I need help figuring out the following.
Let $X$ and $Y$ have the bivariate normal distribution
$$ f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 – 2\rho xy}{2(1-\rho^2)} \right) $$
Show that $X$ and $Z=\frac{Y−ρX}{\sqrt{1−ρ2}}$ are independent standard normal random variables.
For $X$, is it asking to show that $f(x)$ is equal to the density function of the standard normal? If yes, how can I do that with $Z$? And how could I prove they are independent given the information provided? I would really appreciate any help here.
I also have to proof $P(X>0,Y>0) = \frac 14 + \frac{1}{2\pi}\arcsin \rho$. How can I do this using the answer to the above? I'm kind of new trying to understand this concept.
Best Answer
I assume your initial distribution is $$ f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right) $$
Apply the transformation $Y = Z\sqrt{1-\rho^2} + \rho X$
Including the Jacobian determinant $\sqrt{1-\rho^2}$, the pdf for $X,Z$
$$ f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2(1-\rho^2) +2\rho zx\sqrt{1-\rho^2} +\rho^2x^2 - 2\rho (zx\sqrt{1-\rho^2} + \rho x^2)}{2(1-\rho^2)} \right) $$
You can show that this is equivalent to the pdf for independent standard normals.
$$ f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2}{2} \right) $$