50% cutoff on multi-normal distribution I'm doing some visualizations and I'm trying to make outline of an ellipse inside which there is a 50% probability of finding the result given measurement following bivariate normal.
More formally:
Find $q$ s.t. region $U = \{x : f(x) > q\}$ has $\mathbb{P}(U)=0.5$ where $f$ is pdf of bivariate gaussian with some correlation.
Is there a well known formula for $q$? Does this (and similar cutoff values for other distributions) have some established name?
 A: I will provide a solution for arbitrary dimension $n$ and probability $p$ such as $\mathbb{P}(\{x:f(x)>q\})=p$, which gives a simple result for $n=2$. Notation: $x^2 =||x||_2^2,\ |A| = \det A$


*

*Cholesky decomposition of covariance matrix
$$\Sigma = A^TA$$
Note that $|\Sigma| = |A|^2$.

*Substitution
$$x = F(y) = \mu + Ay$$

*Solving for $f(x) \geq q$
$$f(x) = \alpha e^{-\frac12(x-\mu)^T\Sigma^{-1}(x-\mu)} = \alpha e^{-\frac12y^2}, \ \alpha = (2\pi)^{-n/2}|\Sigma|^{-1/2}$$
$$y^2 \leq -2\ln(q/\alpha) =: R^2_q$$

*Calculating $\int_{\{f(x)\geq q\}}f(x)dx$
\begin{align}
p&=\int_{\{f(x)\geq q\}}f(x)dx 
  =\int_{\{y^2\leq R^2_q\}}\alpha e^{-\frac12y^2} |A|dy 
  =\alpha|A|\int_0^{R_q}\left(\int_{S^{n-1}} e^{-\frac12(rs)^2}ds\right)r^{n-1}dr \\
 &=\alpha|A|\mu(S^{n-1})\int_0^{R_q}r^{n-1}e^{-\frac12r^2}dr
  =(2\pi)^{-n/2}\mu(S^{n-1})\int_0^{R_q}r^{n-1}e^{-\frac12r^2}dr
\end{align}
$\mu(S^{n-1})$ being surface measure of unit sphere in $\mathbb{R}^n$.

*Having solved for $R_q$ desired cutoff ellipsis is $F(R_qS^{n-1})$ and $q = \alpha e^{-\frac12R_q^2}$
For $n=2$ it is as follows:
$$p = \frac{2\pi}{2\pi}\int_0^{R_q}re^{-\frac12r^2}dr = [-e^{-\frac12r^2}]_{r=0}^{R_q} = 1 - e^{-R_q^2/2}$$
$$R_q = \sqrt{-2\ln(1 - p)}$$
Cheers!
A: This is not the full answer.
Are you looking for visualization like on the plot below?
One can see on the plot is a 50%-confidence area for a bivariate Normal Copula.

