Question

Let \( X \) be a random variable having Poisson distribution with mean \( \lambda>0 \). Then \( E \left( \frac{1}{X+1} \mid X>0 \right) \) equals:

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

Hint:

- For Poisson distribution, use the PMF to compute the expected value.
- Conditional expectations can be computed by modifying the PMF and normalizing over the desired condition.

Answer 1

The Correct Option is A

1) Using the definition of expectation:
For a random variable \( X \) with a Poisson distribution, the expected value is given by: \[ E \left( \frac{1}{X+1} \mid X>0 \right) = \frac{\sum_{x=1}^{\infty} \frac{1}{x+1} P(X=x)}{P(X>0)} \] Since \( X \sim \text{Poisson}(\lambda) \), the probability mass function (PMF) is: \[ P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!} \] Thus, the sum becomes: \[ E \left( \frac{1}{X+1} \mid X>0 \right) = \frac{\sum_{x=1}^{\infty} \frac{1}{x+1} \frac{\lambda^x e^{-\lambda}}{x!}}{1 - e^{-\lambda}} \] 2) Simplifying the sum:
The sum can be evaluated using known series expansions, and simplifying yields the correct result: \[ E \left( \frac{1}{X+1} \mid X>0 \right) = \frac{1 - e^{-\lambda} - \lambda e^{-\lambda}}{\lambda (1 - e^{-\lambda})} \]