# 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.

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.

• 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. Commented Jun 9, 2022 at 14:45
• Will edit the answer...
– PC1
Commented Jun 9, 2022 at 15:50
• Thanks! But I found the solution and posted it in case it is useful. Commented Jun 9, 2022 at 16:03

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 $$\int_{B(r)} \phi(x) \ \text{d}x \ge 1 - \frac{1}{e} \ge \frac{1}{2}.$$

• 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.
– PC1
Commented Jun 9, 2022 at 16:13