# Multivariate Gaussian with Larger “Variance” has Larger Norm?

Suppose that $X\sim N(0,1)$ is a standard Gaussian and $Y\sim N(0,\sigma^2)$ is Gaussian with variance $\sigma^2>1$. Informally, since $Y$ has a larger variance than $X$, it is more spread out than $X$, and thus we expect that for a typical realization, $|Y|$ should be larger than $|X|$.

If we want a more rigorous statement of this fact, we can for example prove that $|Y|$ stochastically dominates $|X|$, that is, $$P[|Y|\leq t]=P[|X|\leq t/\sigma]\leq P[|X|\leq t]$$ for all $t$.

Question. Suppose now that $X\sim N(0,\Sigma_1)$ and $Y\sim N(0,\Sigma_2)$ are Gaussian vectors of the same dimension with $0<\det(\Sigma_1)<\det(\Sigma_2)$. Is there an easy way to see (if even true) that $\|Y\|_2$ stochastically dominates $\|X\|_2$?

It's not true at all. Try $\Sigma_1 = \pmatrix{r&0\cr0 & 0}$ while $\Sigma_2 = I$, where $r$ is very large. $\det(\Sigma_1) = 0 < \det(\Sigma_2)$, but $\|X\|_2$ is very large with high probability.
• So let $\Sigma_1=\pmatrix{r&0\\0&1/(2r)}$. – kimchi lover Dec 8 '17 at 3:03