Find the density function from a joint density function

Question

I try to solve the following task and I don't know what the correct way to do is.

Let $p\in(0,1)$ and $(X,Y)$ be a pair of random variables with distribution density function $$f(x,y)=\frac{1}{2\pi\sqrt{1-p^2}} \exp\left(-\frac{1}{2(1-p^2)}(x^2-2pxy+y^2)\right)$$ Show that $X$ and $Z=(Y-pX)/\sqrt{1-p^2}$ are independent standard Gauss random variables.

What I've have done so far:

Independence means $\mathbb{E}[XZ]=\mathbb{E}[X]\mathbb{E}[Z]$ meaning if this holds $$\int\int xzf_{X,Z}(x,y)\mathrm{d}x \mathrm{d}y=\int\int xf_X(x,y)\mathrm{d}x \mathrm{d}y \int\int zf_Z(x,y)\mathrm{d}x \mathrm{d}y$$ where $f$ is the density function, then $Z$ and $X$ are independent.

Now here I have great troubles to find the probability density functions of $X$ or $Y$, mainly because of the parameter $p$.

$$\mathbb{E}[XY]=\frac{1}{2\pi\sqrt{1-p^2}}\exp\left(-\frac{1}{2(1-p^2)}\right)\int\int x \exp\left(x^2-2pxy+y^2\right) \mathrm{d}x \mathrm{d}y$$

Let us look at the integral $$\int\int x \exp\left(x^2-2pxy+y^2\right) \mathrm{d}x \mathrm{d}y=\int \exp(y^2)\int x \exp\left(x^2-2pxy\right) \mathrm{d}x \mathrm{d}y$$ Now since this integral is positive, we can apply Fubini-Tonelli.

But this parameter $p$ makes it very difficult to compute the integral, so I'm not even sure, if this is the right / smart way to solve this task.

Thank you for help.

Answer 1

$(X,Z)$ are independent if forany $f,g$ $$\mathbb{E}(f(X)g(Z))=\mathbb{E}(f(X))\mathbb{E}(g(Z))$$

In your case :

$$\mathbb{E}(f(X)g(Z))=\mathbb{E}\left(f(X)g\left(\frac{Y-pX}{\sqrt{1-p^2}}\right)\right)=\int_{\mathbb{R}^2}f(x)g\left(\frac{y-px}{\sqrt{1-p^2}}\right)p(x,y)~dx~dy$$

so this is a change of variable problem, you set : $z=\frac{y-px}{\sqrt{1-p^2}}$ and you get : $y=\sqrt{1-p^2}z+px$ plugging this into $p(x,y)$ gives you : $$p(x,y)=p(x,\sqrt{1-p^2}z+px)=q(x)q(z)$$ where $$q(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$

you then plug it into above equations and this is solved.

Note

in the case of a density $p(x,y)$, $X$ and $Y$ are independent if and only if $p(x,y)=p_X(x)p_Y(y)$ where $p_X$ (resp. $p_Y$) is the density of $X$ (resp. $Y$)

Answer 2

HINT

Note that in the exponential we have $$ x^2-2pxy+y^2=(y-px)^2+(1-p^2)x^2 $$ so that $$ f(x,y)=\frac{1}{2\pi\sqrt{1-p^2}} \mathrm e^{-\frac{(y-px)^2}{2(1-p^2)}-\frac{x^2}{2}}=\frac{1}{\sqrt{2\pi}\sqrt{1-p^2}}\exp\left(-\frac{(y-px)^2}{2(1-p^2)}\right)\times \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right) $$