Let $m$ be a large positive integer and $X=(X\_1,\ldots,X\_m) \sim N(0,I\_m)$. I wish to show that the squared norm of $X$ is much much bigger than the absolute value of any of the $X\_j$'s. For example, [one can show](https://mathoverflow.net/a/369322/78539) that$P(\|X\|^2 \ge \mathcal O(\sqrt{m})|X\_1|) = 1-o(1)$. Recall that $o(1)$ is notation for any quantity which goes to $0$ in the sup-limit $m \to \infty$.


My goal is to improve the $\sqrt{m}$ to something much larger. To this end, let $(u\_m)\_m$ be a sequence of positive numbers.



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> **Question.** What is a sufficient (and ideally, also necessary) condition on the growth of $u\_m$ such that $P(\|X\|^2 \ge u\_m |X\_1|) = 1-o(1)$.
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If $X \sim N(0,I_m)$, what is a necessary and sufficient condition on $u_m > 0$ such that $\lim\sup_{m\to \infty} P(\|X\|^2 \ge u_m|X_1|) = 1$
For real $u\_m>0$, the probability in question is
$$p\_m:=P(\|X\|^2\ge u\_m|X\_1|)=P\Big(\frac{|X\_1|}{\|X\|^2/m}\le\frac m{u\_m}\Big).$$
Passing to a subsequence, without loss of generality
$$\frac{u\_m}m\to c\in[0,\infty]$$
(as $m\to\infty$). By the law of large numbers, $\|X\|^2/m\to1$ in probability. So, by [Slutsky's theorem](https://en.wikipedia.org/wiki/Slutsky%27s_theorem) ,
$$p\_m\to p\_\infty:=P(|X\_1|\le1/c),$$
assuming $1/0:=\infty$ and $1/\infty:=0$. Thus, $p\_m\to1$ iff $c=0$. That is, $P(\|X\|^2\ge u\_m|X\_1|)=1-o(1)$ iff $u\_m=o(m)$.


