Question

Let 𝑋 ∼ 𝑈𝑛𝑖𝑓𝑜𝑟𝑚 8, 20) and 𝑍 ∼ 𝑈𝑛𝑖𝑓𝑜𝑟𝑚(0, 6) be independent random variables. Let 𝑌 = 𝑋 + 𝑍 and 𝑊 = 𝑋 − 𝑍. Then 𝐶𝑜𝑣(𝑌, 𝑊) is ______ (in integer).

Answer 1

Correct Answer: 9

To find: \(\text{Cov}(Y, W)\) 

Given: - \(X \sim Uniform(8,20)\) - \(Z \sim Uniform(0,6)\) - \(X\) and \(Z\) are independent - \(Y = X + Z\) - \(W = X - Z\) We use the covariance formula: \[ \text{Cov}(Y,W) = \text{Cov}(X+Z,\, X-Z) \] Expand using linearity of covariance: \[ \text{Cov}(Y,W)=\text{Cov}(X,X) - \text{Cov}(X,Z) + \text{Cov}(Z,X) - \text{Cov}(Z,Z) \] Since \(X\) and \(Z\) are independent: \[ \text{Cov}(X,Z)=0,\quad \text{Cov}(Z,X)=0 \] So: \[ \text{Cov}(Y,W) = \text{Var}(X) - \text{Var}(Z) \] For a uniform distribution on \([a,b]\): \[ \text{Var}(U) = \frac{(b - a)^2}{12} \] Compute each variance: Variance of X: \[ \text{Var}(X)= \frac{(20 - 8)^2}{12} = \frac{12^2}{12} = \frac{144}{12} = 12 \] Variance of Z: \[ \text{Var}(Z)= \frac{(6 - 0)^2}{12} = \frac{36}{12} = 3 \] Finally: \[ \text{Cov}(Y,W) = 12 - 3 = 9 \]

Final Answer: \(\boxed{9}\)