Level curves of a joint normal distribution I need to plot the level curves of a joint distribution of two standard normal distributions.
I'm trying to proof that level curves are circumferences.
$$p_{X,Y}\left(x,y\right)=\frac{1}{2\pi}\exp\left(\frac{-x^{2}-y^{2}}{2}\right)$$
$$\frac{1}{2\pi}\exp\left(\frac{-x^{2}-y^{2}}{2}\right)=c\Leftrightarrow$$
$$\Leftrightarrow\exp\left(\frac{-x^{2}-y^{2}}{2}\right)=2\pi c\Leftrightarrow$$
$$\Leftrightarrow\ln\left(\exp\left(\frac{-x^{2}-y^{2}}{2}\right)\right)=\ln\left(2\pi c\right)\Leftrightarrow$$
$$\Leftrightarrow\frac{-x^{2}-y^{2}}{2}=\ln\left(2\pi c\right)\Leftrightarrow$$
$$\Leftrightarrow x^{2}+y^{2}=-2\ln\left(2\pi c\right)$$
Is this a correct proof that level lines are circumferences here? I'm a bit confused about the minus in the last equation.

Looks good to me. Regarding the minus: when you wrote
$$\frac{1}{2\pi}\exp\left(\frac{-x^{2}-y^{2}}{2}\right)=c,$$
then you are implicitly assuming that such a $c$ exists. In other words you can not put a $c$ there which is larger than the maximum value of the left side i.e. $\frac{1}{2\pi}\exp\left(\frac{-x^{2}-y^{2}}{2}\right)$. Finally the maximum of this expression is $1/{2\pi}$. Therefore $0<c<\frac{1}{2\pi}$ whichch gives $-2\pi c>0$. Hope this clarifies your doubt.
