If random variable $\xi$ is more peaked than $\eta$ can $P(|\xi| \leqslant |\eta|)$ equals zero?

Solution 1:

Suppose that $P(|\eta|<|\xi|)=1$. Then there exists $t_0>0$ such that $P(|\eta|<t_0<|\xi|)>0$. This can be seen by writing $$ 1=P(|\eta|<|\xi|)=P(\bigcup_{t\in(0,\infty)\cap\mathbb{Q}}\{|\eta|<t<|\xi|\}). $$ For any $t$ we find that $$ P(|\eta|\leq t)\geq P(|\eta|<t)=P(|\xi|\leq|\eta|<t)+P(|\eta|<|\xi|\leq t)+P(|\eta|<t<|\xi|). $$ By choice of $t_0$ this implies that $P(|\eta|\leq t_0)>P(|\eta|<|\xi|\leq t_0)=P(|\xi|\leq t_0)$. This is a contradiction since you assumed that $P(|\eta|\leq t)\leq P(|\xi\leq t)$ for all $t$.

Solution 2:

Your proof is well so far. You could continue for example by considering the moment-generating function. As $|\xi|$ and $|\eta|$ have the same CDF, they also have the same moment-generating function. This leads to $$\mathbb E\exp(t|\xi|)=\mathbb E\exp(t|\eta|)<\mathbb E\exp(t|\xi|), $$ since we can use for the inequality $P(|\xi| > |\eta|) = 1$ and the exponential and the integral are strictly monotone. Thus you have the desired contradiction.