This question examines the properties of independent normal random variables and their linear combinations.
Understanding the Distributions: Let \(X\) and \(Y\) be independent random variables. \(X\) follows a normal distribution \(N(0, \sigma^2_1)\) and \(Y\) follows \(N(0, \sigma^2_2)\).
Independence implies:
Evaluating Independence of Linear Combinations: For two linear combinations \(a_1X + b_1Y\) and \(a_2X + b_2Y\) to be independent, their covariance should be zero.
The covariance is calculated as:
\(cov(a_1X + b_1Y, a_2X + b_2Y) = a_1a_2var(X) + b_1b_2var(Y) + (a_1b_2 + a_2b_1)cov(X, Y)\)
Given that \(cov(X, Y) = 0\), the formula simplifies to:
\(a_1a_2\sigma^2_1 + b_1b_2\sigma^2_2 = 0\)
Analyzing Each Option:
Option 1: \(X + Y\) and \(X - Y\)
Calculate: \(cov(X + Y, X - Y) = var(X) - var(Y) = \sigma^2_1 - \sigma^2_2 \neq 0\). They are not independent.
Option 2: \(2X + Y\) and \(X - Y\) when \(2\sigma^2_1 = \sigma^2_2\)
Calculate: \(cov(2X + Y, X - Y) = 2\sigma^2_1 - \sigma^2_2 = 0\) (since given \(2\sigma^2_1 = \sigma^2_2\))
Thus, they are independent under the condition.
Option 3: \(X + Y\) and \(X - Y\) are identically distributed.
Since they both have the same variance: \((var(X+Y) = \sigma^2_1 + \sigma^2_2 = var(X-Y))\), they are identically distributed.
Option 4: \(X + Y\) and \(2X - Y\) when \(2\sigma^2_1 = \sigma^2_2\)
Calculate: \(cov(X + Y, 2X - Y) = 2\sigma^2_1 - \sigma^2_2 = 0\) (since given \(2\sigma^2_1 = \sigma^2_2\))
Hence, they are independent under the condition.
Conclusion: The true statements are: