What is the correlation between the pairwise differences of 2 bivariate normal random variables?

Question

Given (X,Y) bivariate normal, $U = \frac{X_i - X_j}{\sqrt2\sigma_x}$ and similarly $V = \frac{Y_i - Y_j}{\sqrt2\sigma_y}$ for any two independent pairs $(X_i, Y_i)$ and $(X_j, Y_j)$. Why is this true $\rho(U,V) = \rho(X,Y)$?

Answer 1

Direct calculation: $$ \begin{align*} Cov[U, V] & = \frac {1} {2\sigma_X\sigma_Y}Cov[X_i-X_j, Y_i - Y_j] \\ & = \frac {1} {2\sigma_X\sigma_Y}\left(Cov[X_i, Y_i] + Cov[X_j, Y_j] + 0 + 0\right) \\ & = \frac {1} {2\sigma_X\sigma_Y}\left(2\rho_{XY}\sigma_X\sigma_Y \right) \\ & = \rho_{XY}\end{align*}$$ Also check that $$ Var[U] = \frac {1} {2\sigma_X^2} \left(Var[X_i] + Var[X_j]\right) = \frac {1} {2\sigma_X^2} \left(2\sigma_X^2\right) = 1$$ and similarly for $Var[V] = 1$ Therefore $$ Corr[U, V] = \frac {Cov[U, V]} {\sqrt{Var[U]Var[V]}} = \rho_{XY}$$