# What is the radius of a “Gaussian” sphere such that approx. all the population lie within?

Let $\mathbf{X}\in\Bbb{R}^n$ be a random vector distributed normally, i.e. $\mathbf{X}\sim\mathcal{N}(\mathbf{0},\sigma^2I_n)$, where $\sigma>0$ is the standard deviation and $I_n$ is the identity matrix of order $n$.

In the case of the univariate Gaussian distribution, i.e., $X\sim\mathcal{N}(0,\sigma^2)$, we say that approximately 99.7% of the population lie within three standard deviations.

What would be an appropriate constant in the case of the multivariate "isotropic" Gaussian distribution, i.e., $\mathbf{X}\sim\mathcal{N}(\mathbf{0},\sigma^2I_n)$? That is, what would be the radius of the $n$-sphere in terms of the standard deviation $\sigma$, such that approximately all the population lie within?

P.s.: Please feel free to edit the title (and/or the body of this question). I was not sure how to express what I wanted in the title.

• Maybe, how is $\sigma_n$ expressed in terms of $\sigma ?$ – Narasimham May 29 '16 at 21:31

The distribution of $\|X\|^2/\sigma^2$ is $\chi^2_k$. Since for $k=1$ $$P(|x| \le 3 \sigma) = .9973002$$ I interpret this as the question for which $r$ in the case of $k = 3$ $$P(\|X\| \le r \sigma) = .9973002.$$ This is $r = \sqrt{F^{-1}(.997302)} = 3.76205$ where $F$ is the cumulative distribution function of a $\chi^2_3$ random variable. R command: