# Show that f is density of bivariate normal distribution

Let $$\mathbb{X} := (X_1,X_2)$$ be a random variable with given density function $$f_{\mathbb{X}}(x_1,x_2)= \frac{\sqrt{2}}{\pi}\exp\left(-\frac{3}{2}x_1^2-x_1x_2-\frac{3}{2}x_2^2\right),\space \text{for} (x_1,x_2)\in \mathbb{R^2}$$

Show that $$f$$ is the density of a normal distribution in $$\mathbb{R^2}$$.

At first I thought the joint density can be expressed as the product of the densities, but completing squares gives me a troublesome expression. I then thought that I can identify it with the bivariate normal distribution: $$p(x_1,x_2)=\frac{1}{2 \pi \sigma_1\sigma_2\sqrt{1-p^2}}\exp\left(-\frac{1}{2}\left(\frac{z}{1-p^2}\right)\right)\,,$$ where $$z=\frac{(x_1-\mu_1)^2}{\sigma_1^2}-\frac{2p(x_1-\mu_1)(x_2-\mu_2)}{\sigma_1\sigma_2}+\frac{(x_2-\mu_2)^2}{\sigma_2^2}$$

If you now take out $$-\frac{1}{2}$$ in the exponent of $$f$$ you get $$p=-1$$, $$\mu_1=\mu_2=0$$ and $$\sigma_1=\sigma_2=\sqrt{\frac{1}{3}}$$ and after you multiply $$p(x_1,x_2)$$ with $$\frac{2^{\frac{3}{2}}}{3}$$ you exactly have f. So f is normal distributed in $$\mathbb{R^2}$$. Is this right or one proves this differently?

So $$X_1=X_2 \sim N(0,\frac{1}{3})$$. Because in the next task one is to calculate the density of $$X_1+X_2$$ and $$X_1-X_2$$ and to look whether the are independent or not, but I get the dirac-delta function for the latter one and there I think I might did something wrong. I am thankful for any advice.

HINT 1 : Write the pdf in the form of $$c*e^{\frac{-1}{2}}(x'\Lambda x) \quad$$ Where $$\Lambda$$ is symmetric
HINT 2: Now invert $$\Lambda$$, i.e fine $$\Sigma = \Lambda^{-1}$$ and then find $$A$$ such that $$AA' = \Sigma$$
Then you can argue that if $$Z$$ follows a standard bivariate normal distribution with mean $${0}$$ then $$X=AZ$$ will also follow bivariate normal distribution with appropriate parameters whose pdf will be the one that you wanted to be proved as a bivariate normal's pdf.
Once you show that $$X$$ is a bivariate normal, then showing $$X_1 + X_2$$ to be normal can be done by taking appropriate transformations over $$X$$, i.e choose $$l$$ such that $$l'X = X_1 + X_2$$. In this case $$l' = (1 \quad 1)'$$. You can find the expressions for its parameters under such transformations in this link under the section "Affine Transformation".
Showing Independence between $$X_1 + X_2$$ and $$X_1 - X_2$$ can be done by taking appropriate transformations $$BX = (X_1 + X_2, X_1 - X_2)$$ and showing that the non diagonal elements are $$0$$. I.e Covariance between them is $$0$$ and hence is independent.