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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.

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2 Answers 2

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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.

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  • $\begingroup$ Yes, I couldn't progress from there on. Simple lower bound like $x^T x \le r^2$ does not help with the integral, I get an impossible inequality. $\endgroup$
    – CuriousGuy
    Commented Jun 9, 2022 at 14:45
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    $\begingroup$ Will edit the answer... $\endgroup$
    – PC1
    Commented Jun 9, 2022 at 15:50
  • $\begingroup$ Thanks! But I found the solution and posted it in case it is useful. $\endgroup$
    – CuriousGuy
    Commented Jun 9, 2022 at 16:03
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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 $$ \int_{B(r)} \phi(x) \ \text{d}x \ge 1 - \frac{1}{e} \ge \frac{1}{2}. $$

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    $\begingroup$ The link with the $\chi^2$ distribution is indeed very relevant. If your covariance matrix is not diagonal, you can also look at the Wishart distribution. $\endgroup$
    – PC1
    Commented Jun 9, 2022 at 16:13

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