Question

Let \( X \) be a random variable having the Poisson distribution with mean \( \log_e 2 \). Then \( E\left( e^{(\log_e 3)X} \right) \) equals:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For Poisson-distributed random variables, the moment-generating function is useful for calculating expectations of exponential functions of the random variable.

Answer 1

The Correct Option is D

The Poisson distribution is given by the probability mass function:

\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots \] where \( \lambda = \log_e 2 \) is the mean. We are asked to find \( E\left( e^{(\log_e 3)X} \right) \).

Step 1: Recognizing the moment-generating function of a Poisson random variable.
The moment-generating function (MGF) of a Poisson random variable \( X \) with mean \( \lambda \) is given by:

\[ M_X(t) = E(e^{tX}) = e^{\lambda (e^t - 1)} \] Substituting \( t = \log_e 3 \) and \( \lambda = \log_e 2 \), we get:

\[ E\left( e^{(\log_e 3)X} \right) = e^{\lambda (e^{\log_e 3} - 1)} \] Step 2: Simplifying the expression.
Since \( e^{\log_e 3} = 3 \), the expression becomes:

\[ E\left( e^{(\log_e 3)X} \right) = e^{\log_e 2 (3 - 1)} = e^{2 \log_e 2} \] Step 3: Final calculation.
Using the property of logarithms \( e^{\log_e a} = a \), we find:

\[ e^{2 \log_e 2} = 2^2 = 4 \] Therefore, the correct answer is \( \boxed{4} \).