Question

Let \(X\) be a random variable having the moment generating function

• A. \(16\)
• B. \(4\)
• C. \(6\)
• D. \(8\)

Hint:

For a Poisson distribution, the mean and variance are both equal to the parameter \(\lambda\).

Answer 1

The Correct Option is D

Step 1: Identify the distribution from the MGF.
The given moment generating function is
\[ M_X(t)=e^{2(e^t-1)} \] We know that the MGF of a Poisson random variable with parameter \(\lambda\) is
\[ M_X(t)=e^{\lambda(e^t-1)} \] Comparing, we get
\[ \lambda=2 \] Hence,
\[ X\sim \text{Poisson}(2) \]

Step 2: Find the variance of \(X\).
For a Poisson random variable with parameter \(\lambda\),
\[ Var(X)=\lambda \] Therefore,
\[ Var(X)=2 \]

Step 3: Use variance transformation formula.
For any random variable \(X\),
\[ Var(aX+b)=a^2Var(X) \] Here,
\[ a=2,\qquad b=3 \] Thus,
\[ Var(2X+3)=2^2Var(X) \] \[ =4\times 2 \] \[ =8 \]

Step 4: Final conclusion.
Hence,
\[ \boxed{8} \] Therefore, the correct option is
\[ \boxed{(D)} \]