Question

Let $X$ be a random variable having Poisson(2) distribution. Then $E\left(\dfrac{1}{1+X}\right)$ equals

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

Hint:

When summing over Poisson probabilities with shifted indices, substitution like $y = x+1$ often simplifies the series into an exponential form.

Answer 1

The Correct Option is C

Step 1: Definition of expectation.
For a Poisson random variable with parameter $\lambda = 2$, \[ E\left(\frac{1}{1+X}\right) = \sum_{x=0}^{\infty} \frac{1}{1+x} \cdot P(X = x) = \sum_{x=0}^{\infty} \frac{1}{1+x} \cdot \frac{e^{-2} 2^x}{x!}. \]

Step 2: Simplify the series.
Let $y = x + 1$, then \[ E\left(\frac{1}{1+X}\right) = e^{-2} \sum_{y=1}^{\infty} \frac{2^{y-1}}{y!} = \frac{e^{-2}}{2} \sum_{y=1}^{\infty} \frac{2^y}{y!}. \]

Step 3: Evaluate the exponential series.
\[ \sum_{y=1}^{\infty} \frac{2^y}{y!} = e^2 - 1. \] Hence, \[ E\left(\frac{1}{1+X}\right) = \frac{1}{2}(1 - e^{-2}). \]

Step 4: Conclusion.
Therefore, the required expectation equals $\dfrac{1}{2}(1 - e^{-2})$.