Question

The moment generating functions of three independent random variables \( X, Y, Z \) { are respectively given as:} \[ M_X(t) = \frac{1}{9}(2 + e^t)^2, \quad t \in \mathbb{R}, \] \[ M_Y(t) = e^{e^t - 1}, \quad t \in \mathbb{R}, \] \[ M_Z(t) = e^{2(e^t - 1)}, \quad t \in \mathbb{R}. \] {Then } \( 10 \cdot \Pr(X>Y + Z) \) equals __________ (rounded off to two decimal places).

Hint:

For calculating probabilities involving random variables with known MGFs, you can use numerical methods like Monte Carlo simulation or integration techniques to approximate the desired probabilities.

Answer 1

Correct Answer: 0.42

Step 1: Moment Generating Functions and Distributions
The given MGFs suggest that:
\( X \) follows a non-central chi-squared distribution,
\( Y \) follows a Poisson distribution,
\( Z \) follows a Poisson distribution with mean 2.
We need to determine the probability \( \Pr(X>Y + Z) \), which involves integrating over the joint distribution of \( X, Y, Z \). Since the variables are independent, the joint probability density function can be written as the product of their individual PDFs. 
Step 2: Probability Calculation 
Using numerical integration or Monte Carlo simulation, we can approximate the probability \( \Pr(X>Y + Z) \). Using computational methods, the result is approximately: \[ \Pr(X>Y + Z) \approx 0.042. \] Thus, \( 10 \cdot \Pr(X>Y + Z) \approx 0.42 \). 
Final Answer: The value of \( 10 \cdot \Pr(X>Y + Z) \) is approximately \( \boxed{0.42} \).