Question

Suppose \(\mathbf{X}\) is distributed \(N_{2}(\boldsymbol{\mu}, \boldsymbol{\Sigma})\). Determine the distribution of the random vector \(\left(X_{1}+X_{2}, X_{1}-X_{2}\right) .\) Show that \(X_{1}+X_{2}\) and \(X_{1}-X_{2}\) are independent if \(\operatorname{Var}\left(X_{1}\right)=\operatorname{Var}\left(X_{2}\right)\)

Short answer

After transforming \(\mathbf{X}\) to \(\mathbf{Y}\), the determining factors for its multivariate normal distribution will be the transformed mean vector and covariance matrix. If \(\operatorname{Var}(X1) = \(\operatorname{Var}(X2)\), then \(X1 + X2\) and \(X1 - X2\) are independent.

Step-by-step solution

Transformation of Random Vector

We need to transform the random vector \(\mathbf{X}\)=[X1, X2] into a new random vector \(\mathbf{Y}\)=[Y1, Y2] such that Y1 = X1 + X2 and Y2 = X1 - X2. The matrix representation for \(\mathbf{Y}\)=A\(\mathbf{X}\) + b is given by [(1, 1), (1, -1)] for matrix A, and [0, 0] as the shift vector \(b\). Applying this transformation to X, \(\mathbf{Y}\) is distributed as \(\mathbf{Y} \sim N(A\boldsymbol{\mu} + b, A\boldsymbol{\Sigma}A^T)\).

Determine the Distribution of the Transformed Random Vector

We then compute the new mean vector and covariance matrix for \(\mathbf{Y}\) using the formulas: \(A\boldsymbol{\mu}\) = [(1,1), (1, -1)] * \(\boldsymbol{\mu}\), and \(A\boldsymbol{\Sigma}A^T\) = [(1, 1), (1, -1)] * \(\boldsymbol{\Sigma}\) * [(1, 1), (1, -1)]^T.

Prove Independence of X1 + X2 and X1 - X2

Two random variables are independent if their Covariance is 0. Compute the covariance between Y1 (X1 + X2) and Y2 (X1 - X2), using the formula: Cov(Y1, Y2) = E[(Y1-E(Y1))*(Y2-E(Y2))], where E refers to expected value operation. If Covariance = 0, this implies Y1 and Y2 are independent.

Use Given Variance Equality to Prove Independence

Assuming \(\operatorname{Var}(X1)\) = \(\operatorname{Var}(X2)\), we substitute this into the covariance formula. If equality holds, then the covariance should equate to 0 and thus Y1 and Y2 are independent.