Bivariate normal distribution and law of $Z$

Question

Let $(X,Y)$ a bivariate normal variable $f(x,y)=\frac{1}{2\pi \sqrt{1-\rho^2}}e^{\frac{1}{2(1-\rho^2)}(x^2-2\rho xy+y^2)}$. Let $Z=\frac{Y-\rho X}{\sqrt{1-\rho^2}}$.

• Find the law of $Z$.

• Say if $X$ and $Z$ are independent or not.

• Find $\mathbb{P}(X>0,Y>0)$.

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Already by eye it can be noticed that the mean vector is $\mu=\begin{bmatrix} 0\\ 0\end{bmatrix} $ and the covariance matrix is $\sum=\begin{bmatrix}1 & -\rho\\ -\rho & 1\end{bmatrix}$ . I also know that $Z=X_1+...+X_n\sim N_n(A\mu,A^T\sum A)$, thus $Z$ is a linear of $X$ and $Y$ with $\frac{1}{\sqrt{1-\rho^2}}=const.$. Can I say that $Z\sim N(0,2-2\rho)$?

Answer 1

$(Y|X=x)\sim N(\rho x; 1-\rho ^2)$

Then $Z\sim \Phi$, a standard normal

EDIT: Let's factorize the joint distribution as follows

$$f_{XY}(x,y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2(1-\rho^2)}}\times \frac{1}{\sqrt{2\pi}\sqrt{1-\rho^2}}e^{-\frac{1}{2(1-\rho^2)}[-2\rho xy+y^2]}$$

Now let's complete the square at the esponent by summing $\pm \rho^2x^2$ and obtaining

$$f_{XY}(x,y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2(1-\rho^2)}}\times \frac{1}{\sqrt{2\pi}\sqrt{1-\rho^2}}e^{-\frac{1}{2(1-\rho^2)}[\rho^2x^2-2\rho xy+y^2]}e^{\frac{\rho^2x^2}{2(1-\rho^2)}}$$

That is

$$f_{XY}(x,y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\times \frac{1}{\sqrt{2\pi}\sqrt{1-\rho^2}}e^{-\frac{1}{2(1-\rho^2)}[y-\rho x]^2}$$

$$f_{XY}(x,y)=f_{X}(x)f_{Y|X}(x|y)$$

Where $f_X$ is a standard $\phi$ and $f_{Y|X}$ is a gaussian with mean $\rho x$ and variance $(1-\rho^2)$

Concluding, for evey $x$, your Z is the standardization of the $Y|X$ rv, thus $Z\sim \Phi$