Find $C$ such that
$$Ce^{(-4x^2-xy-4y^2)/2}$$
is a joint probability density of a $2$-variable Gaussian.
If someone could give me a jumping off point, or a process as to how to go about this, I'd really appreciate it.
Find $C$ such that
$$Ce^{(-4x^2-xy-4y^2)/2}$$
is a joint probability density of a $2$-variable Gaussian.
If someone could give me a jumping off point, or a process as to how to go about this, I'd really appreciate it.
For any density $f(x,y)$ you should have $$\int\limits_{\mathbb R^2}f(x,y)\mathrm dx\mathrm dy = 1.$$ From this condition you can find $C$. Recall that $$\int\limits_{\mathbb R}\mathrm e^{-\frac{(y-\mu)^2}2}\mathrm d y = \sqrt{2\pi}$$ for all $\mu$ - you may need it in your calculations.
Hint: you probably know that a Multivariable Gaussian takes the form:
$$ \frac{1}{\sqrt{(2 \pi)^n |\Sigma|}} \exp{ \left(- \frac{({\bf x}-{\bf \mu})^t \Sigma^{-1}({\bf x}-{\bf \mu})}{2} \right)}$$
Try to reduce your given pdf to this, so you can find by inspection the matrix $\Sigma^{-1}$ (it should be easy to see that ${\bf \mu}={\bf 0}$), and from that your are almost done (you don't even need to invert the matrix).