# Upper bound for $\frac{\|x\|_1}{\|x\|_2}$ if each entry of $x\in R^d$ is i.i.d. sampled from Gaussian distribution $N(0,1)$

In the question, $\|x\|_1=\sum_{i=1}^d|x_i|$ with $|\cdot|$ being the absolute value, and $\|x\|_2=\sqrt{\sum_{i=1}^d x_i^2}$.

In general, $\frac{\|x\|_1}{\|x\|_2}\leq \sqrt{d}$ always holds for arbitrary $x\in R^d$.

However, if each entry of $x\in R^d$ is independently and identically sampled from Gaussian distribution $N(0,1)$, then I am interested in if we can get a tighter upper bound for $\frac{\|x\|_1}{\|x\|_2}$ samller than $\sqrt{d}$ with a certain probability? e.g., $P[\frac{\|x\|_1}{\|x\|_2}\leq t]\geq 0.9$ and $t<\sqrt{d}$.

Hope the link helps. https://en.wikipedia.org/wiki/Concentration_inequality

• So are you looking for a function that provides a bound for each probability? – user237392 Apr 24 '16 at 10:56
• @Bey yes, it is right. just similar to bounding the tail probability $P[||X||_1 > t]$ in math.stackexchange.com/questions/33943/… – olivia Apr 24 '16 at 12:05