Constant probability for multivariate normal random variable. I couldn't figure out the following problem.
Consider a multivariate normal random variable in $\mathbb{R}^n$, with the covariance matrix being something like $\sigma^2 \cdot I_n$, where $\sigma \in \mathbb{R}^+$ is a positive value and $I_n$ is the $n$-dimensional identity matrix.
Let $\phi(x)$ be its density distribution, and let $B(r)$ be the euclidean ball centered in the origin of radius $r$.
I have to find a good approximation of the smallest $r$ such that $\int_{B(r)} \phi(x) \ \text{d}x $ is at least a constant, say $1/2$. In particular, I'm interested in the relation between $r$ and $\sigma$. We know that for dimension $n=1$, $r = \sigma$ answers the question. What about dimension $n > 1$?
Many thanks.
 A: The PDF (assuming no mean) is:
$$\phi(x)=\frac{\exp\left(-\frac12x^T\Sigma^{-1}x\right)}{\sqrt{|2\pi\Sigma|}}$$
Where $\Sigma$ is your covariance matrix and $|\Sigma|$ is the determinant of the matrix $\Sigma$. In your case, $\Sigma$ is very simple so we can rewrite the PDF as:
$$\phi(x)=\frac{\exp\left(-\frac{\sigma^2}2x^Tx\right)}{\left(2\pi\sigma^2\right)^\frac n2}$$
So we see that this depends only on $x^Tx$, which is the square of the distance from the point to the origin (i.e.: it lies on the ball of radius $\sqrt{x^Tx}$). This should be enough to solve your problem.
EDIT:
You can rewrite:
$$\phi(\rho)=\frac{\exp\left(-\frac{\sigma^2\rho^2}2\right)}{\left(2\pi\sigma^2\right)^\frac n2}$$
where $\rho=\sqrt{x^Tx}$. Your integral becomes:
$$\int_{B(r)}\phi(x)dx = \int_0^r\phi(\rho)S_n(\rho)d\rho$$
where $S_n(\rho)$ is the area of the $n$-sphere of radius $\rho$. See for example:
https://en.wikipedia.org/wiki/N-sphere#Volume_and_surface_area
Depending on the value of $n$, you will get a closed form of the integral.
A: I found an answer for the problem, which I write here since it can be useful.
The squared radius of a standard normal random $n$-vector $(X_1, \dots, X_n)$ is given by the Chi-squared distribution with degree of freedom $n$, i.e. $\chi_n^2 = \sum_{i=1}^n {X_i^2}$, where $X_i \sim N(0,1)$ for $i = 1, \dots, n$.
Laurent and Massart show a concentration bound for it (page 1325, eq. 4.3): for any positive $x$,
$$
\text{Pr}\left[\chi_n^2 \ge n + 2\sqrt{n x} + 2x \right] \le e^{-x}.
$$
We can take $x=1$, which implies that $\text{Pr}\left[\chi_n^2 \ge n + 2\sqrt{n} + 2 \right] \le e^{-1}$.
In our case, we have $X = \sigma \cdot (X_1, \dots, X_n)$ where $X_i$ is a standard normal r.v. Thus, $\chi_n^2 = \frac{\lvert\lvert X \rvert\rvert_2^2}{\sigma^2}$.
Hence, if $r \ge \sigma \sqrt{5n} \ge \sigma \sqrt{n + 2\sqrt{n} + 2}$, we have
$$
\int_{B(r)} \phi(x) \ \text{d}x  \ge 1 - \frac{1}{e} \ge \frac{1}{2}.
$$
