If $(X,Y)$ is bivariate normal with correlation $\rho$ and $\sigma_X^2 = \sigma_Y^2$, show that $X$ and $Y - \rho X$ are independent

Question

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.

Answer 1

Note that the standard term is "bivariate normal."

One conventional definition of bivariate normal is that a pair $(U,V)$ of random variables is so distributed that every linear combination $aU+bV$ with non-random coefficients $a,b$ is normally distributed.

If the question is how to show that the pair $(X,Y-\rho X)$ has a bivariate normal distribution, then that becomes the question of how to show that for every pair $a,b$ of non-random coefficients, the linear combination $aX+b(Y-\rho X)$ is normally distributed. Then we have $$ aX + b(Y-\rho X) = (a-b\rho)X + bY $$ and that is a linear combination of $X$ and $Y$ with non-random coefficients $a-b\rho$ and $b.$ Since $(X,Y)$ has a bivariate normal distribution, this linear combination has a normal distribution. Thus the pair $(X, Y-\rho X)$ has a bivariate normal distribution.

Answer 2

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 $f_{XZ}(x,z)=f_X(x)f_Z(z)$