Bivariate normal distribution $X$ and $Y$

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.

• I don't think the result is correct unless $X$ and $Y$ are standard normal to start with – Dilip Sarwate Mar 7 '16 at 4:05
• "(since $X$ and $Y$ are independent)" No, they are not. They are correlated random variables. – Dilip Sarwate Mar 7 '16 at 14:22

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.

$$f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2}{2} \right)$$

• thank you very much for your answer. I'm kind of new to this particular topic. How could I show your function is equivalent to the independent standard normals? – user320559 Mar 7 '16 at 10:27
• Just cancel terms and you'll get the expression at the end :). Are you unsure how the expression at the end is the pdf for independent standard normal variables? If so, $f_{XZ} = f_X(x)f_Z(z)$ for independent $X$ and $Z$, and $f_X(x) = (2\pi)^{-1/2} \exp(-x^2/2)$ is the PDF for the standard normal. – D.A.N. Mar 7 '16 at 10:36
• thanks! I also have to proof P(X>0,Y>0)= (1/4) + (1/2pi)(arcsin p). I thought I would have to do the integral from 0 to infinity of density function of the standard normal and multiply it by itseld (since X and Y are independent), but the answer I get is 1/2. Should I be using a different approach for this problem? I will also add this to my original question. – user320559 Mar 7 '16 at 12:04