Question

Let 𝑋₁~Gamma(1,4), 𝑋₂~Gamma(2, 2) and 𝑋₃~Gamma(3, 4) be three independent random variables. If 𝑌=𝑋₁+2𝑋₂+𝑋₃, then 𝐸((\(\frac{y}{4}\))⁴ ) equals ______

Answer 1

Correct Answer: 3024

To solve this problem, we first need to understand the distribution of the random variable \( Y \). Given that \( X_1 \sim \text{Gamma}(1,4) \), \( X_2 \sim \text{Gamma}(2,2) \), and \( X_3 \sim \text{Gamma}(3,4) \), and \( Y = X_1 + 2X_2 + X_3 \), we calculate the mean and variance of \( Y \) using properties of the Gamma distribution and the linearity of expectation. 

1. Mean and variance of each \( X_i \):

• \( \mathbb{E}(X_1) = 1 \times 4 = 4 \), \( \text{Var}(X_1) = 1 \times 4^2 = 16 \)
• \( \mathbb{E}(X_2) = 2 \times 2 = 4 \), \( \text{Var}(X_2) = 2 \times 2^2 = 8 \)
• \( \mathbb{E}(X_3) = 3 \times 4 = 12 \), \( \text{Var}(X_3) = 3 \times 4^2 = 48 \)

2. Mean and variance of \( Y \):

• \( \mathbb{E}(Y) = \mathbb{E}(X_1) + 2 \mathbb{E}(X_2) + \mathbb{E}(X_3) = 4 + 2 \times 4 + 12 = 24 \)
• \( \text{Var}(Y) = \text{Var}(X_1) + 4 \times \text{Var}(X_2) + \text{Var}(X_3) = 16 + 4 \times 8 + 48 = 96 \)

3. Distribution of \( \frac{Y}{4} \):

Using the transformation, if \( Y \sim \text{Gamma}(6,4) \), then \( Z = \frac{Y}{4} \) has shape 6 and scale 1. Thus, \( Z \sim \text{Gamma}(6,1) \).

4. Expected value of \( Z^4 \):

For a gamma-distributed variable \( Z \) with shape \( \alpha = 6 \) and scale \( \beta = 1 \), the expected value \( \mathbb{E}(Z^n) \) is given by:

\[ \mathbb{E}(Z^n) = \frac{\beta^n (\alpha + n - 1)!}{(\alpha - 1)!} \]

So for \( \mathbb{E}(Z^4) \), we have:

\[ \mathbb{E}(Z^4) = 1^4 \times \frac{(6 + 4 - 1)!}{(6 - 1)!} = 24 \times 23 \times 22 \times 21 = 3024 \]

This result is 3024, consistent with the range provided. Thus, the expected value \( \mathbb{E}\left(\left(\frac{Y}{4}\right)^4\right) = 3024 \).