 Solution

For a joint probability density function, independence is equivalent to $f(x,y)=f_X(x)f_Y(y)$ almost everywhere. The displayed density is valid for $|\rho |<1$. Complete the square:Integrating the second normal factor over $y$ givesDividing the joint probability density function by this marginal yields the conditional lawFor $\rho =0$ the joint probability density function factors and the variables are independent. If $\rho \ne 0$, the conditional mean depends on $x$, so independence fails; equivalently their covariance is $\rho$.

Define $Z=(Y-\rho X)/\sqrt{1-{\rho }^2}$. In the change of variables $y=\rho x+\sqrt{1-{\rho }^2}z$, the Jacobian $\sqrt{1-{\rho }^2}$ cancels the remaining density factor, leaving the product of two standard normal densities. Thus $Z$ is standard normal and independent of $X$. One choice isChanging both signs gives the other standardized choice. The density transformation proves independence directly rather than inferring it from zero covariance without a joint-normal justification.