91Ó°ÊÓ

Given a bivariate normal distribution \( \mathbf{X} \sim N(\mathbf{\mu}, \mathbf{\Sigma}) \), we are asked to find the distribution of the transformed random vector \(\mathbf{Y} = (X_{1}+X_{2}, X_{1}-X_{2})\). This is an instance of an affine transformation, and it has an underlying theoretical result that states that an affine transformation of a Gaussian random vector is also Gaussian. The expectation and covariance for \( \mathbf{Y} \) can be obtained using these formulas: \( \mathbf{E}[\mathbf{Y}] = A\mathbf{E}[\mathbf{X}] + \mathbf{b} \) and \( \text{Cov}(\mathbf{Y}) = A\text{Cov}(\mathbf{X})A^{\top} \). Here \( A = [1 1; 1 -1] \) is the matrix parameters for scalar multiplication in the transform and \( \mathbf{b} = \textbf{0} \) because there's no vector addition in the transform.

To get the expectation, multiply matrix A with \( \mathbf{\mu} \), leading to \( [\mu_{1}+\mu_{2}, \mu_{1}-\mu_{2}] \) as the result. To get the covariance matrix, multiply the Cov(\( \mathbf{X} \)) with A on the left and \( A^{\top} \) on the right, it forms a \( 2 \times 2 \) matrix. Remember the-variance-covariance matrix for the bivariate normal distribution is \( \Sigma = [\sigma_{x1}^2, \sigma_{x1,x2}; \sigma_{x2,x1}, \sigma_{x2}^2] \) where \( \sigma_{xi} \) is the standard deviation of \( x_i \) and \( \sigma_{xi, xj} \) is the covariance between \( x_i \) and \( x_j \).

To check if \( X_{1}+X_{2} \) and \( X_{1}-X_{2} \) are independent, examine the covariance matrix of \( \mathbf{Y} \). Two random variables are independent in a multivariate normal distribution if and only if their covariance is 0. Given that \( \operatorname{Var}(X_{1}) = \operatorname{Var}(X_{2}) \) and the covariance \( \sigma_{x1, x2} \) between \( X_1 \) and \( X_2 \) equals to 0. Then you can see from the results in Step 2 that the covariance between \( X_{1}+X_{2} \) and \( X_{1}-X_{2} \) is 0. Hence, \( X_{1}+X_{2} \) and \( X_{1}-X_{2} \) are independent given \( \operatorname{Var}(X_{1}) = \operatorname{Var}(X_{2}) \).