Prove pair (X,Y) has standard normal bivariate distribution

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I'm stuck on this problem and don't really understand how I should attempt it.

Let $Z,W$ be independent standard normal random variables and $=1<p<1$. Check that if $X=Y$, and $Y=pZ+\sqrt{1-p^2}W$ then the pair (X,Y) has standard bivariate normal distribution with parameter $p$.

I understand what it means that $Z,W$ are standard normal, and know the form for standard bivariate, but have no clue how to find the distribution of the pair $(X,Y)$. I'm not really sure what exactly I'm supposed to do to find this distribution.

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I assume you mean if $X = Z$. Are you familiar with the transformation formula for random vectors? If we have a function $g: \mathbb{R}^2 \to \mathbb{R}^2$ that is one-to-one and onto, then if $(X, Y) = g(Z, W)$, where $X = g_1(Z,W)$ and $Y = g_2(Z,W)$ and where the pair $(Z, W)$ has bivariate density $f_{Z,W}(z,w)$. Then we can find the bivariate density $(X,Y)$ with the following formula,

$$f_{X,Y}(x,y) = f_{Z,W}(g^{-1}(x,y)) |J_{g^{-1}}(x,y)| $$

Where $J_{g^{-1}}$ is the Jacobian of the inverse of $g$. In other words,

$$J_{g^{-1}} = \text{det} \begin{bmatrix}\frac{\partial g^{-1}_1}{\partial x} & \frac{\partial g_1^{-1}}{\partial y} \\ \frac{\partial g_{2}^{-1}}{\partial x} & \frac{\partial g_2^{-1}}{\partial y} \end{bmatrix} $$

In our case we have that,

$$(x, y) = (z, \rho z+ \sqrt{1-\rho^2}w) = g(z,w)$$

Finding the inverse we have that,

$$(z,w) = \left(x, \frac{y - \rho x}{\sqrt{1-\rho^2}} \right) = (g_1^{-1}(x,y), g_2^{-1}(x,y) $$

The Jacobian is then given as,

$$J_{g^{-1}} = \text{det} \begin{bmatrix}1 & 0 \\ \frac{-\rho}{\sqrt{1-\rho^2}} & \frac{1}{\sqrt{1-\rho^2}} \end{bmatrix} = \frac{1}{\sqrt{1-\rho^2}} $$

Putting all this together we obtain,

$$f_{X,Y} = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left(-\frac{1}{2}\left(x^2 + \frac{(y-\rho x)^2}{1 - \rho^2} \right) \right) $$

Simplifying we get the standard bivariate form,

$$f_{X,Y}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left(-\frac{1}{2(1-\rho^2)}\left(x^2 -2\rho xy + y^2 \right) \right) $$

Completing the proof. The main thing to take away is the formula for transformations of random vectors and the bivariate standard density.