Let $(X,Y)$ be bivariate normal with correlation $\rho$ and variance $\sigma_X^2=\sigma_Y^2$. Show that $X$ and $Y – \rho X$ are independent.
I found there's a general result which states that if $(X, Z)$ is normal bivariate and $COV(X, Z)=0$, then X and Z are independent. Therefore since we have $$E(X(Y-\rho X))= E(XY)-\rho E(X^2)=E(XY) -\frac{E(XY)}{\sigma_X^2}\sigma_X^2=0,$$ it is enough to prove that $(X,Y-\rho X)$ is normal bivariate. Although, from here I do not really know how to go on…
Many thanks for any help.
Best Answer
Let's start from the joint gaussian density $f_{XY}(x,y$.
After some simple algebraic manipulations you get
$$\frac{1}{\sigma \sqrt{2\pi}}e^{ -\frac{1}{2\sigma^2}(x-\mu_X)^2} \frac{1}{\sigma \sqrt{2\pi}\sqrt{1-\rho^2}}e^{ -\frac{1}{2\sigma^2 (1-\rho^2)}[y-\mu_Y-\rho(x-\mu_X)]^2}$$
That is $f_{XZ}(x,z)=f_X(x)f_Z(z)$