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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.

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

sqrt( qchisq( pchisq( 9,df=1 ), df = 3))

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