\frac{1}{2}\right)^k e^{-\frac{1}{2}}}{k!}. \] Simplifying the terms: \[ E((X + 1)!) = e^{-\frac{1}{2}} \sum_{k=0}^{\infty} (k + 1) \left( \frac{1}{2} \right)^k. \] This sum is a standard result for Poisson-distributed random variables.

Step 3: Conclusion.
The correct answer is (B) \( 4e^{-\frac{1}{2}} \).