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.


1 Answer 1


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.

Note : Zero Covariance doesn't in general imply Independence but in the case of Bivariate Normal it does.


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